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TEST PREP for CQB version 15 The largest and most up-to-date JAA exam questions book

061

EDITION 2012

Contents FO/REWORD ...................................~.............................................................................................................................................................V How to use this book ............................................................................................................................................................................VI JAA exam questions & time limits ........'..............................................................................................................................................VII

01 BASICS OF NAVIGATION

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" 01-01 The solar system ......................................................, ................................................................................................................... 1 01-02 The earth .............................................................~::.~ ..................................................................................................................... 4 01-03 Tim~ and time conversions ..................................................................................................................................................... 21 01-04 Directions ..................:.........................................................................................................................................:....................... 30 01-05 Distance .......................................................................................................................................................................................47

02 MAGNETISM AND COMPASSES 02-01 Knowledge of the principles of direct reading (standby) compass ................................................................................. 61

03 CHARTS 03-01 General properties of miscellaneous types of projections ............................................................................................... 69 03-02 The representation of meridians, parallels, great circles and rhumb lines .................................................................... 80, 03-03 The use of current aeronautical charts .................................................................................................................................83

04 DEAD RECKONING NAVIGATION (DR) 04-01 Basis of dead reckoning .............................................................,........................................................................................... 101 04-02 Use of the navigational computer .......................................................................:............................................................... 103 04-03 The triangle of velocities ................,...................................................................................................................................... 112 04-04 Determination of DR position ......................................................................................:....................................................... 122 04-05 Measurement of DR elements ...........:.................................................................................................................................. 125

05 IN-FLIGHT NAVIGATION 05~01 Use of visual observations and application to in-flight navigation ............................................................................... 131

05-02 Navigation in climb and descent ......................................................................................................................................... 132 05-03 Navigation in cruising flight, use of fixes to revise navigation data .............................................................................. 138 05-04 Flight Log ..................................................................................................................................................................................146

PICTURE SUPPLEMENTS QUESTIONS .................................................................................................................................. 149 PICTURE SUPPLEMENTS EXPLANATIONS ......................................................................................................................... 159

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01 Basics of Navigation

BASICS OF NAVIGATION 01-01 The solar system 1174. Airplane ATPL The term "sidereal" is used:

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to the average apparent time. It is the reference for calculating mean time, zone time, etc.

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A) to describe how two positions of heavenly bodies are located sideways on the sky. B) to describe conditions with reference to the Moon. e) to describe a situation or relationship concerning the stars. D) to describe the time interval between two successive transits of the real apparent Sun at the same meridian.

The plane in which the Earth travels around the Sun is called "Plane

of the Ecliptic".

1196. Airplane ATPL CPL Heli The term "aphelion" is used to describe:

1185. Airplane ATPL CPL Heli When the Sun's declination is northerly:

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A) it is winter on the northern hemisphere. B) the sunrise occurs earlier at southern latitudes than the northern latitudes. e) the daylight period is shorter in the southern hemisphere than the northern. D) midnight Sun may be observed at the South Pole. (Refer to figure 061-EI8) The declination of the Sun is the angle between the rays of the Sun and the plane of the Earth's equator. The Earth's axial tilt is the angle between the Earth's axis and a line perpendicular to the Earth's orbit. The Earth's axial tilt changes gradually over thousands of years, but its current value is about 23°26'. Because this axial tilt is nearly constant, solar declination varies with the seasons and its period is one year. At the solstices, the angle between the rays of the Sun and the plane of the Earth's equator reaches its maximum value of 23°26~ Therefore solar declination = +23°26' at the northern solstice and -23°26' at the southern solstice. At the moment of each equinox, the center of the Sun appears to pass through the celestial equator, and solar declination is 0°. When the solar declination is northerly, the days are longer in the northern hemisphere and shorter in the southern hemisphere. With a southerly solar declination the opposite applies.

(Refer to figure 061-EI6) The astronomer J. Kepler discovered various laws governing the motion of planets around the Sun. These laws are basic to the movement and orbital speeds of the planets. Kepler's 1" Law: The orbit of a planet is elliptical with the Sun at one focus. Kepler's 2 nd Law: The radius vector from the Sun to a planet sweeps out equal areas in equal times. Kepler's 3,d Law: The square of the period of the orbit of a planet is proportional to the cube of that planet's semi major axis. The constant ofproportionality is the same for all planets. Kepler's First Law enunciates that the shape of the orbit of a planet traveling around the Sun is an ellipse, the Sun being positioned at one of the foci of the ellipse. It is known from basic geometry that an ellipse has 2 foci. Because the Sun is one of them => there will be two points along the orbit: one where the planet will be nearest to the Sun - this point is called "perihelion" (early January) and the other one where the planet will be furthest from the Sun which is called "aphelion" (early July). It can be seen from these laws that the orbit of a planet around the Sun is variable, with the orbital speed increasing when the planet is nearest the Sun (Perihelion) and reducing when furthest away from the Sun (Aphelion).

1233. Airplane ATPL The Sun's declination is:

The Sun as seen from Earth is called the true or apparent Sun and appears to cross the Earth from East to West daily. It is, in fact, the rotation of the Earth itself about its own axis which causes this apparent motion of the Sun, leading to the term "apparent" passage of the Sun. The stars and planets are also subject to the same phenomenon.

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For explanation refer to question #1185 on this page.

1270. Airplane ATPL CPL Heli ATPL CPL Assuming mid-latitudes (40 0 to 50 0 N/S). At which time of year is the relationship between the length of day and night, as well as the rate of change of declination of the Sun, changing at the greatest rate?

The mean sun is an imaginary Sun formulated to move eastward along the celestial equator at a rate providing a uniform measure of time equal

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A) the Sun's position relative to the plane of the Equator. B) the distance between the Sun and the horizon. e) the angular distance between the Sun and the celestial North Pole. D) the Sun's position relative to the ecliptic.

1186. Airplane ATPL CPL Heli ATPL CPL As seen from an observer on the surface of the Earth: A) the Sun is in a fixed position relative to the stars. B) the stars will seem to move from West to East during a year. e) the Sun's position relative to the stars is fixed throughout the year. D) the apparent Sun is always in the plane of the ecliptic.

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A) the situation when the distance between the Sun and the Earth is at its longest. B) the relative position between the Earth and the Moon. e) the situation when apparent Sun is passing the plane of the equator. D) the relationship between the length of the day and the length of the night.

A sidereal day is the time interval between two successive transits of a celestial body of the same meridian. The duration of a sidereal day is constant at 23 hrs 56 mins 4,1 seconds, corresponding to the time it takes for the Earth to complete one rotation relative to the stars. Astronomers use sidereal time as a way to keep track of the direction in which their telescopes need to be pointed to view any given star in the night sky. Just as the Sun and Moon appear to rise in the East and set in the West, so do the stars.

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A) Summer solstice and spring equinox. B) Spring equinox and autumn equinox. e) Summer solstice and winter solstice. D) Winter solstice and autumn equinox.

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Aviationexam Test Prep Edition 2012 (Refer to figure 061-EI8) The variation of the length of daylight/night at a given latitude is directly proportional to the variation of the declination of the Sun. Therefore, the rate of change of the length of daylight decreases if the rate of change of declination decreases and increases if the rate of change of declination increases reaching the maximum value when the rate of declination is at its maximum. This situation occurs at the equinoxes (about March the 21" and September the 23,d).

1275. Airplane ATPL CPL Heli ATPL CPL The Sun's declination is on a particular day 12°S. Midnight Sun may this day be observed: A) B) C) D)

North of 78°5. South of 78°5. At 78°5 only. North of 78°N.

(Refer to figure 061-EI3) As shown on the picture, when the Sun's declination is 12°S, it appears on the horizon (tangent to the surface of the Earth) at 78°S on the opposite meridian.Therefore, the Sun is visible all day long South of78°S.

2212. Airplane ATPL CPL Heli ATPL CPL In its path around the Sun, the axis of the Earth has an inclination: A) B) C) D)

varying between zero and 23°27' with the plane of the path. of 66°33' with the plane of the path. varying with the season of the year. of 23°27' with the plane of Equator.

(Refer to figure 061-EI8) The Earth's axis (which passes through the North and the South Pole) is tilted at an angle of 23.5° to line that is perpendicular to the orbital plane. Thus the angle between the Earth's axis and the orbital plane is 90° - 23.5° = 66.5°. Because of this tilt, as the Earth rotates around the Sun, the rays will be perpendicular to the surface in the region between the parallels 23.5° S (Tropic of Capricorn) and 23.5° N (Tropic of Cancer). As it can be seen in the attached figure, the Sun is over the Equator on the 21" of March; As the Earth rotates, the perpendicular rays are moving North so that on the 21" of June the Sun is overhead the 23.5° N parallel. Then moving South so that on the 23,d of September the Sun is again overhead the Equator and on the 21" of December will be overhead the 23.5° S parallel. If the Sun is overhead a certain parallel (the rays are perpendicular to the surface), then a viewer from the ground will see the Sun reaching an altitude of 90° above the horizon. The result is that the daylight period is extended in the far North during the summer months and drastically reduced in the winter months. Above the Latitudes 66.5° N and S the Sun does not rise above the horizon every day during the winter (polar night) and does not set below the horizon every day during the summer (polar day). The plane in which the Earth travels around the Sun is called "Plane

of the Ecliptic'~ The angle between the Plane of the Ecliptic and the Plane of the Equinoctial (plane of the Equator) is equal to the angle between the Earth axis and the perpendicular to the orbital plane (elementary geometry issue). The angle between the Plane of the Ecliptic and the Plane of the Equinoctial = 23.5".

2256. Airplane ATPL CPL Heli ATPL CPL In which months is the difference between apparent noon and mean noon the greatest? A) B) C) D)

November and February. January and July. March and September. June and December.

The Sun as seen from Earth is called the true or apparent sun and appears to cross the Earth from East to West daily. It is, in fact, the rotation of the Earth itself about its own axis which causes this apparent motion of the Sun, leading to the term "apparent" passage of the Sun. The stars and planets are also subject to the same phenomenon. The term "apparent solar time" is used to describe time based upon the Sun as it appears to an observer on Earth. The term "apparent solar day" is the time interval between two successive transits of the real Sun (or other heavenly body) over the same meridian. The apparent solar day starts when the Sun crosses (transits) an observer's ante-meridian - when the Sun

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is at its highest point (zenith) in the sky that day. It finishes with the second transit. Since the speed of the Earth along its elliptical path varies throughout the year, the apparent length ofa day also varies, as do hours, minutes and seconds based on this "day". In addition, as the Earth rotates on its axis in an anticlockwise direction and revolves round the Sun in an anti-clockwise direction, the Sun will appear to have moved in space. The Earth therefore must rotate for more than 360° to produce two successive transits. These problems produced the concept of "mean time" as a solution to the problem.

A mean day is an artificial unit of constant length, based on the average of all apparent solar days over a year. A mean sun is assumed to travel round the Earth in a circular path in the same plane as the Equator at a constant speed. The mean Sun therefore will cross a meridian and cross it again exactly 24 hours (of mean time) later. Because the length of the day is 24 hours and the Sun completes one circuit of the Earth in 24 hours, the rotational speed of the Sun can be expressed in arc of longitude as 15,,!hour W longitude every 4 minutes). This value is used frequently in time calculations and should be remembered. Time reckoned using the mean Sun is called "mean solar time" and is almost equal to the average apparent solar time. The difference in length between the apparent day (based upon the "true" sun) and the mean day (based upon the fictitious "mean" sun) is always less than one minute. The differences are cumulative, however, with the result that the imaginary mean Sun may be ahead or behind the apparent Sun by up to 15 minutes at certain times during the year. The maximum difference between apparent (real) time and Mean Time takes place in mid-November and is approximately 16 minutes. A second maximum difference takes place in mid-February and is approximately 14 minutes. In between these two, the difference is smaller. In November, the real Sun is overhead the observer's meridian at 11:44, while the mean Sun is overhead the meridian, by definition, at 12:00 LMT - i.e. 16 minutes later. In February, the real Sun is overhead the observer's meridian at 12:74, while the mean Sun is overhead the meridian, by definition, at 12:00 LMT - i.e. 14 minutes earlier.

2313. Airplane ATPL CPL Seasons are due to the:

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A) Earth's elliptical orbit around the Sun. B) inclination of the polar axis with the ecliptic plane. C) Earth's rotation on its polar axis. D) variable distance between Earth and Sun. (Refer to figures 061-£15 and 06J-f18) The main cause of the seasons is the inclination (tilt) of the Earth. The Earth's axis (which passes through the North and the South Pole) is tilted at an angle of 23.5° to a line that is perpendicular to the orbital plane. Thus the angle between the Earth's axis and the orbital plane is 90° - 23.5° = 66.5°. To see how the Sun changes its declination during the course ofa year, itis useful to examine its apparent orbit on the celestial sphere. This apparent path is called the "Ecliptic'~ The summer season in the Northern Hemisphere (which is the winter season in the Southern Hemisphere) starts when the Sun is directly overhead the equator (declination zero) at noon on the day of the Vernal Equinox, March 21. Thereafter the Sun is constantly increasing its declination until approximately June 21, the day of the Summer Solstice, when it reaches the northernmost pOint of its path. At this point the Sun is directly overhead (in zenith) at latitude N23°27'. After this, the Sun starts its journey South, crossing the Equator on about September 23, when the summer season in the Southern Hemisphere begins. On December 22 the Sun reaches the lowest point in its southerly path and an observer at 23°27' South will see the Sun directly overhead at local noon. After this date, it begins to move North again to complete the cycle, arriving back three months later overhead the Equator at the Vernal Equinox.

2317. Airplane ATPL CPL The planets move around the Sun: A) B) C) D)

in circular orbits. in elliptical orbits. at constant angular speed. at constant velocity.

For explanation refer to question #1196 on page 1.

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01 Basics of Navigation

4030. Airplane The mean Sun: A) B) C) D)

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is the middle position of the Sun. has a declination equal to the apparent Sun. moves with constant speed along the celestial equator. is only of interest to users of astronomical navigation.

For explanation refer to question #2256 on page 2.

4070. Airplane ATPL CPL Heli ATPL CPL In which two months of the year is the difference between the transit of the apparent Sun and mean Sun across the Greenwich Meridian the greatest?

B) 45° C) 66° D) 23.5" For explanation refer to question #2212 on page 2.

4432. Airplane ATPL CPL Heli ATPL CPL The time it takes for the Earth to complete one orbit around the Sun is: A) B) C) D)

360 days 45 hours 5 minutes 48 seconds. 360 days 5 hours 45 minutes 48 seconds. 365 days 45 hours 48 minutes 5 seconds. 365 days 5 hours 48 minutes 45 seconds.

For explanation refer to question #2256 on page 2.

The Earth completes one orbit of the Sun in a year, taking about 365 and 14 days. A calendar year, however, is exactly 365 days, except for leap years. Because the number of days of the earth's orbit is not whole, adjustments need to be made with an extra day added every four years (leap years) and other adjustments at the end of a period of 100 years so that regular time can be kept on Earth.

4089. Airplane ATPL CPL Heli ATPL CPL At what approximate date is the Earth closest to the Sun (perihelion)?

4534. Airplane ATPL CPL Heli ATPL CPL The angle between the plane of the Ecliptic and the plane of Equator is approximately:

A) B) C) D)

A) B) C) D)

March and September. February and November. June and December. April and August.

A) 27,5° B) 25,3° C) 23,so

End of June. End of March. Beginning of July. Beginning of January.

D) 66,so

For explanation refer to question #1796 on page 1.

For explanation refer to question #2212 on page 2.

4090. Airplane ATPL CPL Heli ATPL CPL At what approximate date is the Earth furthest from the Sun (aphelion)?

4549. Airplane ATPL CPL Heli ATPL CPL The angle between the plane of the Equator and the plane of the Ecliptic is:

A) B) C) D)

A) B) C) D)

Beginning of July. End of December. Beginning of January. End of September.

66,so 23,so 25,3° 65,6°

For explanation refer to question #1796 on page 1.

For explanation refer to question #2212 on page 2.

4219. Airplane ATPL CPL Heli ATPL CPL The direction of the Earth's rotation on its axis is such that:

59808. Airplane ATPL CPL Heli ATPL CPL At what times of the year does the length of the hours of daylight change most rapidly?

A) observed from the point above the North Pole, the rotation is counterclockwise. B) an observer on the surface of the Earth always will face West when observing sunrise. C) any point on the surface of the Earth will move eastward. D) any point on the surface of the Earth will move westward. The Earth is described as rotating from West to East. Ifyou were looking down on the North Pole from space, the Earth would be seen to be rotating anticlockwise. This is why the Sun appears to rise in the East and set in the West to an observer standing on the ground. The Earth then rotates around the Sun in an anti-clockwise direction on an elliptical path.

4326. Airplane ATPL CPL Heli ATPL CPL Which is the highest latitude listed below at which the Sun will rise above the horizon and set every day? A) 62° B) 68° C) 72° For explanation refer to question #2212 on page 2.

4402. Airplane ATPL CPL Heli ATPL CPL What is the highest latitude listed below at which the Sun will reach an altitude of 90° abovethe horizon at some time during the year?

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Spring Equinox and Autumn Equinox. Summer Solstice and Winter Solstice. Spring Equinox and Summer Solstice. Autumn Equinox and Winter Solstice.

For explanation refer to question #1270 on page 1.

230576. Airplane ATPL CPL Heli The declination of the sun is defined as:

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A) The arc from the celestial horizon to the sun measured along a vertical line perpendicular on the horizon. B) The arc along the celestial sphere from zenith to the sun. C) The arc of the meridian of the sun measured from the nearest pole to the sun. D) The angular distance of the sun north or south of the Celestial equator. 230577. Airplane ATPL Which statement is correct?

D) 66.5"

A) 0°

A) B) C) D)

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A) The Sun moves in an elliptical orbit around the Earth. B) The Earth is one of the planets which are all moving in elliptical orbit around the sun. C) The planets move around the sun like all stars of the Solar System. D) The Solar System consists of the Sun, planets and stars.

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230578. Airplane ATPL CPL The first law of Kepler states:

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A) The angular speed of the planet in the orbit around the sun is constant. B) Planets move around the sun in a circular orbit. e) Planets move in elliptic orbits with the sun in one of the foci. D) All planets orbit around the Sun at the same speed. 230581. Airplane ATPL CPL Heli ATPL CPL Which statement regarding the apparent Sun and the mean Sun is correct? A) The mean Sun moves along the ecliptic, the apparent Sun along the celestial equator. B) The apparent Sun is not important for navigation as difference in time with the mean Sun is maximal 4 seconds. e) The apparent Sun is the visible Sun, the mean Sun is a fictitious Sun. D) The apparent Sun is a fictitious Sun coupled to UTe, the mean Sun is related to the local mean time.

230582. Airplane ATPL CPL Heli ATPL CPL Which statement about the orbit of the Earth is correct? A) The orbit of the Earth around the Sun is an ellipse with the Sun at one of the foci. B) The orbit of the Earth around the Sun is an ellipse with the Sun at a point halfway between the two foci. e) The orbit of the Earth is a circle with the Sun at a point next to its centre. D) The orbit of the Earth around the Sun is a circle with the Sun at its centre.

230584. Airplane ATPL CPL Heli ATPL CPL In which statement is the "Mean Sun" best described? A) The mean Sun is a fictitious Sun coinciding each year with the apparent Sun at the Spring Equinox and travelling along the celestial equator at uniform speed. B) The mean Sun is a fictitious Sun coinciding each year with the apparent Sun at the Spring Equinox and travelling along the ecliptic at uniform speed. e) The mean Sun is a fictitious Sun the orbit of which coincides with that of the apparent Sun, but is corrected for mean astronomical and atmospheric refraction. D) The mean Sun is a fictitious Sun the orbit of which coincides with that of the apparent Sun, but is corrected for the mean difference in hour angle. 230585. Airplane ATPL CPL What is meant by "Aphelion"?

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A) The point of the Sun's orbit furthest away from the Earth. B) The point of the Sun's orbit closest to the Earth. C) The point of the Earth's orbit furthest away from the Sun.

D) The point of the Earth's orbit closest to the Sun. For explanation refer to question #1196 on page 1.

230646. Airplane ATPL CPL Heli ATPL CPL The length of the apparent solar day varies continuously throughout a year. This is caused by: A) the fact that the Earth is dosest to the Sun around

the 1" of July. B) ·the fact that the Earth is closest to the Sun around

the 1" of January. C) the tilt of the Earth's axis and the elliptical orbit of the Earth

230583. Airplane ATPL CPL Kepler's second law states that

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A) the radius vector Sun-Earth sweeps out equal areas in equal

time. B) the length of the radius vector Sun-Earth is directly proportional to the square root of its angular speed. , e) the radius vector Sun-Earth moves at constant angular speed. D) the area swept out by the radius vector Sun-Earth per hour increases with increasing length of the radius vector.

around the Sun. D) the equation of time. 230648. Airplane ATPL CPL Hell ATPL CPL An observer is situated on the parallel of 23,5°S. Which statement about the passage of the apparent Sun in relation to this position is correct? A) It passes through the zenith once a year around March 21st. B) It passes through the zenith once a year around December 22nd • e) It passes through the zenith twice a year around June 21" and December 22 nd • D) It passes through the zenith twice a year around March 21 st and September 23rd •

01-02 The earth 1112. Airplane ATPL CPL Heli ATPL CPL If you want to follow a constant true track value: A) you must fly East I West or North I South. B) you must fly a rhumb line. e) you must fly a great circle. D) you, in most cases, will also fly the shortest possible track. (Refer to figures 061-E79, 061-E80 and 061-E81) A greatcirc/e (orthodrome) is a circle on the surface ofa sphere with the same centre and radius as that of the sphere. This applies to the Earth. Great Circles are important in navigation because the shortest distance between two places on the surface of the Earth is the shorter arc of the great circle between those two points. Another way of expressing this is to say that the plane of a great

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circle connecting two places on a sphere (such as the Earth) passes through the centre of the sphere (or Earth). Any great circle on the Earth will therefore divide the Earth into equal parts (bisect it). There can be only one great circle connecting two places on the Earth, unless those two places are diametrically opposite each other (antipodean). Examples are the North and South Pole, 45°N 090 0 W and 45°S 090 0 E etc... In such cases, there is an unlimited number of great circles between the two points. Note: While a great circle track is the shortest distance, the direction of a great circle changes continuously because meridians are not parallel on the Earth. (A flight along the Equator or along a meridian itself is an exception). Check this statement by connecting two places on a globe with a length of string. Measure the angle at each meridian. Unless the two places are due North/South or East/West of each other, the direction between the two places will change

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01 Basics of Navigation from meridian to meridian. Ideally, all aircraft tracks would be great circles, as the distance would be shortest by definition.

C) on the southern hemisphere the meridians run towards

Small circle = any circle on the surface of the Earth which is not a great circle is by definition a small circle. The only small circles which are of navigational

D) the distan!=e, in nautical miles, between two selected merid-

the South Pole. ians will be constant.

interest are parallels of latitude. It is possible to establish an unlimited number ofsmall circles between any two points on Earth.

(Refer to figures 061-E79, 061-E80 and 061-E81) Meridians

A rhumb line (loxodrome) is a line of constant direction on the surface of the Earth or a chart. This is because it crosses each meridian at a constant angle. Meridians are the datums used for measuring true direction. The Equator and Meridians are both Rhumb and great circles by definition. Parallels of Latitude must also be Rhumb lines because they are lines of constant direction since they run East/West. Places on the same parallel of latitude are therefore due East or West of each other. in navigation, a rhumb line is a line crossing all meridians of longitude atthe same angle, i.e. a straight line path derived from a defined initial bearing. That is, upon taking an initial bearing, the pilot proceeds along the line defined

Meridians are straight lines connecting the Poles. A meridian is also a semigreat circle and a meridian and its allied anti-meridian complete a Great Circle. Meridians always indicate the North/South direction as they connect the Poles. True direction is always measured with reference to the local meridian (where the direction is to be measured) because this local meridian will indicate the direction of True North there. A meridian will change direction by 180 0 when it crosses a Pole (a North track or heading will become southerly at the North Pole and vice versa). Meridians cross the Equator at right angles = they are parallel to each other at the Equator (only at the Equator- as Latitude increases, they converge towards the poles).

by it, and continues this line without changing one's true (not magnetic) direction. On a plane surface Rhumb Line would be the shortest distance between two points over short distances, where the curvature of the Earth is not a major factor, and thus can be used for plotting a vehicle, aircraft or ship's course. Over longer distances great circle routes provides the shortest routes. However the inconvenience of having to continuously change bearings while traveling a great circle route makes rhumb line navigation appealing in certain instanc-

The Meridian passing through Greenwich (in London) is known as the Greenwich or Prime Meridian and is now the accepted universal reference datum when measuring Longitude. An aircraft traveling from Greenwich East (or west) for 180 0 0flongitude will arrive at the Greenwich anti-meridian, which is 180 0 from Greenwich (antipodean). Longitude will increase up to the antimeridian and will decrease when passing it. Parallels of latitude

es. Relationship between GC and RL tracks With a few exceptions (eg. Equator, Meridians) the GC tracks are shorter than the RL tracks. GC tracks are situated closer to the poles (higher Latitude)

A circle on the surface of the Earth parallel to the Equator is called a parallel

and RL tracks are situated closer to the Equator (lower Latitude). RL tracks have a constant value, while the GC track values change constantly.

The Equator is the Great Circle which has its plane lying at right angles (nor-

of latitude. Parallels of latitude run in an East/West direction. They are always small circles, with the exception of the Equator, itselfa great circle, which is also the reference' latitude datum (zero degrees north/south - 0 0 N/S).

mal) to the polar axis. Movement along the Equator is East or West as the plane Note: RL and GC trocks are identical if following the Equator due East/West ofthe equator lies in a true East/West direction and divides the Earth into hemi1_ _ _ _--J,lQr1:ff1cking11iDng_anY-mecidiJ:llHiueJ\JDItbfSQutbr~ _ _ _ _ _ _ _ _ _ _ _spheres.

1177. Airplane ATPL CPL Heli ATPL CPL Consider the following statements on longitude:

1200. Airplane ATPL CPL Heli ATPL CPL The poles on the surface ofthe Earth may be defined as:

A) Longitude is stated in degrees up to 360°. B) The value of longitude will never exceed 90°. C) The largest value of longitude is 180°. D) The largest value of change of longitude is 90°. (Refer to figures 061-E79, 061-E80 and 061-E81) Latitude, usually denoted by the Greek letter phi (ip) gives the location of a place on Earth North or South of the Equator. Lines of Latitude are the imaginary horizontal lines shown running east-to-west (or Westro east) on maps (particularly so in the Mercator projection) that run horizontally either North or South of the Equator - Parallels of Latitude. Technically, latitude is an angular measurement in degrees (marked with 0) ranging from 0 0 atthe Equator (low latitude) to 90 0 at the poles (90 0 N or +90 0 for the North Pole and 90 0 S or -90 0 for the South Pole). The latitude is approximately the angle between straight up at the surface (the zenith) and the Sun at an equinox. Longitude, usually identified by the Greek letter lambda (}.), is the geographic coordinate most commonly used in cartography and global navigation for East-West measurement. Constant longitude is represented by lines running from North to South - meridians. The line of longitude (meridian) that passes through the Royal Observatory, Greenwich in England, establishes the meaning of zero degrees of longitude, or the prime meridian. Longitude is expressed as degrees and minutes from the Greenwich Meridian East or West to the Greenwich anti-Meridian. The longitude of Greenwich is 000 0 (east or West- E/W) and the Greenwich Anti-Meridian is at longitude 180 0 E/W. Position on the Earth can be expressed in various ways. The latitude and longitude (always given in that order) of a position is the most common method in navigational practice. This uses a reference system of lines parallel to the Equator (Latitude) and North/South lines - Meridians (Longitude) which can define any point on the Earth, or on a sphere. "Lat and Long" may be written in several ways, e.g. : 25N 128~ 25 OON 128 OO~ N2S00 W12800. Latitude is always written first. When a position is expressed in terms of latitude and longitude, it will be seen that there is only one possible location for each such position.

1197. Airplane ATPL CPL Heli ATPL CPL Consider the following statements on meridians:

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1206 (C)

It is generally known that the Earth rotates around an axis joining the North (NP) and South Poles (SP) - the Polar axis. The North and South Poles provide the datum for directional measurement on a plain sphere and on the surface of the earth.

1204. Airplane ATPL CPL Heli ATPL CPL Consider the following statements on the properties of a great circle:

A) the great circle will maintain their initial true direction. B) the parallels of latitudes are all great circles. C) the great circle running through two positions on the surface ofthe Earth, is the shortest distance between these two posi, tions. D) all answers are correct. For explanation refer to question #1112 on page 4.

1206. Airplane ATPL CPL Heli ATPL CPL Consider the following statements on meridians:

A) any two meridians will form a great circle. B) the meridians are not of equal length. C) the meridians are parallel only at equator. D) any two halves great circle will form a meridians. For explanation refer to question #1197 on this page.

A) all meridians run in true direction from South to North. B) the relative direction between two selected meridians will be constant. 1177 (C)

A) the points where the Earth's axis of rotation cuts the surface ofthe Earth. B) the points on the surface of the Earth where all meridians intersect at right angles. C) the points from where the distance to the equator is equal. D) the point,S at which the vertical lines runs through the centre ofthe Earth.

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Aviationexam Test Prep Edition 2012

1209. Airplane The prime meridian is:

ATPL

Heli

CPL

ATPL

CPL

A) The meridian having the highest value of longitude. B) The meridian 1800 E / W. e) The mid meridian on a chart. D) The meridian running through Greenwich, England. For explanation refer to question #1197 on page 5.

1211. Airplane ATPL CPL Heli ATPL CPL Consider the following statements on the shape of the Earth: A) the diameters of the Earth is the same at all latitudes. B) the longest diameter is between the poles. e) it is slightly flattened at the poles. D) the diameter at the equator is about 60 NM longer than the diameter between the poles. The geographic term for the shape of the Earth is an ellipsoid (a three-dimensional ellipse) or an oblate spheroid (= a sphere which is slenderly flattened at the poles => the equatorial diameter is longer than the polar diameter) as the Earth is not an exact sphere, as it is actually slightly compressed or flattened at the poles. This Hsquashed Hshape is caused by the centrifugal effect of the rotation of the Earth. The navigational or geographic result of this HsquashingH means that the length of a minute of latitude varies very slightly with the position on the Earth. One minute of latitude is equal to one nautical mile, the basic unit of distance in aviation navigation calculations. This compression is usually ignored when making maps, as a Hreduced Earth Hor small scale model of the Earth is used, which treats the Earth as a perfect sphere for practical navigational purposes. Note: The compress/on (i.e. the flattening) of the Earth is about 0,3% (1/300 th) - in other words the polar diameter of the Earth is about 22 NM less than the equatorial diameter.

1212. Airplane ATPL CPL Heli Using latitude and longitude for a place:

ATPL

(Refer to figures 067-E79, Q67-E80 and 067-E87) Position on the Earth can be expressed in various ways. The latitude and longitude (always given in that order) of a position is the most common method in navigational practice. This uses a reference system of lines parallel to the Equator (Latitude) and North/South lines -"'Meridians (Longitude) which can define any point on the Earth, or on a sphere. HLat and LongH may be written in several ways, e.g.: 25N 728~ 25 OON 728 OO~ N2500 W72800. Latitude is always written first. When a position is expressed in terms of latitude and longitude, it will be seen that there is only one possible location for each such position.

1224. Airplane ATPL CPL Heli ATPL CPL Consider the following statements on the great circle and the rhumb line running through the same two positions: A) the rhumb line will in most cases be located closer to the equator than the great circle. B) the great circle will in most cases be shorter of the two. e) the great circle will in most cases run through an area of higher latitude than the rhumb line. D) all statements are correct. For explanation refer to question #1172 on page 4. CPL

Heli

ATPL

CPL

A) the angular difference between the rhumb line and the great circle between two positions, measured at any of the two positions. B) the difference between the rhumb line and the great circle

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(C)

D) the angle at which speech from another person enters the ear. (Refer to figures 067 -E82 and 067 -E83) H Convergency or Hthe angle of convergence = the angle between meridians. Convergency is in fact the angle of inclination between the meridians passing through two points at a given latitude. This angle is zero at the Equator, where meridians are parallel to each other. As meridians leave the Equator and approach the Poles, they converge and the value of convergency reaches a maximum of 7° (between successive meridians) at the Poles, where the meridians are spaced 7° apart. The angle of convergency also represents the difference between the initial and the final Great Circle track (difference between the Great Circle track at the starting and ending points of the specific track between two points). The angle of convergency between two positions (i.e. two meridians) is calculated from the formula: • Convergency = Change in Longitude x sin of Latitude • or: Convergency = GC Initial True Track - GC Final True Track Change in longitude (Ch Long or 0 Long) is the difference in longitude (expressed in 0) between two positions or two meridians. If the positions are at different latitudes, the average value of latitude should be used, which will give an approximate value. Conversion angle is the angle between the Great Circle and Rhumb Line tracks between two places. The Rhumb Line and Great Circle tracks between two points will always be different (except for positions on meridians or along the Equator). The Great Circle track will change direction continuously and the Rhumb Line track will always be a constant direction by definition. It can be seen by drawing that the Great Circle track will always lie closer to the Pole and the Rhumb Line closer to the Equator. Value of Conversion Angle (ca) = l-2 convergency (C). As conversion angleis one halfof convergency, the formula can be written as: • Conversion angle = l-2 Ch Long x sin Latitude (or sin mean Lat). • or: Conversion angle = l-2 (Initial Great Circle - Final Great Circle track).

CPL

A) the distance from this place to another place may be easily calculated. B) the location on the earth's surface of this place is defined. e) the direction from the place to any other place may be easily calculated. D) relative directions to another place may easily be calculated.

1236. Airplane ATPL What is "conversion angle":

directions. e) the angle used to convert from true to compass directions.

1238. Airplane ATPL CPL Heli ATPL CPL In order to fly from position A (10 0 00'N 030 0 00'W) to position B (30 0 00'N 050 0 00'W), maintaining a constant true course, it is necessary to fly: A) the great-circle route. B) the constant average drift route. e) a rhumb line track. D) a straight line plotted on a Lambert chart. For explanation refer to question #7772 on page 4.

1246. Airplane ATPL CPL Heli ATPL CPL The shortest distance between 2 points on the surface ofthe Earth is: A) a great circle. B) the arc of a small circle. e) half the rhumb line distance. D) rhumb line. For explanation refer to question #7772 on page 4.

1248. Airplane ATPL CPL A correct definition of longitude is:

Heli

ATPL

CPL

A) The east-west distance between Greenwich and the place. B) The arc at equator between the Greenwich meridian and the meridian of the place, measured in degrees, minutes and seconds, named East or West. e) The angle between the Greenwich meridian and the meridian of the place. D) The difference between the Greenwich meridian and the meridian of the place, measured at the centre of the Earth. For explanation refer to question #7777 on page 5.

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01 Basics of Navigation

1251. Airplane ATPL CPL Latitude may be defined as:

Heli

ATPL

B) on a North - South track.

CPL

C) on a track which is constantly changing direction.

A) The distance from equator to a place on the surface of the Earth. B) The angle between the plane of the equator and the plane of the parallel of latitude. C) The angular distance measured along a meridian from the eqUator to a parallel of the latitude, measured in degrees, minutes, and seconds and named North or South. D) The displacement of a place from equator.

D) a rhumb line track. For explanation refer to question #1112 on page 4.

2214. Airplane ATPL CPL Heli ATPL CPL The great circle track X => V, measured at "X" is 319° and measured at "V" is 325°. Consider the following statements: A) B) C) D)

For explanation refer to question #1177 on page 5.

2177. Airplane ATPL CPL Any parallel of latitude is a: A) B) C) D)

Heli

ATPL

CPL

great circle. rhumb line. small circle. meridian of tangency.

For explanation refer to question #1112 on page 4.

2178. Airplane ATPL CPL Heli ATPL CPL An aircraft is following the 45°N parallel of latitude. The track followed is a:

For explanation refer to question #1112 on page 4.

A) B) C) D)

ATPL

CPL

Rhumb line. Orthodromic line. Equator. The rhumb line or great circle depending on the chart used.

For explanation refer to question #1112 on page 4.

2206. Airplane ATPL CPL The equator is located:

Heli

ATPL

(Refer to figures 061-E82 and 061-E83) A Great Circle (GC) is a circle on the surface of a sphere with the same centre and radius as that of the sphere. This applies to the Earth. Great Circles are important in navigation because the shortest distance between two places on the surface of the Earth is the shorter arc of the great circle between those two points. Another way of expressing this is to say that the plane of a great circle connecting two places on a sphere (such as the Earth) passes through the centre of the sphere (or Earth). Any great circle on the Earth will therefore .divide the Earth into equal parts (bisect it). There can be only one great circle connecting two places on the Earth unless those two places are diametrically opposite each other (antipodean). Examples are the North and South Pole, 45°N 090 0 W and 45°S 090 0 E etc. .. In such cases, there is an unlimited number of great circles between the two points. While a GC track is the shortest distance, the direction of a GC changes continuously because meridians are not parallel on the Earth (a flight along the Equator or along a meridian itself is an exception). Check this statement by connecting two places on a globe with a length ofstring. Measure the angle at each meridian. Unless the two places are due North/South or East/West of each other, the direction between the two places will change from meridian to meridian. Ideally, all aircraft tracks would be great circles, as the distance would be shortest by definition.

A) constant-heading track. B) rhumb line. C) great circle. D) constant-drift track.

2201. Airplane ATPL CPL Heli Generally what line lies closer to the pole?

southern hemisphere, rhumb line track is 322°. northern hemisphere, rhumb line track is 313°. southern hemisphere, rhumb line track is 331°. northern hemisphere, rhumb line track is 322°.

CPL

The difference between the Initial GC and the RL track is represented by the value of the Conversion Angle: • Conversion angle = ~ Ch Long x sin Latitude (or sin mean Lat). • or: Conversion angle = ~ (Initial Great Circle - Final Great CirCle track). The difference between the Initial GC track and the Final GC track is represented by the value of Convergency: • Convergency = Change in Longitude x sin of Latitude • or: Convergency = GC Initial True Track - GC Final True Track The GC tracks lie closer to the poles than the Rhumb Line (RL) tracks. There are some attributes of GC tracks:

A) on the surface of the Earth, being a circle whose plane is perpendicular to the axis of the Earth and cutting through the centre of the Earth. B) on the surface of the Earth and at right angles to the axis of rotation. C) on the surface as a small circle, horizontal to the axis of rotation. D) on the surface parallel to the magnetic equator. For explanation refer to question #1197 on page 5.

2207. Airplane ATPL CPL Heli ATPL CPL In which occasions does the rhumb line track and the great circle track coincide on the surface of the Earth? A) On East - West tracks in polar areas. B) On high latitude tracks directly East - West. C) On East - West tracks in the northern hemisphere North of the magnetic equator. D) On tracks directly North - South and on East - West tracks along the equator.

Northern Hemisphere: • From the initial to the final point of the GC track, the track value increases in easterly direction and decreases in westerly direction Southern Hemisphere: • From the initial to the final point of the GC track, the track value decreases in easterly direction and increases in westerly direction Now finally let's take a look at this specific question. Since the GC track increases and we are flying in the westerly direction we can say that we are flying in the Southern hemisphere. The difference between the Initial and Final GC tracks equals to the value of convergence. In our case it is 6° (325° - 319°). The difference between the initial or final GC track and the Rhumb Line track equals to the value of conversion angle, which in turn equals to ~ of convergence.ln this case the conversion angle = 3°. The Rhumb line track will be equal to 322° (319° + 3° or 325° - 3°). Another way offinding the RL track, without any calculations of conversion angle, is to find the mean GC track ((319° + 325°) + 2) => 322° => the RL track will be equal to the average GC track => 322°.

2218. Airplane ATPL CPL Heli ATPL CPL How many small circles can be drawn between any two points on a sphere? A) B) C) D)

For explanation refer to question #1112 on page 4.

2209. Airplane ATPL CPL Heli ATPL CPL If you are flying along a parallel of latitude, you are flying:

One. None. An unlimited number. Two.

For explanation refer to question #1112 on page 4.

A) a great circle track. 1251 (C)

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Aviationexam Test Prep Edition 2012

2265. Airplane ATPL CPL Heli ATPL CPL A great circle track joins position A (59°S 141°W) and B (61°S 148°W). How does the true track change when flying from position A to position B?

(Refer to figures 061-E82 and 061-E83) Convergency between two points is the inclination angle between the meridians of the two points measured at their mean latitude. Convergency is equal to the difference between the initial great circle track from one point to another and the final great circle track (both tracks are measured at the meridians of the two points). Convergency is calculated by mUltiplying the change in longitude with the sine of mean latitude of the two points (Convergency Change in Longitude x sin of mean Latitude). Change in longitude (Ch Long or 0 Long) is the difference in longitude (expressed in degrees) between two positions or two meridians. If the positions are at different latitudes, the average value of latitude should be used, which will give an approximate value.

=

In the case of this question, follow a simple calculation: • Convergency = Ch Long x sin mean Lat ·CHLong=r(148°W-141°W} • Mean Lat = 60° ((61° + 59°}.;. 2) • Convergency = r x sin 60° • Convergency = 6° The difference between the initial and final Great Circle track will therefore be 6°. The difference between the initial GC track and the Rhumb Line (RL) track will be of convergency = 3° (= conversion angle). The same applies for the final GC track and the RL track => difference will be 3': Since the case is situated in the S hemisphere, the GC track will be closer to the South Pole => it will gradually increase from the initial to the final waypoint.

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2270. Airplane ATPL CPL Heli ATPL CPL How does the convergence of any two meridians on the Earth change with varying latitude? It changes as cosine of latitude. It changes as sine. of latitude. It increases with decrease of latitude. It is of constant value and does not change with latitude.

2280. Airplane ATPL CPL The convergence of meridians:

Heli

ATPL

CPL

A) is the distance between the meridians in degrees, minutes, and seconds. B) is the angular difference between the meridians. e) is independent of latitude and longitude. D) is greater using rhumb line track than using greater circle. For explanation refer to question #1236 on page 6.

2299. Airplane ATPL CPL The compression factor of the Earth:

Heli

ATPL

For explanation refer to question #1211 on page 6.

Heli

CPL

ATPL

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A) line of variation. B) great circle. e) rhumb line. D) agonic line. For explanation refer to question #7112 on page 4.

4071. Airplane ATPL CPL Heli ATPL CPL When flying on a westerly great circle track in the southern hemisphere you will: A) fly a spiral and finally end up at the South Pole. B) experience an increase in the value of true track. e) always have the rhumb line track between the departure point and the destination to the left of your great circle track. D) experience a decrease in the value of true track. (Refer to figures 061-E82 and 061-E83) Sometimes this particular question may confuse the student. It helps to make a quick sketch which includes two points in the southern hemisphere. We know that we are flying to the West in the southern hemisphere, therefore the shape of our line or path between the two points is going to appear convex to the South pole (concave to the Equator) => the Great Circle track will be situated closer to the South Pole, while the Rhumb Line track will be situated closer to the Equator. For example, if we start out with an initial great circle heading of 230° then the subsequent heading values can only increase due to two factors: the direction of our flight (westerly) and the hemisphere we are flying in.

A) varies by 100 • B) decreases by 60 • e) varies by 40 • D) increases by 50. (Refer to figures 061-E82 and 061-E83) The difference between the Initial or Final GC and the RL track is represented by the value of the Conversion Angle: • Conversion angle = Ch Long x sin Latitude (or sin mean Lat). • or: Conversion angle = (Initial Great Circle - Final Great Circle track).

*

The difference between the Initial GC track and the Final GC track is represented by the value of Convergency: • Convergency = Change in Longitude x sin of Latitude • or: Convergency = GC Initial True Track - GC Final True Track The GC tracks lie closer to the poles than the Rhumb Line (RL) tracks. There are some attributes of GC tracks: Northern Hemisphere: • From the initial to the final point of the GC track, the track value increases in easterly direction and decreases in westerly direction Southern Hemisphere: From the initial to the final pOint of the GC track, the track value decreases in easterly direction and increases in westerly direction

CPL

A) Will always cross equator. B) Has a plane parallel to the Earth's axis of rotation. e) Will also be a rhumb line. D) Has a plane that do not pass through the centre of the Earth.

I

4046. Airplane ATPL CPL Heli ATPL CPL A line which cuts all meridians at the same angle is called a:

*

CPL

A) is so small that it may be ignored when making ordinary maps and charts. B) is about 1:300. e) makes the difference between the polar diameter and the equatorial diameter about 22 NM. D) all answers are correct.

ATPL

For explanation refer to question #1112 on page 4.

4122. Airplane ATPL CPL Heli ATPL CPL An aeroplane flies from A (59°S 142°W) to B (61°S 148°W) with a TAS of 480 kts. The autopilot is engaged and coupled with an Inertial Navigation System in which AB track is active. On route AB, the true track:

For explanatiOn refer to question #1236 on page 6.

4026. Airplane A small circle:

4045. Airplane ATPL CPL Heli ATPL CPL Parallels of latitude, except the equator are: A) both rhumb lines and great circles. B) great circles. e) rhumb lines. D) are neither rhumb lines nor great circles.

A) It increases by 6°. B) It decreases by 60 • C) It increases by 30 • D) It decreases by 30 •

A) B) e) D)

For explanation refer to question #1112 on page 4.

I

Now finally let's take a look at this specific question. Since the point B is located to the West of point A, the track will be in the westward direction. Since both points are in the Southern Hemisphere, the GC track would be arcing down below the RL track, towards the South Pole => the initial GC track at A will be

4026 (0)

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01 Basics of Navigation equal to RL track - the conversion angle => initial GCtrack wifl be lower than RL track and wifl gradually increase until the mid-point between A and B, where the GC track wifl parallel the RL track momentarily (GC track = RL track). From this point on the GC track wifl keep on increasing above the value of RL track up to a maximum value upon arrival at point B (same difference between GC and RL tracks as at point A upon departure). The overall increase in GC track value from departure (point A) up to the arrival (point B) wifl be equal to the value of convergency, that can be easily calculated: • Change in longitude = 6° (148°W - 742°W) • Mean Latitude = 60° ((59° + 67°) .,. 2) • Convergency = Ch Long x sin of Mean Lat • Convergency = 6° x sin 60° • Convergency = 5,79°

4176. Airplane ATPL CPL Heli ATPL CPL An approximate equation for calculation conversion angle is:

For explanation refer to question #7236 on page 6.

4220. Airplane ATPL CPL The Earth may be referred to as:

A) B) C) D)

ATPL

Heli

CPL

ATPL

CPL

are rhumb lines. cut all meridians at the same angle. are great circles. are lines of fixed direction.

4254. Airplane The Earth is:

4128. Airplane ATPL CPL Heli ATPL CPL An approximate equation for·calculating the convergence between two meridians is:

ATPL

CPL

4255. Airplane ATPL CPL Heli ATPL CPL Consider the following statements on rhumb lines:

(Refer to figures 067-E82 and 067 -E83) Do not think of Rhumb Lines only in terms oflines ofparallel. Imagine that you start a flight at the North pole and follow a Rhumb Line of 045° => it means that you wifl have to follow a track that takes you at a 45° angle over each of the meridians that you cross => you wifl end-up following a spiral that wifl eventually lead you towards the South pole. Another way of solving this question is by elimination:

For explanation refer to question #7236 on page 6.

4133. Airplane ATPL CPL Heli ATPL CPL The exact equation for calculating the convergence between two meridians running through two difference positions is: Note: GCTTin = Great Circle True Track Initial GCTTfin = Great Circle True Track Final

Answer 8) is incorrect, because RL tracks can of course cross GC tracks. Answer C) is incorrect, because at the mid-point between the departure and destination the GCtrack has the same value as the RL track - they are parallel for a brief moment. Answer D) is incorrect, because that statement is true only for easterly directions.

4263. Airplane ATPL CPL Heli ATPL CPL A great circle on the Earth running from the North Pole to the South Pole is called:

Convergence = 1/2 x (dlong x sin lat) Convergence = 1/2 x (GCnin + GCnfin) Convergence = GCnin + GCnfin Convergence = GCnin - GCnfin

A) B) C) D)

For explanation refer to question #7236 on page 6.

A) B) C) D)

Heli

CPL

A) most rhumb lines will run as,spirals from the one pole to another. B) a rhumb line will never cross a great circle. C) a rhumb line and a great circle will never have the same true direction for some distance. D) the true direction of a rhumb line on northern hemisphere will increase in true direction, while on southern hemisphere it will decrease.

Convergence = dlat x sin mean latitude Convergence = dlat x cos mean latitude Convergence = dlong x cos mean latitude Convergence = dlong x sin mean latitude

4151. Airplane ATPL CPL Heli The highest value of longitude is found:

ATPL

For explanation refer to question #7277 on page 6.

Convergence =60 x dlong x cos lat Convergence = dlat x sin mean long' Convergence = dlong x sin mean lat Convergence = dlong x cos lat

4132. Airplane ATPL CPL Heli ATPL CPL What is the standard formula for convergence?

A) B) C) D)

CPL

A) a sphere which has a larger polar circumference than equatorial circumference. B) a sphere whose centre is equidistant (the same distance) from the Poles and the Equator. C) considered to be a perfect sphere as far as basic (simple) navigation is concerned. D) none of the above statements is correct.

For explanation refer to question #7236 on page 6.

A) B) C) D)

ATPL

For explanation refer to question #7277 on page 6.

Since radio waves travel in straight lines and follow the shortest distance between the transmitter and receiver, they wifl travel along great circle tracks directly from the radio station to the aircraft and vice versa. All radio bearings are therefore great circle bearings. The direct tracks drawn between radio stations/beacons on Radio-Navigation charts (such as Jeppesen, Aerad) are usually great circle tracks because they are defined by radio bearings, which are used for navigational guidance.

A) B) C) D)

Heli

A) round. B) an oblate spheroid. C) a globe. D) elliptical.

The difference between the INITIAL GC track and the FINAL GC track wifl be 5, 79° (increase).

4123. Airplane Radio bearings:

CA =0,5 x dlong xsin (mean lat) CA = dlong x sin (mean lat) x sin (long) CA = (dlong-dlat) x 0,5 CA =0,5 x dlat x sin (mean lat)

A) B) C) D)

ATPL

CPL

a longitude. a parallel of latitude. a difference of longitude. a meridian.

For explanation refer to question #7797 on page 5.

along equator. close to the poles. close to the prime meridian. at Greenwich anti meridian.

For explanation refer to question #7777 on page 5.

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4265. Airplane A rhumb line is:

ATPL

Heli

CPL

ATPL

CPL

A) the shortest distance between two points on a Polyconic projection. B) a line on the surface of the Earth cutting all meridians at the same angle. C) any straight line on a Lambert projection. D) a line convex to the nearest pole on a Mercator projection. For explanation refer to question #7172 on page 4.

A) the shape of the ecliptic.

B) a great circle on the celestial sphere. C) the shape of the Earth. D) the movement of the Earth around the Sun.

4286. Airplane ATPL CPL Heli ATPL CPL You are flying from A (SOON 100W) to B (S8°N 02°E). What is the convergence between A and B?

For explanation refer to question #1211 on page 6.

Airplane

ATPL

Heli

CPL

ATPL

CPL

A great circle is defined as: A) a circle in any plane on the surface of a sphere. B) a circle on the surface of a sphere, whose plane is cutting

through the centre of the sphere.

perpendicular to the axis of rotation. For explanation refer to question #1112 on page 4.

4275. Airplane ATPL CPL Heli ATPL CPL The following waypoints are ~ntered into an inertial navigation system (INS): 60N 30W 60N20W 60N lOW

The inertial navigation is connected to the automatic pilot on the route WPl - WP2 - WP3. The track change on passing WPT2: A) B) C) D)

a 9 deg increase. a 4 deg decrease. zero. a 9 deg decrease.

(Refer to figure 061-E23) The angle between the Great Circle and Rhumb Line tracks between two places is called the Conversion angle (ca). The Rhumb Line and Great Circle tracks between two points will always be different (except for positions on meridians or along the Equator). The Great Circle track will change direction continuously and the Rhumb Line track will always be a constant direction by definition. • Conversion angle (ca) = Y.l Ch Long x sin Latitude • Conversion angle = Y.l (10° x sin 60°) • Conversion angle = 4,33° Note: The "ca" will be the same at all waypoints since all have the same Lat and there is the same Ch Long (10°) between them. • Great Circle Track arriving at WPT2 = 94,33° • Great Circle Track departing from WPT2 =85,6r • Great Circle Track change at WPT2 = 8,66° decrease (94,33° - 85,6r)

4283. Airplane ATPL CPL Heli ATPL CPL The angle between the true great circle track and the true rhumb line track joining the following points: A (60°5 l6S0W) and B (60°5 l77°E), at the place of departure A, is: A) 7,8° B) 9° C) 15,6° D) 5,2°

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4265 (8)

---------

I

4271 (C)

(Refer to figures 061-E82 and 061-E83) Convergency or "the angle of convergence" = the angle· between meridians. Convergency is in fact the angle of inclination between the meridians passing through two points at a given latitude. This angle is zero at the Equator, where meridians are parallel to each other. As meridians leave the Equator and approach the Poles, they converge and the value of convergency reaches a maximum of 1° (between successive meridians) at the Poles, where the meridians are spaced 1° apart. The angle of convergency also represents the difference between the initial and the final Great Circle track (difference between the Great Circle track at the starting and ending points of the specific track between two points). The angle of convergency between two positions (i.e. two meridians) is calculated from the formula: • Convergency = Change in Longitude x sin ofLatitude • or: Convergency = GC Initial True Track - GC Final True Track Change in longitude (Ch Long or 0 Long) is the difference in longitude (expressed in 0) between two positions or two meridians. If the positions are at different latitudes, the average value of latitude should be used, which will give an approximate value. In the ca~e of this question, the change in Longitude is 12" (2" from 2°E to the Greenwich and then further 10° to 100W). The mean Latitude is 54° (between SOON and 58°N). Now we can use the formula: • Convergency = Change in Longitude x sin of Latitude • Convergency = 12° x sin 54° • Convergency = 9,r

4291. Airplane ATPL CPL Heli ATPL CPL Calculate the 0 lat from 001°lS'N 090000'E to 090°00'5 090000'W: A) 91°15'N B) 88°45'N C) 91°15'5 D) 268°15'N (Refer to figures 061~E79, 061-E80 and 061-E81) Change of latitude (Ch Lat or D Lat) The change of latitude between two places is defined as the smaller arc of meridian between the two latitudes. It is therefore the difference in latitude between the two. It is designated North or South according to the direction of change specified e.g. the Ch Lat between 2rS and 3rS is 10° South. From WS OWE to WN 025°10/, the change of latitude (Ch Lat) will be 30° North. Calculating change of latitude requires the subtraction of one latitude from another, or, if the latitudes are in different hemispheres, the addition of the latitudes. Change of longitude (Ch Long or D Long) Change oflongitude between two places is the smaller arc ofthe Equator measured between the meridians of those places. It is designated East or West ac-

(Refer to figures 061-E82 and 061-E83)

I

A) 6,5" B) 9,7" C) 10,2° D) 6,8°

C) a circle running on the outside ofthe sphere. D) a circle on the surface of the sphere, with its plane running

WPT 1: WPT2: WPT3:

• Conversion angle = Y.l Ch Long x sin Latitude (or sin mean Lat). • or: Conversion angle = Y.l (Initial Great Circle - Final Great Circle track). In the case of this question we have to first calculate the change in Longitude between points A and B. The shortest route will be via the 180° EIW (Greenwich anti-meridian) => 15° from 165°W to the 180° EIW meridian and then further 3° to 17rE = total of 18°. Now we can use the formula: • Conversion angle = Y.l Ch Long x sin Latitude • Conversion angle = Y.l18°x sin 60° • Conversion angle = 7,8°

4271. Airplane ATPL CPL Heli ATPL CPL The term "ellipsoid" may be used to describe:

4273.

Conversion angle is the angle between the Great Circle and Rhumb Line tracks between two places. The Rhumb Line and Great Circle tracks between two points will always be different (except for positions on meridians or along the Equator). The Great Circle track will change direction continuously and the Rhumb Line track will always be a constant direction by definition. It can be seen by drawing that the Great Circle track will always lie closer to the Pole and the Rhumb Line closer to the Equator. Value of Conversion Angie (ca) = Y.l convergency (C). As conversion angle is one half of convergency, the formula can be written as:

I

4273 (8)

I

4275 (0)

I

4283 (A)

I

4286 (8)

I

4291 (C)

I

01 Basics of Navigation cording to the direction of change specified. Change of Longitude (Ch Long) corresponds to change of latitude, but in terms of longitude. For example the Ch Long between 29°E and 31°E is 2°E. Thus from 15°5 OWE to ISW 02S°W, the change of longitude (Ch Long) will be 400W (75° from 01S oE to the Greenwich meridian and then 25° from Greenwich to 02S 0W). There is also a change of latitude, but this does not affect the Ch Long. In the case of this question: • from 0I°1S'N to the Equator =01°15' • from the Equator to 90°00'5 = 90°00' .07°15' + 90°00' = 91°15' • direction is from the North to the South, so the Ch.Lat will be 91°15' South.

4298. Airplane ATPL CPL Heli ATPL CPL What is the initial great circle direction from 45°N 14°12'W to 45°N 12°48'E? A) B) C) D)

86,5° (T). 80,4° (T). 090° (M). 270 0 (M).

(Refer to figures 061-E82 and 061-E83) The difference between the Initial or Final GC and the RL track is represented by the value of the Conversion Angle: • Conversion angle = % Ch Long x sin Latitude (or sin mean Lat). • or: Conversion angle = M! (Initial Great Circle - Final Great Circle track). The difference between the Initial GC track and the Final GC track is represented by the value of Convergency: • Convergency = Change in Longitude x sin of Latitude • or: Convergency = GC Initial True Track - GC Final True Track The GC tracks lie closer to the poles than the Rhumb Line (RL) tracks. There are some attributes of GC tracks: Northern Hemisphere: • From the initia/to the final point of the GCtrack, the track value increases in easterly direction and decreases in westerly direction Southern Hemisphere: • From the initial to the final point of the GC track, the track value decreases in easterly direction and increases in westerly direction

Now finally let's take a look at this specific question. Since the Latitude of both points A and B is the same (4S 0N) and point B is located due East of point A, the Rhumb Line (RL) track would be 090~ Since both points are in the Northern Hemisphere, the GC track would be arcing up above the RL track, towards the North Pole => the initial GC track at A will be equal to RL track - the conversion angle (initial GC track will be smaller than the RL track). The change in Longitude is 2r (74°12' to the Greenwich + further 12°48' to the east). Conversion angle can be calculated using the formula: • Conversion angle = % Ch Long x sin Latitude • Conversion angle = M! 2r x sin 45° • Conversion angle = 9,54° With a RL track of 090° and the conversion angle of 9,5° the initial GC track will be 80,5° (090° - 9,5°).

4306. Airplane ATPL CPL Heli ATPL CPL What is difference in latitude from 30°39'5 20 0 20'E to 45°23'N 400 40'E: A) W44'N B) 76°2'5 C) 76°2'N D) W44'S (Refer to figures 061-E79, 061-E80 and 061-E81) Change of latitude (Ch Lat or D Lat) The change of latitude between two places is defined as the smaller arc of meridian between the two latitudes. It is therefore the difference in latitude between the two. It is designated North or South according to the direction of change specified e.g. the Ch Lat between 2rS and 3rS is 10° South. From ISoS 01S oE to IsoN 02S°W, the change of latitude (Ch Lat) will be 30° North. Calculating change of latitude requires the subtraction of one latitude from another, or, if the latitudes are in different hemispheres, the addition of the latitudes. Change of longitude (Ch Long or D Long) Change of longitude between two places is the smaller arc of the Equator measured between the meridians of those places. It is designated East or West ac-

I

4298 (8)

I

4306 (C)

I

4307 (C)

I

4321 (8)

I

cording to the direction of change specified. Change of Longitude (Ch Long) corresponds to change of latitude, but in terms of longitude. For example the Ch Long between 29°E and 31°E is 2°E. Thus from 15°5 01s oE to IsoN 02S°W, the change of longitude (Ch Long) will be 400W (75° from 01S OE to the Greenwich meridian and then 25° from Greenwich to 02S0W). There is also a change of latitude, but this does not affect the Ch Long. In the case of this question: • from 30 039'S to the Equator =30°39' • from the Equator to 4S023'N =45°23' .30°39' + 45°23' = 76°02' • direction is from the South to the North, so the Ch.Lat will be 76°02' North.

4307. Airplane ATPL CPL Heli ATPL CPL Calculate the 0 long from 001°15'N 090 0 00'E to 001°15'N 015°15'E: A) 74°45'E B) 74°15'E C) 74°45'W D) 105°15'N (Refer to figures 061-E79, 061-E80 and 061-E81) Change of latitude (Ch Lat or D Lat) The change of latitude between two places is defined as the smaller arc of meridian between the two latitudes. It is therefore the difference in latitude between the two. It is designated North or South according to the direction of change specified e.g. the Ch Lat between 2rS and 3rS is 10° South. From WS OWE to IS oN 02S°W, the change of latitude (Ch Lat) will be 30° North. Calculating change of latitude requires the subtraction of one latitude from another, or, if the latitudes are in different hemispheres, the addition of the latitudes. Change of longitude (Ch Long or D Long) Change oflongitude between two places is the smaller arc of the Equator measured between the meridians of those places. It is designated East or West according to the direction of change specified. Change of Longitude (Ch Long) corresponds to change of latitude, but in terms of longitude. For example the Ch Long between 29°E and 31°E is rE. Thus from ISoS 01S 0Eto ISON 02S°W, the change of longitude (Ch Long) will be 400W (75° from 01S OE to the Greenwich meridian and then 25° from Greenwich to 02S 0W). There is also a change of latitude, but this does not affect the Ch Long. In the case of this question, simply deduct 01s o1S'E (75,25°) from 090 0E=> 090°

- 15,25° =74,75° =74°45'. Since we are flying towards the Greenwich meridian (easterly longitude decreases) we are flying westbound, therefore the Ch Long will be "W" => 74°4S'W.

4321. Airplane ATPL CPL Heli ATPL CPL What is the change of longitude between A (45°00'N 163°14'E) and B (31°33'N 157°02'E)? A) 6°12'E B) 6°12'W C) 13°27'W D) 320016'E (Refer to figures 061-E79, 061-E80and 061-E87) Change of latitude (Ch Lat or D Lat) The change of latitude between two places is defined as the smaller arc of meridian between the two latitudes. It is therefore the difference in latitude between the two. It is designated North or South according to the direction of change specified e.g. the Ch Lat between 2rs and 3rS is 70° South. From ISoS OWE to IS oN 02S°W, the change of latitude (Ch Lat) will be 30° North. Calculating change of latitude requires the subtraction of one latitude from another, or, if the latitudes are in different hemispheres, the addition of the latitudes. Change of longitude (Ch Long or D Long) Change of/ongitude between two places is the smaller arc of the Equator measured between the meridians of those places. It is designated East or West according to the direction of change specified. Change of Longitude (Ch Long) corresponds to change of latitude, but in terms of longitude. For example the Ch Long between 29°E and 31°E is 2°E. Thus from ISoS 01S 0Eto ISoN 02S°W, the change of longitude (Ch Long) will be 400W (75° from 01S oE to the Greenwich meridian and then 25° from Greenwich to 02S 0W). There is also a change of latitude, but this does not affect the Ch Long.

In the case of this question: • 163° 14'E = 163,23°E

..

Aviationexam Test Prep Edition 2012 • 15r02'E = 157,03°E .163,23°- 157,03°= 6,2°= 6°12' • Direction is towards the Greenwich meridian = Westerly. Ch Long is therefore 6°12' West.

4331. Airplane ATPL CPL Heli ATPL CPL The automatic flight control system is coupled to the guidance outputs from an inertial navigation system. Which pair of latitudes will give the greatest difference between initial track read-out and the average true course given, in each case, a difference of longitude of 10°? A) 30°5 to 25°5. B) 60 0 N to SooN. C) 30°5 to 30 0 N. 0) 60 0 N to 60 0 N.

in easterly direction and increases in westerly direction Now finally let's take a look at this specific question. Since the Latitude of both points A and B is the same (60 0 N) and point B is located due East of point A, the Rhumb Line (RL) track would be 090~ Since both points are in the Northern Hemisphere, the GC track would be arcing up above the RL track, towards the North Pole => the initial GC track at A will be equal to RL track - the conversion angle and will gradually increase towards the mid-point, where the GC track will be equal to the RL track. From this point on the GC track will increase beyond the RL track until reaching point B where the difference between RL track and GC track will be the greatest (same as at departure point A). The change in Longitude is 1O~ Conversion angle can be calculated using the formula: • Conversion angle = l2 Ch Long x sin Latitude • Conversion angle = l2 10° x sin 60° • Conversion angle = 4,33° With a RL track of 090° and the conversion angle of approx. 4,33° the initial

(Refer to figures 061-E82 and 061-E83) What the JAA is asking in this question is the difference between initial and mean GC track = just a '~AA special way" of asking for the Conversion Ch Long x sin Latitude Angle. Formula for the Conversion Angle = (or sin mean Lot). From the formula we can see that the greater the Latitude, the greater the conversion angle is going to be for a given change oflongitude. When two different latitudes are quotes, we calculate based on their mean latitude. Out of the given answer possibilities, the greatest conversion angle will result from 60 0 N. We can double-check this by calculating the individual cases (Ch.Long in all cases =10° as given by the question):

*

AnswerA: • Mean latitude = 27,5°5 • Conv. angle = Ch Long x sin Latitude • Conv. angle = 10° x sin 27,5° • Conv. angle =2,3°

* *

AnswerB: • Mean latitude = 55°N • Conv. angle = 10° x sin 55° • Conv. angle = 4,1°

*

AnswerC: • Mean latitude = 0° NIS • Conv. angle = 10° x sin 0° • Conv. angle = 0°

*

AnswerD: • Mean latitude = 60 N • Conv. angle = 10° x sin 60° • Conv. angle = 4,33°

GC track will be 85,67" (090° - 4,33°). The INS will follow the GC track, because it is the shortest distance between two points. The GC track will be increasing from 85,67" at point A to 090° (RL track) at a mid-point location = 25°W. The average track between A and the mid-point will be 87,84° ((85,67" + 090°) .;. 2) = average 2,16° difference from the RL track. The RL distance from A to the mid-point = Change in Longitude (in minutes) x cos Latitude => 300 minutes x cos 60° = 150 NM. Now we can apply the 1:60 rule: • TKE = (Dist off-track x 60) .;. Dist along track ·2,16°= (?x 60) .;.150 ·?=2,16.;.60x 150 ·?=5,4NM At the mid-point (25°W) the aircraft will be located 5,4 NM to the North of the RL track. RL track follows the parallel of 60 0 N. We know that 1 NM along a Meridian = 1 minute of Latitude. 5,4 NM will therefore equal 5,4 minutes of Latitude above 60 0 N => the highest Latitude reached will be 60 0 05'24"N.

4338. Airplane ATPL CPL Heli ATPL CPL The maximum difference between geocentric and geodetic latitude occurs at about: A) 90° North and South.

B) 60° North and South. C) 45° North and South. 0) 0° North and South (equator).

0

*

As we can see the greatest value of the conversion angle is reached at latitude 60W

4333. Airplane ATPL CPL Heli ATPL CPL Waypoint 1 is 600N 300W. Waypoint 2 is 60 0N 200W. The aircraft autopilot is coupled to the INS steer. What is the latitude on passing 25°W? 0

A) 60 05'N B) 60 0 11'N

C) 60 0 32'N 0) 59°49'5 (Refer to figures 061-E82 and 061-E83) The difference between the Initial or Final GC and the RL track is represented by the value of the Conversion Angle: • Conversion angle = l2 Ch Long x sin Latitude (or sin mean Lat). • or: Conversion angle = l2 (Initial Great Circle - Final Great Circle track). The difference between the Initial GC track and the Final GC track is represented by the value of Convergency: • Convergency =Change in Longitude x sin of Latitude • or: Convergency = GC Initial True Track - GC Final True Track The GC tracks lie closer to the poles than the Rhumb Line (RL) tracks. There are some attributes of GC tracks:

Northern Hemisphere: • From the initial to the final point of the GC track, the track value increases in easterly direction and decreases in westerly direction Southern Hemisphere: • From the initial to the final point of the GC track, the track value decreases

(Refer to figure 061-EI9) The Geocentric Latitude of a point is the smaller angle between the line joining the centre of the Earth to the Equator and the line joining that point to the centre of the Earth. The Geodetic (or Geographic) Latitude of a point is the smaller angle between the line which is perpendicular to the meridian at that point and the line joining that point to the centre of the Earth, As it can be seen in the attached figure, the maximum difference is at approx 45° NIS and is about 11,6 minutes of arc.

4342. Airplane ATPL CPL Heli ATPL CPL Given: Value for the ellipticity of the Earth is 1/297. Earth's semi-major axis, as measured at the equator, equals 6.378,4 km. What is the semi-minor axis (km) of the Earth at the axis of the Poles? A) 6.356,9 B) 6.378,4 C) 6.367,0 0) 6.399,9 There are two methods to solve these types of questions. In the example #2

a more detailed explanation is presented. Method 1: • 1 .;. 297 = 0,003367003 ·0,003367003 x 6.378,4 = 21,47609428 ·6.378,4 - 21,47609428 = 6356,9 km Method 2: • The equatorial diameter is 6378,4 x 2 = 12756,8 km • Ellipticity 11297 means that the polar diameter is smaller than the equatorial diameter by an amount equal to the equatorial diameter divided by 297.

14331 (D) I 4333 (A) I 4338 (C) I 4342 (A) I

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01 Basics of Navigation • So, the polar diameter = 12756,8 - (12 756,8 -;- 297) = 12713,8 • The semi-minor axis of the Earth is halfof the polar diameter i.e. 12713,8-;2 = 6356,9km

4343. Airplane ATPL CPL Heli ATPL CPL The diameter of the Earth is approximately: A) 18.500 km B) 6.350 km C) 12.700 km D) 40.000km Remember that the relationship between distance and angular change of Longitude, where 1 minute of Longitude = 1 NM (or 1° of Longitude =60 NM) is valid only at the Equator (0° NIS). Therefore, the circumference of the Earth at the Equator is approx. 21 600 NM (360° x 60 NM).

As far as the Radius of the Earth, you can either remember a value of approx. 12700 km, oryou can calculate it from the circumference: • Circumference =2lTR (. .. R =radius; IT =3,14) .21600NM =2 XlTX R ·21600NM=6,28xR ·R=3439,49NM Now just convert NM to km: • 1 NM = 1,852 km ·3 439,49 NM =approx. 6370 km If a radius 12740km.

= 6 370 km, then the diameter = twice this value => 6370 x 2 =

4349. Airplane ATPL CPL Heli ATPL CPL The automatic flight control system (AFCS) in an aircraft is coupled to the guidance outputs from an inertial navigation system (INS) and the aircraft is flying from waypoint No. 2 (60°00'5 070 000'W) to No. 3 (60°00'5 OSooOO'W). Comparing the initial track (0T) at 070 000'W and the final track (0T) at OSooOO'W, the difference between them is that the initial track is approximately: A) 9° greater than the final one. B) 5° greater than the final one. C) 9° less than the final one. D) 5° less than the final one. (Refer to figures 061-E82 and 061-E83) The difference between the Initial or Final GC and the RL track is represented by the value of the Conversion Angle: • Conversion angle = ~ Ch Long x sin Latitude (or sin mean Lat). • or: Conversion angle = ~ (initial Great Circle - Final Great Circle track). The difference between the Initial GC track and the Final GC track is represented by the value of Convergency: • Convergency = Change in Longitude x sin ofLatitude • or: Convergency = GC Initial True Track - GC Final True Track Now let's take a look at this specific question. We do not need to draw any sketches in this case in order to visualize the relationship between the RL and GC tracks. The question is asking for the difference in the initial GC track and the final GC track => we know that this is represented by the value of the Convergency. The change in longitude in our case = 10°. Simply use the convergency formula: • Convergency = Change in Longitude x sin of Latitude • Convergency = 10° x sin 60° • Convergency =8,66° =approx. 9° The difference between the initial and final GC tracks will be approximately 9°.

4369. Airplane ATPL CPL Heli ATPL CPL You are flying from A (SOON 100W) to B (SBON 02°E). If initial great circle track is 047°(T} what is final great circle track? A) B) C) D)

57° 52° 43° 29°

(Refer to figures 061-E82 and 061-E83) The difference between the Initial or Final GC and the RL track is represented by the value of the Conversion Angle:

• Conversion angle = ~ Ch Long x sin Latitude (or sin mean Lat). • or: Conversion angle = ~ (initial Great Circle - Final Great Circle track). The difference between the Initial GC track and the Final GC track is represented by the value of Convergency: • Convergency = Change in Longitude x sin of Latitude • or: Convergency = GC Initial True Track - GC Final True Track The GC tracks lie closer to the poles than the Rhumb Line (RL) tracks. There are some attributes ofGCtracks: Northern Hemisphere: • From the initial to the final point of the GC track, the track value increases in easterly direction and decreases in westerly direction Southern Hemisphere: • From the initial to the final point of the GC track, the track value decreases in easterly direction and increases in westerly direction In the case of this question, the change in Longitude is 12° (r from 2°E to the Greenwich and then further 10° to 10 0 W). The mean Latitude is 54° (between 50 0 N and 58 ON). Now we can use the formula: • Convergency = Change in Longitude x sin of Latitude • Convergency = 12° x sin 54° • Convergency = 9,r Since we are flying Eastbound, initially the GC track will be lower than the RL track and will gradually increase towards the RL track. Exactly mid-point between A and B the GC track will be parallel with the RL track => GC and RL tracks will be identical. From this point on, towards point B the GC track will be becoming greater than RL track, reaching the maximum difference at point B. The difference between the initial GC track and the final GC track equals to the value of convergency. We have calculated that convergency = 9,r = approximately 10°. If the initial GC track is smaller than RL track, then the final GC track has to be greater than RL track => the GC track will increase. Initial GC track of 04r + convergency of 10° =final GC track of 057~

4397. Airplane ATPL CPL Heli ATPL CPL Given: The coordinates ofthe heliport at Issy les Moulineaux are: 4BoSO'N 002°16,S'E The coordinates of the antipodes are: A) 41°10'5177°43,5'W B) 48°50'5 17J043,5'E C) 48°50'5 177°43,5'W 0,) 41°lO'5177°43,5'E Antipodal points are on the same great circle and are diametrically opposite, i.e. they lie on meridian and anti-meridian and their latitudes are of equal value and bpposite sign. The anti-meridian of E002°16.5 is W17r43.5 (original Longitude + 180°).

4410. Airplane ATPL CPL Heli ATPL CPL What is the convergence at SooOO'N between the meridians 10S000'W and 14S000'W on the Earth? A) 32,1° B) 40,0° C) 30,6° D) 50,0° (Refer to figures 061-E82 and 061-E83) Convergency or "the angle of convergence" = the angle between meridians. Convergency is in fact the angle of inclination between the meridians passing through two points at a given latitude. This angle is zero at the Equator, where meridians are parallel to each other. As meridians leave the Equator and approach the Poles, they converge and the value of convergency reaches a maximum of 1° (between successive meridians) at the Poles, where the meridians are spaced 1° apart. The angle of convergency also represents the difference between the initial and the final Great Circle track (difference between the Great Circle track at the starting and ending points of the specific track between two points). The angle of convergency between two positions (i.e. two meridians) is calculated from the formula: • Convergency = Change in Longitude x sin of Latitude • or: Convergency = GC Initial True Track - GC Final True Track Change in longitude (Ch Long or D Long) is the difference in longitude (expressed in 0) between two positions or two meridians. If the positions are at different latitudes, the average value of latitude should be used,

I 4343 (C) I 4349 (C) I 4369 (A) I 4397(C) I 4410 (C) I

EI

Aviationexam Test Prep Edition 2012 • Change in Latitude = 3°30' = 210 minutes • 1 minute along Meridian = 1 NM ·210 minutes along Meridian = 210 NM • DistanceB=210NM

which will give an approximate value. In case of this question the solution is very easy: • Latitude = 50 0N • Ch. Longitude =40° (145°W - 105°W) • Convergency = Change in Longitude x sin of Latitude • Convergency =40° x sin 50° • Convergency =30,6°

4413. Airplane Conversion angle is:

ATPL

Heli

CPL

ATPL

We can see that the two distances are almost equal (we do not have to use trigonometry in this case) = what we have here is a right-angle isosceles triangle =the other two internal angles will be equal =angle a will be 45~ From this information we can easily determine the Rhumb Line track from A to B => simply add 45° to 270° to get a track of 315°.

CPL

4501. Airplane ATPL CPL Heli ATPL CPL An aircraft passes position A (60 0 00'N 1200 00'W) on route to position 8 (60 0 00'N 1400 30'W). What is the great circle track on departure from A?

A) convergency. B) 4 times convergency. C) twice convergency. 0) 0,5 convergency. For explanation refer to question #1236 on page 6.

4420. Airplane ATPL CPL Heli ATPL CPL What is the change in longitude (in degrees and minutes) from A (45°N 165°30'E) to 8 (45°N 155°40'W)? A) 38°05'E B) 38°50'W-

C) 38°05'W 0) 38°50'E (Refer to figures 061-E79, 061-E80 and 061-E81) Change of/atitude (Ch Lator D Lot) The change of latitude between two places is defined as the smaller arc of meridian between the two latitudes. It is therefore the difference in latitude between the two. It is designated North or South according to the direction of change specified e.g. the Ch Lat between 2rS and 3rS is 10° South. From WS OWE to WN 025°W; the change of latitude (Ch Lat) will be 30° North. Calculating change of latitude requires the subtraction of one latitude from another, or, if the latitudes are in different hemispheres, the addition of the latitudes. Change of longitude (Ch Long or D Long) Change of/ongitude between two places is the smaller arc of the Equator measured between the meridians of those places. It is designated East or West according to the direction of change specified. Change of Longitude (Ch Long) corresponds to change of latitude, but in terms of longitude. For example the Ch Long between 29°E and 31°E is 2°E. Thus from 15°S 015°Eto 15°N 025°W; the change of longitude (Ch Long) will be 400W (15° from OWE to the Greenwich meridian and then 25° from Greenwich to 025°W). There is also a change of latitude, but this does not affect the Ch Long.

In the case of this question you have to realize that the shortest change of longitude between these two points will be via the Greenwich anti-meridian (180° EIW) and not via the Greenwich meridian (0° EIW): • from 165°30'E to the 180 0EIW meridian = 14°30' • from the 180 0E/W meridian to 155°40'W =24°20' • 14°30' + 24°20' =38°50' • direction is from an Easterly Longitude towards the Greenwich anti-meridian (180 0EIW), therefore the direction of flight is Eastbound. CH. Long is therefore going to be 38°50' E.

4477. Airplane ATPL CPL Heli ATPL CPL The rhumb line track between position A (45°00'N, 010 0 00'W) and position 8 (48°30'N, 015°00'W) is approximately: A) 345°

B) 300° C) 330° 0) 315° (Refer to figure 061-E50) Start by drawing a sketch of the situation described by the question. To solve the question we need to first find the distances along the parallel and meridian and then use trigonometry (if needed): Let's start by finding the distance A =distance along the parallel of 45°N: • Change in Longitude = 5° = 300 minutes • Distance = Change in Longitude (in minutes) x cos Latitude • Distance = 300 minutes x cos 45° • Distance A = 212 NM

4413 (0)

I

4420 (0)

I

4477 (0)

I

4501 (C)

(Refer to figures 061-E82 and 061-E83) The difference between the Initial or Final GC and the RL track is represented by the value of the Conversion Angle: • Conversion angle = ~ Ch Long x sin Latitude (or sin mean Lat). • or: Conversion angle = ~ (Initial Great Circle - Final Great Circle track). The difference between the Initial GC track and the Final GC track is represented by the value of Convergency: • Convergency =Change in Longitude x sin of Latitude • or: Convergency = GC Initial True Track - GC Final True Track The GC tracks lie closer to the poles than the Rhumb Line (RL) tracks. There are some attributes of GC tracks: Northern Hemisphere: • From the initial to the final point of the GC track, the track value increases in easterly direction and decreases in westerly direction Southern Hemisphere: • From the initial to the final point of the GC track, the track value decreases in easterly direction and increases in westerly direction

Now finally let's take a look at this specific question. Since the Latitude of both points A and B is the same (60 0N) and point B is located due West of point A, the Rhumb Line (RL) track would be 270~ Since both points are in the Northern Hemisphere, the GC track would be arcing up above the RL track, towards the North Pole => the initial GC track at A will be equal to RL track + the conversion angle. Change in Longitude in this case = 20,5° (140 030'W - 1200W). Conversion angle can be calculated using the formula: • Conversion angle = ~ Ch Long x sin Latitude • Conversion angle = ~ 20,5° x sin 60° • Conversion angle =8,9° =approx. 9° With a RL track of 270° and the conversion angle of approx. 9° the initial GC track will be 279° (270° + 9°).

4504. Airplane ATPL CPL Heli ATPL CPL { An aircraft travels from point A to point 8, using the autopilot \ connected to the aircraft's inertial system. The coordinates of A (45°5 010 0 W) and 8 (45°5 030 0 W) have been entered. The true course of the aircraft on its arrival at 8, to the nearest degree, is:

A) 277 0 B) 284° C) 263 0 0) 270 0 (Refer to figures 061-E82 and 061-E83) The difference between the Initial or Final GC and the RL track is represented by the value of the Conversion Angle: • Conversion angle = ~ Ch Long x sin Latitude (or sin mean Lat). • or: Conversion angle = ~ (Initial Great Circle - Final Great Circle track). The difference between the Initial GC track and the Final GC track is represented by the value of Convergency: • Convergency =Change in Longitude x sin of Latitude :or: Convergency =GC Initial True Track - GC Final True Track

Second step will be to find the distance B = distance along the meridian 015°W:

I

A) 261° B) 288° C) 279° 0) 270°

I

The GC tracks lie closer to the poles than the Rhumb Line (RL) tracks. There are some attributes of GC tracks:

4504 (A)

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01 Basics of Navigation Northern Hemisphere: • From the initial to the final point of the GC track, the track value increases in easterly direction and decreases in westerly direction Southern Hemisphere: • From the initial to the final point of the GC track, the track value decreases in easterly direction and increases in westerly direction Now finally let's take a look at this specific question. Since the Latitude of both points A and B is the same (45°5) and point B is located due West of point A, the Rhumb Line (RL) track would be 270°. Since both points are in the Southern Hemisphere, the GC track would be arcing down below the RL track, towards the South Pole => the final GC track upon arrival at B will be equal to RL track + the conversion angle (final GC track will be greater than the RL track). The change in Longitude in this case = 20°. Conversion angle can be calculated using the formula: • Conversion angle = JI2 Ch Long x sin Latitude • Conversion angle = JI2 20° x sin 45° • Conversion angle = r With a RL track of270° and the conversion angle ofapprox. rthe final GC track will be 27r (270° + 7").

4513. Airplane ATPL CPL Heli ATPL CPL The automatic flight control system (AFCS) in an aircraft is coupled to the guidance outputs from an inertial navigation system (INS). The aircraft is flying between inserted waypoints No.3 (55°00'N020000'W) and No.4 (55°00'N 030 000'W). With OSRTK/STS selected on the COU, to the nearest whole degree, the initial track read-out from waypoint No.3 will be: A) B) C) D)

274

0

2660 270 0

The difference between the Initial GC track and the Final GC track is represented by the value of Convergency: • Convergency = Change in Longitude x sin of Latitude • or: Convergency = GC Initial True Track - GC Final True Track The GC tracks lie closer to the poles than the Rhumb Line (RL) tracks. There are some attributes of GC tracks: Northern Hemisphere: • From the initial to the final point of the GC track, the track value increases in easterly direction and decreases in westerly direction Southern Hemisphere: • From the initial to the final point of the GC track, the track value decreases in easterly direction and increases in westerly direction Now finally let's take a look at this specific question. Since the Latitude of both points 3 and 4 is the same (55°N) and point 4 is located due West of point 3, the Rhumb Line (RL) track would be 270°. Since both points are in the Northern Hemisphere, the GC track would be arcing up above the RL track, towards the North Pole => the initial GCtrack at A will be equal to RL track + the conversion angle (GC track will be greater than the RL track). Change in Longitude in our case = 10~ Conversion angle can be calculated using the formula: • Conversion angle = JI2 Ch Long x sin Latitude • Conversion angle = JI2 10° x sin 55° • Conversion angle = 4° With a RL track of 270° and the conversion angle of 4° the initial GC track will be 274° (270° + 4°). .

4517. Airplane ATPL CPL Heli ATPL CPL You are flying from A (30 0S 200E) to B (30 0S 200WJ:"What is the RL track from A to B? A) 250 0 (T) B) 270 0 C) 290 0

m m

A) 260 0 (T) B) 270 0 (T) C) 290 0 (T) D) 300 0 (T) (Refer to figures 061-E82 and 061-E83) The difference between the Initial or Final GC and the RL track is represented by the value of the Conversion Angle: • Conversion angle = JI2 Ch Long x sin Latitude (or sin mean Lat). • or: Conversion angle = JI2 (Initial Great Circle - Final Great Circle track). The difference between the Initial GC track and the Final GC track is represented by the value of Convergency: • Convergency =Change in Longitude x sin of Latitude • or: Convergency = GC Initial True Track - GC Final True Track The GC tracks lie closer to the poles than the Rhumb Line (RL) tracks. There are some attributes of GC tracks: Northern Hemisphere: • From the initial to the final point of the GC track, the track value increases in easterly direction and decreases in westerly direction

(Refer to figures 061-E82 and 061-E83)

I

4517 (8)

I

4518 (A)

Now finally let's take a look at this specific question. Since the Latitude of both points A and B is the same (30"5) and point B is located due West of point A, the Rhumb Line (RL) track would be 270~ Since both points are in the Southern Hemisphere, the GC track would be arcing down below the RL track, towards the South Pole: • the INITIAL GC track upon departure from A will be equal to RL track - conversion angle (GC track will be smaller than RL track) • the FINAL GC track at arrival to B will be equal to RL track + the conversion angle (GC track will be higher than RL track). The change in Longitude between points A and B is 40° (from 200E to 200W). Conversion angle can be calculated using the formula: • Conversion angle = JI2 Ch Long x sin Latitude • Conversion angle = JI2 40° x sin 30° • Conversion angle = 10° With a RL track of270° and the conversion angle of 100the INITIAL GC track will be 260° (270° - 10°) and the FINAL GC track will be 280° (270° + 10°).

4519. Airplane ATPL CPL Heli ATPL CPL You are flying from A (30 0S 200E) to B (300S 20 0W). What is the final GC track?

m m

A) 250 0 B) 270 0 (T) C) 280 0 D) 300 0 (T)

For explanation refer to question #4518 on this page.

4528. Airplane ATPL CPL Heli ATPL CPL The initial great circle track from A to B is 080° and the rhumb line track is 083°. What is the initial great circle track from B to A and in which hemisphere are the two positions located? A) B) C) D)

2660 260 0 260 0 2660

and in the Northern Hemisphere. and in the Southern Hemisphere. and in the Northern Hemisphere. and in the Southern Hemisphere.

(Refer to figures 061-E22, 061-E82 and 061-EB3) A Great Circle (GC) is a circle on the surface of a sphere with the same centre and radius as that of the sphere. This applies to the Earth. Great Circles are important in navigation because the shortest distance between two places on the surface of the Earth is the shorter arc of the great circle between those

D) 3000 (T)

4513 (8)

4518. Airplane ATPL CPL Heli ATPL CPL You are flying from A (30 0S 200E) to B (30 0S 200W). What is the initial GC track?

Southern Hemisphere: • From the initial to the final point of the GC track, the track value decreases in easterly direction and increases in westerly direction

278 0

(Refer to figures 061-E82 and 061-E83) The difference between the Initial or Final GC and the RL track is represented by the value of the Conversion Angle: • Conversion angle = JI2 Ch Long x sin Latitude (or sin mean Lat). • or: Conversion angle = JI2 (Initial Great Circle - Final Great Circle track).

I

Since the Latitude of both points A and B is the same (30°5) and point B is located due West ofpoint A, the Rhumb Line (RL) track would be 270°. Do not get the RL track confused with the Great Circle (GC) track. RL track is constant track, whereas the GC track changes.

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4519 (C)

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4528 (A)

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Aviationexam Test Prep Edition 2012 two points. Another way of expressing this is to say that the plane of a great circle connecting two places on a sphere (such as the Earth) passes through the centre of the sphere (or Earth). Any great circle on the Earth will therefore divide the Earth into equal parts (bisect it). There can be only one great circle connecting two places on the Earth, unless those two places are diametrically opposite each Other (antipodean). Examples are the North and South Pole, 45°N 090 0W and 45°S 090 0E etc. .. In such cases, there is an unlimited number of great circles between the two points. While a GC track is the shortest distance, the direction of a GC changes continuously because meridians are not parallel on the Earth (a flight along the Equator or along a meridian itself is an exception). Check this statement by connecting two places on a globe with a length of string. Measure the angle at each meridian. Unless the two places are due North/South or East/West of each other, the direction between the two places will change from meridian to meridian. Ideally, all aircraft tracks would be great circles, as the distance would be shortest by definition.

but this time along the Equator. In this case the distance would be 21.600 NM. To cover this distance in 22,5 hrs the speed would have to be 960 kts = double the original speed. Even without any calculations of time, if you take a look at the distances, the distance along the Equator is exactly twice the distance at 60 0N => hence also the speed would double if the time is to remain constant.

4564. Airplane ATPL CPL Heli ATPL CPL The great circle bearing of position B from position A in the Northern Hemisphere is 040°. If the conversion angle is 4°, what is the great circle bearing of A from B1 A) 228° B) 212° C) 220° D) 224°

The difference between the Initial GC and the RL track is represented by the value of the Conversion Angle: • Conversion angle = 1'.2 Ch Long x sin Latitude (or sin mean Lat). • or: Conversion angle 1'.2 (Initial Great Circle - Findl Great Circle track).

(Refer to figures 061-E21, 061-E82 and 061-E83) The difference between the Initial or Final GC and the RL track is represented by the value of the Conversion Angle: • Conversion angle = 1'.2 Ch Long x sin Latitude (or sin mean Lat). • or: Conversion angle = 1'.2 (Initial Great Circle - Final Great Circle track).

The difference between the Initial GC track and the Final GC track is represented by the value of Convergency: • Convergency = Change in Longitude x sin of Latitude • or: Convergency = GC Initial True Track - GC Final True Track

The difference between the Initial GC track and the Final GC track is represented by the value ofConvergency: • Convergency = Change in Longitude x sin of Latitude • or: Convergency = GC Initial True Track - GC Final True Track

The GC tracks lie closer to the poles than the Rhumb Line (RL) tracks. There are some attributes of GC tracks:

The GC tracks lie closer to the poles than the Rhumb Line (RL) tracks. There are some attributes of GC tracks:

Northern Hemisphere: • From the initial to the final point of the GC track, the track value increases

Northern Hemisphere:

=

in easterly direction and decreases in westerly direction

Southern Hemisphere: From the initial to the final point of the GC track, the track value decreas-

es in easterly direction and increases in westerly direction

4539. Airplane ATPL CPL Heli ATPL CPL If an aeroplane was to circle around the Earth following parallel600N at a ground speed of 480 kts.ln order to circle around the Earth along the equator in the same amount of time, it should fly at a ground speed of: 550 kts 240 kts 960 kts 480kts

Remember that the relationship between distance and angular change ofLongitude, where 1 minute ofLongitude = 1 NM (or 1°ofLongitude = 60 NM) is valid only at the Equator (0° N/S). Therefore, the circumference of the Earth at the Equator is approx. 21.600 NM (360° x 60 NM}.ln order to obtain the distance corresponding to 1° change in Longitude at any other Latitude, we use the formula below: • When following a line of Parallel (Rhumb Line), each 1° change of Longitude change represents a distance of 1° x 60 x cos Latitude. The circumference of the Earth at 60 0N can easily be calculated using the formula above: • Circumference =360° x 60 x cos Latitude • Circumference =360 0x 60 x cos 60° • Circumference = 10.800 NM In the first instance the aircraft is flying along the parallel 60 0N and completes a circle around the Earth. We now know that this distance is 10.800 NM and the speed is 480 kts. The time needed to complete this circle would be 22,5 hrs.ln the second instance, the aeroplane has to circle the Earth as weI/,

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4539 (C)

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4564 (A)

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4611 (8)

Southern Hemisphere: • From the initial to the final point of the GC track, the track value decreases in easterly direction and increases in westerly direction

Now finally let's take a look at this specific question. The question states that the initial GC track from A to B is 080° and the RL track is 083° => the difference is 3° => this is our conversion angle. The RL is track is constant all the way from A to B = 083°. The reciprocal RL track from B to A would be 263~ Now it helps to draw a sketch => you will see that the GC track curves up above the RL track and this identifies a Northern Hemisphere. If we take the RL track from B to A (263°) and add the conversion angle (3°) we will obtain the initial GC track from B to A => 266°. Another way to solve this question is to take the conversion angle (3°) and mUltiply it by 2 to obtain the angle of convergency of 6° => this is the difference between the initial and final GC track. If the initial GC track is 080° and the RL track is 083~ then the final GC track will be 086° (RL track is in the middle of these values). Now just find the reciprocal of 086° to obtain the result of266°.

A) S) C) D)

• From the initial to the final point of the GC track, the track value increases in easterly direction and decreases in westerly direction

I

Now finally let's take a look at this specific question. Since the Latitude of both points A and B lies in the Northern Hemisphere we can say that the GC track would be arcing up above the RL track, towards the North Pole. If the initial GC track from point A to point B is 040° and the conversion angle is given as 4~ the Rhumb Line (RL) track from pOint A to point B will be 044° (040° + 4°). The RL track from point B to point A will be the reciprocal = 224°. When flying from B to A the GC track will be higher than the RL track and the difference between these two will be again 4° (conversion angle) => 224° RL track + 4° ca will give us an initial GC track from B to A of 228~

4611. Airplane ATPL CPL Heli ATPL CPL Position A is at latitude 33°45'N and position B is at latitude 14°25'N. What is the change in latitude between A and B1 A) B) C) D)

48°10' 19°20' 23°45' 76°15'

(Refer to figures 061-E79, 061-E80 and 061-E81)

Change of latitude (Ch Lat or D Lat) The change of latitude between two places is defined as the smaller arc of meridian between the two latitudes. It is therefore the difference in latitude between the two. It is designated North or South according to the direction of change specified e.g. the Ch Lat between 2JOS and 3JOS is 10° South. From WS OWE to WN 025°W, the change of latitude (Ch Lat) will be 30° North. Calculating change of latitude requires the subtraction of one latitude from another, or, if the latitudes are in different hemispheres, the addition

of the latitudes. Change of longitude (Ch Long or D Long) Change oflongitude between two places is the smaller arc of the Equator mea7 sured between the meridians of those places. It is designated East or West according to the direction of change specified. Change of Longitude (Ch Long) corresponds to change of latitude, but in terms of longitude. For example the Ch Long between 29°E and 31°E is 2°E. Thus from 15°S 015°E to 15°N 025°W, the change of longitude (Ch Long) will be 400W (15° from 015°E to the Greenwich meridian and then 25° from Greenwich to 025°W). There is also a change of latitude, but this does not affect the Ch Long. In the case of this question, simply deduct the two values: • 33°45'N - 14°25'N = 19°20' • direction is from the North to the South, so the Ch.Lat will be 19°20' South.

01 Basics of Navigation

4628. Airplane ATPL CPL Heli ATPL CPL The circumference ofthe parallel of latitude at 60 0 N is approximately: A) 10.800 NM B) 18.706 NM C) 20.000 NM D) 34.641 NM Remember that the relationship between distance and angular change ofLongitude, where 1 minute of Longitude 1 NM (or 1° of Longitude 60 NM) is valid only at the Equator (0° N/S). Therefore, the circumference of the Earth at the Equator is approx. 21.600 NM (360° x 60 NM).ln order to obtain the distance corresponding to 1° change in Longitude at any other Latitude, we use the formula below:

=

=

• When following a line of Parallel (Rhumb Line), each 1° change of Longitude change represents a distance of 1° x 60 x cos Latitude. The circumference of the Earth at 60 0N can easily be calculated using the formula above: 0 • Circumference =360 x 60 x cos Latitude 0 • Circumference =360 x 60 x cos 60° • Circumference = 10.800 NM.

are important in navigation because the shortest distance between two places on the surface of the Earth is the shorter arc of the great circle between those two points. Another way of expressing this is to say that the plane of a great circle connecting two places on a sphere (such as the Earth) passes through the centre of the sphere (or Earth). Any great circle on the Earth will therefore divide the Earth into equal parts (bisect it). There can be only one great circle connecting two places on the Earth unless those two places are diametrically opposite each other (antipodean). Examples are the North and South Pole, 45°N 090 0W and 45°S 090 0E etc. .. In such cases, there is an unlimited number of great circles between the two points. While a GC track is the shortest distance, the direction of a GC changes continuously because meridians are not parallel on the Earth (a flight along the Equator or along a meridian itself is an exception). Check this statement by connecting two places on a globe with a length ofstring. Measure the angle at each meridian. Unless the two places are due North/South or East/West of each other, the direction between the two places will change from meridian to meridian. Ideally, all aircraft tracks would be great circles, as the distance would be shortest by definition. The difference between the Initial GC and the RL track is represented by the value of the Conversion Angle: • Conversion angle = Ch Long x sin Latitude (or sin mean Lat). • or: Conversion angle = (Initial Great Circle - Final Great Circle track).

*

4667. Airplane ATPL CPL Heli ATPL CPL How many NM are equivalent to 1° of arc of latitude:

*

The difference between the Initial GC track and the Final GC track is represented by the value of Convergency: • Convergency = Change in Longitude x sin of Latitude • or: Convergency = GC Initial True Track - GC Final True Track

A) 1 NM B) 15 NM C) 60NM D) 600NM

The GC tracks lie closer to the poles than the Rhumb Line (RL) tracks. There are some attributes of GC tracks:

If we assume that the Earth is a perfect sphere, then the following applies: • When following a Great Circle in the North/South direction each 1° change of Latitude represents a distance of 60 NM (1 minute of Latitude = 1 NM). • When following a Great Circle in the East/West direction each 1° change of Longitude represents a distance of 1° x 60 x cos Latitude. However, in reality the Earth is an ellipsoid, therefore 1 minute of an arc of a Great Circle varies very slightly depending of the position on the Earth. For practical navigation purposes, however, these errors (or variations) can be disregarded and in this respect we treat the Earth as if it were a perfect sphere.

4717. Airplane ATPL CPL Heli ATPL CPL You are flying from A (30°5 20° E) to B (30°5 20 0 W). At what longitude will the GC track equal the RL track? A) B) C) D)

A Great Circle (GC) is a circle on the surface of a sphere with the same centre and radius as that of the sphere. This applies to the Earth. Great Circles

100E 100W OOE/W 200W

(Refer to figures 061-E82 and 061-E83) The Latitude of both points A and B is identical- 30 0S. Point B is located due West of point A. The Rhumb Line (RL) track will therefore be 270~ Since both points A and B are located in the Southern Hemistphere, the Great Circle (GC) track will be curved down below the RL track. towards the South Pole. Since we are flying Westbound, initially the GC track will be lower than the RL track and will gradually increase towards the RL track. Exactly mid-point between A and B the GC track will be parallel with the RL track => GC and RL tracks will be identical. The mid-point between A and B is exactly at OOE/W (Greenwich meridian). From this point on, towards point B the GC track will be greater than RL track, reaching the maximum difference at point B.

Northern Hemisphere: • From the initial to the final point of the GC track, the track value increases in easterly direction and decreases in westerly direction Southern Hemisphere: • From the initial to the final point of the GC track, the track value decreases in easterly direction and increases in westerly direction Now finally let's take a look at this specific question. The initial GC track at point P is 095° and we are located in the Southern Hemisphere. Now it would help to draw a sketch => you would see that the initial GC track is greater than the Rhumb Line (RL). The difference between GC and RL track = the conversion Therefore, the RL track will be 095° - = 088°. angle =

r.

r

4730. Airplane ATPL CPL Heli ATPL CPL The circumference of the Earth is approximately: A) B) C) D)

43.200 NM 10.800 NM 21.600 NM 5.400 NM

For explanation refer to question #4343 on page 13.

4821. Airplane ATPL CPL Heli ATPL CPL As the INS position of the departure aerodrome, coordinates 35°32,7'N 139°46,3'W are input instead of 35°32,7'N 139°46,3'E. When the aircraft subsequently passes point 52°N 1800W, the longitude value show on the INS will be: A) 080027,4'W B) 099°32,6'W

4726. Given:

Airplane

ATPL

Heli

CPL

ATPL

C) 099°32,6'E

CPL

D) 080 0 27,4'E

=

(Refer to figure 061-E17) Refer to the illustration to visualize the situation. The correct (actual) position of the aircraft is 139°46,3'E. An incorrect IRS position entry is made and the position 139°46,3'W is entered by mistake. As the aircraft departs and flies toward the 1800E/W meridian the actual change in Longitude will be 40013'40'~ The IRS position will experience the same magnitude and direction of Longitude change, but it will be taken from the IRS position => to solve the question simply deduct 40°13'40" from the IRS position (139°46'20"'W) to obtain the result of 99°32'40''w.

Great circle from P to Q measured at P 095° Southern hemisphere Conversion angle..!, - Q =7° What is the rhumb line track P - Q? A) B) C) D)

081° 102° 088° 109°

(Refer to figures 061-E82 and 061-E83)

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4628 (A)

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4667 (C)

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4717 (C)

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4726 (C)

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4730 (C)

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4821 (8)

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Aviationexam Test Prep Edition 2012

Airplane

4823.

ATPL

Heli

CPL

ATPL

the Rhumb Line (RL) track would be 270~ Since both points are in the Northern Hemisphere, ·the GC track would be arcing up above the RL track, towards the North Pole => the initial GC track at A will be equal to RL track + the conversion angle. Conversion angle can be calculated using the formula: • Conversion angle = J.2 Ch Long x sin Latitude • on version angle = J.2 9° x sin 45° • Conversion angle = 3, 18°

CPL

Given: 55°N OOOOE 54°N 0100E

PointA: PointB:

The average true course of the great circle is 100°. The true course ofthe rhumb line at point A is:

With a RL track of 270° and the conversion angle of approx. 3° the initial GC track will be 273° (270° + 3°).

A) 100° B) 096 C) 1040 D) 1070 0

4864.

(Refer to figures 061-E82 and 061-E83) This is actually a tricky, yet very simple question. You just have to visualize (or draw a sketch) the relationship between a Great Circle (GC) and Rhumb Line (RL) tracks. In the Northern Hemisphere the value of the GC track is initially lower than the value of RL track and the GC track gradually increases until the mid-point, where the GC and RL tracks are parallel. From this point on the GC track continues to increase over the RL track until the final waypoint point. The average GC track will be equal to the RL track mid-point between the two waypoints.ln our case it will be 1OO~ We know that the RL track is constant, therefore we can say the RL track for the entire flight from point A to point B will be 700~ GC track will change, but this question asks about the RL track and not the GC track. The result is therefore 100~ without the need for any calculation. 4836. Airplane ATPL CPL Heli ATPL CPL What is the rhumb line (RL) direction from 45°N 14°12'W to 45°N 12°48'E?

A) 270 0 (T) B)

Airplane

ATPL

CPL

Heli

ATPL

CPL

Given:

090 0 (T)

Waypoint1: Waypoint2:

60 05030 0W 60 0S0200W

What will be the approximate latitude shown on the display unit of an inertial navigation system at longitude 025°W? A) 060°11 '5. B) 059°49'5.

C) 060°00'5. D) 060°06'5. (Refer to figures 061-E82 and 061-E83) The difference between the Initial or Final GC and the RL track is represented by the value of the Conversion Angle: • Conversion angle = J.2 Ch Long x sin Latitude (or sin mean Lat). • or: Conversion angle = J.2 (Initial Great Circle - Final Great Circle track). The difference between the Initial GC track and the Final GC track is represented by the value of Convergency: • Convergency =Change in Longitude x sin of Latitude • or: Convergency = GC Initial True Track - GC Final True Track The GC tracks lie closer to the poles than the Rhumb Line (RL) tracks. There are some attributes of GC tracks:

C) 090 0 (M) D) 270 0 (M) (Refer to figures 061-E79, 061-E80 and 061-E81) A rhumb line is a line ofconstant direction on the surface of the Earth or a chart. This is because it crosses each meridian at a constant angle. Meridians are the datums used for measuring true direction. The Equator and Meridians are both Rhumb Lines and Great Circles by definition. Parallels of Latitude must also be Rhumb lines because they are lines of constant direction since they run East/West. Places 'on the same parallel of latitude are therefore due East or West of each other. In the case of this question, we would be flying due East along the same parallel (45°N) => we would follow a track 090° True. 4838. Airplane ATPL CPL Heli ATPL CPL The great circle track measured at A (45°00'N 010000'W) from A to B (45°00'N 019°00'W) is approximately:

A) 270 0

B) 090 0 C) 273 0 D) 093 0

(Refer to figures 061-E82 and 061-E83) The difference between the Initial or Final GC and the RL track is represented by the value of the Conversion Angle: • Conversion angle = J.2 Ch Long x sin Latitude (or sin mean Lat). • or: Conversion angle = J.2 (Initial Great Circle - Final Great Circle track). The difference between the Initial GC track and the Final GC track is represented by the value of Convergency: • Convergency = Change in Longitude x sin of Latitude • or: Convergency = GC Initial True Track - GC Final True Track The GC tracks lie closer to the poles than the Rhumb Line (RL) tracks. There are some attributes of GC tracks: Northern Hemisphere:

• From the initial to the final point of the GC track, the track value increases in easterly direction and decreases in westerly direction Southern Hemisphere:

• From the initial to the final point of the GC track, the track value decreases in easterly direction and increases in westerly direction Now finally let's take a look at this specific question. Since the Latitude of both points A and B is the same (45°N) and point B is located due West of point A,

I

4823 (A)

I

4836 (8)

I

4838 (C)

I

4864 (0)

I

Northern Hemisphere:

• From the initial to the final point of the GC track, the track value increases in easterly direction and decreases in westerly direction Southern Hemisphere:

• From the initial to the final point of the GC track, the track value decreases in easterly direction and increases in westerly direction Now finally let's take a look at this specific question. Since the Latitude of both points 1 and 2 is the same (60°5) and point 2 is located due East of point 1, the Rhumb Line (RL) track would be 090°. Since both points are in the Southern Hemisphere, the GC track would be arcing down below the RL track, towards the South Pole => the initial GC track at point 1 will be equal to RL track + the conversion angle and will gradually decrease towards the mid-point, where the GC track will be equal to the RL track. From this point on the GC track will decrease below the RL track until reaching point 2 where the difference between RL track and GC track will be the greatest (same as at departure point 1). The change in Longitude is 10~ Conversion angle can be calculated using the formula: • Conversion angle = J.2 Ch Long x sin Latitude • Conversion angle = J.2 70° x sin 60° • Conversion angle =4,33° With a RL track of 090° and the conversion angle of approx. 4,33° the initial GC track will be 94,33° (090° + 4,33°). The INS will follow the GC track, because it is the shortest distance between two points. The GC track will be decreasing from 94,33° at point 1 to 090° (RL track) at a mid-point location = 25°W. The average track between point 1 and the mid-point will be 92,16° ((94,33° + 090°) .;. 2) = average 2,16° difference from the RL track. The RL distance from point 1 to the mid-point = Change in Longitude (in minutes) x cos Latitude => 300 minutes x cos 60° = 150 NM. Now we can apply the 1:60 rule: • TKE = (Dist off-track x 60).;. Dist along track ·2,16°= (?x 60).;. 150 ·?=2,16.;.60x 150 ·?=5,4NM At the mid-point (25°W) the aircraft will be located 5,4 NM to the South of the RL track. RL track follows the parallel of600s. We know that 1 NM along a Meridian = 1 minute of Latitude. 5,4 NM will therefore equal 5,4 minutes of Latitude South of60°5 => the approximate Latitude reached at point 025°W will be 60°05'24"5 => approximately 60°05' 5.

01 Basics of Navigation

4868. Airplane ATPL CPL Heli ATPL CPL In the Northern Hemisphere the rhumb line track from position A to B is 230°, the convergence is 6° and the difference in longitude is 10°. What is the initial rhumb line track from B

to A? A) 050° B) 053° C) 056° D) 047° (Refer to figures 061-E82 and 061-E83) A rhumb line is a line ofconstant direction on the surface of the Earth or a chart. This is because it crosses each meridian at a constant angle. Meridians are the datums used for measuring true direction. The Equator and Meridians are both Rhumb Lines and Great Circles by definition. Parallels of Latitude must also be Rhumb lines because they are lines of constant direction since they run East/West. Places on the same parallel of latitude are therefore due East or West of each other. The solution of this question is very easy - do not get tricked by the JAA into performing a complex calculation. JAA is trying to confuse you by providing a lot of extra and unnecessary information. Rhumb Line is a constant track line => all the way from A to 8 the track will be 230°. Therefore, the track from 8 to A will be a simple reciprocal of 230° => 230° - 180° = 050°. It is as simple as that.

4872. Airplane ATPL CPL Heli ATPL CPL The great circle bearing from A (70 0 S 030 0 W) to B (70 0 S 0600E) is approximately: A) 090° (T) B) 048° (T) C) 132°(T) D) 312° (T) (Refer to figures 061-E82 and 061-E83) The difference between the Initial or Final GC and the RL track is represented by the value of the Conversion Angle: • Conversion angle = Jo:2 Ch Long x sin Latitude (or sin mean Lat). • or: Conversion angle = Jo:2 (Initial Great Circle - Final Great Circle track). The difference between the Initial GCtrack and the Final GC track is represented by the value of Convergency: • Convergency =Change in Longitude x sin of Latitude • or: Convergency = GC Initial True Track - GC Final True Track The GC tracks lie closer to the poles than the Rhumb Line (RL) tracks. There are some attributes of GC tracks: Northern Hemisphere:

• From the initial to the final point of the GC track, the track value increases in easterly direction and decreases in westerly direction Southern Hemisphere:

• From the initial to the final point of the GC track, the track value decreases in easterly direction and increases in westerly direction Now finally let's take a look at this specific question. Since the Latitude of both points A and 8 is the same (70°5) and point 8 is located due East of point A the Rhumb Line (RL) track would be 090~ Since both points are in the Southern Hemisphere, the GC track would be arcing down below the RL track, towards the South Pole => the initial GC track at A will be equal to RL track + the conversion angle (GC track will be higher than the RL track). The change in Longitude .between A and 8 is 90° (30° from 30 0W to the Greenwich and then further 60° to 60 0E). Conversion angle can be calculated using the formula: • Conversion angle = Jo:2 Ch Long x sin Latitude • Conversion angle = Jo:2 90° x sin 70° • Conversion angle = 42° With a RL track of 090° and the conversion angle of approx. 42° the initial GC track from A to 8 will be 132° (090° + 4n.

4903. Given:

Airplane

ATPL

Heli

CPL

ATPL

Position B:

W~at is the positiori of B?

4868 (A)

I

4872 (e)

I

4903 (C)

Ifyou are an overachiever then the calculation is as follows: Draw a sketch with the starting point and your track line. Now draw a horizontal line from the starting point ':4" and a vertical line from the finishing point "8" => you now have a right angle triangle with an internal angle of 30° at point A. We know that the longest arm of the triangle is 200 NM => we can easily calculate the remaining two arms of the triangle: • cos 30° =Distance 1 + 200 NM • Distance 1 =cos300x200NM • Distance 1 = 173,2 NM • sin 30° = Distance 2 + 200 NM • Distance 2 = sin 30° x 200 NM • Distance 2 = 100NM Now that we know the distances along the parallel and meridian, we can easily calculate the new co-ordinates: • 1 minute of an arc along the Equator equals 1 NM • Distance of 173,2 NM = 173,2 minutes • 173,2 minutes = approx. 2°53' Westbound • Longitude of 8 will be 1000E minus 2°53' => 9rOl'E When following a Great Circle in the North/South direction (along a Meridian) each 1°change ofLatitude represents a distance of 60 NM (1 minute ofLatitude =INM). • 1 minute of an arc along the Meridian equals 1 NM • Distance of 100 NM = 100 minutes • 100 minutes = approx. 1°40' Southbound • Latitude of 8 will be OOooN/S minus 1°40' => 1°40'S The co-ordinates ofposition 8 will be 001°40'5 09rOl'E.

4905. Airplane ATPL CPL Heli ATPL CPL Position A is located on the equator at longitude 1300 00'E. Position B is located 100 NM from A on a bearing of 225° (T). The coordinates of position Bare: ' A) 01°11'N 128°49'E B) 01°11'5128°49'E C) 01°11'N 131°11'E D) 01°11'5131°11'E (Refer to figure 061-E41) You actually don't need to do any calculations. Take a deep breath, have a sip of water and look at the question and available answers one more time. Hopefully, you found the correct answer. Time management is essential to completing these tests in the given amount of time. Don't waste valuable time doing unnecessary work! You are located at the Equator at 1000E and the track ofyour flight is 240° = to the South West (SW). Out of the possible answers only one meets the SW requirement with respect to the starting positionA. Ifyou are an overachiever then the calculation is as follows: Draw a sketch with the starting point and your track line. Now draw a horizontal line from the starting point "A" and a vertical line from the finishing point "8" => you now have a right angle triangle with an internal angle of 45° at point A. We know that the longest arm of the triangle is 100 NM => we can easily calculate the remaining two arms of the triangle: • cos 45°= Distance 1 + 100NM • Distance 1= cos 45° x 100NM • Distance 1 =70,71 NM

Now that we know the distances along the parallel and meridian, we can easily calculate the new co-ordinates: • 1 minute of an arc along the Equator equals 1 NM

A) 01°40'5101°40'E

I

(Refer to figure 061-E39) You actually don't need to do any calculations. Take a deep breath, have a sip of water and look at the question and available answers one more time. Hopefully, you found the correct answer. Time management is essential to completing these tests in the given amount of time. Don't waste valuable time doing unnecessary work! You are located at the Equator at 100 0E and the track ofyour flight is 240° = to the South West (SW). Out of the possible answers only one meets the SW requirement with respect to the starting positionA.

• sin 45°= Distance 2 + 100 NM • Distance 2 = sin 45° x 100 NM • Distance 2 =70,71 NM

OOON 1000E 240 0(T), 200 NM from A

Position A:

CPL

B) 01°40'N 097°07'E C) 01 °40'5 097"07'E D) 01°40'N 101°40'E

I

4905 (8)

I

Aviationexam Test Prep Edition 2012 • Distance of 70,77 NM = 70,71 minutes ·70,71 minutes = approx. 1°11' Westbound • LongitudeofB will be 130 0 Eminus 1°11'=> 128°49'E

230609. Airplane ATPL CPL Heli ATPL CPL Geodetic latitude and geocentric latitude coincide

When following a Great Circle in the North/South direction (along a Meridian) each 1°change ofLatitude represents a distance of60 NM (1 minute ofLatitude = 1 NM). • 1 minute of an arc along the Meridian equals 1 NM • Distance of 70,71 NM = 70,71 minutes ·70,71 minutes =approx. 1°11' Southbound • Latitude of B will be OOooN/S minus 1°11' => 1°11'S The co-ordinates ofposition B will be 001°11'5 128°49'E.

4914. Airplane ATPL CPL Heli ATPL CPL An aircraft at position OooOO'N/S 163°27'W flies a track of 225° (T) for 70 NM. What is its new position? A) B) C) D)

00049'N 162°38'W 00049'5162°38'W 00049'N 164°16'W 00049'5164°16'W

This is a fairly simple question - there is no need for any calculation - just use the process of elimination of incorrect answers. The starting position is at the Equator and the direction of flight is to the South-West (225° T) - therefore the final position is going to be in the Southern Hemisphere => answers A) and C) are incorrect. The Longitude of the starting point is 163°27'W and as we stated above, the direction of flight is SW => the longitude of the final position has to be further to the West of the starting position => that makes answer B) incorrect. The only answer that meets the criteria that the final position has to be to the SW of the starting position is answer D) - the correct one.

230580. Airplane ATPL CPL Heli ATPL CPL The main reason for the occurrence of seasons on earth is A) B) C) D)

the elliptical form of the orbit of the earth around the sun. the distance between the sun and the earth. the length of the day as stated by the second law of Kepler. the inclination of the earth axis with regard to the plane of the ecliptic.

230592. Airplane ATPL CPL Heli ATPL CPL Given Waypoint 1600S 0300W and Waypoint 260 0S 0200W, which have been inserted into an INS connected to an autopilot, what will be the approximate Latitude shown on the display at longitude 025°W? A) B) C) D)

only on the equator. at 45~N/5. only at the Poles. at the Poles and on the equator.

230610. Airplane ATPL CPL Heli ATPL CPL What is the correct definition of latitude of a position on the Earth? A) Latitude is the angle between the plane of the ecliptic and the parallel of the position. B) Latitude is the angle between the Earth's rotational axis and the line from the centre of the earth to the position. C) Latitude is the angle between the plane of the equator and the line from the centre of the earth to the position. D) Latitude is the angle between the plane of the Prime Meridian and the plane ofthe meridian ofthe position. 230611. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-11) Which figure in the Appendix represents the geographic latitude of position P, which is situated above the surface of the ellipsoid? A) B) C) D)

Figure C. Figure B. Figure A. Figure D.

230612. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-11) Which figure in the Appendix represents the geocentric latitude of position P, which is situated above the surface of the ellipsoid? A) B) C) D)

Figure B. Figure C. Figure D. FigureA.

230613. Airplane ATPL CPL Heli ATPL CPL On the Earth's ellipsoid one degree of latitude near the equ<;itor is:

060°11'5 059°49'5 060°00'5 060°06'5

Airplane ATPL CPL Heli ATPL CPL 230605. A great circle intersects the equator in 030 0W with a great circle direction of 035°(T). An aircraft tracking the great circle will reach the maximum Northern/Southern latitude in position A) (55°N, 0600E) B) (55°5, 0600E) C) (35°5, 1200W) D) (35°N, 1200W) 230606. Airplane ATPL CPL Position A (56°00.0'S, 163°57.2'E) Position B (56°00.0'S, 171°47.4'W) For the route from A to B the:

= =

A) B) C) D)

A) B) C) D)

Heli

ATPL

CPL

A) B) C) D)

more than 60 NM but less than 61 NM. more than 60 NM. 60NM. less than 60 NM.

Airplane ATPL CPL Heli ATPL CPL 230626. An aircraft follows a great circle in the Northern Hemisphere. At a certain moment the aircraft is in the position on the great circle where the great circle direction is 2700(T). Continuing on the great circle the: A) B) C) D)

track angle will track angle will track angle will track angle will

decrease and the latitude will increase. increase and the latitude will decrease. increase and the latitude will increase. decrease and the latitude will decrease.

rhumb line distance is 1206,6 NM. rhumb line distance is 1455,4 NM. great circle direction at B is 080.7° great circle direction at B is 100.1°.

1 4914 (D) 1230580 (D) 1230592 (0) 1230605 (A) 1230606 (C) 1230609 (0) 1230610 (C) 1230611 (C) 1230612 (A) 1230613 (0) 1 1230626 (D) 1

~~~~~------~---------------------------

01 Basics of Navigation

230627. Airplane ATPL CPL Heli ATPL CPL Which definition describes best the notion "Poles i,? A) The Poles are the points of intersection between the surface of the earth and the extended line joining the star Polaris with the centre of the Earth. B) The Poles are the points of intersection between the earth's axis and the surface of the Earth. C) The Poles are the points on the surface of the Earth where gravity acts under an angle of exactly 90°. D) The Poles are the points on the surface of the Earth which have the same distance to all points of the ecliptic. 230629. Airplane ATPL CPL Heli ATPL CPL An aircraft is in the position (86°N, 020 0E). When following a rhumb line track of 085°(T) it will: A) follow a small circle which lies to the North of the parallel of 86°N. B) follow a line which lies at first to the North of the parallel of 86°N but after having passed a DL of 180° to the South of it. C) fly to the north via an arbitrary line. D) fly via a spiral to the North Pole. 230630. Airplane ATPL CPL Heli Which definition of the equator is correct?

ATPL

CPL

A) The equator is a greatcircle with its plane parallel to the Earth rotational axis. B) The equator is a small circle, the plane is parallel to the Earth rotational axiS, C) The equator isa greatcircle with its plane perpendicular to the Earth rotational axis. D) The equator is a small circle with its plane perpendicular to the Earth rotational axis. 230632. Airplane ATPL CPL Heli ATPL CPL Which statement about meridians is correct? A) A meridian and its anti-meridian form a complete great circle. B) The plane of a meridian is parallel to the equator. C) A meridian is a complete great circle of 360°. D) All meridians-are parallel to the Meridian of Greenwich.

230639. Airplane ATPL CPL Position A (30 0 00.0'N, 175°23.2'W) Position B (30 0 00.0'N, 173°48.1'E) For the route from A to B the

Heli

ATPL

CPL

= =

A) great circle direction at A is 275.4°. B) great circle direction at B is 092]0. C) rhumb line distance is 578NM. D) rhumb line distance is 648.7 NM. 230650. Airplane ATPL CPL Heli ATPL CPL The reason that the solar day lasts longer than the sidereal day isthat A) the orbital track of the earth around the sun is an ellipse. B) both the direction of rotation of the Earth around its axis and its orbital rotation around the sun are the same. C) the sun has an own movement through the universe opposite to the movement of the earth due to the gravitational forces of celestial bodies. D) the orbital speed ofthe earth around its axis diminishes slowly. 230671. Airplane ATPL CPL Which statement is true?

Heli

ATPL

CPL

A) Civil twilight at the equator lasts longer than at 600N or 600S because the radius of the equator is larger than the radius of the 600parallel. B) The declination of the Sun and the latitude of the observer will affect the duration of civil twilight. C) Only the declination of the Sun will affect the duration of civil twilight. D) The duration of the civil twilight on 21 st of March and on 23,d of September is equal at all places on Earth independent of latitude. 231094. Airplane ATPL CPL Heli ATPL CPL An aircraft flies from waypoint 7 (63°00' N, 073°00'W) to waypoint 8 (62°00' N, 073°00' W). The aircraft position is (62°00' N, 073°10'W). The cross track distance in relation to the planned track is: A) 8,8 NM L B) 4,7NMR C) 8,8 NM R D) 4,7 NM L

01-03 Time and time conversions 1126. Airplane ATPL CPL Heli ATPL CPL The countries having a standard time slow on UTe: A) will often have an earlier standard date than the UTC date. B) will generally be located at western longitudes. C) will often experience sunrise earlier than the sunrise occurs at the Greenwich meridian. D) will generally be located at eastern longitudes. (Refer to figures 067-E707, 067-E702, 067-E703 and 067-E704) The English grammar used in many of these questions is extremely poor. The terms "slow and fast" have no place in this question. What the question is attempting to demonstrate is that if you are located West of the Greenwich line then you will be behind in time in relation to the time at Greenwich. For example, ifyou are in Iceland and it is midnight in Greenwich, England then you know the time in Iceland is going to be before midnight.

Converting UTC into LMT and vice versa is straightforward, using the fixed relationship between longitude and the time of 75° difference per hour. Any difference in LMT between places is caused by their difference in longitude. When Local Mean Time at Greenwich is 72:00 (72:00 UTC), Local Mean Time at longitude 07 SoW will be one hour earlier (77:00 LMT - the Sun has not yet reached the meridian of 07S 0 W), but UTC at the position is stiff 72:00. A good, reliable and straight-forward rule to use when converting to and from UTC and or LMT is: • Longitude East - UTC least (smaller) - standard time fast on UTC • Longitude West - UTC best (greater) - standard time slow on UTC

1230627 (8) 1230629 (D) 1230630 (C) 1230632 (A) 1230639 (C) 1230650 (8) 1230671 (8) 1231094 (8) 1 1126 (8)

1

Aviationexam Test Prep Edition 2012

1131. Airplane ATPL CPL Heli ATPL CPL A day at a place as measured in local mean time starts:

1159. Airplane UTe stands for:

A) when the mean Sun transits the meridian of the place in question. B) when the mean Sun transits the Greenwich meridian. C) when the mean Sun transits the anti meridian of the place in question. D) when the mean Sun transits the 180 E/W meridian. (Refer to figures 067-E107, 067-E702, 067-E103 and 067-E704) Time at a place needs to be related directly to the position (i.e. the longitude) of the Sun in order to follow the regular routines of daily life. These routines require that a start to the day is made at or near sunrise and that the middle of the day will be coincident with the Sun passing overhead the local meridian. Similarly, the time for sleep is when the Sun is over the horizon (at a different longitude West of the observer). Local Mean Time (LMT) is the time at a position (i.e. longitude) relative to Sun position. When the mean Sun transits the antimeridian of a position, the local day starts at 00:00 hours LMT. The Sun transits the actual meridian of the position at local midday - 72:00 LMT. Crossing the anti-meridian again occurs at 24:00 LMT, the end of the day. Simultaneously, 00:00 signifies the start of another day. It will be clear that all places on any given meridian have the identical LMT as time is regulated by the position of the Sun in relation to the local meridian. This is not the case for positions on Earth at different longitudes. The Local Mean Time at a position is required when calculating precise sunrise/sunset or twilight times at a particular longitude. Even though Local Standard Time (LST) may be the same at two places, the time difference between the LMTs of the places at different longitudes will be equal to their difference in/ongitude divided by 75° per hour (the average rotational speed of the sun). For example, Lyon and Bordeaux both keep French standard time. As one place is West of the other, however, the times of sunrise and sunset will differ by several minutes. Longitude can be converted into time at the rate of 75° of longitude per hour which is the average rate of motion of the Sun across the sky: ·7 hour of the Sun movement across the sky = 75° of longitude (or 7° in 4 minutes).

1157. Airplane ATPL CPL Heli ATPL CPL What is the difference between UTe and GMT? A) GMT is valid only at the Greenwich meridian and is of no use at other longitudes. B) UTC is adjusted abruptly at announced time and is hence not practical to use. C) UTC is slightly more accurate than GMT, but the difference between the two is so small that it has no importance in everyday navigation of aircraft. D) All answers above are correct. (Refer to figures 067-E107, 067-E102, 067-E103 and 067-E704) Greenwich Mean Time (GMT) is local time on the Greenwich meridian based on the hypothetical mean Sun. As the Earth's orbit is elliptical and its axis is tilted, the actual pOSition of the Sun against the background ofstars appears slightly ahead or behind the expected position. The accumulated timing error varies through the year in a smoothly periodic manner by up to 74 minutes slow in February to 76 minutes fast in November. The use of a hypothetical mean Sun removes this effect. GMT was used generally until 7925. At that time the reference point was changed from noon to midnight and it was recommended that, to avoid confusion, the term "Universal Time" should be used for the "new GMT'~ Universal co-ordinated time (UTe) is a time standard more accurately calculated than the local mean time atthe Greenwich meridian (Greenwich Mean Time), but in practice it amounts to the same thing. UTC is regulated against International Atomic Time. Flight arrival and departure times are usually given in as local times (Local Standard Time) for passenger convenience. In civil aviation, however, all operational times are given using a common time datum UTC - for air safety and to prevent ambiguity. Position reporting and estimating is always made using UTe. While this can result in apparent anomalies if UTC is compared with local time at the operating base, it is essential and is followed throughout civil aviation by all operators and agencies.

1131 (C)

I

1157 (C)

I

1159 (A)

I

lEI -------------

1182 (C)

I

A) B) C) D)

ATPL

CPL

Heli

ATPL

CPL

Universal Time Co-ordinated. Universal Time Coefficient. Universal Time Constant. Universal Time Compensated.

(Refer to figures 067-E107, 067-E102, 067-E103 and 067-E704) Universal co-ordinated time (UTC) is a time standard more accurately calculated than the local mean time at the Greenwich meridian (Greenwich Mean Time), but in practice it amounts to the same thing. UTe is regulated against International Atomic Time. Flight arrival and departure times are usually given in as local times (Local Standard Time) for passenger convenience. In civil aviation, however, all operational times are given using a common time datum UTC - for air safety and to prevent ambiguity. Position reporting and estimating is always made using UTe. While this can result in apparent anomalies if UTC is compared with local time at the operating base, it is essential and is followed throughout civil aviation by all operators and agencies.

1182. Airplane Standard time is:

ATPL

CPL

Heli

ATPL

CPL

A) the time which is accepted and used as a standard for the whole world. B) the time most frequently used for air navigation. C) the time enforced by the legal authority to be used in a country or an area. D) the time used at a particular meridian. (Refer to figures 067-E707, 067-E702, 067-E703 and 067-E704) Local Standard Time (LST) is a convenient system where everywhere in a country, or in a time zone in a country keeps the same time as everywhere else. If Local Mean Time were used throughout a country, particularly one with a large East/West spread, life would be very difficult if not impossible (except for places north/south of each other) as all LMTs would be different. The system to counteract this problem is called Local Standard Time. Standard time is the time kept by a country. It is decided by the government of that country alone. It can also be the time in a particular zone of a country with a largechangeoffongitude. Usually, itis theLMTofthe centre ofthecountry adjusted to a convenient figure ahead or behind UTe. The value chosen will usually reflect the passing of the Sun'at midday.

1199. Airplane ATPL CPL Heli ATPL CPL Some standard times may differ from UTe by other times than whole hours, because: A) some areas have limited communication with neighboring areas, which does not call for coordinated standard times. B) the political authorities have emphasized the importance ofthe sunlight period in a particular position. C) it has been considered highly desirable that the sunlight period of the day is balanced around noon, standard time. D) all answers are correct. Have a look at a website that details the time in different cities around the world. You will notice that Afghanistan is 30 minutes different from the whole hour. Even stranger is the land of Nepal. The time there is set 75 minutes different from the whole hour. I doubt there is anyone reason thqt certain countries do this action and I also doubt that the reasons for doing this are solely contained as the three reasons mentioned above.

1203. Airplane A day is defined as:

ATPL

CPL

Heli

ATPL

CPL

A) the period from morning to evening. B) the period from sunrise to sunset. C) the period in which day flying is authorized. D) the period elapsed between two successive transits of a heavenly body. For explanation refer to question #2256 on page 2.

1199 (D)

I

1203 (D)

I

01 Basics of Navigation

1216. Airplane ATPL CPL Heli ATPL CPL When the length of the day is measured with reference to the passage of the apparent Sun:

when the light is refracted from the atmosphere of the Earth and thus giving a certain quantity of illumination (the amount of illumination experienced makes it possible to carry out daytime tasks without the need of additional artificial lighting).

A) the length of the days, as indicated by our watches, will be exactlyequal. B) the length of the day will vary with the latitude of the observer. e) the length of the day will be the same once every month. 0) the length of the day will vary in the course of the year.

The duration of twilight is affected by the following factors: a) The declination of the Sun, and therefore by implication the time ofyear. When the Sun is in observer's hemisphere the duration of twilight increases as the Sun's declination increases. b) The observer's latitude. As latitude increases so the duration of twilight increases. At high latitudes, it is possible for civil twilight to exist all night. c) The observer's altitude. Although only MSL tables are provided for the course and the examination, there are tables available at 1000 ft intervals up to 50,000 ft above MSL. The high altitude tables show that twilight commences earlier and ends later than at MSL but has a shorter duration. This is because of the decreased atmosphere and decreased scattering effect with increased altitude.

For explanation refer to question #2256 on page 2.

1230. Airplane ATPL CPL Heli ATPL CPL The length of a apparent solar day is not constant because: A) the Earth's speed of revolution in its orbit varies continuously, due to the orbit being elliptical. B) the Earth's speed of rotation is not the same at all latitudes. e) the Sun's declination is not constant. 0) the plane of the Ecliptic and the plane of the Equator are inclined to each other.

1243. Airplane ATPL CPL Heli ATPL CPL By the term "transit" of a heavenly body it is understood that: A) the body is moving. B) the body is passing the meridian of the observer or another specified meridian. e) the body is passing the anti meridian of the observer. 0) the body is at the same celestial meridian as another body.

For explanation refer to question #2256 on page 2.

1237. Airplane ATPL CPL Heli ATPL CPL Which of the following alternatives is correct when you cross the international date line? A) The date will increase if you are crossing on a westerly heading. -B) The date will increase if you are crossing on a easterly heading. e) The date will always be the same. 0) If you are crossing from westerly longitude to easterly long itude the date will remain the same. (Refer to figures 061-EI1 and 061-E14) The local date of a place always changes at 24:00 LMTILST.lt will also change at longitude 180 0EIW (the International Date Line) when the Sun crosses the Greenwich (Prime) meridian. The loL follows the general direction of the Greenwich anti-meridian. When an aircraft crosses the line, the local date for passengers and crewcranges.lfthe flight is from the western hemisphere to the eastern hemispFmie (say from 1700W to 1700E) on the 4 October, the date will change to 5 October when crossing the Greenwich anti-meridian. If the flight is from 1700E to 1700W and the flight departs on the 5 October, the date will change to 4 October when crossing the Greenwich anti-meridian. It is perhaps fortunate for some countries that the Greenwich ante-meridian runs mainly through the Pacific Ocean with a few zig-zags to keep all islands in a group under the parent country or the main island. This avoids a country having two different days in different parts of the state. For example, many Pacific South sea islands are part of the New Zealand time system for mutual convenience. Note: It may be useful to think of the Earth as stationary and the Sun as orbiting westwards instead of the Sun as stationary and the Earth as orbiting eastwards like in reality. See the attached figure - it can be seen that the date will increase .ifyou are crossing on a westerly heading.

1239. Airplane ATPL CPL Morning civil twilight begins when: A) B) e) 0)

Heli

ATPL

CPL

For explanation refer to question #2256 on page 2.

1259. Airplane ATPL CPL Civil twilight is defined by: A) B) e) 0)

Sun's altitude is 12° below the celestial horizon. Sun's altitude is 18° below the celestial horizon. Sun's upper edge tangential to horizon. Sun's altitude is 6° below the celestial horizon.

1268. Airplane ATPL CPL The "Equation of time":

A) should be prepared to increase your date by 1. B) should increase your date by an extra date at the first midnight you experience. e) should be prepared to decrease your date by 1. 0) should not change date at the first midnight you experience.

2227. Airplane ATPL CPL Heli ATPL CPL Observed from a position on the surface ofthe Earth the heavenly bodies seem to:

I

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1243 (8)

CPL

1288. Airplane ATPL CPL Heli ATPL CPL When approaching the International Date Line from the East, you:

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1237 (A)

ATPL

When the apparent Sun is ahead of or behind the mean Sun, apparent time is faster or slower than mean civil time. The difference between apparent and civil time (or between the apparent solar day and the mean solar day) at any instant is called the equation of time. It is used to convert civil time at any instant to apparent time and vice versa. The equation of time varies from +14 minutes in February to -16 minutes in November.

A) B) e) 0)

I

Heli

A) states the difference in time of transit of the mean Sun and the apparent Sun any particular day. B) states the difference between celestial time an apparent time. e) is used to calculate mean time when standard time is known. 0) is used when calculating the difference between UTe and LMT.

Civil twilight is the value used in civil aviation. Morning civil twilight starts when the Sun is 6° below the horizon and lasts until sunrise, when the upper limb of the Sun rises to the horizon of the observer. Evening civil twilight is of course the reverse condition. The true horizon is the tangent to the Earth's surface. The twilight is that period of time before sunrise and after sunset

1230 (A)

CPL

For explanation refer to question #1237 on this page.

(Refer to figures 061-El05 and 061-El06) There are three types of twilight: • Civil twilight, when the Sun is 6° below the horizon • Nautical twilight, when the Sun is 12° below the horizon • Astronomical twilight, when the Sun is 18° below the horizon.

I

ATPL

For explanation refer to question #1239 on this page.

the sun's upper edge is tangential to the celestial horizon. the centre of the Sun is 12° below the celestial horizon. the centre of the Sun is 18° below the celestial horizon. the centre of the Sun is 6° below the celestial horizon.

1216 (0)

Heli

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1259 (0)

not change their relative positions on the sky. move from West to East on the Southern Hemisphere. move from East to West on the Northern Hemisphere. move from East to West.

1268 (A)

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1288 (A)

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2227 (0)

lEI

Aviationexam Test Prep Edition 2012 The Earth is described as rotating from West to East. If you were looking down on the North Pole from space, the Earth would be seen to be rotating anticlockwise. This is why the Sun appears to rise in the East and set in the West to an observer standing on the ground. The Earth then rotates around the Sun in an anti-clockwise direction on an elliptical path.

2228. Airplane ATPL CPL Heli ATPL CPL What is the definition of "morning civil twilight"? A) The period in the morning, just before sunrise. B) Morning civil twilight is the period in the morning from the centre of the Sun is 6° below the horizon until the upper limb of the Sun appears at the horizon. C) The period when the Sun, in the morning, has its centre between 6° under the horizon and the horizon. D) The period of semi-darkness in morning. For explanation refer to question #1239 on page 23.

2262. Airplane ATPL CPL Heli ATPL CPL What is the meaning of the term standard time? A) B) C) D)

It is the time zone system applicable only in the USA. It is a expression for local mean time. It is another term for UTe. It is the time set by the legal authorities for a country or part of a country.

For explanation refer to question #1182 on page 22.

2277. Airplane ATPL CPL Heli Daylight Saving Time (Summer Time):

ATPL

CPL

A) is used in some countries. B) is used to extend the sunlight period in the evening. C) is introduced by setting the standard time forward by one hour. D) all answers are correct. Daylight saving time (DST; also summer time in British English) is the convention of advancing clocks so that afternoons have more daylight and mornings have less. Typically clocks are adjusted forward one hour near the start of spring and are adjusted backward in autumn. Modern DST was first proposed in 1907 by William Willett. Many countries have used it since then; details vary by location and change occasionally. If you ever forget which way to set your clock, just remember the old saying, "Spring forward / Fall back!"

2287. Airplane ATPL CPL The "duration of twilight":

Heli

ATPL

CPL

A) will in the period around the equinoxes increase as you approach the equator from North or South. B) is generally longer in positions at high latitudes than in positions at lower latitudes. C) is independent of the Sun's declination, and only depends on the observers latitude and longitude., D) is longer in the morning than in the evening because ofthe refraction in the atmosphere. For explanation refer to question #1239 on page 23.

2316. Airplane ATPL CPL Heli ATPL CPL Consider the following statements on sunrise and sunset: A) at equator sunrise and sunset occur at quite regular times throughout the year. B) in May sunrise occurs later at 45°00'N than at 45°00'S. C) in November sunset occurs earlier at 45°00'S than at 45°00'N. D) in July the period of sunlight is longer at 15°00'S than at 15°00'N. (Refer to figures 061-E105, 061-El06 and 061-E18) The Earth's axis (which passes through the North and the South Pole) is tilted at an angle of 23.5 0 to line that is perpendicular to the orbital plane. Thus the angle between the Earth's axis and the orbital plane is 90 0 - 23.5 0 = 66.5~ Because of this tilt, as the Earth rotates around the Sun, the rays will be

I

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2262 (0)

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perpendicular to the surface in the region between the parallels 23.5 0 Sand 23.5 0 N. The Sun is over the Equator on the 21" of March - as the Earth rotates, the perpendicular rays are moving North so that on the 21" of June the Sun is overhead the 23.5°N parallel. Then moving South so that on the 23,d ofSeptember the Sun is again overhead the Equator and on the 21" of December will be overhead the 23.5 0 S parallel. The result is that the daylight period is extended in the far North during the summer months and drastically reduced in the winter months. If you are at a location near the equator, for example Ecuador, then you would have a fairly consistent sunrise and sunset at relatively similar times throughout the entire year - approximately ibround 06:00/18:00 LMT every day plus or minus a few minutes. '

2903. Airplane ATPL CPL Heli ATPL CPL "Mean time" has been introduced in order to: A) compensate for the irregularities of the speed of rotation of the Earth around it's axis. B) have one fixed time to be used within the border of a country. C) introduce a constant measurement of time, independent of the daily variations in the movement of the Sun as observed from the Earth. D) save energy as the problem of adjusting our watches entering every leap year is eliminated. For explanation refer to question #2256 on page 2.

4096. Airplane ATPL CPL Heli ATPL CPL The main reason that day and night, throughouttheyear, have different duration, is due to the: A) B) C) D)

inclination of the ecliptic to the equator. Earth's rotation. relative speed of the Sun along the ecliptic. gravitational effect of the Sun and Moon on the speed of rotation of the Earth.

For explanation refer to question #2316 on this page.

4173. Airplane ATPL CPL Atmospheric refraction:

Heli

ATPL

CPL

A) causes the sunrise and the sunset to occur earlier. B) causes the sunrise and the sunset to occur later. C) causes the sunrise to occur later and the sunset to occur earlier. D) cause the sunrise to occur earlier and the sunset to occur later. (Refer to figures 061-E105 and 061-E106) , Sunrise or sunset occurs when the upper edge of the Sun is on the observer's hb" rizon. Certain atmospheric conditions can affect the times ofsunrise and sunset and the length of twilight. If the air is clear, the Sun will rise or set quickly. When the air is thick with dust or water particles, it is occasionally possible to see the Sun after it has "official/y" set or before the official time of sunrise because of refraction of the sun's rays. The amount of refraction will determine how early or late'a sunrise or sunset will be and how long a sunrise or sunset will last.

4179. Airplane ATPL CPL Heli ATPL CPL Times of sunrise and sunset is in the Air Almanac only given for one particular time in every 24 hour period. These dC!,~ are accurate: A) enough to be used for all longitudes, when calculating light conditions. B) but may call for an adjustment if the observe is at a high altitude. C) only for the places on the Greenwich meridian. D) all answers are correct. (Refer to figures 061-El05 and 061-E106) The times are given for a particular latitude and date. This is because the Sun rises (and sets) at the same LMT for all places on the same latitude. The times of sunrise and sunset at any place will not change greatly over successive days, except at high latitudes. The times tabulated in the tables for specific latitudes may therefore be taken as the same for aI/longitudes. The LMTfor a particular date and place (latitude) is extracted and then converted to the time standard

2316 (A)

I 2903 (C) I 4096 (A) I

4173 (0)

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4179 (0)

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01 Basics of Navigation

(LSTILMT) required. Sunrise, sunset and twilight calculations are time calculations based on the LMT ofsunrise, sunset or twilight.

4190. Airplane "Apparent Time" is:

ATPL

Heli

CPL

ATPL

4288. Airplane ATPL CPL Heli ATPL CPL If the mean Sun moves 121°30' along the equator, that equals: A) 20 hours 10 minutes. B) 9 hours 15 minutes. C) 6 hours 20 minutes. D) 8 hours 06 minutes.

CPL

A) the correct time, as it appears on our watches and clocks. B) based on the time of transit of the apparent Sun. C) the time that is apparent to everybody: in other words the official time. D) the average time, calculated from the movements of the Sun and the stars.

(Refer to figures 061-E101, 061-El02, 061-El03 and 061-El04) The Earth rotates around its axis in 24 hours. There are 360° of Longitude => the Sun appears to be traveling across the sky at a rate of 15° of Longitude in 1hour => 1° in 4 minutes. A Longitude change of 121,5° will therefore be covered in 8 hrs 6 min (121,5° x 4 min = 486 min = 8,1 hrs = 8 hrs 6 min).

For explanation refer to question #2256 on page 2.

4194. Airplane ATPL CPL Heli The International Oate Line is located: A) B) C) D)

ATPL

4305. Airplane ATPL CPL Heli ATPL CPL 5 hrs 20 min 20 sec corresponds to a longitude difference of:

CPL

at the apparent sun's anti meridian. at the Greenwich meridian. at the 180° E/W meridian, or in the vicinity ofthis meridian. at latitudes on the 180° E/W meridian.

For explanation refer to question #1237 on page 23.

4197. Airplane ATPL CPL Heli ATPL CPL Consider the following statements on sunset: A) for positions at the same longitude, sunset will occur simultaneously at all latitudes. B) at sunset the centre of the Sun is at the observers horizon. C) sunset is the time when the observer at sea level sees the last part of the Sun disappear below the horizon. D) night-flying regulations start at the time of sunset.

A) 75°00' B) 78°45' C) 80°05' D) 81°10'

(Refer to figures 061-E101, 061-E102, 061-£103 and 061-El04) The Earth rotates around its spin axis in a day, i.e. it rotates 360° in 24 hours. Therefore 15°oflongitude= 1 hour 1°of longitude =4 minutes of time 5 hours => 20 minutes => 20 seconds =>

Airplane

ATPL

Heli

CPL

ATPL

4324. Airplane ATPL CPL Civil twilight occurs between:

CPL

ciallighting. C) are the periods when an observer is illuminated by both di-

rect and indirect rays of light from the Sun. D) are the periods before sunrise and after sunset when the light

4215. Airplane ATPL CPL Heli ATPL CPL The times given for sunrise, sunset, morning and evening twilight in the Air Almanac: are given in Standard Time. are given in UTC. must always be corrected for atmospheric refraction. are given in LMT.

For explanation refer to question #4179 on page 24.

~he duration

ATPL

Heli

CPL

ATPL

CPL

I I

D) 50000'S010 035'E.

(Refer to figures 061-El0l, 061-El02, 061-E103 and 061-EI04) The difference between 19:41 and 21:13 is 1 hr 32 min = 92 minutes. The Sun appears to be traveling across the sky at a rate of 1° of longitude in 4 minute (15° in 1 hour). 92 minutes..,. 4 = 23° of longitude. Note that the sunset is listed as 21:13 urc and it happened at 19:41 LMT => obviously our position is to the West of Greenwich. .

A) B) C) D)

of civil twilight is the time:

For explanation refer to question #1239 on page 23.

4190 (8) 4535 (A)

4194 (C)

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4197 (C)

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CPL

A) 50°00'5 023°00'W. B) 50°00'5 010 035'W. C) 50°00'5 022°00'E.

4535. Airplane ATPL CPL When the time is 20:00 UTC, it is:

A) between sunset and when the centre of the Sun is 12° below the true horizon. B) agreed by the international aeronautical authorities which is 12 minutes. C) needed by the Sun to move from the apparent height of 0° to the apparent height of 6°. D) between sunset and when the centre of the Sun is 6° below the true horizon.

I I

ATPL

4379. Airplane ATPL CPL Heli ATPL CPL On 4th February the Air Almanac lists 19:41 as the time of sunset at 50°00'5. An observer registers sunset at 21:13 UTC this day. What is the observers position? "

For explanation refer to question #1239 on page 23.

Airplane

Heli

For explanation refer to question #1239 on page 23.

is lower than when the Sun is above the horizon.

4240.

20..,. 60

A) sunset and 6° below the horizon. B) 6° and 12° below the horizon. C) 12° and 18° below the horizon. D) sunrise and sunset.

A) are periods when it is nearly dark. B) are periods when it is too dark to move around without artifi-

A) B) C) D)

300 minutes 20 minutes 0,333 minutes 320,33 minutes

320,33 minutes..,. 4 = 80,08° = 80°05' of Longitude.

For explanation refer to question #4173 on page 24.

4210. Twilight:

5 x 60

I

Heli

ATPL

CPL

14:00 LMT at 90° West. 24:00 LMT at 120° West. 12:00 LMT at 60° East. 08:00 LMT at the Prime meridian.

(Refer to figures 061-El0l, 061-E102, 061-E103 and 061-EI04) We are concerned with the longitude in this question. We have a location with a Longitude of900W => we can use the standard of1° ofLongitude = 4 minutes ofSun travel across the sky. 90 0x 4 minutes = 360 minutes. 360..,. 60 = 6 hours. 20:00 - 6 = 14:00. Final answer is 14:00 local mean time. We deduct the 6 hrs from the UTe because the position is Westerly. We would add 6 hours if the position was 90°f.

4240 (0)

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4324 (A)

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Aviationexam Test Prep Edition 2012 4550. Airplane ATPL CPL Heli ATPL CPL What is the local mean time, position 65°25'N 123°45'W at 22:00 UTC? A) 13:45 B) 22:00 C) 06:15 D) 08:15 (Refer to figures 061-EI0l, 061-EI02, 061-E103 and 061-EI04) We are concerned with the longitude in this question. Your first step is to convert the 45 minutes into something we can work with => 0,75°. So we have 123,75° and we can use the standard of 1° of Longitude = 4 minutes of Sun travel across the sky. 123,75° x 4 minutes =495 minutes. 495 + 60 =8,25 hours. 22:00 - 8,25 = 13,75 => 13:45. Final answer is 13:45 local mean time. So, we are basically talking about somewhere in the western part of North America.

4565. Airplane ATPL CPL Heli ATPL CPL (Refer to figures 061-01 and 061-02) What is the Standard Time in Hawaii when it is 06:00 ST on the 16th July in Queensland, Australia? A) 10:00 5T 15th July. B) 20:00 5T 15th July. C) 10:00 5T 16'h July. D) 10:00 5T 17th July. (Refer to figures 061-E101, 061-EI02, 061-EI03 and 061-EI04) It helps to know a little geography when doing these questions. In particular where the Prime Meridian and International Date Line are located in relation to Australia and the Hawaiian Islands. ·In Queensland at 06:00 ST on 16 July = 20:00 UTC on 15 July (note from the picture that Queensland is 10 hours ahead of GMT). ·In Hawaii at 20:00 UTe on 15 July = 10:00 ST on 15 July (note from the picture that Hawaii is 10 hours behind GMT).

4623. Airplane ATPL CPL Heli ATPL CPL The relationship between the mean Sun's movement along the equator and Mean time is: A) 10 of arc equals 4 minutes of time. B) 1800 of arc equals 12 hours of time. C) 5 hours of time equals 75 0 of arc. D) all answers are correct.

4671. Airplane ATPL CPL Heli ATPL CPL In 8 hours and 8 minutes the mean Sun has moved how many degrees (0) along the celestial equator? A) 180 B) 1480

C) 1220 D) 560 (Refer to figures 061-EI0l, 061-EI02, 061-E103 and 061-E104) The Earth rotates around its axis in 24 hours. There are 360° of Longitude => the Sun appears to be traveling across the sky at a rate of 15° of Longitude in 1 hour => 1° in 4 minutes. 8 hours and 8 minutes = 488 minutes of time. 488 + 4 = 122" of/ongitude.

120 kts 12:32UTC 12:47UTC 12:50 UTC

TAS: ATAX: ETAY: ATAY:

What is ETA at Z? A) B) C) D)

1257 UTC 1302 UTC 1300 UTC 1303 UTC

CPL

What we need to do is find the actual ground speed (GS) for the first segment and then use this GS to calculate the time estimate for the second segment. • Time it took to fly X - Y is 18 min (12:50 - 12:32) = 0,3 hrs • Distance =Rate x Time ·30NM =? kts x 0,3 hrs .? kts=30NM + 0,3 hrs .?= 100kts Now that we know the actual GS we can easily calculate the estimated time of arrival (ETA) at Z: • Distance = Rate x Time ·20NM = 100kts x ?hrs • ?hrs =20NM + 100kts .? = 0,2 hrs = 12 minutes If the Actual Time ofArrival at "Y" was 12:50, then 12 minutes later we should arrive at "Z" => 13:02.

4686. Airplane ATPL CPL Heli ATPL CPL (Refer to figures 061-01 and 061-02) An aircraft takes off from Guam at 23:00 Standard Time on 30 April local date. After a flight of 11 hrs 15 min it lands at Los Angeles (California). What is the Standard Time and local date of arrival (assume summer time rules apply)? A) 17:15 on 30 April. B) 12:15 on 1 May. C) 13:15 on 1 May. D) 16:15 on 30 April. (Refer to figures 061-E101, 061-E102, 061-EI03 and 061-EI04) Guam is 23:00 ST on 30 April = 13:00 UTC on 30 April (Guam is 10 hours ahead of UTC). An 11 hour and 15 minute flight puts us in Los Angeles at 24:15 UTe on 30 April (24:15 is actually at 00:15 UTC on 1" May). Los Angles, California is listed as 8 hours behind UTe, but we need to reduce this by 1hour due to daylight savings time. So, we need to subtract 7 hours from the UTe time of 00:15. We will land at 17:15 local time on April30rh •

4687. Airplane ATPL CPL Heli When the time is 14:00 LMT at 90° West, it is:

For explanation refer to question #1131 on page 22.

4681. Airplane ATPL (Refer to figure 061-09) Given:

Time abbreviations: • Estimated Time of Departure = ETD • Actual Time of Departure =ATD • Estimated Time ofArrival = ETA • Actual Time ofArrival = ATA

Heli

ATPL

CPL

ATPL

CPL

A) 14:00 LMT at 90 0 East. B) 12:00 LMT at 1200 West. C) 10:00 LMT at 60 0 West. D) 06:00 LMT at the Prime meridian. (Refer to figures 061-E101, 061-E102, 061-E103 and 061-EI04) The difference in longitude is 120° - 90° = 30°. We know that 1° of longitude equals 4 minutes of time (of Sun traveling across the sky). 30° x 4 minutes = 120 minutes of time = 2 hours. It may be useful to think of the Earth as stationary and the Sun as orbiting westwards instead of the Sun as stationary and the Earth as orbiting eastwards like in reality. So, if the LMT at900Wis 14:00 the time at 1200W will be 14:00 minus 2 hrs, because the Sun still need to travel to this place => 12:00 LMT.

4689. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-01) When it is 10:00 Standard Time in Kuwait, the Standard Time in Algeria is: A) 07:00 B) 12:00 C) 13:00 D) 08:00 (Refer to figures 061-EI0l, 061-EI02, 061-E103 and 061-E104) Start by taking the time in sunny Kuwait (you can start with Algeria, but forthis example we will start with Kuwait). The question states that it is 10:00 standard in Kuwait. Looking at the chart, we see that Kuwait is 3 hours ahead of GMT. So, subtracting three hours from local Kuwait time tells us that the UTe is 07:00. Looking once more at the chartinforms us that Algeria is 1 hour ahead ofGMT. Well, if we know that it is currently 07:00 UTe, then it is 08:00 Standard Time

I 4550 (A) I 4565 (A) I 4623 (D) I 4671 (C) 14681 (8) I 4686 (A) 14687(8) I 4689 (D) I

01 Basics of Navigation in beautiful Algeria. Kuwait 70:00 ST = 07:00 UTC Algeria 07:00 UTC = 08:00 ST

4701. Airplane ATPL CPL Heli ATPL CPL In the Air Almanac the highest time difference listed for difference between UTC and Standard time is maximum: A) B) C) 0)

13 hours. 12 hours. 6 hours. 24 hours.

You would think that since a day is 24 hour long, you cannot find a country having a standard time of UTC +/- 73. But when you add Daylight Saving Time, you may have a standard time of UTC +/- 73. This is the case of the extreme East of Russia out on the Kamchatka peninsula, near Alaska, USA.

4716. Airplane ATPL CPL Heli ATPL CPL How much time does it take for the mean Sun to move from meridian 145°15'E to meridian 023°45'W? A) B) C) 0)

4720. Airplane ATPL CPL Heli ATPL CPL (Refer to figures 061-01 and 061-02) At 12:00 Standard Time on the 10th July in Queensland, Australia, what is the Standard Time in Hawaii, USA? 12:00 ST 10July. 10:00 ST 10 July. 16:00 ST 09 July. 02:00 ST 10 July.

(Refer to figures 067-E707, 067-E102, 067-E703 and 067-£104) It helps to know a little geography when doing these questions. In particular where the Prime Meridian and International Date Line are located in relation to Australia and the Hawaiian Islands. ·~/n Queensland at 72:00 ST on 70 July = 02:00 UTC on 10 July (according to the chart, Queensland is 10 hours ahead of GMT). ·In Hawaii at 02:00 UTC on 10 July = 76:00 ST on 09 July (according to the chart, Hawaii is 70 hours behind GMT).

7 hours 56 minutes and starts at 09:00 UTC. 1 hour 4 minutes and starts-at 09:00 UTC. 1 hour 4 minutes and starts at 07:56 LMT. 8 hours 36 minutes and starts at 07:56 UTC.

(Refer to figures 067-E705 and 067-E706) The Air Almanac states the beginning of morning twilight (when the centre of the Sun's disc is 6° below the horizon) is at 07:56. The civil twilight ends with the sunrise (when the upper edge of the Sun's disc is co-incident with the horizon) = at 09:00. The difference between 07:56 and 09:00 is 7 hour 4 minutes = the duration of twilight.

I

4701 (A)

I

4716 (8)

I

Kuwait 07:00 ST = 04:00 UTC Algeria 04:00 UTC =05:00 ST

4720 (C)

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4780 (C)

A) B) C) 0)

I

4820 (A)

2324UTC 0724UTC 1552 UTC 0738 UTC

(Refer to figures 067-E105 and 067-E706) The sunrise/sunset times listed in the Almanac are the UTC time the sunrise/ sunset occurs at the Greenwich meridian. However, these times can be taken without much error to be the LMT times of sunrise/sunset occurs at any other meridian. ·In the left column find the closest two parallels - in our case 50 0 N and 45°N => in the first row at the top find the date or use the two closest dates for interpolation. • Read the sunrise information for 50 0 N (07:40) and for 45°N (07:27) for the first day (Dec 4th). • Interpolate between 07:40 and 07:27 to obtain the time applicable for 49°N =>07:36. • Repeat the same steps for Dec l'h => find the interpolated value for 49°N of 07:39. • Now we have the sunrise on Dec 4th (07:36) and on Dec l'h (07:39) => interpolate to obtain the sunrise on Dec 6th => 07:38. Now we have to convert the time of 07:38 LMT to the UTC => the Longitude is 723°30'W = 723,5~ We know that the Sun travels across the sky at a speed of 7° in 4 minutes => 723,5° will therefore take 494 minutes = 8 hrs 74 minutes. Therefore, the sunrise at 723°30Woccurs at 75:52 UTC (07:38 + 08:74 = 75:52UTC).

A) B) C) 0)

66°00'N 07:56 09:00 15:28 16:32

The duration of morning twilight at 66°00'N is: A) B) C) 0)

(Refer to figures 067-E107, 067-E102, 067-E103 and 067-E104) Start by taking the time in sunny Kuwait (you can start with Algeria, but for this example we will start with Kuwait). The question states that it is OZ·OO standard in Kuwait. Looking at the chart, we see that Kuwait is 3 hours ahead of GMT. So, subtracting three hours from local Kuwait time tells us that the UTC is 04:00. Looking once more at the chart informs us that Algeria is 7 hour ahead ofGMT. Well, if we know that it is currently 04:00 UTe, then it is 05:00 Standard Time in beautiful Algeria.

4859. Airplane ATPL CPL Heli ATPL CPL The local mean time at longitude 095°20'W, at 0000 UTC, is:

4780. Airplane ATPL CPL Heli ATPL CPL For 1st February the Air Almanac lists the following data: Latitude: Morning civil twilight: Sl!nrise: Sunset: Evening civil twilight:

A) 05:00 B) 09:00 C) 12:00 0) 03:00

4854. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-03) What is the UTC time of sunrise in Vancouver, British Columbia, Canada (49°N 123°30'W) on the 6th December?

8 hours 06 minutes. 11 hours 16 minutes. 9 hours 41 minutes. 6 hours 15 minutes.

(Refer to figures 067-E707, 067-E102, 067-E103 and 067-E704) The Earth rotates around its axis in 24 hours. There are 360° of Longitude => the Sun appears to be traveling across the sky at a rate of 75° of Longitude in 7 hour => 7° in 4 minutes. A Longitude change from 745°75'E (745,25°E) to 023°45'W (023,75°W) is 769° (745,25° + 23,75°). A Longitude change of 769° will therefore be covered in 77 hrs 76 min (769° x 4 min = 676 min = 71,26 hrs = 77 hrs 76 min).

A) B) C) 0)

4820. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-01) If it is 07:00 hours Standard Time in Kuwait, what is the Standard Time in Algeria?

I

17:38:40 same day. 06:21 :20 same day. 17:38:40 previous day. 06:21 :20 previous day.

(Refer to figures 067-E107, 067-£702, 067-£103 and 067-E104) We are concerned with the longitude in this question. Your first step is to convert the 20 minutes of Longitude into something we can work with => 0,3333°. So we have 90,33° and we can use the standard of 7° of Longitude =4 minutes ofSun travel across the sky. 95,33"x 4 minutes = 387,32 minutes. 387,32 + 60 = 6,34hours. 00:00-6,35= 77,65=> 77:39 previous day. Final answer is 77:3910cal mean time. It is the previous day because the location is West of the Greenwich meridian an the time at Greenwich is 00:00 (the new day is just starting).

4854 (C)

I

4859 (C)

I

Aviationexam Test Prep Edition 2012

4863. Airplane ATPL CPL Heli ATPL CPL 0 On 27 Feb, at 52°10'5 040 00'E, the sunrise is at 02:30 UTC. On the same day, at 52°10'5 035°00'W, the sunrise is at: A) 02:30 UTC B) 05:10UTC e) 07:30 UTC D) 21:30UTC (Refer to figures 067-£707, 067-E702, 067-E703 and 067-E704) The difference in longitude is 35° + 40° = 75~ We know that 7° of longitude equals 4 minutes of time (of Sun traveling across the sky). 75° x 4 minutes = 300 minutes of time = 5 hours. It may be useful to think of the Earth as stationary and the Sun as orbiting westwards instead of the Sun as stationary and the Earth as orbiting eastwards like in reality. So, the Sun will rise at 02:30 UTC at 040 0 E and after 5 hours of traveling westwards will rise at 035°W at 07:30 UTC.

4884. Airplane ATPL CPL Heli ATPL CPL On the 27th of February, at 52°5 and 0400E, the sunrise is at 02:43 UTC. On the same day, at 52°5 and 035°W, the sunrise is at: A) 21:43 UTC B) 02:43 UTC C) 07:43 UTC D) 05:23 UTC (Refer to figures 067 -E707, 067 -E702, 067-E703 and 067 -E7 04) The difference in longitude is 35° + 40° = 75°. We know that 7° of longitude equals 4 minutes of time (of Sun traveling across the sky). 75° x 4 minutes = 300 minutes of time = 5 hours. It may be useful to think of the Earth as stationary and the Sun as orbiting westwards instead of the Sun as stationary and the Earth as orbiting eastwards like in reality. So, the Sun will rise at 02:43 UTC at 040 0 E and after 5 hours of traveling westwards will rise at 035°W at 07:43 UTC.

4908. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-03) The UTC of sunrise on 6 December at WINNIPEG (Canada) (49°50'N 097°30'W) is: A) B) e) D)

09:30 01:13 22:30 14:13

A) The declination ofthe sun and the latitude ofthe observer. B) The inclination of the axis of the earth and the longitude ofthe observer. e) The latitude and the longitude of the observer. D) The date and the longitude of the observer. 230645. Airplane ATPL CPL Heli ATPL CPL Which statement is correct about the apparent solar day? A) The duration of the apparent solar day is constant throughout a year due to the constant rotational speed of the earth around its axis. B) The apparent solar day is the period between two successive transits of the true sun through the same meridian. e) The duration of the apparent solar day is constant throughout a year due to the constant velocity of the earth in its orbit around the sun. D) The apparent solar day is the period between two successive transits of the mean sun through the same meridian. 230647. Airplane ATPL CPL Heli ATPL CPL In a sunrise/sunset table given for the 28th of June at a certain latitude, sunrise is gven as 0239 and sunset is given as 2127. What is the latitude? A) 55°S B) 60 0 N e) 80 0 N D) 00 0 N/S 230649. Airplane ATPL CPL Heli ATPL CPL When proceeding, on a given date, along a parallel towards the East, the moment of sunrise will occur one hour earlier every 15° difference in longitude when it is expressed in: A) LAT (Local Apparent Time) B) Standard Time e) UTC D) LMT

(Refer to figures 067-E705 and 067-E706) The sunrise/sunset times listed in the Almanac are the UTe time the sunrise/ sunset occurs at the Greenwich meridian. However, these times can be taken without much error to be the LMT times of sunrise/sunset occurs at any other meridian. ·In the left column find the closest parallel to 49°50'N - in our case 50 0 N (in this case our Lat is so close to the tabulated one that we do not need to interpolate). • Read the sunrise information for 50 0 N on Dec 4th (07:40) and for Dec l'h (07:43). ·Interpolate between 07:40 and 07:43 to obtain the time applicable for Dec 6th => 07:42. Now we have to convert the time of 07:42 LMT to the UTe => the Longitude is 9r30'W = 97,5°. We know that the Sun travels across the sky at a speed of 7° in 4 minutes => 97,5° will therefore take 390 minutes = 6 hrs 30 minutes. Therefore, the sunrise at 9r30'W occurs at 74:72 UTC (07:42 + 06:30 = 74:72 UTC).

230567. Airplane ATPL CPL Heli ATPL CPL At (54°N, 0200W) the sun rises on November 28th a1 09:01 UTC. At (44°N, 0200W) the sun will rise: A) B) e) D)

230579. Airplane ATPL CPL Heli ATPL CPL The time interval between sunrise and sunset is dependent on:

230655. Airplane ATPL CPL Heli At 0000 Local Mean Time of an observer:

ATPL

CPL

_

A) the mean Sun is in transit with the observer's anti-meridian. B) the mean Sun is in transit with the observer's meridian. e) the apparent Sun is in transit with the observer's meridian: D) the apparent Sun is in transit with the observer's anti-meridian. 230658. Airplane ATPL CPL Heli ATPL CPL (Refer to figures 061-01 and 061-02) When it is 0600 Standard Time in Queensland (Australia) the Standard Time in Hawaii (US) is: A) B) e) D)

10:00 12:00 02:00 06:00

at 07:41 LMT. earlier since the latter position lies further South. later since the latter position lies further South. also at 09:01 UTe since both positions are situated on the same meridian.

1 4863 (C)

1 4884 (e)

1 4908 (D) 1230567 (8) 1230579 (A) 1230645 (8) 1230647 (8) 1230649 (C) 1230655 (A) 1230658 (A) 1

01 Basics of Navigation 230660. Airplane ATPL CPL Heli ATPL CPL An aircraft departs from Schiphol airport and flies to Santa Cruz in Bolivia (South America) via Miami in Florida. The de~ parture time (off blocks) is 07:45 ST at the 10th of November, taxi time before take off at Schiphol i,s 25 minutes. The flight time to Miami over the Atlantic Ocean is 09:20. The total taxi time in Miami to and from the gate is 25 minutes. The time spend at the gate is 02:40. From Miami to Santa Cruz the airborne time is 06:30. Calculate the time and date of touch down in Santa Cruz in ST Bolivia if the difference between ST and UTC is 5 hours. A) 09:05 11th Nov. B) 22:05 10th Nov. C) 21:05 10th Nov. 0) 07:05 11th Nov. 230661.

Airplane

ATPL CPL

Heli

ATPL

CPL

Which statement about ST is true? A) Standard time is determined by the government of the appropriate state and does not necessarily follow the borders of 15° wide longitude zones. B) In all cases the standard times at Western longitudes are slow on and at Eastern longitudes fast on UTe. C) Standard time is the time that is determined by division of the longitude by 15 and rounding off the answer to the nearest integer. 0) The standard time at 125° W is UTC - 08:20. 230663.

Airplane

ATPL

CPL

Heli

ATPL

CPL

Standard time for some areas is listed in the Air Almanac as UTC +13 instead of UTC -11. The reason forthis is: A) the sense of Earth rotation. B) the setup of the sunrise/sunset tables. C) keeping the same date as the political and or economical entity to which they belong. 0) the fact that they are keeping daylight saving time. 230668.

Airplane

ATPL

CPL

Heli

ATPL

CPL

The time difference in Local Mean Time between sunset at positions A (SOON, 1200E) and B (50 0S, 1200E) on the 21st of Novemberis:

-
Airplane

ATPL

CPL

Heli

ATPL

CPL

Which statement about the duration of daylight is true? A) Close to the solstices the influence of latitude on the duration of daylight is at its smallest. B) Close to the equinoxes the influence of latitude on the duration of daylight is at its smallest. C) In sUl:nmer the length of the period of daylight decreases with increasing latitude. 0) On September 10th the duration of daylight is longer on the Southern Hemisphere than on the Northern Hemisphere.

230670. Airplane ATPL CPL Heli ATPL CPL The SRISS table for the 23,d of February at latitude 400N gives: SR=06:44 SS= 17:44 At 12:00 Central European Time (UTC+l) at 400N: A) the Sun sets at 116°E. B) the Sun rises at 64°W. C) the Sun rises at 79°W. 0) the Sun sets at 86°E. 230672. Airplane ATPL CPL ' Heli ATPL CPL Mu'a, Tonga Islands, is situated at (21°11'S, 175°07'W) In the Air Almanac the standard time of Tonga Islands is listed as UT<;+13. For August 21 st the sunrise table in the Air Almanac shows: 200S:06:18 30 0S:06:28 What is the Standard Time of sunrise at Mu'a? A) 06:59 on August 21't. B) 07:39 on August 22 nd • C) 06:59 on August 22 nd • 0) 07:39 on August 21". Airplane ATPL CPL Heli ATPL CPL 230673. Position "Elephant Point" is situated at (58°00'N, 135°30'W). Standard time for this location is listed in the Air Almanac as UTC -8. If sunset occurs at 00:57 UTC on 21 st Janl!ary, what is the time of Sunset in LMT? A) 08:57 on January 21't. B) 15:55 on January 20 th . C) 09:59 on January 21't. 0) 16:57 on January 20 th . 230674.

Airplane

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CPL

Heli

ATPL

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(Refer to figure 061-04) " The UTC of Morning Civil Twilight at (66°48'N, 095°26'W) on 27th of January is? A) 0927 UTC B) 0814 UTC C) 1541 UTC 0) 1436 UTC 230675. Airplane ATPL CPL Heli ATPL CPL '(Refer to figure 061-04) The UTC of Sunrise at (66°48'N, 095°26'W) on 27th of January is? A) B) C) 0)

1541 UTC 1549 UTC 0814 UTC 0927 urc

230676.

Airplane

ATPL

CPL

Heli

ATPL

CPL

(Refer to figure 061-04) What is the duration of morning Civil Twilight at (66°48'N, 095°26'W) on 27th of January? A) 09:27 B) 08:14 C) 01:02 0) 01:13

1230660 (C) 1230661 (A) 1230663 (e) 1230668 (D) 1230669 (8) 1230670 (8) 1230672 (C) 1230673 (8) 1230674 (D) 1230675 (8) 1 1230676 (D) 1 i! "

j

., lEI

Aviationexam Test Prep Edition 2012

231112. Airplane ATPL CPL Heli ATPL CPL An aircraft follows a radial to a VOR/DME station. At 10:00 the DME reads 120 NM. At 10:03 the DME reads 105 NM. The estimated time overhead the VOR/DME station is: A) B) e) D)

231113. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-18) Two consecutive waypoints of a flight plan are Stornoway VOR/DME (N58°12,4', W006°11,0') and Glasgow VOR/DME (N55°52,2', W 004°26,7'). During the flight the Actual Time Over Stornoway is 11:15 UTe and the Estimated Time Over Glasgow is 11:38 UTe. At 11:21 UTe the fix ofthe aircraft is exactly over reporting point RONAR. What is the Revised UTe over Glasgow,.based on this last fix?

10:21:00 10:24:00 10:18 10:27:00

A) B) C) D)

11:36 11:38 11:34 11:33

01-04 Directions 1128. Airplane ATPL CPL Heli ATPL CPL The directive force acting on a compass needle in an aircraft: A) will always pull the north-seeking end down at southern magnetic latitudes and up at northern magnetic latitudes. B) will be higher at high magnetic latitudes than near the magnetic equator. e) is the resultant magnetic force in the horizontal plane in the position where the compass is installed. D) will be canceled by the compass compensation system. (Refer to figure 061-E20) The Earth can be visualized with a large magnet at the centre, roughly aligned with the geographical NIS axis. The magnetic poles of this "magnet" lie close to, but are not exactly at the geographic poles. A magnetic field is associated with the Earth and it surrounds the Earth with the field concentrated at the magnetic poles. The field has a magnetic equator and magnetic meridians. The- total magnetic field (T) can be resolved into horizontal (H) and vertical (Z) components. The magnetic field can be shown as "lines offorce" which run from pole to pole, just as in a magnet. These lines offorce are concentrated above the magneticpoles, pointing downwards to the pole. The area near the magnetic pole, where the horizontal component H is least is known as the "6 micro tesla zone". Teslas are a measurement ofmagnetic force and are usually in units of micro or nano teslas.

If a magnet is freely suspended in the Earth magnetic field it will be under the influence of the total magnetic field T and will align itself with that field and point to magnetic North. Direction on Earth is measured in the horizontal plane. A compass is constructed with a magnetised compass needle (i.e. a small magnet) which itself is freely suspended in the horizontal plane. The needle will indicate magnetic direction in that plane. Thus a compass needle aligns itself with the Earth horizontal component (H) of the Earth's magnetic field. This field is known as the directive force. With increasing magnetic latitude, the needle will "dip" according to the direction of the field as the value ofH decreases, becoming almost zero at the poles. At the pole, where the value ofZ is maximum, the needle will tend to point vertically down towards the pole, making it ineffective for navigation. At the magnetic Equator, the value of H is maximum and a freely suspended compass needle in the earth's magnetic field will be horizontal to the Earth, exactly in line with component H. The attached figure depicts the Earth's magnetic field. The direction

of the Earth's magnetic field is parallel to the surface at the magnetic equator, and perpendicular to the surface at the Magnetic North. The horizontal component is therefore at its maximum at the magnetic equator, and decreases to become null at the Magnetic North. The horizontal component of the magnetic field will indicate the direction of the Magnetic Poles at a place. The angle of dip is the angle between the Earth's magnetic field and the horizontal component of the earth's magnetic field. Aircraft compasses are designed to sense the horizontal component (H) of the magnetic field, therefore they are most sensitive around the Magnetic equator (mid-way between the Magnetic North and the Magnetic South poles) and least sensitive around the Magnetic poles. The magnetic equator is a line on the surface of the Earth where the angle of dip is zero.

1231112 (8) 1231113 (A) 1 1128 (C)

1 1135 (D)

The value of magnetic dip (also referred to as the magnetic inclination) is 0° at the magnetic equator and constantly increases towards the magnetic poles, where it reaches its maximum value of 90° (magnet is vertical). As it can be seen the "H" weakens with increasing distance from the magnetic equator, reaching zero at the poles. The magnetic meridian is an imaginary line connecting the Magnetic South and North poles and can be taken as the magnetic force lines along the surface of the Earth. That is, a compass needle will be parallel to the magnetic meridian. Remember that the magnetic compass depends solely on the horizontal component "H" for its directional properties and is therefore no use for navigation close to the magnetic poles where "H" is too weak.

1135. Airplane ATPL CPL Heli ATPL CPL The horizontal component of the Earth's magnetic field: A) is approximately the same at all magnetic latitudes less than 60°. B) weakens with increasing distance from the magnetic poles. e) weakens with increasing distance from the nearer magnetic pole. D) is approximately the same at magnetic latitudes SOON and SooS. For explanation refer to question #7728 on this page.

1136. Airplane ATPL CPL Heli ATPL CPL The horizontal component of the Earth's magnetic field: A) weakens with increasing distance from the nearer magnetic pole. B) weakens with increasing distance from the magnetic poles. e) is stronger closer to the magnetic equator. D) is approximately the same at all magnetic latitudes less than 60 0 • For explanation refer to question #1128 on this page.

1152. Airplane ATPL CPL Heli ATPL CPL The angle between True North and Magnetic North is known as: A) deviation. B) variation. e) alignment error. D) dip. (Refer to figures 061-E9S, 061-E96 and 067-E97) The deviation is the angle between the Compass North and the Magnetic North. It is expressed as how many degrees East or West is the Compass North situated from the Magnetic North. If Compass North lies to the West ofMagnetic North the deviation is westerly (or negative). Another way to memorize this is when the compass needle points to the West of magnetic North,

1 1136 (C)

1 1152 (8)

1

---.-~----------

-------._----------- _.

~-

--~~---..

------

._----

01 Basics of Navigation the deviation is "west': When the needle points to the East of the magnetic North, deviation is "east': If Corripass North lies to the East of Magnetic North the deviation is easterly (or positive). Deviation applied to magnetic heading gives compass heading. Next time you are in your airplane look underneath the direct reading compass and you will notice the compass correction card. The deviations will be listed on the card. .

1165. Airplane ATPL CPL Heli ATPL CPL A negative (westerly) magnetic variation signifies that:

A) True North is East of Magnetic North. B) True North is West of Magnetic North. C) compass North is East of Magnetic North. D) compass North is West of Magnetic North.

Variation is a correction applied to magnetic to obtain true heading. The Magnetic Poles are in continuous movement around the geographic poles. This movement is monitored and it can be predicted. Since Magnetic Poles are not situated at the True Poles, the direction of a magnetic compass needle will vary according to the aircraft position. The compass needle will only give an approximate idea of the direction of Magnetic North at a place. The difference between True North and Magnetic North (the true and magnetic meridians) at a place is called magnetic variation. When the magnetic North Pole lies to the East of the True Meridian, variation is easterly. When the magnetic North Pole lies to the West of the True Meridian, variation is westerly. The magnetic variation has maximum value of 180° on the line (arc of a Great Circle) that joins the True North with Magnetic North (or the True South with the Magnetic South).

For explanation refer to question #1154 on this page.

1167.

A) B) C) D)

ATPL

Heli

CPL

ATPL

CPL

Airplane

ATPL

Heli

ATPL

1190. Airplane ATPL CPL Heli ATPL CPL A magnetic compass will be most effective at:

CPL

dip is zero. variation is zero. deviation is zero. the isogonal is an agonic line.

A) a position roughly halfway between the magnetic poles. B) the South magnetic pole. C) the North magnetic pole. D) the equator.

For explanation refer to question #1128 on page 30.

For explanation refer to question #1128 on page 30.

1156. Airplane ATPL CPL Heli ATPL CPL If the True HOG is 165° and the variation is -3, what is the magnetic heading?

1194. Airplane ATPL Heli ATPL Isogrives are lines that connect positions that have:

A) B) C) D)

A) Variation is negative (Westerly), therefore magnetic heading

is 168°. B) Variation is negative (Westerly), therefore magnetic heading is 162°. C) Variation is negative (Easterly), therefore magnetic heading is 162°. D) Variation is negative (Easterly), therefore magnetic heading is 168°. (Refer to figure 061-E97) The magnetic variation is the angle between the True North and the Magnetic North. It is expressed as - at how many degrees East(+) or West(-) is the Magnetic North situated from the True North.

=

Magnetic Heading (MH) TH + Wvariation (or - E variation) True Heading (TH) = MH - W variation (or + E variation)

the same horizontal magnetic field strength. the same grivation. the same variation. 0° magnetic dip.

(Refer to figures 061-E93 and 061-E94) Grid values must often be converted into magnetic values (and vice versa) when a magnetic compass is used. The difference between Grid and Magnetic directions is GRIVATION (Grid variation). Grivation is the algebraic sum of convergency and variation. Lines joining places of equal grivation are known as Isogrivs. On most grid navigation charts, the isogrivs are already plotted.lsogonals are also normally printed on charts as well as lines joining positions with the same convergency. On a Lambert's chart these latter may be assumed to be meridians, which will give only a small error. At or near the pole, the same assumption applies for polar stereographic and transverse mercator charts. It takes a long time to calculate and plot isogrivs and it is laborious. Pilots are not usually required to do it.

Variation of-3 = 3°W MH= 165°+3°W MH=368°

I

1154(C)

I

1155(A)

I

1156(A)

CPL

(Refer to figures 061-E93 and 061-E94) Great circle tracks are always preferred because of distance savings. When flying near the poles however, convergency changes so rapidly that it is almost impossible to fly great circle tracks. Instead, a technique known as "Grid Navigation" is applied, using a grid for measuring directions. The advantage of measuring direction with respect to Grid North is that direction is the same everywhere on the chart. This technique gives a practical means of flying great circle tracks. Savings can also be made when flying great circle tracks over long distances at lower latitudes with the grid technique. In very high latitudes magnetic compasses are useless and a directional gyro is used as a heading reference, utilizing specialized techniques.

• When the Magnetic North Pole lies to the East of the True Meridian, variation is Easterly (positive). • When the Magnetic North Pole lies to the West of the True Meridian, variation is Westerly (negative).

A) B) C) D)

ATPL

A) to make the system of latitude and longitude available on a gridded map. B) to provide a system for directions where a great circle has a constant direction, even if its true direction varies. C) make a chart covering high latitudes that has the same qualities as the equatorial Mercator chart. D) minimise the errors introduced when making calculations involving variation.

(Refer to figure 061-E96) . The Earth's magnetic poles do not have the same position as the geographic poles. The difference between True North and Magnetic North (the true and magnetic meridians) at a place is called magnetic variation = rel="nofollow"> in other words it is the angular difference between the geographical meridian and the magnetic meridian running through the same position. The maximum value of magnetic variation is 180·.

ATPL

Heli

The purpose of establishing a grid is:

A) variation will never exceed 90°. B) variation will always increase when the total strength of the terrestrial magnetic field increases. C) the variation is East when True North seems to be located West of Magnetic North. D) the largest values of variation are found along the anti meridians of the magnetic poles.

Heli

CPL

For explanation refer to question #1152 on page 30.

Consider the following statements on magnetic variation:

1155. Airplane ATPL CPL At the magnetic equator:

ATPL

magnetic course. true heading. compass heading. magnetic track.

1171.

Airplane

1154.

Airplane

Deviation applied to magnetic heading gives:

I

1165(A)

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1167(C)

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1171 (8)

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1190(A)

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1194(8)

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Aviationexam Test Prep Edition 2012 1205. Airplane ATPL CPL Heli ATPL CPL Consider the following statements on a freely suspended magnetic needle in the terrestrial magnetic field:

1232. Airplane ATPL CPL Heli ATPL CPL In the areas close to the magnetic poles compasses are not of any use in air navigation, mainly because:

A) the needle will align itself along the magnetic meridian. B) the needle will align itself with the direction of the magnetic lines of force. C) the needle will not be influenced by the magnetic inclination (dip). D) all answers above are correct.

A) the field strength of the Earth's magnetic field is at it's weakest in this area. B) the distance from the magnetic equator is too long. C) the horizontal component of the Earth's magnetic field is too weak. D) the inclination is insufficient in these areas.

(Refer to figure 061-E20) The attached figure depicts the Earth's magnetic field. The angle of dip is the angle between the Earth's magnetic field and the horizontal component of the earth's magnetic field. Aircraft compasses are designed to sense the horizontal component (H) of the magnetic field, therefore they are most sensitive around the Magnetic equator. The magnetic equator is a line on the surface of the Earth where the angle of dip is zero. From the magnetic equator towards the magnetic poles the dip is constantly increasing. Thus reaching a dip of 90° (magnet is vertical) at the magnetic poles. As it can be seen the "H" weakens with increasing distance from the magnetic equator, reaching zero at the poles. The magnetic meridian is an imaginary line connecting the Magnetic South and North poles and can be taken as the magnetic force lines along the surface of the Earth. That is, a compass needle will be parallel to the magnetic meridian.

1210. Airplane ATPL CPL Heli ATPL CPL As you move from a lower to a higher southern magnetic latitude, the characteristics of the terrestrial magnetic field will change: A) the horizontal component of the field will increase and the inclination will decrease. B) the magnetic lines of force will spread further apart. C) the magnetic meridian will become more and more vertical. D) the inclination will increase and the vertical component of the field will increase. For explanation refer to question #1128 on page 30.

1220. Airplane ATPL CPL Heli Referring to the Earth's magnetic field:

ATPL

CPL

A) the inclination is 90 0 at the geographical poles. B) the inclination decreases with increased geographical latitude. C) the dip is maximum at the magnetic equator. D) the inclination is 90 0 at the magnetic poles. For explanation refer to question #1128 on page 30.

1225. Airplane ATPL CPL Heli ATPL CPL A simple magnet is surrounded by a magnetic field having the following properties: A) the magnetic lines of force are straight lines which will not have the characteristics to bend. B) the field's direction is from the magnets red pole to the magnets blue pole. C) a small number of the magnetic lines of force leaving one of the poles will pass out through space and will never return to this magnet. D) the direction of the field is from the magnets blue pole to the magnets red pole. (Refer to figure 061-E20) The maximum strength of a magnet is normally found at each of its poles (hence the term "pole strength'? however, in the case of the Earth the maximum strength has been found to be at two locations near to each pole => the North and South magnetic poles. In terms of magnetism the Earth behaves as if a huge permanent magnet were situated near the center producing a magnetic field over the surface, with a Blue pole in the North and a Red pole in the South. In 2001, the North Magnetic Pole was determined by the Geological Survey ofCanada to lie near Ellesmere Island in northern Canada at 81°18W 110048'W, In 2009, it was moving toward Russia at almost 40 miles (64 km) per year due to magnetic changes in the Earth's core.

I

Ell

1205 (8)

I

1210 (0)

I

1220 (0)

I

1225 (8)

I

For explanation refer to question #1128 on page 30.

1234. Airplane ATPL CPL Heli The magnetic meridian in a position is:

ATPL

CPL

A) the direction of the great circle running from the magnetic South Pole to the Magnetic North Pole, measured in the position. B) the direction in the Earth's magnetic lines of force in the position. C) the horizontal direction of the Earth's magnetic field in that position, toward the Magnetic North Pole. D) the direction in which a freely suspended magnet will point in that position. For explanation refer to question #1205 on this page.

1235. Airplane "True North" is:

ATPL

Heli

CPL

ATPL

CPL

A) the direction along a meridian. B) the direction along the meridian, toward the North Pole when on the northern hemisphere and toward the South Pole when on the southern hemisphere. C) is in any direction out from the True North Pole. D) the direction along any meridian toward the True North Pole. (Refer to figures 061-E77 and 061-E78) True direction is measured with respect to True North (the direction of the Geographic North Pole from a place). The direction of True North is indicated by the meridian at the actual place where the measurement is made since meridians always run in a North/South direction. Usually navigational maps and charts are based on True directions.

1244. Airplane ATPL CPL Heli ATPL CPL Which of the following statements concerning the Earth's magnetic field is completely correct? A) Dip is the angle between total magnetic field and ve-rti~al field component. B) The blue pole ofthe Earth's magnetic field is situated in North Canada. C) At the Earth's magnetic equator, the inclination varies depending on whether the geograhic equator is North or South of the magnetic equator. D) The Earth's magnetic field can be classified as transient semipermanent or permanent. For explanation refer to question #1225 on this page.

1249. Airplane ATPL CPL Heli ATPL CPL Which of these is a correct statement about the Earth's magnetic field? A) It acts as though there is a large blue magnetic pole in northern Canada. B) The angle of dip is the angle between the vertical and the total magnetic force. C) It may be temporary, transient, or permanent. D) It has no effect on aircraft deviation. For explanation refer to question #1225 on this page.

1232 (C)

I

1234 (C)

I

1235 (0)

I

1~44

(8)

I

1249 (A)

I

01 Basics of Navigation

2181. Airplane ATPL CPL Heli ATPL (PL The Earth can be considered as being a magnet with the:

1250. Airplane ATPL CPL Heli ATPL CPL What is the definition of magnetic variation? A) The angle between the direction indicated by a compass and Magnetic North. B) The angle between True North and compass North. C) The angle between Magnetic North and True North. D) The angle between magnetic heading and Magnetic North.

A) blue pole near the North Pole of the Earth and the direction of the magnetic force pointing straight up from the Earth's surface. B) red pole near the North Pole of the Earth and the direction of the magnetic force pointing straight down to the Earth's surface. C) blue pole near the North Pole of the Earth and the direction of the magnetic force pointing straight down to the Earth's surface. D) red pole near the North Pole of the Earth and the direction of the magnetic force pointing straight up from the Earth's surface.

For explanation refer to question #7154 on page 31.

1252. Airplane What is deviation? A) B) C) D)

ATPL

Heli

CPL

ATPL

CPL

The angle between Magnetic North and compass North. The angle between Magnetic North and True North. The angle between True North and compass North. The angle between True North and Magnetic North.

For explanation refer to question #1225 on page 32.

For explanation refer to question #1152 on page 30.

1256. Airplane ATPL CPL Heli ATPL CPL If the CH = 220°, Variation = 12°E, Deviation = 2°W, what is the corresponding TH? A) B) C) D)

A) inversely proportional to the horizontal component of the Earth's magnetic field. B) proportional to the horizontal component of the Earth's magnetic field. C) inversely proportional to the vertical component of the Earth's magnetic field. D) inversely proportional to the vertical and horizontal components of the Earth's magnetic field.

TH =234° TH =206° TH =230° TH = 210°

(Refer to figure 061-E97) When applying the values of Deviation and Variation using the above mnemonic we usually start off with the True heading => we work from the right to the left using the "CDMVT" mnemonic. When applying the Deviation and Variation values we have to apply it in the correct sense - either with a "plus" or a "minus" sign. Remember the mnemonic "West is Best, East is Least" = in other words West = plus sign, East = minus sign. This mnemonic works when starting with True heading and working your way towards the left though the Magnetic heading to the Compass heading. When working the other way around = starting with the Compass heading and wishing to obtain the Magnetic heading or the True heading, you have to reverse this logic => West = minus sign and East = plus sign. C D M V T 220° - 2°W => 218° + 12°E => ? True Heading (?) = 230° Note: Remember that you have to work your way from the left (from CH) to the right (to TH).

1313. Airplane An isogonal is a line:

ATPL

Heli

CPL

ATPL

(Refer to figure 067-E96) Magnetic variation = the difference between True North and Magnetic North (the true and magnetic meridians) at a place. When the magnetic North Pole lies to the East of the True Meridian, variation is easterly. When the magnetic North Pole lies to the West of the True Meridian, variation is westerly. Unes on a map or chart joining positions with equal magnetic variation are termed 'sogona's (Isogonic lines). Isogonals are shown as broken lines on the chart, with the value of variation marked alongside. A line joining places with zero variation is termed an Agonic line. When variation is zero (i.e. heading to Magnetic North would be the same as heading to True north), magnetic and true directions are the same. One of the Agonic lines is currently running through Western Europe and the other through the USA. The Aclinic is the magnetic equator or the line of 0° dip. It jOins all points where the angle of dip is zero. 'soclina's are lines of equal magnetic dip.

1250 (C)

I

1252 (A)

I

1256 (C)

I

1313 (0)

I

For explanation refer to question #1128 on page 30.

2189. Airplane ATPL CPL An agonic line is a line that connects: A) B) C) D)

2181 (C)

I

Heli

ATPL

CPL

positions that have the same variation. positions that have 0° variation. points of equal magnetic dip. points of equal magnetic horizontal field strength.

For explanation refer to question #1313 on this page.

2193. Airplane ATPL CPL Heli ATPL (PL The magnetic force causing compass deviation will be a force in direction: A) B) C) D)

CPL

A) running through all positions having the same magnetic Inclination. B) running through all positions having the same magnetic longitude. C) on the surface of the Earth, running through all positions having the same magnetic latitude. D) running through all positions having the same variation.

I

2187. Airplane ATPL CPL Heli ATPL CPL The sensitivity of a direct reading magnetic compass is:

parallel to the magnetic meridian. perpendicular to the aircraft lateral axis. perpendicular to the compass needle. perpendicular to the aircraft longitudinal axis.

For explanation refer to question #1128 on page 30.

2216. Airplane ATPL CPL Isogonals converge at the: A) B) C) D)

Heli

ATPL

(PL

magnetic equator. north and South geographic and magnetic poles. north magnetic pole only. north and South magnetic poles only.

(Refer to figures 061-E09 and 061-E96) Magnetic variation = the difference between True North and Magnetic North (the true and magnetic meridians) at a place. When the magnetic North Pole lies to the East of the True Meridian, variation is easterly. When the magnetic North Pole lies to the West of the True Meridian, variation is westerly. Unes on a map or chart joining positions with equal magnetic variation are termed 'sogona's (Isogonic lines). Isogonals are shown as broken lines on the chart, with the value of variation marked alongside. The question on where they converge is a tricky one - they actually do converge at both Geographical poles as well as at both Magnetic poles. To verify this you can take a look at any Polar High Altitude enroute chart and see the isogonal lines on the chart => they indeed do converge at both the Magnetic North pole as well as the Geographic North pole.

2187 (8)

I

2189 (8)

I

2193 (C)

I

2216 (8)

I

EI

Aviationexam Test Prep Edition 2012 2229. Airplane ATPL The following values are given:

Grid track: Grid convergence: Variation: Deviation:

Heli

ATPL

192

2236.

B) MH 125 C) MH 145 D) MH 223 (Refer to figures 067-£93,067-£94 and 067-E97) With a Grid track (GT) of 792° and the Convergency (C) of 48°W we can calculate the True Track (TT): Use the mnemonic "Convergence West, True Track best": 792° (GT) + 48° (C) =240° (TT) Now that we have the ITand variation of 700E we can easily calculate the Magnetic Track (MT): Use the mnemonic "Variation East, Magnetic least": 240° (IT) - 70 0 E (Var) =230° (MT) Knowing the MT we can calculate the Magnetic Heading (MH) using the information on the Wind Correction Angle (WCA).lfwe experience a crosswind from the left, we have to apply a WCA to the left in order to maintain our desired magnetic track. Our magnetic heading in this case will be equal to the magnetic track decreased by the amount of the left WCA => MT of230°WCA of 9° =Magnetic Heading of227 ° Note: disregard the Deviation information - we would use the Deviation info only if the question asked for the Compass Heading.

Another way to calculate is to use the CDMVTCG mnemonic (Compass Heading / Deviation / Magnetic Heading / Variation / True Heading / Convergency / Grid Heading):

?1" <=

T

700E

-

C

G

1" <= 48°W + 792°

True Heading (1) = 240° Magnetic Heading (11) 230·

=

Note: Remember that you have to work your way from the right (from Grid heading) to the left (to Magnetic Heading). Now the last step is to apply the WCA (see the text above) => 230° - 9° =227 ~

2232. Airplane ATPL Heli Preparing a chart for use of grid means:

ATPL

CPL

ATPL

CPL

For explanation refer to question #7373 on page 33.

A) MH 221

V

Heli

A) compass deviation. B) magnetic variation. C) pressure. D) wind velocity.

48W 10E 2W

Find magnetic heading when WCA is 9L.

M

2234. Airplane ATPL CPL Isogonals are lines of equal:

ATPL

A) pre-calculating grid directions for all positions marked on the chart. B) mark the chart with lines and values for grid latitude and longitude. C) selecting suitable grid tracks, position line and bearings. D) selecting a meridian on the chart and drawing lines on the chart, parallel to the meridian selected. (Refer to figures 067-E93 and 067-E94) Preparing a chart for use of grid means a series of equally-spaced, parallel lines is drawn on the chart parallel to a reference (datum) meridian. These parallel lines may be considered as "false" meridians. The datum meridian is chosen for convenience as the datum meridian for that particular chart. It will often be the central meridian of the chart or sector covered for the flight. It may be the Greenwich Meridian. The space between meridians is arbitrary, but in order to make measuring distance straightforward, 60 NM spacing is preferred. A series of "false parallels" can also be drawn perpendicular to the false meridians which forms the GRID. If the same spacing between the parallels is used, then the grid is square. The chart now has two reference directions: grid North and true North. Grid North is a constant direction but true North varies as convergency of the meridians.

Airplane

ATPL

CPL

Heli

Complete the following statement regarding magnetic variation. The charted values of magnetic variation on Earth normally change annually due to: A) a reducing field strength causing numerical values at alilocations to decrease. B) magnetic pole movement causing numerical values at all 10cations to increase. C) magnetic pole movement causing numerical values at all 10cations to increase or decrease. D) an increasing field strength causing numerical values at all locations to increase. (Refer to figure 067 -E96) The Earth's magnetic poles do not have the same position as the geographic poles. The Magnetic Poles are in slight, but continuous movement around the geographic poles. This movement is monitored and it can be predicted. The approximate rate of movement is 7° in 5 years. Since Magnetic Poles are not situated at the True (Geographical) Poles, the direction of a magnetic compass needle will vary according to the aircraft position. The difference between True North and Magnetic North (the true and magnetic meridians) at a place is called magnetic variation => in other words it is the angular difference between the geographical meridian and the magnetic meridian running through the same pOSition. • When the magnetic North Pole lies to the East of the True Meridian, variation is Easterly (positive). • When the magnetic North Pole lies to the West of the True Meridian, variation is Westerly (negative). As stated, the location of the Magnetic poles will change continuously and therefore also the value of magnetic variation at any given place will change slowly over a period of time. Depending on the specific place the change con be an increase or a decrease in magnetic variation value. The rate of change should always be printed on a chart, either with arrows showing the amount and direction of the annual movement, or in a table. The chart date is usually shown, so the precise value of variation can be calculated for any year.

2237.

Airplane

ATPL

CPL

Heli

ATPL

CPL

'

,

When a magnetized compass needle is freely suspended in the Earth's magnetic filed, when free from extraneous magnetic influence, it will align itself with: A) B) C) D)

True North. Magnetic North. absolute North. relative North.

(Refer to figure 067-E95) When a magnetized compass needle is freely suspended in the magnetic field of the Earth, it will align itself along the magnetic field. The direction of the Earth's magnetic field is parallel to the surface at the magnetic equator, and perpendicular to the surface at the Magnetic North. The horizontal component is therefore at its maximum at the magnetic equator, and decreases to become null at the Magnetic North. The horizontal component of the magnetic field will indicate the direction of the Magnetic Poles at a place. In doing so, the needle aligns itself along the direction of the magnetic meridian at that point. A magnetic meridian gives the direction in the horizontal plane of Magnetic North. Therefore, without any external magnetic influence, the compass needle will align itself so that it points toward the Magnetic North. However, when it is subject to an external magnetic influence such as from a magnetized material or electrical equipment it will align itself with the so called Compass North, which is not the same as the Magnetic North. The compass North can be looked at as a "virtual" magnetic North when the needle is subject to external magnetic fields other than the Earth's magnetic field.

I 2229(A) I 2232(0) I 2234 (B) I 2236(C) I 2237(B) I

------------------

01 Basics of Navigation 2238. Airplane ATPL CPL Heli ATPL CPL When a magnetized compass needle is freely suspended in the Earth's magnetic field; and affected by extraneous magnetic influence, it will align itself with: A) B) C) D)

For explanation refer to question #1225 on page 32.

2298. Airplane ATPL CPL The compass needle marked red:

2239. Airplane ATPL CPL Heli ATPL CPL Compass deviation is defined as the angle between: True North and Magnetic North. Magnetic North and compass North. True North and compass North. the horizontal and the total intensity of the Earth's magnetic field.

2240. Airplane ATPL CPL Heli ATPL CPL The angular difference between the geographical meridian and the magnetic meridian running through the same position is named:

2301. Airplane ATPL CPL The value of magnetic variation: A) B) C) D)

2271. Airplane ATPL CPL Heli ATPL CPL The lines on a chart joining places of equal magnetic dip are called: A) aclinic lines. B) isogonals. C) isoclinals. D) agonic lines. For explanation refer to question #1373 on page 33.

2272. Airplane ATPL CPL Heli ATPL CPL The North and South magnetic poles are the only positions on the Earth's surface where: .~'

a freely suspended compass needle will stand horizontal. isogonals converge. a freely suspended compass needle will stand vertical. the value of magnetic variation equals 90 0 •

A) B) C) D)

isogrives isoclines isogonals isotachs

I I

ATPL

2312. Airplane ATPL (PL Heli ATPL CPL The value of magnetic variation on a chart changes with time. This is due to: A) B) C) D)

movement of the magnetic poles, causing an increase. increase in the magnetic field, causing an increase. reduction in the magnetic field, causing a decrease. movement of the magnetic poles, which can cause either an increase or a decrease.

For explanation refer to question #2236 on page 34.

A) B) C) D)

For explanation refer to question #1373 on page 33.

A) B) C) D)

Heli

(Refer to figures 061-E93 and 061-E94) The difference between true and grid directions on the chart is grid convergency (or convergence). The value of convergency is that between the local meridian, where true direction is measured and the datum meridian on which the chart grid is based. If the direction of true North at the local meridian is West of grid North, convergency is West and vice versa. Therefore, in the northern hemisphere, if the local meridian itselfis West of the datum meridian, then convergency is East (true North is East of grid north}. If a meridian is East of the datum meridian, convergency is West. In the southern hemisphere the picture changes. If the local meridian is West of the grid datum, convergency is West (a sketch always resolves uncertainty).

2314. Airplane ATPL CPL Where is a compass most effective?

2276. Airplane ATPL CPL Heli ATPL CPL The total magnetic force of the terrestrial magnetic field:

CPL

A) named easterly when Grid North appears to be East of True North. B) independent of longitude, depending only of latitude. C) 1 on an equatorial Mercator chart. D) the difference in direction between Grid North and True North.

For explanation refer to question #1128 on page 30.

2273. Airplane ATPL CPL Heli ATPL CPL The lines on the Earth's surface that join points of equal magnetic variation are called:

ATPL

varies between maximum values of 45°E and 45°W. is a maximum of 1800 • is always 00 at the magnetic equator. is never greater than 90 0 •

2308. Airplane ATPL "Grid convergence" is:

variation. magnetic correction. inclination. deviation.

Heli

For explanation refer to question #1154 on page 31.

For explanation refer to question #2236 on page 34.

A) B) C) D)

CPL

(Refer to figure 061-E20) The north-seeking pole of a magnet is termed red and the south-seeking pole is termed blue. Poles of magnets repel and unlike poles attract, (red repels red, red attracts blue, blue repels blue, blue attracts red). On the Earth the North magnetic pole is considered by convention to be blue and the South magnetic pole red.

For explanation refer to question #1152 on page 30.

A) B) C) D)

ATPL

A) will be repelled by terrestrial Magnetic North Pole. B) is called "The Northseeking Pole". C) will align itself in the reciprocal direction of the terrestrial magnetic field. D) is the North Pole of the compass needle.

True North. Magnetic North. compass North. relative North.

For explanation refer to question #2237 on page 34.

A) B) C) D)

Heli

Heli

ATPL

CPL

About midway between the earths magnetic poles. In the region of the magnetic South Pole. In the region ofthe Magnetic North Pole. On the geographic equator.

For explanation refer to question #1128 on page 30.

is horizontal in all positions on the surface of the Earth. is vertical at the magnetic equator. is strongest at the magnetic poles. will be stronger at higher altitudes because the attenuation is less at high altitudes.

2238(C) 12239(8) 2312 (0) I 2314 (A)

I I

2240 (A)

I

2271 (C)

I

2272 (C)

I

2273(C)

I

2276 (C)

I

2298(8)

I

2301 (8) 12308(0)

I

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Aviationexam Test Prep Edition 2012

2318. Airplane ATPL CPL Directions are stated:

Heli

ATPL

CPL

A) as a reference direction and a number of degrees. B) in degrees with reference to True North when plotted with reference to the latitude/longitude grid on a chart. e) in degrees in a 360 system, starting out clockwise from the reference direction. D) all answers are correct.

4088. Airplane ATPL Isogrives on a chart indicate lines of:

Heli

A) zero magnetic variation. B) equal magnetic dip. e) equal horizontal directive force. D) equal grivation. For explanation refer to question #1194 on page 31.

(Refer to figures 061-E77 and 061-E78) Direction may be measured in various ways. In all cases the direction To or From a position will be measured with reference to a particulardatum, usually North. All directions are measured clockwise in degrees from 0° - 360°. There are 4 main types of direction referred to in Navigation: • Degrees True (0T) • Degrees Magnetic (OM) • Degrees Compass (OC) • Degrees Relative (OR or oRe!.)

4093. Airplane Grid convergence is:

4028. Airplane ATPL CPL The value of variation:

For explanation refer to question #2308 on page 35.

Heli

ATPL

CPL

ATPL

A) grid convergence and variation. B) grid convergence and deviation. e) grid convergence, variation and deviation. D) variation and deviation.

The rule for applying Grivation is the same as that for variation, substituting "Grivation" for" Variation": • "Grivation East - Magnetic Least" • "Grivation West- Magnetic Best':

Heli

ATPL

A) GH is 156° and MH is 103°. B) GH is 103° and MH is 159°. C) _grid convergence is 46W and variation is 10E. D) grid convergence is 58W and deviation is 2E. (Refer to figures 061-E93, 061-E94 and 061-E96) Grivation = Convergence + Variation. Westerly Grivation = negative value; Easterly Grivation =positive value (just like variation). Grid heading = Magnetic heading ± Grivation. Grid heading =Magnetic heading - W grivation (or + E grivation) 103° (GH) = 159° (MH) - 56° (Grivation)

4064. Airplane The agonic line:

ATPL

CPL

Heli

ATPL

A) easterly for positions East of the grid datum meridian on the northern hemisphere. B) easterly in all positions having westerly longitude. C) easterly in all positions having easterly longitude. D) westerly for positions East of the grid datum meridian on the northern hemisphere.

For explanation refer to question #1205 on page 32.

(Refer to figures 061-E93, 061-E94 and 061-E96) Grid values must often be converted into magnetic values (and vice versa) when a magnetic compass is used. The difference between Grid and Magnetic directions is GRIVATION (Grid variation). Grivation is the algebraic sum of convergency and variation. Therefore: • Grid =Magnetic ± Variation ± Convergency. • Grid =Magnetic ± Grivation.

4049. Airplane ATPL Grivation is 56W when:

Heli

A) directly with the horizontal component of the Earth's magnetic field. B) directly with the vertical component of the Earth's magnetic field. e) inversely with both vertical and horizontal components of the Earth's magnetic field. D) inversely with the horizontal component of the Earth's magnetic field.

For explanation refer to question #1152 on page 30.

Heli

ATPL

4105. Airplane ATPL CPL Heli ATPL CPL The force acting on the needle of a direct reading compass varies:

A) is zero at the magnetic equator. B) has a maximum value of 180°. e) has a maximum value of 45 E or 45 W. D) cannot exceed 90°.

4042. Airplane ATPL "Grivation" is the sum of:

ATPL

ATPL

CPL

A) is midway between the Magnetic North and South Poles. B) follows the geographic equator. e) is the shorter distance between the respective true and Magnetic North and South Poles. D) follows separate paths out of the North polar regions, one currently running through western Europe and the other through the USA.

4112. Airplane ATPL CPL Heli ATPL CPL The dip angle in the terrestrial magnetic field is given by the following equation: A) dip = sin·'(H/T) B) dip = cos·, (H/T) e) dip = 12 xT/H D) dip=T/Z (Refer to figure 061-E20) The attached figure depicts the Earth's magnetic field. The direction of the Earth's magnetic field is parallel to the surface at the magnetic equator, and perpendicular to the surface at the Magnetic North. The horizonta7t1jm~ ponent is therefore at its maximum at the magnetic equator, and decreases to become null at the Magnetic North. The horizontal component of the magnetic field will indicate the direction of the Magnetic Poles at a place. The angle of dip is the angle between the Earth's magnetic field and the horizontal component of the earth's magnetic field. Aircraft compasses are designed-to sense the horizontal component (H) of the magnetic field, therefore they are most sensitive around the Magnetic equator (mid-way between the Magnetic North and the Magnetic South poles) and least sensitive around the Magnetic poles. The magnetic equator is a line on the surface of the Earth where the angle of dip is zero. The value of magnetic dip (also referred to as the magnetic inclination) is 0° at the magnetic equator and constantly increases towards the magnetic poles, where it reaches its maximum value of 9 0° (magnet is vertical). As it can be seen the "H" weakens with increasing distance from the magnetic equator, reaching zero at the poles. The magnetic meridian is an imaginary line connecting the Magnetic South and North poles and can be taken as the magnetic force lines along the surface of the Earth. That is, a compass needle will be parallel to the magnetic meridian. The main theme of this questiorns not the formula (you will probablinever again use it during your aviation career), but that the value of dip is directly related to its north/south distance in relation to the poles and the equator.

For explanation refer to question #1313 on page 33.

I 2318 (0) I 4028 (8) I 4042 (A) I 4049 (8) I 4064 (0) I 4088 (0) I 4093 (0) I 4105 (A) I 4112 (8) I

- - - - _.. _ - - -.. _-_.--_..._

....

01 Basics of Navigation 4119. Airplane ATPL CPL Heli ATPL CPL When an aircraft is moved to a place of lower magnetic latitude:

A) the deviation values will decrease because the horizontal component of the terrestrial field is becoming stronger. B) deviation values will change considerably and irregularly. C) the deviation values will become more southerly. D) the deviation values will increase. (Refer to figure 067-E20) Deviation is solely a compass error which varies from aircraft to aircraft. It varies in many ways, such as with heading and latitude. Deviation is caused by localised magnetic fields in the aircraft which are originate in different ways, such as magnetic fields introduced into the aircraft structure during manufacture and electrical fields produced by electrical and radio equipment. We know that the sensitivity of the compass is directly proportional to the horizontal component of the Earth's magnetic field, referred to as "H". "H" is strongest at the Magnetic Equator and decreases with an increase of Magnetic Latitude and eventually becomes zero at the Magnetic Poles. We have to realize that the stronger the "W is the more difficult it is for the local magnetic fields to deflect the compass needle and thus to cause the compass deviation. Therefore, as we move close to the Magnetic Equator (decrease our Magnetic Latitude), the force "H" becomes stronger and the deviation decreases.

Airplane ATPL CPL Heli ATPL CPL If compass HOG is 340 0 and deviation +3, what is magnetic heading? 4120.

A) Deviation is 343°. B) Deviation is 343°. C) Deviation is 337°. D) Deviation is 337°.

than the magnetic heading. D) the compass needle seems to be pointing at a pole located West of the magnetic pole. (Refer to figures 067-E95 and 067-E97) Easterly Deviation as well as Easterly Variation is a positive value (+ °E). Westerly Deviation as well as Westerly Variation is a negative value (- OW). Magnetic heading = Compass heading + Deviation Compass heading = Magnetic heading - Deviation When deviation is Westerly it is actually a negative value => using the above relationship we would deduct a Westerly deviation from the magnetic heading to obtain the compass heading. When deviation is Easterly it is actually a positive value => using the above relationship we would add an Easterly deviation to the magnetic heading to obtain the compass heading. Example: Magnetic heading = 700° If Compass heading = 720° • in this case deviation is -20° = 200W .700° MH = 720·CH + (-20· Dev) If Compass heading =080° • in this case deviation is +20° =200E • 700° MH = 080° CH + (+20· Dev) 4145. Airplane ATPL CPL Heli ATPL CPL The angle between Magnetic North and compass North is called:

A) magnetic variation. B) compass error. C) compass deviation. D) alignment error.

is positive (Easterly), therefore magnetic heading is positive (Westerly), therefore magnetic heading is positive (Easterly), therefore magnetic heading is positive (Westerly), therefore magnetic heading

For explanation refer to question #7 752 on page 30. 4181. Airplane ATPL CPL Heli ATPL CPL In a particular position the horizontal and the total strength ofthe terrestrial magnetic field are the same. This position is:

(Refer to figure 067 -E97) The compass deviation is the angle between the Compass North and the Magnetic North. It is expressed as how many degrees East(+} or West(-} is the Compass North situated from the Magnetic North. Compass Heading (CH) = MH + W deviation (or - Edeviation) Magnetic Heading (MH) = CH - W deviation (or + Edeviation)

A) B) C) D)

at one of the magnetic poles. between the geographic and the magnetic pole. at magnetic latitude 45°. at the magnetic equator.

For explanation refer to question #7728 on page 30.

Deviation of+3 =3°E 4189. Airplane ATPL CPL Heli ATPL CPL The deviation will change with a change in aircraft heading:

MH = 340° + 3°E MH =343°

A) because the undesired magnetic pole is moved relative to the direction of the Earth's magnetic field. B) because the relative direction of the magnetic meridian , is changed. C) because the strength of the undesired hard magnetism is changed. D) because the magnetic inclination will change whefle"(,er the heading is changed.

Airplane ATPL CPL Heli ATPL CPL At a specific location, the value of magnetic variation: 4121.

A) B) C) D)

depends on the true heading. depends on the type of compass installed. depends on the magnetic heading. varies slowly over time.

For explanation refer to question #2236 on page 34.

Airplane ATPL CPL Heli If variation is expressed as Westerly, then: 4141.

ATPL

(Refer to figures 067-E709 and 067-E770) Deviation is produced by the iron and steel components inside the airplane. It is the angle between the local magnetic meridian and the direction in which the compass magnets are positioned. This question is trying to inform us that the amount ofdeviation will change due to the relationship of the magnet with the earth's magnetic field.

CPL

A) True North is West of Magnetic North. B) compass North is West of Magnetic North. C) True North is East of Magnetic North. D)· Magnetic North is West of compass North.

4221. Airplane ATPL CPL Heli What is a line of equal magnetic variation?

For explanation refer to question #7754 on page 37.

A) B) C) D)

Airplane ATPL CPL Heli ATPL CPL The deviation of a compass is describe as +4. This means that: 4144.

A) the deviation may be described as westerly. B) compass heading will always be different by 4 degrees from true heading. C) the compass heading will have a lower number in degrees 4119 (A)

I

4120 (A)

I

4121 (D)

I

4141 (C)

I

4144 (C)

I

An isocline. An isogonal. An isogriv. An isovar.

For explanation refer to question #7373 on page 33.

4145 (C)

I

4181 (D)

I

4189 (A)

I

4221 (8)

I

ATPL

CPL

_-

Aviationexam Test Prep Edition 2012 4226. Airplane ATPL CPL Heli ATPL CPL Which of the following statements concerning Earth magnetism is completely correct? A) An isogonal is a line which connects places with the same magnetic variation; the agonic line is the line of zero magnetic dip. B) An isogonal is a line which connects places with the same magnetic variation; the aclinic is the line of zero magnetic dip. e) An isogonal is a line which connects places of equal dip; the aclinic is the line of zero magnetic dip. D) An isogonal is a line which connects places with the same magnetic variation; the aclinic connects places with the same magnetic field strength. For explanation refer to question #1313 on page 33.

4231.

Airplane

ATPL

Heli

CPL

ATPL

CPL

A line running through positions where the magnetic and the true meridians are parallel, is called: A) B) e) D)

an agonic line. an aclinic line. the magnetic equator. the zero-variation meridian.

C) a correction to be added to compass heading to obtain mag-

netic heading. D) an error to be added to compass heading to obtain magnetic heading. Fo~ explanation

4284.

refer to question #4144 on page 37.

Airplane

ATPL

CPL

Heli

ATPL

CPL

What is the dip angle at the South magnetic pole? A) 0°

B) 90° e) 180°

D) 64° For explanation refer to question #1128 on page 30.

4296. Airplane ATPL Heli ATPL (Refer to figure 061-10) Assume a North polar stereographic chart whose grid is aligned with the Greenwich meridian. An aircraft flies from the Geographic North Pole for a distance·of 480 NM along the 1100E meridian, then follows a grid track of 1540 for a distance of 300 NM. Its position is now approximately: A) 70015'N 080 0E B) 80 000'N 080 0E e) 78°45'N 087°E

For explanation refer to question #1313 on page 33.

D) 79°15'N 074°E 4244. Airplane ATPL CPL Heli ATPL CPL A line drawn on a chart which joins all points where the value of magnetic variation is zero is called an: A) isogonal. B) aclinic line. e) agonic line. D) isotach.

First of all we need to find the position after the aircraft travels 480 NM along the meridian 1100Efrom the North Pole southbound. We know that 1 NM along a meridian = 1 minute ofLatitude change.480 NM =480 minutes of Latitude = 8°. The new position will therefore be 90° - 8° => 82°N 110°E.

For explanation refer to question #1313 on page 33.

4250.

Airplane

ATPL

Heli

CPL

ATPL

Now we need to convert the Grid track to a True Track. In the Northem Hemisphere the chart convergency is equal to the longitude, but with sign reversed (for example chart convergency at 110 0E is 110 0W). Therefore, traveling from a point 82"N 1100E the convergency is 1100W => True direction will be Grid direction + 110° => True direction is 154°G + 110° = 264°T. We plot a line 264° True from the position 82°N 110°E.

CPL

When is the magnetic compass most effective? A) In the region of the magnetic South Pole. B) About midway between the magnetic poles. e) In the region of the Magnetic North Pole. D) On the geographic equator.

Last step is to apply the distance of 300 NM to this line. 1 NM = 1 minute of Latitude change => 300 NM = 300 minutes = 5°. Measure the distance that is represented by 5° on the chart along a meridian (use the 90° E meridian as it is easiest) => apply the same distance to the line you have .drawn from the point 82W 1100E and note the final position => it will be afrProx;mately 80 0N080°E.

For explanation refer to question #1128 on page 30.

4267.

Airplane

ATPL

Heli

CPL

ATPL

CPL

An isogonal is: A) B) e) D)

4430. Airplane ATPL Heli ATPL The initial straight track from A (75°N 60 0 E) to B (75°N 60 0 W) on a polar stereographic chart is:

a line of equal wind speed. a line of equal magnetic deviation. a line of zero magnetic variation. a line of equal magnetic variation.

A) B) e) D)

For explanation refer to question #1313 on page 33.

4268.

Airplane

ATPL

Heli

CPL

ATPL

CPL

A) a line of zero magnetic deviation. B) a line of equal magnetic deviation. C) a line of zero magnetic variation. D) a line of equal magnetic variation. For explanation refer to question #1313 on page 33.

Airplane

ATPL

Heli

CPL

ATPL

030 0 360 0 060° 330 0

(Refer to figures 061-E93 and 061-E94) This is another Polar Gridded Chart problem. The Datum Meridian (Grid North) is OOE/W. The chart convergency is the angle between the Grid North and the True North. It is expressed as - at how many degrees East(+) or West() the True North situated from the Grid North. In the Northem Hemisphere the chart convergency is equal to the longitude, but with sign reversed (for example chart convergency at 60 0E is 60 0W). ·IfChart convergency is 60°W, then Grid direction =True direction - 60°. • From A (75°N 060 0E) to B (75°N 060 0W) the Grid direction is a straight line and the initial track at A is 270 0G. • True direction = 270 0G + 60° = 330 0T

The agonic line is:

4272.

(Refer to figures 061-E29, 061-E93 and 061-E94) Please note that this is not primarily a calculation question, but rather plotting question. The only calculation involved is the conversion of distance to degrees of Latitude change.

CPL

Deviation is: A) an error to be added to magnetic headings. B) a correction to be added to magnetic heading to obtain compass heading.

I

EI

4226 (8)

I

4231 (A)

I

4244 (C)

I

4250 (8)

I

4267 (0)

I

4268 (C)

I

4272 (C)

I

4284 (8)

I

4296 (8)

I

4430 (0)·

I

01 Basics of Navigation

4443. Airplane ATPL CPL Given: True track: 352° 11 oW Variation: _5° Deviation: 100L Drift: Calculate the compass heading.

Heli

ATPL

CPL

When applying the values of Deviation and Variation using the above mnemonic we usually start off with the True heading => we work from the right to the left using the above "CDMVT" mnemonic. When applying the Deviation and Variation values we have to apply it in the correct sense - either with a "plus" or a "minus" sign. Remember the mnemonic "West is Best, East is Least" = in other words West = plus sign, East = minus sign. This mnemonic works when starting with True heading and working your way towards the left though the Magnetic heading to the Compass heading. When working the other way around = startIng with the Compass heading and wishing to obtain the Magnetic heading or the True heading, you have to reverse this logic => West = minus sign and East = plus sign.

A) 018 0 B) 348 0 C) 358 0

D) 008 0 (Refer to figures 061-E55 and 061-E97) Let's start with some definitions: • Heoding = the direction of the fore and aft axis of the aircraft. • Track =the line (direction) which the aircraft tracks over the ground. • Drift = the angle between the Heading and the Track. It is expressed as how many degrees left or right the actual track is situated from the heading (see the attached figure). Based on on the information above we can determine the heading. If True Track =352° and the Drift = 100L then the True Heading (TH) will be oor (352° + 10°). Now that we have the value ofTH we can calculate the Magnetic Heading (MH) and the Compass Heading (CH) using the mnemonic "CDMVT" = Can Dead Men Vote Twice? It stands for Compass Heading; Deviation; Magnetic Heading; Variation; True Heading. Easterly Deviation as well as Easterly Variation is a positive value (+ °E). Westerly Deviation as well as Westerly Variation is a negative value (- OW). When applying the values of Deviation and Variation using the above mnemonic we usually start off with the True heading => we work from the right to the left using the above "CDMVT" mnemonic. When applying the Deviation and Variation values we have to apply it in the correct sense - either with a "plus" or a "minus" sign. Remember the mnemonic "West is Best, East is Least" = in other words West = plus sign, East = minus sign. This mnemonic works when starting with True heading and working your way towards the left though the Magnetic heading to the Compass heading. When working the other way around = starting with the Compass heading and wishing to obtain the Magnetic heading or the True heading, you have to reverse this ' logic => West = minus sign and East = plus sign.

C

D

V

M

? <= 5°W + ?1" <= 11°W + Magnetic Heading (??) = 013° Compass Heading (?) 018°

002°

4445. Airplane ATPL CPL Given: True Track: Variation: Deviation: Drift: Calculate the Compass heading:

Heli

ATPL

CPL

Based on the information above we can determine the heading. If True Track = 352° and the Drift = 100R then the True Heading (TH) will be 342° (35r - 10°). Now that we have the value ofTH we can calculate the Magnetic Heading (MH) and the Compass Heading (CH) using the mnemonic "CDMVT" = Can Dead Men Vote Twice? It stands for Compass Heading; Deviation; Magn~ic Heading; Variation; True Heading.

I

4450 (A)

T

?1" <= 11°W + 34r

= =

Magnetic Heading (??) 353° Compass Heading (?) 358° Note: Remember that you have to work your way from the right (from TH) to the left (to CH).

4450. Airplane ATPL Heli ATPL Grid heading is 299°, grid convergency is 55° West and magnetic variation is 90° West. What is the corresponding magnetic heading? A) B) C) D)

0840

3340 1540 2640

(Refer to figures 061-E93, 061-E94 and 061-E97) With a Grid Heading (GH) of299 0 and the Convergency (C) of55°W we can calculate the True Heading (TH): Use the mnemonic "Convergence West, True Track best": 299 0 (GH) + 550 (C) = 3540 (TH) Now that we have the TH and variation of 90 0W we can easily calculate the Magnetic Heading (MH): Use the mnemonic "Variation West, Magnetic best": 354 0 (TH) + 90 0 (Var) = 084 0 (MH)

V

T

C

0 90 W + 1" <= 55°W True Heading (?) = 354· Magnetic Heading (??) 084°

?1" <=

G

+ 299°

=

Note: Remember that you have to work your way from the right (from Grid heading) to the left (to Magnetic Heading).

A) 030 0 (T) B) 330 0 (T) C) 1500 (T) D) 210 0 (T)

(Refer to figures 061-E55 and 061-E97) Let's start with some definitions: • Heading = the direction of the fore and aft axis of the aircraft. • Track = the line (direction) which the aircraft tracks over the ground. • Drift = the angle between the Heading and the Track. It is expressed as how many degrees left or right the actual track is situated from the heading (see the attached figure).

4445 (C)

+

4451. Airplane ATPL Heli ATPL On a polar stereographic chart, the initial great circle course from A 700N 060 0W to B 700N 0600E is approximately:

A) 078 0 (C) B) 3460 (C) C) 358 0 (C) D) 025 0 (C)

I

V

M

? <= 5°W

M

Note: Remember that you have to work your way from the right (from TH) to the left (to CH).

4443 (A)

D

C

Another way to calculate is to use the CDMVTCG mnemonic (Compass Heading I Deviation I Magnetic Heading I Variation I True Heading I Convergency I Grid Heading):

T

=

I

Easterly Deviation as well as Easterly Variation is a positive value (+ °E). Westerly Deviation as well as Westerly Variation is a negative value (- OW).

I

4451 (A)

I

(Refer to figures 061-E93 and 061-E94) This is yet another Polar Gridded Chart problem. The Datum Meridian (Grid North) is OOEIW. The chart convergency is the angle between the Grid North and the True North. It is expressed as - at how many degrees East(+} or West(-} the True North is situated from the Grid North. In the Northern Hemisphere the chart convergency is equal to the longitude, but with sign reversed (for example chart convergency at 60 0W is 60 0E). ·IfChart convergency is 60 0E, then Grid direction = True direction + 60~ • From A (looN 060 0W) to B (70 0N 060 0E) the Grid direction is a straight line and the initial track at A is 090 0G. 0 • True direction = 090 0G - 60° = 030 T

Aviationexam Test Prep Edition 2012 4456. Airplane ATPL CPL Heli ATPL CPL Route A (44°N 026°E) to B (46°N 024°E) forms an angle of 35° with longitude 026°E. Average magnetic variation between A and B is 3°E. What is the average magnetic course from A to B? A} 322° B) 328°

C) 032° D) 038° (Refer to figure 061-E45) Draw a simple sketch and the question will become very easy. Determine the True heading of 325° and then apply the value of variation to convert the True Heading into Magnetic Heading. When converting from True to Magneticthe mnemonic "Westis Beast, Eastis Least" is applicable => in our case we simply deduct the 3° ofEasterly variation to obtain a Magnetic Heading of322°.

4460. Given:

Airplane

True track Drift Variation Deviation

ATPL

Heli

CPL

ATPL

CPL

300° 8°R 100W _4°

D) 278° (Refer to figures 061-E55 and 061-E97) Let's start with some definitions: • Heading =the direction of the fore and aft axis of the aircraft. • Track = the line (direction) which the aircraft tracks over the ground. • Drift = the angle between the Heading and the Track. It is expressed as how many degrees left or right the actual track is situated from the heading (see the attached figure). Based on on the information above we can determine the heading. 1fTrue Track =300° and the Drift = BOR then the True Heading (TH) will be 292° (300° - BO). Now that we have the value ofTH we can calculate the Magnetic Heading (MH) and the Compass Heading (CH) using the mnemonic "CDMVT" = Can Dead Men Vote Twice? It stands for Compass Heading; Deviation; Magnetic Heading; Variation; True Heading. Easterly Deviation as well as Easterly Variation is a positive value (+ °E). Westerly Deviation as well as Westerly Variation is a negative value (- OW). When applying the values of Deviation and Variation using the above mnemonic we usually start off with the True heading => we work from the right to the left using the above "CDMVT" mnemonic. When applying the Deviation and Variation values we have to apply it in the correct sense - either with a "plus" or a "minus" sign. Remember the mnemonic "West is Best, East is Least" = in other words West = plus sign, East = minus sign. This mnemonic works when starting with True heading and working your way towards the left though the Magnetic heading to the Compass heading. When working the other way around = starting with the Compass heading and wishing to obtain the Magnetic heading or the True heading, you have to reverse this logic => West = minus sign and East = plus sign.

V

T

? <= 4°W +

??" <= 10 0 W + 292° Magnetic Heading (m = 302° Compass Heading (1)

D

M

V

2°E - 319° <= 4°W Compass Heading (1) = 317"

?

<=

T + 315°

Note: Remember that you have to work your way from the right (from TH) to the left (to CH).

Airplane

ATPL

Heli

CPL

ATPL

CPL

4°E

If the aircraft is flying a compass heading of 270, the true and magnetic headings are:

B) 322° C) 294°

M

C

Variation: Deviation:

A) 306°

D

When applying the values of Deviation and Variation using the mnemonic "CDMVT" we usually start off with the True heading => we work from the right to the left. When applying the Deviation and Variation values we have to apply it in the correct sense - either with a "plus" or a "minus" sign. Remember the mnemonic "West is Best, East is Least" = in other words West = plus sign, East = minus sign. This mnemonic works when starting with True heading and working your way towards the left though the Magnetic heading to the Compass heading. When working the other way around =starting with the Compass heading and wishing to obtain the Magnetic heading or the True heading, you have to reverse this logic => West = minus sign and East = plus sign.

4525. Given:

Calculate the compass heading?

C

Men Vote Twice? It stands for Compass Heading; Deviation; Magnetic Heading; Variation; True Heading.

=306°

Note: Remember that you have to work your way from the right (from TH) to the left (to CH).

4466. Airplane ATPL CPL Heli ATPL CPL In still air, you wish to fly a true of 315°. Variation is 4°W. Deviation is 2°E. What compass heading should you fly? A) 321 B) 313 C) 317 D) 309 (Refer to figure 061-E97)

A very useful mnemonic for the actual calculation exists: CDMVT = Can Dead

A) 274°(T); 267°(M). B) 267°(T); 274°(M). C) 277°(T); 281°(M). D) 263°(T); 259°(M). (Refer to figure 061-E97)

A very useful mnemonic for the actual calculation exists: CDMVT = Can Dead Men Vote Twice? It stands for Compass Heading; Deviation; Magnetic Heading; Variation, True Heading. When applying the values of Deviation and Variation using the above mnemonic we usually start off with the True heading => we work from the right to the left. When applying the Deviation and Variation values we have to apply it in the correct sense - either with a "plus" or a "minus" sign. Remember the mnemonic "West is Best, East is Least" =in other words West =plus sign, East = minus sign. This mnemonic works when starting with True heading and working your way towards the left though the Magnetic heading to the Compass heading. When working the other way around = starting with the Compass heading and wishing to obtain the Magnetic heading or the True heading, you have to reverse this logic => West = minus sign and East = plus sign.

C D M 270° + 4°E => ?" Magnetic Heading (?) 274° True Heading (11) = 267"

=

V

rw

T

=> ??"

Note: Remember that you have to work your way from the left (from CH) to the right (to TH).

4537. Given:

Airplane

True track: Drift: Variation: Compass HDG:

ATPL

CPL

Heli

ATPL

CPL

245° 5° right 3°E 242°

Calculate the magnetic heading. A) 247" B) 243°

C) 237° D) 253° (Refer to figures 061-E55 and 061-E97) Let's start with some definitions: • Heading = the direction of the fore and aft axis of the aircraft. • Track = the line (direction) which the aircraft tracks over the ground. • Drift = the angle between the Heading and the Track. It is expressed as how many degrees left or right the actual track is situated from the heading (see the attached figure).

I 4456 (A) I 4460 (A) I 4466 (C) I 4525(8) I 4537 (C) I

01 Basics of Navigation Based on on the information above we can determine the heading. If True Track = 245° and the Drift = 5°R then the True Heading (TH) will be 240° (245° - 5°). Now that we have the value ofTH we can calculate the Magnetic Heading (MH) using the value of Variation using the mnemonic "CDMVT" = Can Dead Men Vote Twice? It stands for Compass Heading; Deviation; Magnetic Heading; Variation; True Heading. Easterly Deviation as well as Easterly Variation is a positive value (+ °E). Westerly Deviation as well as Westerly Variation is a negative value (- OW). When applying the values of Deviation and Variation using the above mnemonic we usually start off with the True heading => we work from the right to the left using the above "CDMVT" mnemonic. When applying the Deviation and Variation values we have to apply it in the correct sense - either with a "plus" or a "minus" sign. Remember the mnemonic "West is Best, East is Least" = in other words West = plus sign, East = minus sign. This mnemonic works when starting with True heading and working your way towards the left though the Magnetic heading to the Compass heading. When working the other way around = starting with the Compass heading and wishing to obtain the Magnetic heading or the True heading, you have to reverse this logic => West = minus sign and East = plus sign.

C

D

M

T

V

242° x r <= 3°E - 240° Magnetic Heading (?) 23r (240° TH - 3°E variation)

=

Note: You can completely disregard the compass heading - it is there just to confuse you - you could use it to determine the magnetic heading only ifyou were given the Deviation value.

4544. Airplane ATPL CPL Heli ATPL CPL If the compass heading is 265°, variation is 33°W and deviation is 3°E, what is the true heading? A) 229° B) 235 0 C) 301 0 D) 295 0 (Refer to figure 067-E97)

A very useful mnemonic for the actual calculation exists: CDMVT = Can Dead Men Vote Twice? It stands for Compass Heading; Deviation; Magnetic Heading; Variation, True Heading. When applying the values of Deviation and Variation using the above mnemonic we usually start off with the True heading => we work from the right to the left. When applying the Deviation and Variation values we have to apply it in the correct sense - either with a "plus" or a "minus" sign. Remember the mnemonic "West is Best, East is Least" = in other words West = plus sign, East = minus sign. This mnemonic works when starting with True heading and working your way towards the left though the Magnetic heading to the Compass heading. When working the other way around =starting with the Compass heading and wishing to obtain the Magnetic heading or the True heading, you have to reverse this logic => West = minus sign Ulid East = plus sign.

C D M V 265° + 3°E => r - 33°W Magnetic Heading (?) 268° True Heading (7?) = 235°

=

T

=>

?r

Note: Remember that you have to work your way from the left (from CH) to the right (to TH).

4551. Airplane ATPL Heli ATPL Given: A polar stereographic chart (Northern hemisphere) whose grid is aligned with the zero meridian. Grid track: 344° Longitude: 115°00'W Calculate the true course. A} 099 0 (T)

B) 229 0 (T)

C) 279°(T) D) 049 0 (T) (Refer to figures 067-E93 and 067-E94) The Datum Meridian (Grid North) is OOEIW. The chart convergency is the angle between the Grid North and the True North. It is expressed as - at how many

I

4544 (8)

I

4551 (8)

I

4553 (8)

I

4558 (8)

I

degrees East(+) or West(-) is the True North situated from the Grid North. In the Northern Hemisphere the chart convergency is equal to the longitude, but with sign reversed (for example chart convergency at 775°W is 775°E). ·If Chart convergency is 775°E, then Grid direction = True direction + 775° • True direction = Grid direction - 775° • True direction =344°G - 775° =229°T

4553. Airplane ATPL Heli ATPL On a polar stereographic projection chart showing the South Pole, a straight line joins position A (70°5 065°E) to position B (70°5 025°W). The true course on departure from position A is approximately: A) 250° B) 225 0 C) 1350 D) 3W (Refer to figures 067-E21, 067-E93 and 067-E94) The Datum Meridian (Grid North) is OOEIW. In the Northern Hemisphere the chart convergency is equal to the longitude, but with sign reversed. Refer to the attached figure: ·If Chart convergency is 065°E, then Grid direction = True direction + 065° • True direction = Grid direction - 065° ·25° + 65° = 90° and the two sides that compose this angle are radii of the circle and thus, equal. • The green triangle is a right-angled isosceles one. The angle a7 =a2 =45° • Theangle{3= 780°-45°-25°= 770° 0 0 • The Grid direction from A to B is 780°+ (3 = 780°+ 770 =290 G 0 • From A to 8 the True direction =290 G - 065° =225°T

4558. Airplane ATPL CPL Given: True track: Drift: Compass heading: Deviation: Calculate the variation.

Heli

ATPL

CPL

A) 25°W B) 21°W

C) SoW D) 9°W (Refer to figures 067-E55 and 067-E97) Let's start with some definitions: • Heading = the direction of the fore and aft axis of the aircraft. • Track = the line (direction) which the aircraft tracks over the ground. • Drift =the angle between the Heading and the Track. It is expressed as how many degrees left or right the actual track is situated from the heading (see the attached figure). Based on on the information above we can determine the heading. If True Track = 780° and the Drift = 8°R then the True Heading (TH) will be 77r (780° - 8°). With the value of CH and Deviation we can calculate the Magnetic Heading. Then, knowing the MH and TH we can determine the Variation. All of this using the mnemonic "CDMVT" = Can Dead Men Vote Twice? It stands for Compass Heading; Deviation; Magnetic Heading; Variation; True Heading. Easterly Deviation as well as Easterly Variation is a positive value (+ °E). Westerly Deviation as well as Westerly Variation is a negative value (- OW). When applying the values of Deviation and Variation using the above mnemonic we usually start off with the True heading => we work from the right to the left using the above "CDMVT" mnemonic. When applying the Deviation and Variation values we have to apply it in the correct sense - either with a "plus" or a "minus" sign. Remember the mnemonic "West is Best, East is Least" = in other words West = plus sign, East = minus sign. This mnemonic works when starting with True heading and working your way towards the left though the Magnetic heading to the Compass heading. When working the other way around = starting with the Compass heading and wishing to obtain the Magnetic heading or the True heading, you have to reverse this logic => West = minus sign and East = plus sign.

C

D

M

V

T

795° - rw => ?r <= ? +77r Magnetic Heading (??) 193° (795°CH - rw deviation) Variation (?) =21°W(793°MH- 772°TH)

=

Aviationexam Test Prep Edition 2012 Note: Remember that you have to work from both ends towards the value of Variation.

4569. Airplane Given: Aircraft position:

Heli

ATPL

ATPL

C

M

D

242° <= ? + 231" Deviation (?) = SoW

800 00,0'S 1400 00,0'E 025° (G)

Aircraft tracking: If the grid is aligned with the Greenwich Anti-Meridian, the true track is: A) 245° (T) B) 205° (T) C) 165° (T) 0) 065° (T) (Refer to figures 061-E93 and 061-E94) The Grid track = 025°G, the Datum Meridian (Grid North) being 180 0E/W. If we change the Datum Meridian to OOooEIIN, the Grid track = 025° +180° = 205°. In the Southern Hemisphere the chart convergency is equal to the longitude if a standard grid is used (for example chart convergency at 1400E is 140 0E). If Chart convergency is 1400E, then Grid direction = True direction + 140°. True direction = Grid direction - 140 205°G - 140°= 065°T

=

4584. Airplane ATPL CPL Heli ATPL CPL In a particular position the total strength of the terrestrial magnetic field is 5 nanotesla. The inclination is 55°. What is the strength of the horizontal component in this position? A) 4,09 nanotesla. B) 2,87 nanotesla. C) 6,05 nanotesla. 0) 1,25 nanotesla.

V

T

<= 3°E - 240°

Note: Remember that you have to work your way from the right (from TH) to the left (to CH).

4596. Airplane ATPL CPL Heli ATPL CPL The Magnetic North Pole seems to rotate around the geographical North Pole. A rate ofthis movement is approximately: A) B) C) 0)

1° in 1 year. 1° in 11 years. 1° in 5 years. 1° in 9 years.

For explanation refer to question #2236 on page 34.

4619. Airplane ATPL CPL Heli ATPL CPL Given: True track: Magnetic variation: Drift angle: What is the magnetic heading required to maintain the given track? A) 190° B) 194° C) 204°

Teslas are a measurement of magnetic force and are usually in units of micro or nano teslas. The formula is as follows: Strength =Tesla x cos dip. 5 x cos 55°= 2,87 nanotesla.

4595. Airplane Given: True track: Drift: Variation: Compass heading:

the left though the Magnetic heading to the Compass heading. When working the other way around = starting with the Compass heading and wishing to obtain the Magnetic heading or the True heading, you have to reverse this logic => West = minus sign and East = plus sign.

ATPL

Heli

CPL

ATPL

CPL

245° 5° right 3°E 242°

0) 180° (Refer to figures 061-E55 and 061-E96) We can start by converting the True Track into Magnetic Track => when converting from True to Magnetic, we can use the mnemonic "West is Best, East H is Least => we add Westerly variation and subtract Easterly variation. In our case the variation is Easterly, so the Magnetic track will equal to 192° (True Track) - 1"E (variation) = 185° Magnetic Track. Now that we have our required magnetic track we need to think about the drift. With a drift 5°L it means that our aircraft is being blown off to the left of course => our actual ground track is 5° to the left of our heading. Therefore, to follow an actual track of 185° magnetic, our magnetic heading needs to be 190~

Calculate the deviation.

4624. Airplane ATPL CPL Heli ATPL CPL . What is the value of magnetic dip at the South magnetic pole?

A) WE B) l°E

A) 360°

C) 5°E

B)

0) SoW (Refer to figures 061-E55 and 061-E97) Let's start with some definitions: • Heading = the direction of the fore and aft axis of the aircraft. • Track = the line (direction) which the aircraft tracks over the ground. • Drift = the angle between the Heading and the Track. It is expressed as how many degrees left or right the actual track is situated from the heading (see the attached figure). Based on on the information above we can determine the heading. If True Track = 245° and the Drift = 5°R then the True Heading (TH) will be 240° (245° - 5°). Now that we have the value ofTH we can calculate the Magnetic Heading (MH) and the Compass Heading (CH) using the mnemonic "CDMVT" = Can Dead Men Vote Twice? It stands for Compass Heading; Deviation; Magnetic Heading; Variation, True Heading. Easterly Deviation as well as Easterly Variation is a positive value (+ °E). Westerly Deviation as well as Westerly Variation is a negative value (- OW). When applying the values of Deviation and Variation using the above mnemonic we usually start off with the True heading => we work from the right to .the left using the above "CDMVT" mnemonic. When applying the Deviation and Variation values we have to apply it in the correct sense - either with a "plus" or a "minus" sign. Remember the mnemonic HWest is Best, East H is Least = in other words West = plus sign, East = minus sign. This mnemonic works when starting with True heading and working your way towards

I

4569 (D)

I

4584 (8)

I

4595 (D)

I

4596 (e)

I

180°

C) 090° 0) 0° For explanation refer to question #1128 on page 30.

4625. Airplane ATPL CPL Heli ATPL CPL What is the maximum possible value of dip angle? A) B) C) 0)

66° 180° 90° 45°

For explanation refer to question #1128 on page 30.

4639. Airplane ATPL CPL Heli ATPL CPL Given magnetic heading 075°, variation 4°W, drift angle 12°R, relative bearing to the station 270°. What is the true bearing of the aircraft from the station? A) B) C) 0)

4619 (A)

149° 173° 169° 161°

I

4624 (C)

I

4625 (C)

I

4639 (D)

I

01 Basics of Navigation (Refer to figure 061-E97) Magnetic Bearing (MB) to the station = Magnetic Heading (MH) Bearing (RB). MB to the station = 075° + 270° MB to the station = 345° MB from the station = 165° (345° - 180°)

• Track = the line (direction) which the aircraft tracks over the ground. • Drift = the angle between the Heading and the Track. It is expressed as how many degrees left or right the actual track is situated from the heading (see the attached figure).

+ Relative

We need the true bearing, so we need to subtract the 4°W variation to get the final result of 161°. Note: In this type of question we need to disregard the drift angle.

4663. Airplane ATPL Heli ATPL A pilot navigates from A to B on 700N on a polar stereographic chart. A is at 60 0W, B is at 600E. The initial track at A is: A) 030° B) 150° C) 350° D) 210° (Refer to figures 061-E93 and 061-E94) This is a Polar Gridded Chart problem. The Datum Meridian (Grid North) is OOE/w. The chart convergency is the angle between the Grid North and the True North. It is expressed as - at how many degrees East(+) or West(-) is the True North situated from the Grid North. In the Northern Hemisphere the chart convergency is equal to the longitude, but with sign reversed (for example chart convergency at 60 0 W is 60 0 E). ·If Chart convergency is 60 0 E, then Grid direction =True direction + 60°. • From A (70 0 N 060 0 W) to B (70 0 N 060 0 E) the Grid direction is a straight line and the initial track at A is 090 0 G. 0 • True direction = 090 G - 60° = 030°T.

4672. Airplane ATPL CPL Heli ATPL CPL Given true heading 066°, variation 4°W, drift angle 12°R, relative bearing to the station 070°. What is the true bearing of the aircraft from the station?

Based on on the information above we can determine the heading. If True Track = 070° and the Drift = 10 0 R then the True Heading (TH) will be 060° (070 0 - 10°). Now that we have the value ofTH we can calculate the Magnetic Heading (MH) and the Compass Heading (CH) using the mnemonic "CDMVT" = Can Dead Men Vote Twice? It stands for Compass Heading; Deviation; Magnetic Heading; Variation; True Heading. Easterly Deviation as well as Easterly Variation is a positive value (+ °E). Westerly Deviation as well as Westerly Variation is a negative value (- OW). When applying the values of Deviation and Variation using the above mnemonic we usually start off with the True heading => we work from the right to the left using the above "CDMVT" mnemonic. When applying the Deviation and Variation values we have to apply it in the correct sense - either with a "plus" or a "minus" sign. Remember the mnemonic "West is Best, East is Least" = in other words West = plus sign, East = minus sign. This mnemonic works when starting with True heading and working your way towards the left though the Magnetic heading to the Compass heading. When working the other way around = starting with the Compass heading and wishing to obtain the Magnetic heading or the True heading, you have to reverse this logic => West = minus sign and East =plus sign.

? <= I°E

4817. Airplane ATPL CPL Heli Isogonic lines connect positions that have:

ATPL

CPL

A) 000 0 B) 090 0

4B50. Airplane ATPL Heli ATPL On a polar stereographic map, a straight line is drawn from position A (70 0 N 102°W) to position B (BOON 006°E). The point of highest latitude along this line occurs at longitude 035°W. What is the initial straight-line track angle from A to B, measured atA?

the same variation.

A) 049 0 B) 077° C) 229°

the same elevation. the same angle of magnetic dip.

4831. Airplane Given: True track: Variation: Deviation: Drift:

D)

ATPL

Heli

CPL

ATPL

CPL

A) 1000

091 089 0 D) 101° 0

(Refer to figures 061-E55 and 061-E97) Let's start with some definitions: • Heading = the direction of the fore and aft axis of the aircraft.

I

4672 (0)

I

4817 (A)

I

4831 (C)

I

4837 (C)

023 0

(Refer to figure 061-E12) It will help to draw a sketch of the situation. You will notice that at the point where the straight line intersects the meridian 35°W the line is perpendicular to the 35°W meridian - a 90° angle. Now we basically have a right-angle triangle between point A, this intersecting point at 35°W and the North Pole. We can easily determine the internal angle of this triangle at the North Pole: lOr - 35° = 67~ Since we know two of the internal angles of the triangle (090° 0 and 06r) we can easily determine the third angle => 180° - 90 - 6r = 023°. The True direction from A will be 023°.

Calculate the compass heading.

4663 (A)

0

(Refer to figures 061-E77 and 061-E78) In the question figure the True Bearing ofpoint B from point A is 090°. The True bearing ofpoint A from point B is 270°. Do not get confused - the question asks about the bearing "from B to A".

0° variation.

For explanation refer to question #1313 on page 33.

I

= =

T 060°

D) 360°

Note: In this type of question we need to disregard the drift angle. We can also disregard the value of magnetic variation if we work with the formula 'TH + RB = TB'~

B) C)

0

4837. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-14) What is the true bearing of point A from point B?

0

True Bearing (TB) to the station =Relative Bearing (RB) + True Heading (TH). TB to the station = 070 0 + 066° TB to the station = 1360 TB from the station = 376° (136° + 180°)

A) B) C) D)

?r <= 30 W +

Note: Remember that you have to work your way from the right (from TH) to the left (to CH).

C) 270

312 D) 316°

V

M -

Magnetic Heading (??) 090 0 Compass Heading (?) 089°

A) 136° B) 148° C)

D

C

I

4850 (0)

I

Aviationexam Test Prep Edition 2012

4862. Airplane Given the following:

ATPL

CPL

Heli

ATPL

CPL

Magnetic heading: Magnetic variation: Drift angle:

ic works when starting with True heading and working your way towards the left though the Magnetic heading to the Compass heading. When working the other way around = starting with the Compass heading and wishing to obtain the Magnetic heading or the True heading, you have to reverse this logic => West = minus sign and East =plus sign.

C ?? <=

D M V 4°E - r <= 3rW Magnetic Heading (?) = 037" Compass Heading (??) = 033 0

What is the true track? A) 048 0 B) 0640 C) 0560 D) 072 0

T

+

005°

Note: Remember that you have to work your way from the right (from TH) to the left (to CH).

(Refer to figures 061-E55 and 061-E97) Let's start with some definitions: • Heading the direction of the fore and aft axis of the aircraft. • Track = the line (direction) which the aircraft tracks over the ground. • Drift =the angle between the Heading and the Track. It is expressed as how many degrees left or right the actual track is situated from the heading (see the attached figure).

=

When applying the values of Deviation and Variation using the above mnemonic we usually start off with the True heading => we work from the right to the left using the above "CDMVT" mnemonic. When applying the Deviation and Variation values we have to apply it in the correct sense - either with a "plus" or a "minus" sign. Remember the mnemonic "West is Best, East is Least" = in other words West = plus sign, East = minus sign. This mnemonic works when starting with True heading and working your way towards the left though the Magnetic heading to the Compass heading. When working the other way around = starting with the Compass heading and wishing to obtain the Magnetic heading or the True heading, you have to reverse this logic => West =minus sign and East =plus sign. Based on on the information above we can easily determine the true track. First of all we will obtain the true heading: TH = 060° (MH) - 80W (Var) TH=05r Second step is to convert the True Heading into the True Track using the drift information. Drift to the right means that our heading will be less than our track by the amount of the drift => our true track will be equal to the heading + the drift angle => 052° + 4° => 056°. Note: Easterly Deviation as well as Easterly Variation is a positive value (+ °E). Westerly Deviation as well as Westerly Variation is a negative value (- OW).

4878. Given:

Airplane

ATPL

CPL

Heli

ATPL

CPL

4880. Airplane ATPL Heli ATPL A straight line is drawn on a North polar stereographic chart joining Point A (70 0 N 060 0W) to Point B (700N 0600E). What is the initial track direction (going to the east) ofthe line at A? A) 090 0 (T)

B) 030 (T) 0

C) 1200 (T) D) 3300 (T) For explanation refer to question #4451 on page 39.

4900. Airplane ATPL Heli ATPL Two positions plotted on a polar stereographic chart, A (80 0 N 000°) and B (70 0 N 102°W) are joined by a straight line whose highest latitude is reached at 035°W. At point B, the true course is: A) 247 0

023 C) 203 D) 305

B)

0 0 0

(Refer to figure 061-El0) It will help to draw a sketch of the situation. You will notice that at the point where the straight line intersects the meridian 35°W the line is perpendicular to the 35°W meridian - a 90° angle. Now we basically have a right-angle triangle between point B, this intersecting point at 35°W and the North Pole. We can easily determine the internal angle of this triangle at the North Pole: lor - 35° = 67". Since we know two of the internal angles of the triangle (090° and 061") we can easily determine the third angle => 180° - 90° - 67" = 623~ The True direction upon arrival to B will be 023° + 180° = 203°.

4950. Airplane ATPL CPL Heli ATPL CPL When the Magnetic North Pole is East of the True North Pole, the variation is:

True track: Drift: Variation: Deviation:

4°E

What is the compass heading?

A)

+ and easterly.

B) - and easterly.

C)+ andwesterly.

A) 0-07 0 B) 033 0 C) 359 0 D) 33]0

D) - and westerly. For explanation refer to question #1154 on page 31.

(Refer to figures 061-E55 and 061-E97) Let's start with some definitions: • Heading = the direction of the fore and aft axis of the aircraft. • Track = the line (direction) which the aircraft tracks over the ground. • Drift = the angle between the Heading and the Track. It is expressed as how many degrees left or right the actual track is situated from the heading (see the attached figure). Based on on the information above we can determine the heading. If True Track = 348° and the Drift = 17"L then the True Heading (TH) will be 005° (348° + 11"). Now that we have the value ofTH we can calculate the Magnetic Heading (MH) and the Compass Heading (CH) using the mnemonic "CDMVT" = Can Dead Men Vote Twice? It stands for Compass Heading; Deviation; Magnetic Heading; Variation; True Heading.

4953. Airplane ATPL CPL Heli ATPL CPL When the Magnetic North Pole is West of the True North Pole variation is: A) + and easterly. B) - and easterly. C) - and westerly. D) + and westerly. For explanation refer to question #1154 on page 31.

When applying the values of Deviation and Variation using the above mnemonic we usually start off with the True heading => we work from the right to the left using the above "CDMVT" mnemonic. When applying the Deviation and Variation values we have to apply it in the correct sense - either with a "plus" or a "minus" sign. Remember the mnemonic "West is Best, East is Least" = in other words West =plus sign, East = minus sign. This mnemon-

I 4862 (C) 14878(8) 14880(8) I 4900 (C) I 4950 (A) I 4953 (C) I

----------------~--~

---_.

01 Basics of Navigation

230631. Given:

Airplane

ATPL

CPL

Heli

ATPL

CPL

A (56°N , l45°E) B (57°N, l65°W) WI1at is the difference in longitude between A and B? A) 020° B) 130° C) 050° D) OW

A) the compass north at that position corrected for variation. B) the isogonal to the magnetic North Pole. C) the direction of the horizontal component of the Earth's magnetic field at that position. D) the great circle between the position and the magnetic North Pole. 230687. Airplane ATPL CPL Heli ATPL CPL The horizontal component of the Earth's magnetic field: A) is minimum at the magnetic equator. B) is very small close to the magnetic poles. C) is maximum at the magnetic poles. D) increases with an increase of the magnetic latitude. 230688. Airplane ATPL CPL Heli ATPL CPL The direction of Magnetic North at a certain position coincides with the direction of: the isogonic line to the Magnetic North Pole. the horizontal component of the Earth's magnetic field. the great circle to the magnetic North Pole. the isoclinic line to the Magnetic North Pole.

230689. Airplane ATPL CPL Near the magnetic pole:

Heli

ATPL

CPL

A) the angle of dip is minimal whereby a freely suspended compass needle will be almost fully vertically aligned. B) the vertical component of the earth's magnetic field is too small to permit the use of a magnetic compass. C) the angle of dip is maximal whereby a freely suspended compass needle will be almost fully horizontally aligned. D) the horizontal component of the earth's magnetic field is too small to permit the use of a magnetic compass. 230690. Airplane ATPL CPL Heli ATPL CPL The long term periodic change in the Earth's Magnetic Field A) is caused by sunspot activity. B) affects mainly the compass deviation. C) is reflected in the slow movement ofthe magnetic poles. D) is caused by the westerly movement of the geographic North Pole. 230691. Airplane The Directive Force:

A) angle of dip. B) vertical component of the Earth's magnetic field. C) total magnetic force of the Earth's magnetic field. D) directive force. 230697. Airplane ATPL CPL Heli ATPL CPL Which of the following variables affect deviation?

230686. Airplane ATPL CPL Heli ATPL CPL The direction "Magnetic North" at a position on the Earth is:

A) B) C) D)

230692. Airplane ATPL CPL Heli ATPL CPL With an increase in magnetic latitude there will be a decrease in the:

ATPL

CPL

Heli

ATPL

CPL

A) is zero over the geographical poles. B) is about twice as strong on 60 0N/S as on the Equator. C) is the component of the Earth's magnetic field which aligns the compass needle. D) equals the vertical component of the Earth's magnetic field.

1) Magnetic latitude. 2) Aircraft heading. 3) Aircraft altitude. 4) Aircraft electronic equipment. A) 2,3 B) 1,2,4 C) 1,2,3 D) 3,4

230704. Airplane ATPL CPL Heli A definition of a Magnetic Track angle is:

ATPL

CPL

A) The direction of a line referenced to the isogonic line to the Magnetic North pole. Compass North. B) The direction of a line referenced to Magnetic North. C) The direction ofthe longitudinal axis of an aircraft referenced to Compass North. D) The direction ofthe longitudinal axis of an aircraft referenced to Magnetic North. 230705. Given:

Airplane

ATPL

Compass Heading: True Track: Drift Angle: Deviation:

CPL

Heli

ATPL

CPL

233° 256° 100R _3°

What is the variation? A) B) C) D)

36°E 200W WE 100W

230706. Airplane ATPL CPL Heli ATPL CPL A VOR is situated at position (74°N, 094°W); local variation is 50 0W. A Polar Stereographic chart supplied with a Greenwich grid is used for navigation. To proceed along (magnetic) radial 238 inbound an aircraft has to follow a Grid Track of A) B) C) D)

283° 013° 103° 193°

230707. Airplane ATPL CPL Heli ATPL CPL A route is flown from (80°5, 1000W) to (80°5, l400E).At l80 0 E/W the Grid Track (GT) and True Track (TT) on a Polar Stereographic chart, whose grid is aligned with the Greenwich meridian, are respectively: A) 1100(G) and 2900(T) B) 2500(G) and 0700(T) C) 0700(G) and 250°(T)

D) 290 0(G) and 1100(T)

1230631 (C) 1230686 (C) 1230687 (8) 1230688 (8) 1230689 (0) 1230690 (C) 1230691 (C) 1230692 (0) 1230697 (8) 1230704 (8) 1 1230705 (C) 1230706 (C) 1230707 (A) 1

Aviationexam Test Prep Edition 2012 Airplane ATPL CPL Heli ATPL CPL 230708. A route is flown from (85°S, 1OOOE) to (85°S, 1400W). At 1800E/W the Grid Track (GT) and True Track (TT) on a Polar Stereographic chart, whose grid is aligned with the Greenwich meridian, are respectively: A) 290 0(G) and 110°(T). B) 1100(G) and 290°(T). C) 250 0(G) and 070°(T). D) 070 0(G) and 250°(T).

230709. Airplane ATPL CPL Heli ATPL CPL A route is flown from (80 0S, 1000W) to (80 0S, 1400E). At 1600W the Grid Track (GT) and True Track (TT) on a Polar Stereographic chart with a grid orientated on the 180° meridian are respectively: A) B) C) D)

110~(G) and 270°(T). 290 0(G) and 270°(T). 2700(G) and 290°(T). 2700(G) and 110°(T).

230710. Airplane ATPL CPL Heli ATPL CPL A route is flown from (85°S, 1000E) to (85°S, 1400W). At 1600E the Grid Track (GT) and True track (TT) on a Polar Stereographic chart with a grid orientated on the 180~ meridian are respectively: A) B) C) D)

090 0(G) and 070°(T). 070 0(G) and 090°(T). 250 0(G) and 090°(T). 090 0(G) and 250°(T).

230711. Airplane ATPL CPL Heli ATPL CPL Thule VOR is located at (76°32'N, 68°15'W). A Polar Stereographic chart with the grid aligned with the Greenwich meridian is to be used. The local variation is 75°W. Which grid track must be maintained to track radial 210(M) inbound? A) B) C) D)

285°(G) 203°(G) 323°(G) 023°(G)

230712. Airplane ATPL CPL Heli ATPL CPL A route is drawn from (75°00'N, 060000'E) to (75°00'N, 030 000'W) on a Polar Stereographic chart with the grid aligned with the Greenwich meridian. The Grid Track (GT) is: A) B) C) D)

225°(G) 255°(G) 315°(G) 285°(G)

230714. Airplane ATPL CPL Heli ATPL CPL Route A - B is drawn on a Polar Stereographic chart with the grid aligned with the Greenwich meridian. The true track of the straight line at A is 060°. When passing the meridian 1000E, the true track is 090°. The grid track of this route on the chart is: A) B) C) D)

350 0(G) 060 0 (G)

1300 (G) 010 0 (G)

230715. Airplane ATPL CPL Heli ATPL CPL Route A - B is drawn on a Polar Stereographic chart with the grid aligned with the Greenwich meridian. The true track of the straight line at A (75°S, 0100W) is 080°. What is the Grid Track when passing the meridian of 050 0E? A) B) C) D)

1100(G) 330 0(G) 090 0(G) 070 0(G)

230716. Airplane ATPL CPL Heli ATPL CPL Route A - B is drawn on a Southern Polar Stereographic chart whose grid is aligned with the Greenwich meridian. The true track of the straight line at A is 120°. When passing the meridian of 1000E the true track is 090°. The grid track of this route on the chart is: A) B) C) D)

1200(G) 1900(G) 350 0(G) 030 0(G)

230728. Airplane ATPL CPL Heli ATPL CPL Consider the positions (OooN/S, OOooE/W) and (OooN/S, 1800E/W) on the ellipsoid. Which statement about the distances between these positions is correct?

A) The route via the North Pole is shorter than the route along the equator. B) The route via the equator is shorter than the route via the South Pole. C) The route via the South pole is shorter than the. ro.ute via the North Pole. D) The route via either pole and the route via the equator are of equal length. 230729. Airplane ATPL CPL Heli ATPL CPL The maximum difference in distance when proceeding along the great circle between two positions, in stead of the rhumb line, will occur:

230713. Airplane ATPL CPL Heli ATPL CPL Route A - B is drawn on a Polar Stereographic chart with the grid aligned with the Greenwich meridian. The True Track of the straight line at A (75°N, 0100W) is 080°. What is the Grid Track when passing the meridian 050 0E?

A) B) C) D)

on North-South tracks at low latitudes. on North-South tracks at high latitudes. on East-West tracks at low latitudes. on East-West tracks at high latitudes.

A) 1500(G) B) 1100(G)

C) 070 0 (G)

D) 090 0(G)

1230708 (C) 1230709 (8) 1230710 (8) 1230711 (0) 1230712 (8) 1230713 (0) 1230714 (A) 1230715 (0) 1230716 (8) 1230728 (A) 1 1230729 (0) 1

01 Basics of Navigation

01-05 Distance 1201. Airplane ATPL CPL Heli ATPL CPL You have calculated point of no return (PNR) on a flight, having all negative WCs in the flight plan. During the flight you experience that the WIV is stronger but coming from the same direction as in the flight plan. Consider the following statements: A) the PNR will not change because the neitherTA5 nor fuel flow has changed. B) the PNR will, if recalculated, move toward the no-wind PNR. C) you will arrive at the PNR at a later time than flight planned. D) a recalculated PNR will move toward the place of departure. (Refer to figure 067-E108) Negative Wind Component (WC) means a headwind. In still air conditions the PSR would be located furthest from the departure airport. However, in reality we almost never have absolutely still air. Therefore, if we experience any kind of wind component - irrespective whether it isihead or tail wind, the distance from departure airport to PSR will be smaller (PSR will move closer to the departure airport => you will have to turn back sooner ifit becomes necessary). The greater the wind, the greater the distance by which the PSR moves closer to the departure airport.

1245. Airplane ATPL CPL A Nautical mile is defined as:

Heli

ATPL

CPL

A) the length of a 1 minute arc, measured anywhere on the surface of the Earth. B) the average length of a l' arc of longitude and a l' arc of latitude. C) 1 852 meters. D) the average length of a 1 minute arc of a meridian. In aviation we can have the horizontal distance expressed in various units - most frequently in 3 types (according to ICAO Annex 5): • Nautical Miles (NM) for long distances • Kilometres (km) for long distances • Meters (m) for short distances

2254. Airplane ATPL CPL Heli ATPL CPL In international aviation the following units shall be used for horizontal distance: A) B) C) D)

For explanation refer to question #7245 on this page.

2304. Airplane ATPL CPL Heli ATPL CPL From the departure point, the distance to the point of equal time is: A) proportional to the sum of ground speed out and ground speed back. B) inversely proportional to the sum of ground speed out and ground speed back. C) inversely proportional to the total distance to go. D) inversely proportional to ground speed back. (Refer to figure 067-E707) Point of Equal Time = (Distance x GS Home).;. (GS Out + GS Home) In still air conditions the PET would be located in the middle between point A and point B. However, in reality we almost never have absolutely still air. Therefore, if we experience a headwind on a flight from A to B, the PET will move closer to the destination (closer to B) - into the wind. If we experience a tailwind on a flight from A to B, the PET will move closer to the departure (closer to A). In both cases, the stronger the wind, the greater the PET move away from the mid-point towards the particular direction. Now with the above formula and information on wind effects let's examine all of the answer possibilities: • Answer A) is incorrect => take a look at the PETformula and you will realize that the greater the value of (GS Out + GS Home) the shorter the Distance to PET => it is inversely proportional to the sum of GS Out and GS Home. • Answer B) is correct => as mentioned above - the greater the value of (GS Out + GS Home) the shorter the distance to PET and vice versa => therefore we can say that the distance to PET is inversely proportional to the sum of GS Out and GS Home. However, this statement is rather incomplete and pulled out of the context of the formula (it assumes that GS Home remains constant) - take this answer for what it's worth - nevertheless the JAA believes it is correct.

When dealing with height or altitude, the SI units defined by leAO are: .Meters(m) ·Feet(ft)

As far as the definitions of the units go, they are typically based on the spherical model of the Earth (even though the Earth is not a perfect sphere, for navigation purposes it is considered a sphere as the error is within the tolerances). With thisrespect, 7 Nautical Mile (NM) is defined as 7 minute of an arc of a Meridian. 7 kilometer was defined as '110000 th part of a Meridian from the Equator to the Pole (90°= 5 400 minutes =5 400NM = 70000km).

2215. Airplane ATPL CPL Heli ATPL CPL An aircraft leaves OON/S 45°W and flies due South for 10 hours at a speed of 540 kts. What is its position? A) South Pole. B) North Pole.

• Answer C) is incorrect => distance to PET is directly proportional to the total distance of the flight => the longer the distance between the two airports the longer the distance to the PET. • Answer D) is incorrect => distance to PET is directly proportional to the value of GS Home => the higher the GS Home speed, the longer the distance from the starting point to the PET and vice versa.

2319. Airplane ATPL CPL Consider the following statements:

D) 45°5. (Refer to figure 067-E100) To solve this question we will need to use the following principle: • When following a Great Circle in the NorthlSouth direction (along a Meridian) each 7° change of Latitude represents a distance of 60 NM (7 minute of Latitude = 7 NM).

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1201 (D)

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1245 (D)

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2215 (A)

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2254 (C)

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2304 (8)

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Heli

ATPL

CPL

A) the exact length of a l' of arc is shorter at high altitude than at sea level, when the arc is observed from the centre ofthe Earth. B) in any position on the surface of the Earth, the length of l' of arc east/west is equal to the length of l' of arc north/south in the same position on a perfect sphere. C) the exact length of a l' of Great Circle arc varies a little from position to position because the Earth radius vary. D) all statements are correct.

C) 300 5.

Now in the case of this question, we first need to determine the distance that we will cover in 70 hours offlight time. Ata speed of540 kts we will cover a distance of 5.400 NM. If 7 minute of Latitude change = 7 NM then 5.400 NM will represent a Latitude change of 5.400 minutes => 90° of Latitude change. If we have started at the Equator (OWlS) and traveled 900 S then the new position will be at the South Pole (hope you have packed a warm coat).

metres, statute miles and nautical miles. kilometres, feet and nautical miles. metres, kilometres and nautical miles. kilometres, statute miles and nautical miles.

For explanation refer to question #4667 on page 77.

2319 (C)

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Aviationexam Test Prep Edition 2012

4059. Airplane ATPL CPL Heli ATPL CPL When dealing with heights and altitudes in international aviation, we use the following units: A) metre and foot. B) foot, kilometre and decimals of nautical mile. C) foot and yard. 0) all answers are correct.

ATPL

CPL

A) distances will vary, dependent on the latitude.

B) distances will vary, dependent on their directions. C) a 1 minute arc of a Great Circle measured on the surface of the Earth will be equally long wherever it is measured. 0) all answers are correct. For explanation refer to question #4667 on page 1Z

4101. Airplane ATPL CPL Heli ATPL CPL The term IIdeparturel i used in navigation also has the following meaning: B) distance north/south. C) angular distance along a meridian.

0) angular distance along a parallel of latitude. (Refer to figures 061-E79, 061-E80 and 061-E81) The term "Departure" in navigation can be defined as the Earth distance (in NM) along a parallel of Latitude - in other words, a change of longitude along a parallel of latitude expressed in nautical miles. When traveling due East or West along a parallel of latitude, the distance covered is called "Departure': The term was taken from sailing techniques invented several hundred years ago by maritime navigators; who would often sail North or South to the latitude of a position, then alter course onto due East or West, as appropriate and then sail to destination by making good the particular latitude. They were able to measure latitude and keep their course by celestial observations of the Pole Star in the northern hemisphere. The advantage of the method was that with the Mercator chart, plotting position along the parallel was simplified. The technique was known a.~ "parallel sailing': The standard formula to calculate the value of "Departure" in Nautical Miles (NM) is:

• Departure = Change in Longitude (in minutes) x cos Latitude The diameter of the Earth is greatest at the Equator. As the Latitude increases North or South from the Equator, the diameter decreases and together with it the circumference of the Earth. For example the length of the parallels at 60 0 N and S is only 50% of the length of the Equator. Therefore, the distance between each 1° meridian gets progressively smaller and smaller as the Latitude increases. 1°ofLongitude change (from one meridian to anoth0 er - e.g. from 10 W to 11 OW) at the Equator equals a ground distance of 60 NM, but the same 1° Longitude change (e.g.from 10 0 W to 11 OW) at 60W represents a distance of30 NM only.

4187. Airplane ATPL CPL Heli ATPL CPL Why do we normally overlook the descend phase when calculating point of equal time (PET)? A) Because there are so many uncertain factors in the descend

phase. B) Because we never know what kind of descend clearance we will get from ATe. C) Because the descend will have an equal effect, whatever destination we decide to proceed to. 0l/Because the W/V during the descend is not known / in academic situations. (Refer to figure 061-EI07) When solving the PET calculations we do not make any specific adjustments for descent speeds because we will have to make a descend either to destination (if we decide to continue) or back at the departure (if we decide to return from the PET) => either way we will have to make a descent and the PET calculation does not take into account any descent specifics. PET calculation

4059 (A)

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4100 (C)

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4101 (A)

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4187 (C)

For explanation refer to question #4101 on this page.

4276. Airplane ATPL CPL IIKiiometer l i is defined as:

Heli

ATPL

CPL

A) the mean length of a 1/40000 part of the equator.

B) a 1/10 000 part of the meridian length from equator to the pole. C) 0,621 NM. 0) 0,454SM. For explanation refer to question #1245 on page 4Z

A) distance in direction east/west, given in nautical miles.

I

4198. Airplane ATPL CPL Heli ATPL CPL Consider the following statements on IIdeparturel l: A) as the difference of longitude increases, the departure is constant if the latitude is constant. B) as the latitude increases, the departure between two meridians decreases. C) departure may be calculated using the equation: departure=Sin Lat. x sin Long. 0) departure is independent of difference of longitude.

For explanation refer to question #1245 on page 4Z

4100. Airplane ATPL CPL Heli Assuming the Earth being a perfect sphere:

assumes that a descent at one airport will be the same as a descent at another airport.

I

4292. Airplane ATPL CPL Heli ATPL CPL An aeroplane is flying at TAS 180 kts on a track of 090°. The WIV is 045°/50 kts. How far can the aeroplane fly out from its base and return in one hour? \. A) 56 NM

B) 88NM C) 85 NM

0) 176 NM (Refer to figure 061-EI08) We can solve this question using the concept of the Point of Safe Return. The PSR (Point of Safe Return), or sometimes referred to as the PNR (Point of No Return), is defined as the furthest point along track (on a flight from point A to point B) that an aircraft can fly to and still return backtci point A and retain the minimum required fuel reserves upon landing. To find the PSR we will use the following variables and formula: . • Endur = Endurance in hrs (flight time to having only reserve fuel available) = 1 hr • GS Out = Ground Speed Out • GS Home = Ground Speed back to base Time from bl!1se to PSR = (Endur x GS Home) .{- (GS Out + GS Home) Now we need to calculate the Ground Speeds (GS) based on our TAS and the wind. Use your flight computer to calculate the GS: TAS = 180 kts;course = 090° and wind = 045° / 50 kts. You should come-up with a GS of approx. 141 kts (GS Out). For the return back to base the GS will be 212 kts (GS Home). With this information we can now proceed with the actual PSR calculation: Time to PSR = (1 x 212) 7 (141 + 212) Time to PSR = (212) 7 (353) Time to PSR = 0,6 hrs => 36 minutes. Now to find the distance from the departure airport to the PSR we simply take the time to fly to PSRand mUltiply it by the GS Out=> 0,6hrsx 141 kts =85 NM.

4304. Airplane ATPL CPL Heli ATPL CPL A flight is planned from A (37°00,0'N OOooOO,O'E/W) to B (46°00,0'N OOOoOO,O'E/W). The distance in kilometres from A to B is approximately: A) 540 km

B) 794km C) 1.000km 0) 1.771 km (Refer to figure 061-E100) To solve this question we will need to use the following principle: • When following a Great Circle in the North/South direction (along a Meridian) each 1° change of Latitude represents a distance of 60 NM (1 minute of Latitude = 1 NM).

4198 (8)

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4276 (8)

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4292 (C)

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4304 (C)

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01 Basics of Navigation Now in the case of this question, we first need to determine the change of Latitude and then use the formula above. To determine the Ch.Lat simply deduct the two values since they are in the same hemisphere => 46"N - 3rN = 9° = 540 minutes. If 1 minute of Latitude equals 1 NM then 540 minutes of Latitude wifl represent a distance of 540 NM. 1 NM equals 1,852 km => 540 NM equals 1.000km.

South for 3.600 NM East for 3.600 NM North for 3.600 NM West for 3.600 NM The final position of the aircraft is: A) 59°00'N 090 0 00'W B) 60 0 00'N 0900 00'W C) 60 0 00'N 030 0 00'E 0 D) 59°00'N 060 00'W

4312. Airplane ATPL CPL Heli ATPL CPL The distance along a meridian between 63°55'N and 13°47'S is: A) 3.008 NM

B) 7.702 NM C) 5.008NM D) 4.662 NM (Refer to figure 061-El00) To solve this question we wifl need to use the following principle: • When following a Great Circle in the North/South direction (along a Meridian) each 1° change of Latitude represents a distance of 60 NM (1 minute of Latitude = 1 NM). In case of our question, we have to first find the total difference in Latitude: • from 63°55'N to the Equator = 63°55 minutes = 3 835 minutes ((63° x 60) + 55) • from the Equator to 13°47'5 = 13°47 minutes =827 minutes ((13° x 60) + 47) • total change in Latitude distance = 4 662 minutes

(Refer to figure 061-E100) To solve this question we will need to use two principles: • When following a Great Circle in the North/South direction (along a Meridian) each 1° change of Latitude represents a distance of 60 NM (1 minute ofLatitude = 1 NM). • When following a Great Circle in the East/West direction (along a Parallel) each 1° change ofLongitude represents a distance of1 ° x 60 x cos Latitude. Now in the case of this question, let's solve it step-by-step, by calculating each of the legs: Leg 1 = 3.600 NM south: ·3.600 NM along a Meridian = 3.600 minutes = 60° Southbound ·If we started at 060 0 N 30 0 W we will end up at OOooN/S 30 0 W

If 1 minute of Latitude change = 1 NM then 4.662 minutes wifl represent a distance of 4.662 NM.

Leg 2 = 3.600 NM east: • Distance along the Equator = 1 NM per 1 minutes of Longitude ·3.600 NM along a the Equator =3.600 minutes =60° Eastbound .If we started at OOooN/S 30 0 W we will end up at OOOoN/S 30 0 E

4329. Airplane ATPL CPL Heli ATPL CPL Determine the distance between points A (45°00,0'N 010 0 00,0'W) and B (45°00,0'N 005°00,0'E):

Leg 3 = 3.600 NM north: ·3.600 NM along a Meridian =3.600 minutes =60° Northbound .If we started at OOooN/S 30 0 E we wifl end up at 060 0 N 30 0 E

A) B) C) D)

300NM 636,4NM 900NM 212,1 NM

(Refer to figure 061-E100) To solve this question we will need to use the following principle: • When following a line of Parallel (Rhumb Line), each 1° change of Longitude change represents a distance of 1° x 60 x cos Latitude. In case of our question, the difference in Longitude = 15° (10° to the Greenwich from the 10 0 00'W position and additional 5° from the Greenwich to the 005°00'E position). 15° = 900 minutes.

Leg 4 = 3.600 NM west: ·3.600 NM along a Parallel at 060 0 N = ? • Distance = Change in Longitude (in minutes) x cos Latitude ·3.600 NM = ? minutes x cos 60° • ? minutes = 3.600 NM + cos 60° .? =Z200 minutes = 120° Westbound .If we started at 060 0 N 30 0 E we will end up at 060 0 N 90°W. The final position wifl therefore be 060 0 N 90°W.

4336. Airplane ATPL CPL Heli ATPL CPL An aircraft at position 60 0 00'N 005°12'W flies 165 km due West. The aircraft's new position is:

Now we can use the formula: • Distance = Change in Longitude (in minutes) x cos Latitude • Distance = 900 minutes x cos 45° • Distance = 636,4 NM

A) 60 0 00'N 002°14'E B) 600 00'N 008°10'W C) 60 0 00'N 001°10'E D) 60 0 00'N 001°10'W

4330. Airplane ATPL CPL Heli ATPL CPL The distance between the parallels of latitude 17°23'S and 23°59'N is: A) B) C) D)

4.122 NM 636 NM 2.473 NM 2.482 NM

In case of our question: • 1 NM = 1,852 km • 165 km =89,09 NM

(Refer to figure 061-E100) To solve thi~ qllPstion wp will nppd tn 1I5f:' the fol/owing principle: • When following a Great Circle in the North/South direction (along a Meridian) each 1° change of Latitude represents a distance of 60 NM (1 minute of Latitude = 1 NM). Now in the case of this question, we first need to determine the change of Latitude and then use the formula above: • From lr23'StotheEquator= lr23' • From the Equator to 23°59'N =23°59' • Total Ch.Lat= lr23' + 23°59' = 41°22' ·41°22'=2.482 minutes ·If 1 minute = 1 NM, then 2.482 minutes = 2.482 NM.

4335. Airplane ATPL CPL Heli ATPL CPL What is the final position after the following rhumb line tracks and distances have been followed from position 60 0 00'N 0300 00'W?

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4312 (0)

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4329 (8)

(Refer to figure 061-E100) To solve this question we will need to use the following principle: • When following a line of Parallel (Rhumb Line), each 1° change of Longitude change represents a distance of 1° x 60 x cos Latitude.

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4330 (0)

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4335 (8)

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4336 (8)

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Now we can use the formula: • Distance = Change in Longitude (in minutes) x cos Latitude ·89,09 NM ~ ? minutes x cos 60° • ? minutes = 89,09 NM + cos 60° .? = 178,18 minutes = 2°58'11" Westbound ·If we started at 005°12'W (5,2"W) and travel 2,9r westbound, we will end up at 8, lrW = 008°1O,2'W.

4337. Airplane ATPL CPL Heli ATPL CPL 0 An aircraft at position 60 00'N 005°22'W flies 165 km due East. What is the new position? A) B) C) D)

60 0 00'N 008°20'E 60 0 00'N 002°24'W 600 00'N 001°08'E 60 0 00'N 00l°08'W

(Refer to figure 061-E100) . To solve this question we will need to use the following principle:

4337 (8)

I

Aviationexam Test Prep Edition 2012 • When following a line of Parallel (Rhumb Line), each 1° change of Longitude change represents a distance of lOx 60 x cos Latitude. In case of our question: • 1 NM = 1,852 km • 165km=89,09NM Now we can use the formula: • Distance = Change in Longitude (in minutes) x cos Latitude ·89,09 NM = ? minutes x cos 60° • ? minutes = 89,09 NM .,. cos 60° • ? = 178,18 minutes = 2°58'11" Eastbound ·If we started at 60 0 N 005°22'W we will end up at 60 0 N 002°23'50"W => approximately 60 0 N 002°24'W.

4358. Airplane ATPL CPL Heli ATPL CPL Two points A and Bare 1.000 NM apart. TAS is 490 kts. On the flight between A and B the wind component is -20 kts. On the return leg between B and A, the wind component is +40 kts. What distance from A, along the route A to B, is the point of equal time (PET)? A) 470NM

B) 530NM C) 455 NM 0) 500 NM

In case of this question, we need to fly at the same speed at which the Sun is moving across the sky. The question states that the Sun is moving ata speed of 15° change ofLongitude in 1 hour. We just Hplug Hthis information into the formula listed above to get the correct result. 15° of Longitude change = 900 minutes of Longitude change (15° x 60 minutes). • Distance = Change in Longitude (in minutes) x cos Latitude • Distance = 900 minutes x cos 60° • Distance = 450 NM The Sun covers a distance of 450 NM in 1 hour. To keep-up with the Sun we would need to maintain a speed of 450 kts.

4372. Given: Position A: Position B: Position C:

Airplane

ATPL

CPL

Heli

ATPL

CPL

60 0N0200W 60 0N 021°W 59°N0200W

What are, respectively, the distances from A to B and from A toC?

(Refer to figure 061-El07) Dist =Distance between A and B GS Out =Groundspeed to destination GS Home = Groundspeed back home

A) B) C) 0)

Dist to PET = (Dist x GS Home) .,. (GS Out + GS Home) Time to PET = Distance to PET.,. GS Out First step of our calculation is going to involve the determination of Ground Speeds: • GS Out = 470 kts (490 kts - 20 kts headwind) • GS Home = 530 kts (490 kts + 40 kts tailwind) Distance from Departure to Destination is quoted by the question as 1.000 NM. Now we have all of the variables that we need to calculate the PET: Dist to PET = (1.000 x 530) .,. (470 + 530) Dist to PET = (530.000) .,. (1.000) Dist to PET = 530 NM Note: Remember that when a wind component has a negative (-) prefix, it means a headwind, whereas a positive (+) prefix means a tailwind. A good mnemonic is to realize that when you want to get to your destination a headwind is bad for you =negative; whereas a tailwind is good for you =positive.

4370. Airplane ATPL CPL Heli ATPL CPL What is the rhumb line distance, in nautical miles, between two positions on latitude 60 0N, that are separated by 10° of longitude? A) B) C) 0)

0) 780 kts (Refer to figure 061-E100) To solve this question we will need to use the following principle: • When following a line of Parallel (Rhumb Line), each 1° change of Longitude change represents a distance of 1° x 60 x cos Latitude.

300NM 520NM 600 NM 866 NM

60 NM and 30 NM. 52 NM and 60 NM. 30 NM and 60 NM. 60 NM and 52 NM.

(Refer to figure 061-El00) To solve this question we will need to use two principles: • When following a Great Circle in the North/South direction (along a Meridian) each 1° change of Latitude represents a distance of 60 NM (1 minute of Latitude = 1 NM). • When following a Great Circle in the East/West direction (along a Parallel) each 1° change of Longitude represents a distance of 1° x 60 x cos Latitude. Now in the case of this question: From A to B. Notice that this leg will be along the same line of parallel (60 0 N) => we will fly 1° (60 minutes) to the west: • Distance = Change in Longitude (in minutes) x cos Latitude • Distance = 60 minutes x cos 60° • Distance = 30 NM From A to C. Notice that this leg will be along the same meridian (20 0 W) => we will fly 1° (60 minutes) to the south: • 1 minute along a Meridian = 1 NM • 1° along a Meridian = 60 minutes ·60 minutes along a Meridian = 60 NM Note: be careful - the question is asking about the distances from A to B and then from A to C!!! Not for A to B and the from B to C.

4373. Airplane ATPL CPL Heli ATPL CPL What is the length of one degree of longitude at latitude 60° South?

(Refer to figure 061-El00) To solve this question we will need to use the foilowing principle: • When following a line of Parallel (Rhumb Line), each 1° change of Longitude change represents a distance of 1° x 60 x cos Latitude. In case of our question, the difference in Longitude = 10° = 600 minutes. Now we can use the formula: • Distance =Change in Longitude (in minutes) x cos Latitude • Distance = 600 minutes x cos 60° • Distance = 300 NM

4371. Airplane ATPL CPL Heli ATPL CPL The Sun moves from East to West at a speed of 15° longitude an hour. What ground speed will give you the opportunity to observe the Sun due South at all times at 60000'N?

A) 30 NM B) 52 NM C) 60NM 0) 90NM (Refer to figure 061-E100) To solve this question we will need to use the following principle: • When following a line of Parallel (Rhumb Line), each 1° change of Longitude change represents a distance of 1° x 60 x cos Latitude. In case of our question, the difference in Longitude = 1° = 60 minutes: • Distance =Change in Longitude (in minutes) x cos Latitude • Distance = 60 minutes x cos 60° • Distance =30 NM

A) 300 kts B) 520 kts C) 450 kts

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4358 (8)

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4370 (A)

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4371 (C)

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4372 (C)

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4373 (A)

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01 Basics of Navigation

4380. Airplane ATPL CPL Heli How many feet are equivalent to 9,5 km?

ATPL

• When following a line of Parallel (Rhumb Line), each 1° change of Longitude change represents a distance of 1° x 60 x cos Latitude.

CPL

Now we can use the formula: • Distance = Change in Longitude (in minutes) x cos Latitude ·280 NM = ? minutes x cos 45° • ? minutes = 280 NM .;. cos 45° • ? = 396 minutes =6°36' Eastbound

A) 31.160 ft B) 50.160 ft C) 57.760ft D) 9.500ft

If the position of B is 45°N 45° 15'E and the position of A is to the West of B, displaced by 6°36~ we can easily determine the co-ordinates ofA as 45°N 38°39'E.

7 m = 3,2808 ft = 39,37 inches 7 km = 7000 m = 3.280 ft

Therefore, 9,5 km =31160ft(9,5 km x 3280ft).

4383. Airplane ATPL CPL How many feet are there in 1 SM?

Heli

ATPL

4400. Airplane ATPL CPL How long is 25 km at 60 0 00'N?

CPL

A) B) C) D)

A) 3.280ft B) 5.280ft C) 6.080ft

1inch = 2,54 cm 1 m = 3,2808 ft = 39,37 inches 1 km = 1.000 m = 3.280 ft 1 ft= 0,3048 m = 12 inches

A) B) C) D)

ATPL

CPL

A) 53°20'N 172°38'E B) 45°00'N 172°38'E C) 53°20'N 169°22'W

3.280ft 5.280ft 6.080ft 1.000ft

D) 45°00'N 169°22'W

4385. Airplane ATPL CPL How many feet are there in a km?

(Refer to figure 061-E700) To solve this question we will need to use two principles: • When following a Great Circle in the North/South direction (along a Meridian) each 1° change of Latitude represents a distance of 60 NM (/ min,ute of Latitude = 1 NM). • When following a Great Circle in the East/West direction (along a Parallel) each 1° change ofLongitude represents a distance of 1° x 60 x cos Latitude.

Heli

ATPL

CPL

3.280ft 5.280ft 6.080ft 1.000ft

Now in the case of this question, let's solve it step-by-step, by calculating each of the legs: Leg 1 = 2.950 NM north: ·2.950 NM along a Meridian = 2.950 minutes = 49,16° Northbound • If we started at 004°70'5 (-4,16°) we will end up at045°N (-4,16°+49,16°) • The co-ordinates of after this leg will be 45°N 178°22'W

For explanation refer to question #4383 on this page.

4389. _. Airplane ATPL CPL 1 nautical mile equals: A) B) C) D)

Heli

ATPL

CPL

Leg 2 = 382 NM west ·382 NM along a Parallel at 045°N = ? • Distance = Change in Longitude (in minutes) x cos Latitude ·382 NM = ? minutes x cos 45° • ? minutes =382 NM.;. cos 45° .? = 540,23 minutes =approx. 9,23° Westbound ·If we started at 178°22'W (/78,36°W) and continue West, we will reach the 180° E/W meridian in 1,64° (180° - 778,36°). That will leave us to travel 7,59° further West to a final location ofBat 172,41°E (180° -7,59°) = 172°25'E.

1 852 meters. 5 280 feet. 0,896 statute mile. 3 081 yards.

1 NM = 1,852 km = 1,1507 SM = 6080 ft 1 km = 7000m=3280ft

If 1 km = 1000 m and if 1 NM = 1,852 km, then 1 NM = 1852 m.

4394. Given:

Airplane

Position A: Position B: Distance A-B: B is to the East of A

ATPL

Heli

CPL

ATPL

The co-ordinates ofB will be 045°N 172°25'E.lfwe rounded-up the angular distance in leg 1 to yo we would obtain the result of 172°38'E (9° - 1,64 => 7,36° ... 180°-7,36°= 172,64°= 172"38').

CPL

4403. Airplane ATPL CPL Heli ATPL CPL A is at 55°00'N 151°00'W and Bat 45°00'N 162°53'W. What is the departure?

45°N, ?OE 45°N,45°15'E 280NM

A) B) C) D)

Determine longitude of position A. A) B) C) D)

38°39'E 49°57'E 51°51'E 40 0 33'E

4380 (A)

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4383 (8)

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4384 (C)

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4385 (A)

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4389 (A)

584NM 458 NM 546 NM 409 NM

(Refer to figure 061-E700) The term "Departure" in navigation can be defined as the Earth distance along a parallel of Latitude. The standard formula to calculate the value of "Departure" in Nautical Miles (NM) is:

(Refer to figure 061-El00) To solve this question we will need to use the following principle:

I

40,2 statute miles. 46,3 nautical mile. 13,5 nautical mile. 27,0 nautical mile.

4401. Airplane ATPL CPL Heli ATPL CPL An aircraft departs from position A (04°10'S 178°22'W) and flies northward following the meridian for 2.950 NM. It then flies westward along the parallel of latitude for 382 NM to position B. The coordinates of position Bare?

For explanation refer to question #4383 on this page.

A) B) C) D)

CPL

1NM = 1,852 km = 1,7507 SM = 6.080 ft 25 km = 13,5 NM (25.;. 1,852)

1 cm = 0,3937 inches

Heli

ATPL

Do not get tricked by this question => 1 NM always equals 1,852 km, much in the same way as 1 km consists of 1 000 meters, regardless of the position on the Earth.

D) 1.000 ft

Summary of conversion factors for distance calculations: 1 NM= 1,852km= 1,1507SM=6.080ft 1SM = 1,609 km =0,8689 NM =5.280 ft

4384. Airplane ATPL CPL How many feet are there in a NM?

Heli

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4394 (A)

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4400 (C)

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4401 (8)

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4403 (8)

Aviationexam Test Prep Edition 2012 • Departure

=Change in Longitude (in minutes) x cos Latitude

In the case of this question, where the Latitudes are different, we will use the mean Latitude. Mean latitude of 55°N and 45°N will be SOON ((55° + 45°) + 2). The change in Longitude is W53' (762°53'W - 151°W) = 713 minutes. We are now ready to use the Departure formula: • Distance = Change in Longitude (in minutes) x cos Latitude • Distance = 713 x cos 50° • Distance =458,3 NM

4404. Airplane ATPL CPL Heli ATPL CPL The rhumb line distance between points A (60 0 00'N 002°30'E) and B (60 0 00'N 007°30'W) is: A) 150 NM B) 450 NM e) 600 NM

D) 300 NM (Refer to figure 061-E100) To solve this question we will need to use the following principle: • When following a line of Parallel (Rhumb Line), each 1° change of Longitude change represents a distance of 1° x 60 x cos Latitude.

Dist to PET = (7.086.900) + (670) Dist to PET = 1.622 NM If we now wanted to determine the time it will take us to reach the PET from our departure, we simply divide the distance to PET by our groundspeed (GS Out) => 1.622 NM + 370 kts = 4,38 hrs => 4 hrs 23 minutes => 263 minutes.

4560. Airplane ATPL CPL Heli ATPL CPL An aircraft at latitude 06°12,0'5 tracks 000° (T) for 1.667 km. On completion of the flight the latitude will be: A) 21°12,0'5 B) 21°12,5'N C) 08°48,0'N D) 09°14,0'N (Refer to figure 061-EI00) To solve this question we will need to use the following principle: • When following a Great Circle in the North/South direction (along a Meridian) each 1° change of Latitude represents a distance of 60 NM (7 minute of Latitude = 1 NM).

In case of our question, the difference in Longitude = 10° (2"30' to the Greenwich from the 2°30'E position and additional 7"30' from the Greenwich to the 7"30'W position). 10° = 600 minutes.

Now in the case of this question, we first need to determine the distance that we will cover in 3 hours of flight time. At a speed of 444 km/h we will cover a distance of 7.332 km: • 1 NM = 1,852 km ·1667 km = 900,1 NM

Now we can use the formula: • Distance = Change in Longitude (in minutes) x cos Latitude • Distance = 600 minutes x cos 60° • Distance =300 NM

If 1 minute of Latitude change = 1 NM then 900, t NM will represent a Latitude change of 900, 1 minutes => approximately 15° ofLatitude change. If we have started at 06°12'5 (-6,2°) and traveled 15° northbound then the new position will be 8,8°N (-6,2" + W) = 08°48'N.

4409. Airplane ATPL CPL Heli ATPL CPL An aircraft departing A (40 0 00'N 080 0 00'E) flies a constant true track of 270° at a ground speed of 120 kts. What are the coordinates of the position reached in 6 hrs?

4561. Given:

A) B) e) D)

40 0 00'N 068°1 O'E. 40 0 00'N 064°20'E. 40 0 00'N 070 0 30'E. 40 0 00'N 060 0 00'E.

Airplane

Distance A to B: Ground speed OUT: Ground speed BACK:

ATPL

CPL

Heli

ATPL

CPL

2.346NM 365 kts 480 kts

The time from A to the point of equal time (PET) between A and B is:

(Refer to figure 061-E100) To solve this question we will need to use the following principle: • When following a line of Parallel (Rhumb Line), each 1° change of Longitude change represents a distance of lOx 60 x cos Latitude.

A) 167 min. B) 219 min. e) 260 min. D) 197 min.

Now in the case of this question, we first need to determine the distance that we will cover in 6 hours of flight time. At a speed of 120 kts we will cover a distance of 720 NM. Now we can use the formula: • Distance = Change in Longitude (in minutes) x cos Latitude • 720 NM = ? minutes x cos 40° • ? minutes = 720 NM + cos 40° .? =939,9 minutes = 15,66° Westbound ·If we started at 40W 080 0E we will end up at 400N 64,34°E (080 0E - 15,66°) => approx. 400N 64°20'E.

(Refer to figure 061-E107) Dist =Distance between A and B =2.346 NM GS Out =Groundspeed to destination =365 kts GS Home = Groundspeed back home = 480 kts

4503. Given:

If we now wanted to determine the time it will take us to reach the PET from our departure, we simply divide the distance to PET by our groundspeed (GS Out) => 7.333 NM + 365 kts = 3,65 hrs => 3 hrs 39 minutes => 219 minutes.

Airplane

Distance A to B: Ground speed OUT: Ground speed BACK:

ATPL

CPL

Heli

ATPL

CPL

3.623 NM 370 kts 300 kts

The time from A to the point of equal time (PET) between A and B is: A) B) e) D)

Dist to PET = (Dist x GS Home) + (GS Out + GS Home) Time to PET = Distance to PET + GS Out Dist to PET = (2.346 x 480) + (365 + 480) Dist to PET = (7.126.080) + (845) Dist to PET = 7.333 NM

4566. Airplane ATPL CPL Heli ATPL CPL The distance from A to B is 2.368 NM. If outbound ground speed is 365 kts and homebound ground speed is 480 kts and safe endurance is 8 hrs 30 min, what is the time to th,e PNR? A) 290 minutes. B) 209 minutes. e) 219 minutes. D) 190 minutes.

323 min. 288 min. 263 min. 238 min.

(Refer to figure061-E107) Dist =Distance between A and B =3.623 NM GS Out = Groundspeed to destination =370 kts GS Home =Groundspeed back home =300 kts Dist to PET = (Dist x GS Home) + (GS Out + GS Home) Time to PET = Distance to PET + GS Out Dist to PET = (3.623 x 300) + (370 + 300)

(Refer to figure 061-E108) The PSR (Point of Safe Return), or sometimes referred to as the PNR (Point of No Return), is defined as the furthest point along track (on a flight from point A to point B) that an aircraft can fly to and still return back to point A and retain the minimum required fuel reserves upon landing. To find the PSR we will use the following variables and formula: • Endur Endurance in hrs (flight time to having only reserve fuel available) = 8,5 hrs

=

I 4404 (D) I 4409 (8) I 4503 (C) I 4560 (C) I 4561 (8) I 4566 (A) I

EI

01 Basics of Navigation • GS Out = Ground Speed Out (from A to B) = 365 kts • GS Home =Ground Speed back Home (from PSR back to A) =480 kts Time from A to PSR = (Endur x GS Home) -;- (GS Out + GS Home) With the information stated in the question we can now proceed with the actual PSR calculation:

Time to PSR = (5.550) -;- (945) Time to PSR = 5,B7 hrs => 5 hrs 52 minutes => 352 minutes. Now to find the distance from the departure airport to the PSR we simply take the time to fly to PSR and multiply it by the GS Out => 5,B7 hrs x 390 kts = 2.2B9NM.

Time to PSR = (B,5 x 4BO) -;- (365 + 4BO) Time to PSR = (4.0BO) -;- (B45) Time to PSR = 4,B3 hrs => 4 hrs 50 minutes => 290 minutes.

Note 1: Remember that when a wind component has a negative (-) prefix, it means a headwind, whereas a positive (+) prefix means a tailwind. A good mnemonic is to realize that when you want to get to your destination a head~ wind is bad for you =negative; whereas a tailwind is good for you =positive.

4570. Airplane ATPL CPL Heli ATPL CPL An aircraft has a TAS of 300 knots and is over a stretch of water between 2 airfields 500 NM apart. If the wind component is 60 knots head, what is the distance from the first airfield to the critical point?

Note 2: Do not confuse PSR (Point of Safe Return) with PET (Point of Equal Time). PET = the point at which the time to return back to departure is equal to the time to continue to the destination.

A) 250 NM B) 200 NM C) 300NM D) 280 NM (Refer to figure 061-E107) The Point of Equal Time (PET) or sometimes referred to as the Critical Point (CP) is a point along the route offlight from which it will take the same time to either continue to the destination airport or to return to the departure airport.

4581. Airplane ATPL CPL Heli ATPL CPL An aircraft leaves point A (75°N 50 0 W) and flies due North. At the North Pole it flies due South along the meridian of 65°50'E unit reaches 75°N (point B). What is the total distance covered? A) B) C) D)

1.650 NM 2.000NM 2.175 NM 1.800 NM

(Refer to figure 061-E100) To solve this question we will need to use the following principle: • When following a Great Circle in the North/South direction (along a Meridian) each 1° change of Latitude represents a distance of 60 NM (1 minute of Latitude = 1 NM).

Dist =Distance between two airports =500 NM GS Out =Groundspeed to destination " GS Home =Groundspeed back home Dist to PET = (Dist x GS Home) -;- (GS Out + GS Home) Time to PET =Distance to PET -;- GS Out First step of our calculation is going to involve the determination of Ground Speeds: • GS Out = 240 kts (300 kts - 60 kts headwind) • GS Home = 360 kts (300 kts + 60 kts tailwind)

Now in the case of this question, we have started at the position 75°N 50 0 W and traveled to the North Pole = 15° up. At the North Pole we have turned to the right and traveled back to 75°N 65°50'E = 15° down => total of30° along a Meridian. 30° = I.BOO minutes. If 1 minute of Latitude change = 1 NM then I.BOO minutes of Latitude will represent a distance of I.BOO NM.

Distance from Departure to Destination is quoted by the question as 500 NM. Now we have all of the variables that we need to calculate the PET: Dist to PET = (500 x 360) -;- (240 + 360) Distto PET = (1BO.OOO) -;- (600) Dist to PET = 300 NM

4617. Given:

4578. Airplane ATPL CPL Heli ATPL CPL An aircraft was over Q at 13:20 hours flying direct to R. Given: Distance Q to R: True airspeed: Mean wind component OUT: Mean wind component BACK: Safe endurance:

3.016 NM 480 kts -90 kts +75 kts 10:00 hrs

The distance from Q to the point of safe return (PSR) is: A) 2.370 NM B) 2.290NM C) 1.310 NM D) 1.510 NM (Refer to figure 067-£10B) The PSR (Point of Safe Return), or sometimes referred to as the PNR (Point of No Return), is defined as the furthest point along track (on a flight from point A to point B) that an aircraft can fly to and still return back to point Aand retain the minimum required fuel reserves upon landing. To find the PSR we will use the following variables and formula: • Endur = Endurance in hrs (flight time to having only reserve fuel available). = 10hrs • GS Out =Ground Speed Out (from Q to R) • GS Home = Ground Speed back Home (from PSR back to Q) Time from Q to PSR = (Endur x GS Home) -;- (GS Out + GS Home) Now we need to calculate the Ground Speeds based on our TAS and the wind. Note that when a wind component has a negative (-) prefix, it means a headwind, whereas a positive (+) prefix means a tailwind. • GS Out = 390 kts (4BO kts TAS - 90 kts headwind); • GS Home = 555 kts (4BO kts TAS + 75 kts tailwind).

Airplane

ATPL

CPL

Distance A to B: Wind component A to B: Wind component B to A: TAS:

Heli

ATPL

CPL

360NM -15 kts +15 kts 180 kts

What is the distance from the equal-time-point to B? A) B) C) D)

170 NM

195 NM 180 NM 165 NM

(Refer to figure 061-El07) Dist = Distance between A and B GS Out =Groundspeed to destination GS Home = Groundspeed back home Dist to PET = (Dist x GS Home) -;- (GS Out + GS Home) Time to PET =Distance to PET -;- GS Out First step of our calculation is going to involve the determination of Ground Speeds: • GS Out = 165 kts (1BO kts - 15 kts headwind) • GS Home = 195 Ab (1BO kb + 15 kls tailwind) Distance from Departure to Destination is quoted by the question as 360 NM. Now we have all of the variables that we need to calculate the PET: Dist to PET = (360 x 195) -;- (165 + 195) Dist to PET = (70.200) -;- (360) Dist to PET = 195 NM Be careful- the question asks about the distance from the PET to point B => 360 NM - 195 NM = 165 NM. Note: Remember that when a wind component has a negative (-) prefix, it means a headwind, whereas a positive (+) prefix means a tailwind. A good mnemonic is to realize that when you want to get to your destination a headwind is bad for you = negative; whereas a tailwind is good for you =positive.

With this information we can now proceed with the actual PSR calculation: Time to PSR = (10 x 555) -;- (390 + 555)

I

4570 (C)

I

4578 (8)

I

4581 (D)

I

4617 (D)

I

EI

Aviationexam Test Prep Edition 2012

4633. Airplane ATPL CPL Heli ATPL CPL An aircraft was over Q at 1320 hours flying direct to R. Given:

If the aircraft left the departure point at 14:35 then 2 hrs 22 minutes later equals

Distance Q to R: True airspeed: Mean wind component OUT: Mean wind component BACK:

Note: Remember that when a wind component has a negative (-) prefix, it means a headwind, whereas a positive (+) prefix means a tailwind. A good mnemonic is to realize that when you want to get to your destination a headwind is bad for you =negative; whereas a tailwind is good for you =positive.

3.016 NM 480 kts -90 kts +75 kts

The ETA for reaching the point of equal time (PET) between Q and Ris:

to 16:57.

4647. Given:

Airplane

ATPL

Distance A to B: Ground speed OUT: Ground speed BACK:

A) 18:20 B) 17:56 e) 17:52 D) 17:42

Heli

CPL

ATPL

CPL

2.484NM 420 kts 500 kts

The time from A to the point of equal time (PET) between A and B is:

(Refer to figure 061-E107) Dist = Distance between Q and R GS Out = Groundspeed to destination GS Home = Groundspeed back home

A) B) e) D)

Dist to PET = (Dist x GS Home) -i- (GS Out + GS Home) Time to PET = Distance to PET -i- GS Out First step of our calculation is going to involve the determination of Ground Speeds: • GS Out = 390 kts (480 kts - 90 kts headwind) • GS Home =555 kts (480 kts + 75 kts tailwind) Distance from Departure to Destination is quoted by the question as 3.016 NM. Now we have all of the variables that we need to calculate the PET: Dist to PET = (3.016 x 555) -i- (390 + 555) Dist to PET = (1.673.880) -i- (945) DisttoPET= 1.771 NM

173 min. 163 min. 193 min. 183 min.

(Refer to figure 061-E107) Dist = Distance between A and B = 2.484NM GS Out =Groundspeed to destination =420 kts GS Home =Groundspeed back home =500 kts Dist to PET = (Dist x GS Home) -i- (GS Out + GS Home) Time to PET = Distance to PET -i- GS Out Dist to PET = (2.484 x 500) -i- (420 + 500) Dist to PET = (1.242.000) -i- (920) Dist to PET = 1.350 NM

If we now wanted to determine the time it will take us to reach the PET from our departure, we simply divide the distance to PET by our groundspeed (GS Out) => 1.771 NM -i- 390 kts = 4,54 hrs => 4 hrs 32 minutes => 272 minutes. If the aircraft left the departure point at 13:20 then 4 hrs 32 minutes later equals

If we now wanted to determine the time it will take us to reach the PETfrom our departure, we simply divide the distance to PET by our groundspeed (GS Out) => 1.350 NM -i- 420 kts = 3,21 hrs => 3 hrs 13 minutes => 193 minutes.

to 17:52.

4674. Airplane ATPL CPL Heli ATPL CPL The departure between positions 60 0 N 1600 E and 60 0 N "x" is 900 NM. What is the longitude of "x" when flying eastbound?

Note: remember that when a wind component has a negative (-) prefix, it means a headwind, whereas a positive (+) prefix means a tailwind. A good mnemonic is to realize that when you want to get to your destination a headwind is bad for you = negative; whereas a tailwind is good for you = positive.

4634. Airplane ATPL CPL Heli ATPL CPL An aircraft was over A at 14:35 hours flying direct to B. Given: Distance A to B: True airspeed: Mean wind component OUT: Mean wind component BACK:

2.900 NM 470 kts +55 kts -75 kts

The ETA for reaching the point of equal time (PET) between A and B is": A) B) e) D)

17:21 17:44 18:46 16:57

(Refer to figure 061-E107) Dist =Distance between A and B GS Out = Groundspeed to destination GS HUllle = GlUuIIJ~JJeeJ vaLl\. flullle

Distance from Departure to Destination is quoted by the question as 2.900 NM. Now we have all of the variables that we need to calculate the PET: Dist to PET = (2.900 x 395) -i- (525 + 395) Dist to PET = (1.145.500) -i- (920) DisttoPET= 1.245NM If we now wanted to determine the time it will take us to reach the PET from our departure, we simply divide the distance to PET by our groundspeed (GS Out) => 1.245 NM -i- 525 kts = 2,37 hrs => 2 hrs 22 minutes => 142 minutes.

I

4634 (D)

(Refer to figure 061-E100) The term "Departure" in navigation can be defined as the Earth distance (in NM) along a parallel of Latitude - in other words, a change of longitude along a parallel of latitude expressed in nautical miles. When traveling due East or West along a parallel of latitude, the distance covered is called "Departure': The term was taken from sailing techniques invented several hundred years ago by maritime navigators, who would often sail North or South to the latitude of a position, then alter course onto due East or West, as appropriate and then sail to destination by making good the particular latitude. They were able to measure latitude and keep their course by celestial observations of the Pole Star in the northern hemisphere. The advantage of the method was that with the Mercator chart, plotting position along the parallel was simplified. The technique was known as "parallel sailing".

• Departure = Change in Longitude (in minute$) x r.t)$ Latitude

First step of our calculation is going to involve the determination of Ground Speeds: • GS Out =525 kts (470 kts + 55 kts tailwind) • GS Home =395 kts (470 kts - 75 kts headwind)

4633 (e)

1700W 1400w 145°E 175°E

The standard formula to calculate the value of "Departure" in Nautical Miles (NM) is:

Dist to PET = (Dist x GS Home) -i- (GS Out + GS Home) Time to PET =Distance to PET -i- GS Out

I

A) B) e) D)

I

4647 (C)

I

4674 (A)

I

In the case of this question: ·900 NM = Change in Longitude (minutes) x cos 60 0N • Change in Longitude (minutes) = 900 NM -i- cos 60 0N • Change in Longitude = 1.800 minutes • 1.800 minutes = 30° (1.800 -i- 60) If we started at 1600E and flying eastbound, then the final point will be 1700W (160 0E + 30°).

01 Basics of Navigation

4682. Given:

Airplane

ATPL

Heli

CPL

ATPL

C) 106 min.

CPL

D) 102 min.

Distance A to B: Ground speed OUT: Ground speed BACK:

1.973 NM 430 kts 385 kts

The time from A to the point of equal time (PET) between A and B is: A) 145 min. B) 130min. e) 162 min. D) 181 min.

(Refer to figure 061-El07) Dist =Distance between Q and R = 1.160 NM GS Out =Groundspeed to destination =435 kts GS Home = Groundspeed back home = 385 kts Dist to PET = (Dist x GS Home) .;. (GS Out + GS Home) Time to PET = Distance to PET.;. GS Out Dist to PET = (1.760 x 385).;. (435 + 385) Dist to PET = (677.600) .;. (820) Dist to PET = 826 NM If we now wanted to determine the time it will take us to reach the PET from our departure, we simply divide the distance to PET by our groundspeed (GS Out) => 826 NM.;. 435 kts = 1,9 hrs => 1 hr 54 minutes => 114 minutes.

(Refer to figure 061-E107) Dist =Distance between A and B = 1,973 NM GS Out =Groundspeed to destination =430 kts GS Home =Groundspeed back home =385 kts

4732. Airplane ATPL CPL Heli ATPL CPL The great circle distance between position A (59°34,"N 008°08,4'E) and B (30 0 25,9'N 171°51,6'W) is:

Dist to PET = (Dist x GS Home) .;. (GS Out + GS Home) Time to PET = Distance to PET.;. GS Out Dist to PET = (1.973 x 385).;. (430 + 385) Dist to PET = (759.605).;. (815) Dist to PET = 932 NM If we now wanted to determine the time it will take us to reach the PET from our departure, we simply divide the distance to PET by our groundspeed (GS Out) => 932 NM.;. 430 kts = 2, 17 hrs => 2 hrs 10 minutes => 130 minutes.

4710. Airplane ATPL CPL Heli ATPL CPL An aircraft takes-off from an airport 2 hours before sunset. The pilot flies a track of 090° (T), W/V 130°/20 kts, TAS 100 kts. In order to return to the point of departure before sunset, the furthest distance which may be traveled is: A) 97 NM

(Refer to figure 061-El00) Look at this question very carefully. Notice anything unique about the longitudes of positions A and B? 008°08.4'E is the anti-meridian of 171°51.6'W. The great circle track from A to B is over the North Pole. One minute of latitude along a meridian = 1 NM. The difference in latitude = (90° - 59°34.1) + (90°30°25.9') = 90° = 5400 minutes = 5 400 NM.

4736. Airplane ATPL CPL Heli An arc of 1 minute of a meridian equals:

B) 115 NM e) 105 NM

A) 1 NM B) 10 km

D) 84NM

e) 1 SM

(Refer to figure 061-El08) We can solve this question using the concept of the Point of Safe Return. The PSR (Point ofSafe Return), or sometimes referred to as thePNR (Point of No Return), is defined as the furthest point along track (on a flight from point A to point B) that an aircraft can fly to and still return back to point A and retain the minimum required fuel reserves upon landing. To find the PSR we will use the following variables and formula: • Endur = Endurance in hrs (flight time to having only reserve fuel available) =2hrs • GS Out =Ground Speed Out • GS Home =Ground Speed backto base Time from airport to PSR = (Endur x GS Home).;. (GS Out + GS Home) Now we need to calculate the Ground Speeds (GS) based on our TAS and the wind. Use your flight computer to calculate the GS: TAS = 100 kts; course = 090° and wind = 130° / 20 kts. You should come-up with a GS of approx. 84 kts (GS Out). For the return back to the airport the GS will be 114 kts (GS Home). With this information we can now proceed with the actual PSR calculation: Time to PSR = (2 x 114).;. (84 + 114) Time to PSR = (228) .;. (198) Time to PSR = 1,15 hrs => 69 minutes. Now to find the distance from the departure airport to the PSR we simply take the time to fly to PSR and mUltiply it by the GS Out => 1,15 hrs x 84 kts = 96,6NM.

4711. Given:

Airplane

ATPL

Heli

CPL

ATPL

CPL

ATPL

CPL

D) 1 km For explanation refer to question #4667 on page 17.

4744. Given:

Airplane

ATPL

Distance A to B: Mean ground speed OUT: Mean ground speed BACK: Safe endurance:

CPL

Heli 2.484 NM 420 kts 500 kts 08 hrs 30 min

The distance from A to the point of safe return (PSR) ~.'is: A) 1.908 NM B) 1.940 NM e) 1.736 NM D) 1.630 NM (Refer to figure 061-E108) The PSR (Point of Safe Return), or sometimes referred to as the PNR (Point of No Return), is defined as the furthest point along track (on a flight from point A to point B) that an aircraft can fly to and still return back to point A and retain the minimum required fuel reserves upon landing. To find the PSR we will use the following variables and formula: • Endur =Endurance in hrs (flight time to having only reserve fuel available) =8,5hrs • GS Out =Ground Speed OLit (from A to B) =420 kts • GS Home =Ground Speed back Home (from PSR back to A) =500 kts

CPL

With the information stated in the question we can now proceed with the actual PSR calculation:

1760NM 435 kts 385 kts

The time from Q to the point of equal time (PET) between Q and Ris: A) 110min. B) 114 min. 4682 (8)

ATPL

Time from A to PSR = (Endur x GS Home) .;. (GS Out + GS Home)

Distance Q to R: Ground speed OUT: Ground speed BACK:

I

A) 5.400 NM B) 10.800km e) 2.700 NM D) 10.800 NM

I

4710 (A)

I

4711 (8)

I

4732 (A)

I

4736 (A)

I

Time to PSR = (8,5 x 500).;. (420 + 500) Time to PSR = (4.250) .;. (920) Time to PSR = 4,62 hrs => 4 hrs 37 minutes => 277 minutes. If we now wanted to determine the distance to the PST from our departure (point A), we simply multiply the time to PSR by our groundspeed (GS Out) => 4,62 hrs x 420 kts = 1.940 NM.

4744 (8)

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Aviationexam Test Prep Edition 2012

4745. Given:

Airplane

ATPL

Distance Q to R: Ground speed OUT: Ground speed BACK: Safe endurance:

Heli

CPL

ATPL

• GS Out = Ground Speed Out (from A to B) =430 kts • GS Home = Ground Speed back Home (from PSR back to A) =385 kts

CPL

Time from A to PSR = (Endur x GS Home) + (GS Out + GS Home)

1.760 NM 435 kts 385 kts 9 hrs

With the information stated in the question we can now proceed with the actual PSR calculation:

The distance from Q to the point of safe return (PSR) between Qand Ris: A) B) C) D)

1.313 NM 1.838 NM 1.467 NM 1.642 NM

• Endur = Endurance in hrs (flight time to having only reserve fuel available) =9hrs • GS Out = Ground Speed Out (from A to B) = 435 kts • GS Home = Ground Speed back Home (from PSR back to A) =385 kts Time from A to PSR = (Endur x GS Home) + (GS Out + GS Home) With the information stated in the question we can now proceed with the actual PSR calculation: Time to PSR = (9 x 385) + (435 + 385) Time to PSR = (3.465) + (820) Time to PSR = 4,23 hrs => 4 hrs 74 minutes => 254 minutes. If we now wanted to determine the distance to the PST from our departure (point A), we simply mUltiply the time to PSR by our groundspeed (GS Out) => 4,23 hrs x 435 kts = 7.840 NM.

4747. Airplane ATPL CPL Heli ATPL CPL An aircraft flies a great circle track from 56°N 070 0 W to 62°N 11 OOE. The total distance traveled is:

(Refer to figure 067-E700) The first thing you need to realize is that the meridian 070 0W is the anti-meridian of 770°E. Therefore, the Great Circle track from 56°N 070 0W to 62°N 7700E passes over the North Pole and is represented by the meridian itself. One minute ofLatitude along a meridian = TNM (7" ofLatitude =60 NM). We simply find the number of degrees/minutes between these two positions => 34° from 56°N to the pole and additional 28° from the pole to 62"N => total of 62° of Latitude /3.720 minutes => 3.720 NM.

Airplane

ATPL

Distance A to B: Ground speed OUT: Ground speed BACK: Safe endurance:

Heli

CPL

ATPL

1.973 NM 430 kts 385 kts 7 hrs 20 min

2.900 NM 470 kts +55 kts -75 kts 9 hrs 30 min

The distance from A to the point of safe return (PSR) is: A) 2.844NM B) 1.611 NM C) 1.759 NM D) 2.141 NM (Refer to figure 067-E708) The PSR (Point ofSafe Return), or sometimes referred to as the PNR (Point of No Return), is defined as the furthest point along track (on a flight from point A to point B) that an aircraft can fly to and still return back to point A and retain the minimum required fuel reserves upon landing. To find the PSR we will use the following variables and formula: • Endur = Endurance in hrs (flight time to having only reserve fuel available) =9,5hrs • GS Out = Ground Speed Out (from A to B) • GS Home = Ground Speed back Home (from PSR back to A) Time from Q to PSR = (Endur x GS Home) + (GS Out + GS Home)

• Endur = Endurance in hrs (flight time to having only reserve fuel available) = 7,33 hrs

I

4747 (0)

I

4748 (0)

Now to find the distance from the departure airport to the PSR we simply take the time to fly to PSR and mUltiply it by the GS Out => 4,08 hrs x 525 kts = 2.742NM.

4755. Airplane ATPL CPL Heli ATPL CPL For a distance of 1.860 NM between Q and R, a ground speed OUT of 385 kts, a ground speed BACK of 465 kts and an endurance of 8 hrs (excluding reserves) the distance from Q to the point of safe return (PSR) is:

(Refer to figure 067-E708) The PSR (Point of Safe Return), or sometimes referred to as the PNR (Point of No Return), is defined as the furthest point along track (on a flight from point A to point B) thatan aircraft can fly to and still return back to point A and retain the minimum required fuel reserves upon landing. To find the PSR we will use the following variables and formula:

4745 (8)

With this information we can now proceed with the actual PSR calculation: Time to PSR = (9,5 x 395) + (395 + 525) Time to PSR = (3.752,5) + (920) Time to PSR = 4,08 hrs => 4 hrs 05 minutes => 245 minutes.

Note 2: Do not confuse PSR (Point of Safe Return) with PET (Point of Equal Time). PET = the point at which the time to return back to departure is equal to the time to continue to the destination.

1.664 NM 1.698 NM 1.422 NM 1.490 NM

I

Distance A to B: True airspeed: Mean wind component OUT: Mean wind component BACK: Safe endurance:

Note 7: Remember that when a wind component has a negative (-) prefix, it means a headwind, whereas a positive (+) prefix means a tailwind. A good mnemonic is to realize that when you want to get to your destination a headwind is bad for you =negative; whereas a tailwind is good for you =positive.

CPL

The distance from A to the point of safe return (PSR) is: A) B) C) D)

4753. Airplane ATPL CPL Heli ATPL CPL An aircraft was over A at 14:35 hours flying direct to B. Given:

Now we need to calculate the Ground Speeds based on our TAS and the wind. Note that when a wind component has a negative (-) prefix, it means a headwind, whereas a positive (+) prefix means a tailwind. • GS Out = 525 kts (470 kts TAS + 55 kts tailwind); • GS Home = 395 kts (470 kts TAS - 75 kts headwind).

2.040 NM 1.788 NM 5.420 NM 3.720 NM

4748. Given:

If we now wanted to determine the distance to the PST from our departure (point A), we simply mUltiply the time to PSR by our groundspeed (GS Out) => 3,46 hrs x 430 kts 7.488 NM.

=

(Refer to figure 067-E708) The PSR (Point ofSafe Return), or sometimes referred to as the PNR (Point of No Return), is defined as the furthest point along track (on a flight from pOint A to point B) that an aircraft can fly to and still return back to point A and retain the minimum required fuel reserves upon landing. To find the PSR we will use the following variables and formula:

A) B) C) D)

Time to PSR = (7,33 x 385) + (430 + 385) Time to PSR = (2.822) + (875) Time to PSR = 3,46 hrs => 3 hrs 28 minutes => 208 minutes.

I

4753 (0)

I

A) B) C) D)

930 NM 1.532 NM 1.685 NM 1.865 NM

(Refer to figure 067-E708) The PSR (Point of Safe Return), or sometimes referred to as the PNR (Point of No Return), is defined as the furthest point along track (on a flight from point Q

4755 (C)

I

01 Basics of Navigation

to point R) that an aircraft can fly to and still return back to point A and retain the minimum required fuel reserves upon landing. To find the PSR we will use the following variables and formula: • Endur = Endurance in hrs (flight time to having only reserve fuel available) =8hrs • GS Out = Ground Speed Out (from 0 to R) = 385 kts • GS Home =Ground Speed back Home (from PSR back to 0) =465 kts Time from A to PSR = (Endur x GS Home) + (GS Out + GS Home) With the information stated in the question we can now proceed with the actual PSR calculation:

If we now wanted to determine the distance to the PST from our departure (point 0), we simply mUltiply the time to PSR by our groundspeed (GS Out) => 4,38 hrs x 385 kts = 1.686 NM.

4756. Airplane ATPL CPL Heli ATPL CPL An aircraft has a TAS of 300 kts and a safe endurance of 10 hrs. If the wind component on the outbound leg is 50 kts head, what is the distance to the point of safe endurance? A) 1.500 NM B) 1.458 NM e) 1.544NM D) 1.622 NM (Refer to figure 061-EI08) The PSR (Point ofSafe Return), or sometimes referred to as the PNR (Point of No Return), is defined as the furthest point along track (on a flight from point A to point B) that an aircraft can fly to and still return back to point A and retain the minimum required fuel reserves upon landing. To find the PSR we will use the following variables and formula: • Endur = Endurance in hrs (flight time to having only reserve fuel available) = 10hrs • GS Out = Ground Speed Out (from A to B) • GS Home = Ground Speed back Home (from PSR back to A) Time from 0 to PSR = (Endur x GS Home) + (GS Out + GS Home) Now we need to calculate the Ground Speeds based on our TAS and the wind: • GS Out = 250 kts (300 kts TAS - 50 kts headwind); • GS Home = 350 kts (300 kts TAS + 50 kts tailwind). With this information we can now proceed with the actual PSR calculation: Time to PSR = (10 x 350) + (250 + 350) Time to PSR = (3.500) + (600) Time to PSR = 5,83 hrs => 5 hrs 50 minutes => 350 minutes. Now to find the distance from the departure airport to the PSR we simply take the time to fly to PSR and mUltiply it by the GS Out => 5,83 hrs x 250 kts = 1.457,5NM.

Airplane

ATPL

Heli

CPL

ATPL

CPL

(Refer to figure 061-El08) The PSR (Point ofSafe Return), or sometimes referred to as the PNR (Point of No Return), is defined as the furthest point along track (on a flight from point A to point B) that an aircraft can fly to and still return back to point A and retain the minimum required fuel reserves upon landing. To find the PSR we will use the following variables and formula: • Endur = Endurance in hrs (flight time to having only reserve fuel available) =6,5hrs • GS Out =Ground Speed Out (from A to B) =380 kts • GS Home =Ground Speed back Home (from PSR back to A) =440 kts

4762 (C)

I

4764 (8)

I

4776 (A)

(Refer to figure 061-EI00) If two points are located on the same line of Longitude (same Meridian) it is quite easy to determine the distance between these two points. We can assume that 1 minute of Latitude = 1 NM Waf Latitude =60 NM). Just calculate the difference in Latitude and apply this rule. In the case of this question we have a total 9° of difference => 9° x 60 NM = 540NM. 1 NM = 1,852 km 540 NM = 1.000 km.

4776. Airplane ATPL CPL Heli ATPL CPL The distance between point of departure and destination is 340 NM and wind velocity in the whole area is 100°/25 kts. TAS is 140 kts, true track is 135° and safe endurance 3 hrs 10 min. How long will it take to reach the point of safe return? A) 1 hrs 49 min. B) 1 hrs 37 min. e) 1 hrs 21 min. D) 5 hrs 30 min. (Refer to figure 061-El08) The PSR (Paint of Safe Return), or sometimes referred to as the PNR (Point of No Return), is defined as the furthest point along track (on a flight from point A to point B) that an aircraft can fly to and still return back to point A and retain the minimum required fuel reserves upon landing. To find the PSR we will use the following variables and formula: • Endur = Endurance in hrs (flight time to having only reserve fuel available) = 3,166 hrs • GS Out == Ground Speed Out • GS Home = Ground Speed back to base

With this information we can now proceed with the actual PSR calculation:

D) 1.536 NM.

I

A) 1.222 km B) 1.000 km C) 540 km D) 804km

Time toPSR = (3,166 x 160)+(119+ 160) Time to PSR = (507) + (279) Time to PSR = 1,82hrs=> 109minutes=> 1 hr49 min.

A) 889 NM. B) 1.030 NM. e) 1.326 NM.

4756 (8)

If we now wanted to determine the distance to the PST from our departure (point A), we simply multiply the time to PSR by our groundspeed (GS Out) => 3,49 hrs x 380 kts = 1.326 NM.

Time from airport to PSR = (Endur x GS Home) + (GS Out + GS Home)

The distance from A to the point of safe return (PSR) is:

I

Time to PSR = (6,5 x 440) + (380 + 440) Time to PSR = (2.860) + (820) Time to PSR = 3,49 hrs => 3 hrs 29 minutes => 209 minutes.

Now we need to calculate the Ground Speeds (GS) based on our TAS and the wind. Use your flight computer to calculate the GS: TAS = 140 kts; course = 135° and wind = 100° I 25 kts. You should come-up with a GS of approx. 119 kts (GS Out). For the return back to the airport the GS will be 160 kts (GS Home).

1.920 NM 380 kts 440 kts 6 hrs 30 min

Distance A to B: Ground speed OUT: Ground speed BACK: Safe endurance:

With the information stated in the question we can now proceed with the actual PSR calculation:

4764. Airplane ATPL CPL Heli ATPL CPL A flight is to be made from A 49°5 1800 E/W to B 58°5 1800 E/W. The distance from A to B is approximately:

Time to PSR = (8 x 465) + (385 + 465) Time to PSR = (3.720) + (850) Time to PSR = 4,38 hrs => 4 hrs 23 minutes => 263 minutes.

4762. Given:

Time from A to PSR = (Endur x GS Home) + (GS Out + GS Home)

I

4778 (C)

I

4778. Airplane ATPL CPL Heli ATPL CPL What is the time required to travel along the parallel of latitude 60 0 N between meridians 010 0 E and 030 0 W at a groundspeed of 480 kts? A) 1 hr 45 min. B) 1 hr 15 min. e) 2 hrs 30 min. D) 5 hrs 00 min. (Refer to figure 061-El00) To solve this question we will need to use the following principle:

Aviationexam Test Prep Edition 2012 • When following a line of Parallel (Rhumb Line), each 1° change of Longitude change represents a distance of 1° x 60 x cos Latitude. In case of our question, the difference in Longitude =40° (10° to the Greenwich from the 100E position and additional 30° from the Greenwich to the 30 0W position). 40° =2.400 minutes. Now we can use the formula: • Distance = Change in Longitude (in minutes) x cos Latitude • Distance = 2.400 minutes x cos 60° • Distance = 1.200 NM We know the distance and the speed. We can therefore easily calculate the time required to cover this distance: • Distance = Rate x Time • 1.200 NM = 480 kts x ? hrs .? hrs = 1.200 NM"," 480 kts • ? = 2,5 hrs (2 hrs 30 minutes)

4803. Airplane ATPL CPL Heli ATPL CPL The International Nautical Mile defined by ICAO is equivalent to: A) 1.582 m B) 1.652 C) 1.852 D) 1.962

m m m

A) 08°29'N B) 06°01'N C) 02°45'5 D) 03°50'5 (Refer to figure 061-El00) To solve this question we will need to use the following principle: • When following a Great Circle in the North/South direction (along a Meridian) each 1° change of Latitude represents a distance of 60 NM (1 minute of Latitude = 1 NM). Now in the case of this question, we first need to convert the distance from km toNM: • 1 NM = 1,852 km ·685 km =369,9 NM If 1 minute of Latitude change = 1 NM then 369,9 NM will represent a Latitude change of 369,9 minutes => 6,16° of Latitude change. If we have started at 002°20'N (2,33°) and travelled 6,16° northbound (360°T) then the new position will be => 2,33° + 6,16° = 8,49° => 008°29'N.

4858. Airplane ATPL CPL Heli ATPL CPL An aircraft departs a point 04°00'N 1700 00'W and flies 600 NM South, followed by 600 NM East, then 600 NM North, then 600 NM West. What is its final position?

1 NM = 1,852 km = 1,1507 SM = 6080 ft 1 km = 1000 m =3280 ft If 1 km = 1000 m and if 1 NM = 1,852 km, then 1 NM = 1852 m.

4819. Airplane ATPL CPL Heli ATPL CPL At what approximate latitude is the length of one minute of arc along a meridian equal to one NM (1852 m) correct? A) 45°

B) 0° C) 90° D) 30° Read this question carefully - it is asking about one minute of arc of a MERIDIAN and not the Parallel of Latitude. When performing most navigation calculations in this subject we assume that 1 minute along a Meridian equals 1 NM (1852 mY. For this to be true the Earth woiJld have to be a perfect sphere and for simple navigation calculations we assume this. However in reality the Earth is not a perfect sphere - it is slightly flattened at the poles. For this very reason a definition of 1 NM with respect to a 1 minute arc of a Meridian varies depending on the Latitude - it is not a constant length at the surface of the Earth but gradually lengthens with increasing distance from the equator. At the Equator, 1 NM = 1843 m while at the poles 1 NM = 1861 m. Today the internationally accepted definition of 1 NM is a length of 1852 m which would correspond to 1 minute of arc of a Meridian measured at a latitude of approximately 45°.

4825. Airplane ATPL CPL Heli ATPL CPL What is the longitude of a position 6 NM to the East of 58°42'N 094°00'W?

A) B) C) D)

4857. Airplane ATPL CPL Heli ATPL CPL An aircraft at latitude 02°20'N tracks 360° (T) for 685 kilometres. What is its latitude at the end of the flight?

093°53,1'W 093°54,0'W 093°48,5'W 094°12,0'W

(Refer to figure 061-El00) To solve this question we will need to use the following principle: • When following a line of Parallel (Rhumb Line), each 1° change of Longitude change represents a distance of 1° x 60 x cos Latitude. In case of our question we can directly use the formula above: • Distance = Change in Longitude (in minutes) x cos Latitude • 58°42' =58,7" • 6 NM = ? minutes x cos 58,7" • ? minutes = 6 NM "'" cos 58,7" • ? = 11,5 minutes Eastbound ·If we started at 58°42'N 094°00'W we will end up at 58°42'N 093°48'30"W (093°48,5'W)

A) B) C) D)

0

04°00'N 170 00'W 06°00'5170 00'W 04°00'N 169°58,1'W 04°00'N 1700 01,8'W 0

(Refer to figure 061-El00) To solve this question we will need to use two principles: • When following a Great Circle in the North/South direction (along a Meridian) each 1° change of Latitude represents a distance of 60 NM (1 minute of Latitude = 1 NM). • When following a Great Circle in the East/West direction (along a Parallel) each 1°change of Longitude represents a distance of 1° x 60 x cos Latitude. Now in the case of this question, let's solve it step-by-step, by calculating each of the legs: Leg 1 =600 NM south: ·600 NM along a Meridian = 600 minutes = 10° Southbound ·lfwestartedat004°N 170 0Wwewillendupat006°S 1700W Leg 2 = 600 NM east: ·600 NM along a Parallel at 006°S = ? • Distance = Change in Longitude (in minutes) x cos Latitude ·600 NM = ? minutes x cos 6° • ? minutes = 600 NM "'" cos 6° • ? =603,3 minutes = 10°03,3' Eastbound ·If we started at 006°S 1700W we will end up at 006°S 159°56,l'W Leg 3 = 600 NM north: ·600 NM along a Meridian = 600 minutes = 10° Northbound ·If we started at 006°S 159°56,l'W we will end up at 004°N 159°56,l'W Leg 4 = 600 NM west: ·600 NM along a Parallel at 004°N = ? • Distance = Change in Longitude (in minutes) x cos Latitude ·600 NM = ? minutes x cos 4· • ? minutes = 600 NM "'" cos 4° • ? =601,47 minutes = 10°01,47' Westbound ·If we started at 004"N 159°56,l'W we will end up at 004°N 169°57, 17'w. The final position will therefore be 004°N 169°57, Il'W => 004°N 169°5l'1O"W.

4870. Airplane ATPL CPL Heli ATPL CPL An aircraft at latitude 02°20'N tracks 180e (T) for 685 km • On completion of the flight the latitude will be:

A) B) C) D)

03°50'5 04°10'5 04°30'5 09°05'5

(Refer to figure 061-E100)

I

4803 (C)

I

4819 (A)

I

4825 (C)

I

4857 (A)

I

4858 (C)

I

4870 (A)

I

01 Basics of Navigation To solve this question we will need to use the following principle: • When following a Great Circle in the North/South direction (along a Meridian) each 1° change of Latitude represents a distance of 60 NM (1 minute ofLatitude = 1 NM).

we will cover in 1,5 hours of flight time. At a speed of 890 km/h we will cover a distance of 1.335 km: • 1 NM = 1,852 km • 1.335 km = 720,8 NM

Now in the case of this question, we first need to convert the distance from km toNM: • 1 NM = 1,852 km ·685 km = 369,9 NM

If 1 minute of Latitude change = 1 NM then 720,8 NM will represent a Latitude change of 720,8 minutes => 12,01° => approximately 12° of Latitude change. If we have started at 10°5 and traveled 12°N then the new position will be orN.

If 1 minute of Latitude change = 1 NM then 369,9 NM will represent a Latitude change of 369,9 minutes => 6,16° of Latitude change. If we have started at 002°20W (2,33°) and travelled 6,16° southbound (180°T) then the new position will be =>2,33°-6,16°=-3,83°=> 003°50'5.

4881. Airplane ATPL CPL Heli ATPL CPL An aircraft is at lOON and is flying South at 444 km/h. After 3 hours the latitude is:

4871. Airplane ATPL CPL Heli ATPL CPL An aircraft flies the following rhumb line tracks and distances from position 04°00'N 030 0 00'W: 600 NM South, then 600 NM East, then 600 NM North, then 600 NMWest. The final position of the aircraft is: A) 04°00'N 029°S8'W. B) 04°00'N 030 0 02'W. C) 04°00'N 030 0 00'W. 0) 03°S8'N 030 0 02'W. (Refer to figure 061-EI00) To solve this question we will need to use two principles: • When following a Great Circle in the North/South direction (along a Meridian) each 1° change of Latitude represents a distance of 60 NM (1 minute of Latitude = 1 NM). • When following a Great Circle in the East/West direction (along a Parallel) each 1°change ofLongitude represents a distance of 1° x 60 x cos Latitude. Now in the case of this question, let's solve it step-by-step, by calculating each of the legs: Leg 1 = 600 NM south: ·600 NM along a Meridian = 600 minutes = 10° Southbound ·If we started at 004°N 030 0 W we will end up at 006°5 030 0 W Leg 2 = 600 NM east: ·600 NM along a Parallel at 006°5 = ? • Distance =Change in Longitude (in minutes) x cos Latitude ·600 NM =? minutes x cos 6° • ? minutes = 600 NM + cos 6° .? = 603,3 minutes = 10,055° = 10°03,3' Eastbound ·If we started at 006°5 030 0 W we will end up at 006°5 019°56,l'W (30 0 W 10,055°) Leg 3 = 600 NM north: ·600 NM along a Meridian =600 minutes = 10° Northbound ·If we started at 006°5 019°56,l'W we will end up at 004°N 019°56,l'W Leg 4 = 600 NM west: ·600 NM along a Parallel at 004°N = ? • Distance = Change in Longitude (in minutes) x cos Latitude ·600 NM = ? minutes x cos 4° • ? minutes =600 NM + cos 4° .? =601,47 minutes = 10,0245°= 10°01,47' Westbound ·If we started at 004°N 019°56,l'W we will end up at 004°N 029°58,17'W (19,945°+ 10,0245°). The final position will therefore be 004°N 029°57, 17'W => 004°N 029°57'1O"W.

4876. Airplane ATPL CPL Heli ATPL CPL An aircraft at latitlJde 10°5 flies North at a GS of 890 km/h. What will its latitude be after 1,5 hrs? A) 22°00'N B) 03°S0'N C) 02°00'N 0) 12°1S'N (Refer to figure 061-EI00) To solve this question we will need to use the following principle: • When following a Great Circle in the North/South direction (along a Meridian) each 1° change of Latitude represents a distance of 60 NM (1 minute of Latitude = 1 NM). Now in the case of this question, we first need to determine the distance that

I

4871 (A)

I

4876 (C)

I

4881 (C)

I

4882 (8)

I

4901 (0)

I

A) 10°5 B) 02°N C) 02°5 0) OONiS (Refer to figure 061-EI00) To solve this question we will need to use the following principle: • When following a Great Circle in the North/South direction (along a Meridian) each 1° change of Latitude represents a distance of 60 NM (1 minute of Latitude = 1 NM). Now in the case of this question, we first need to determine the distance that we will cover in 3 hours of flight time. At a speed of 444 km/h we will cover a distance of 1.332 km: • 1 NM = 1,852 km • 1.332 km = 179,2 NM If 1 minute of Latitude change = 1 NM then 179,2 NM will represent a Latitude change of 179,2 minutes => 11°59,2' => approximately 12° of Latitude change. If we have started at lOON and travelled 12°5 then the new position will be ors.

4882. Airplane ATPL CPL Heli ATPL CPL An aircraft at latitude 10°5 flies North at a groundspeed of 444 km/h. What will be its latitude after 3 hrs? A) 02°00'5 B) 02°00'N C) 22°00'N 0) 22°00'5 (Refer to figure 061-El00) To solve this question we will need to use the following principle: • When following a Great Circle in the North/South direction (along a Meridian) each 1° change of Latitude represents a distance of 60 NM (1 minute of Latitude = 1 NM). Now in the case of this question, we first need to determine the distance that we will cover in 3 hours of flight time. At a speed of 444 km/h we will cover a distance of 1.332 km: • 1 NM = 1,852 km • 1.332 km = 179,2 NM If 1 minute of Latitude change = 1 NM then 179,2 NM will represent a Latitude change of719,2 minutes => 11°59,2' => approximately 12° ofLatitude change. If we have started at 10°5 and traveled 12° northbound then the new position willbe02"N.

4901. Airplane ATPL CPL Heli ATPL CPL You start from P (70 0 00'N 015°00'E) and fly westward along the parallel of latitude for 2 hours at ground speed 220 kts. What is your position after two hours flight? A) 006°44'W B) 021°26'W C) 00r40'E 0) 006°26'W (Refer to figure 061-El00) To solve this question we will need to use the following principle: • When following a line of Parallel (Rhumb Line), each 1° change of Longitude change represents a distance of 1° x 60 x cos Latitude. Now in the case of this question, we first need to determine the distance that we will cover in 2 hours of flight time. At a speed of220 kts we will cover a distance of 440 NM. Now we can use the formula: • Distance = Change in Longitude (in minutes) x cos Latitude ·440 NM = ? minutes x cos 70° • ? minutes = 440 NM + cos 70°

Aviationexam Test Prep Edition 2012 .? = 1.286,5 minutes = 21,44° Westbound • If we started at 700N o15°E we will end up at 700N 006°26'W => 15°E - 21,44° = -6,44° => the negative sign means that we have crossed the Greenwich meridian and the final Longitude is 6,44°W = 6°26'W.

4919. Airplane ATPL CPL Heli ATPL CPL An aircraft at position 27°00'N 170000'W travels 3.000 km on a track of 180° (T), then 3.000 km on a track of 090° (T), then 3.000 km on a track of 000° (T), then 3.000 km on a track of 270° (T). What is its final position? A) 2rOO'N 170 00'W B) OOoOO'N/S 170000'W C) 2rOO'N 173°18'W D) 2rOO'N 143°00'W 0

(Refer to figure 061-E100) To solve this question we will need to use two principles: • When fallowing a Great Circle in the North/South direction (along a Meridian) each 1° change ofLatitude represents distance of60 NM (1 minute ofLatitude =lNM). • When following a Great Circle in the East/West direction (along a Parallel) each 1°change of Longitude represents a distance of 1° x 60 x cos Latitude.

a

Now in the case of this question, let's solve it step-by-step, by calculating each of the legs: 1 NM = 1,852 km 3.000 km =approx. 1.620 NM

300 (T) B) 000° (T) C) 45° (T) D) 60 (T) A)

0

(Refer ta figures 061-EI00, 061-E79, 067-£80 and 061-E81) To solve this question we will need to use the following principle: • When following a Great Circle in the North/South direction (along a Meridian) each 1° change of Latitude represents a distance of 60 NM (1 minute ofLatitude = 1 NM). Now in the case of this question, we first need to determine the distance that we will cover in 10 hours of flight time. At a speed of 540 kts we will cover a distance of 5.400 NM. If 1 minute of Latitude change = 1 NM then 5.400 NM will represent a Latitude change of 5.400 minutes => 90° of Latitude change. If we have started at the Equator (ooN/S) and traveled 90°5 then the new position will be at the South Pole (hope you have packed a warm coat). From the South Pole all directions are 360 0T (= Due North = 360° / 000° True).

4944. Airplane ATPL CPL Heli ATPL CPL How many centimeters are equivalent to 36,25 inches? A) 92,08 em

Leg 1 = 1.620 NM south: • 1.620 NM along a Meridian = 1.620 minutes =2r Southbound • If we started at 02rN 1700Wwe will end up at the Equator (OW/S) 170°W. Leg 2 = 1.620 NM east: • 1 NM = 1 minute along the Equator ·1.620 NM = 1.620 minutes along the Equator ·1.620 minutes =2r Eastbound ·If we started at OOOON/S 1700W we will end up at OOooN/S 143°W Leg 3 = 1.620 NM north: • 1.620 NM along a Meridian = 1.620 minutes =2r Northbound ·If we started at OOooN/S 14rw we will end up at 02rN 14rW Leg 4 = 1.620 NM west: • 1.620 NM along a Parallel at 02rN = ? • Distance =Change in Longitude (in minutes) x cos Latitude • 1.620 NM = ? minutes x cos 2r .? minutes = 1.620 NM + cos 2r .? = 1.818,17 minutes =30,30° =30°18' Westbaund ·If we started at 02rN 143°W we will end up at 02rN 173°18'w.

The final position will therefore be 02rN 173°18'W.

4925. Airplane ATPL CPL Heli ATPL CPL An aircraft at position 60 0 N 005°W tracks 090° (T) for 315 km. On completion of the flight the longitude will be: A) 002°10'W B) 000015'E C) 000040'E D) 005°15'E (Refer to figure 061-E100) To solve this question we will need to use the following principle: • When following a line of Parallel (Hhumb Line), each 1" change ot Longitude change represents a distance of 1° x 60 x cos Latitude. In case of our question: • 1 NM = 1,852 km ·315km= 170NM Now we can use the formula: • Distance = Change in Longitude (in minutes) x cos Latitude • 170 NM =? minutes x cos 60° .? minutes = 170 NM + cos 60° • ? =340 minutes =5,66° =5°40' Eastbound ·If we started at 60 0N 005°W we will end up at 60 0N 000040'E.

1 4919 (e)

4927. Airplane ATPL CPL Heli ATPL CPL An aircraft leaves OONIS 45°W and flies due South for 10 hours at a speed of 540 kts. What is its position as a true bearing from the South Pole?

1 4925 (e)

1 4927 (8)

B) C) D)

0,014 m 14,27 em 11,05 em

1 m =3,2808 ft =39,37 inches 1 inch = 2,54 cm

Therefore, 36,25 inches = 92,08 cm.

230628. Airplane ATPL CPL Heli ATPL CPL Two places on the parallel of 47°5 lie 757,8 km apart. Calculate the difference in longitude. A)

4°39'

B) 9°19' C) 10°00' D) 4°51'

230640. Airplane ATPL CPL Heli ATPL CPL On an oblate spheroid representing the earth's shape: A) 1 minute of are along the meridian at OON/S measures the same distance as 1 minute of arc at 90 0 N/S. B) 1 minute of arc along the equator measures the same distance as 1 minute of arc along the meridian at a latitude of 45°N/S. C) 1 minute of arc along the equator measures a greater distance than 1 minute of arc along the meridian at a latitude of 45°N/S. D) 1 minute of arc along the meridian at low latitudes measures a greater distance than 1 minute of arc along the meridian at high latitudes.

230641. Airplane ATPL CPL Heli ATPL Position A is (31°00'5, 176°17'W) Rhumb line track (T) from A to B is 270°. Initial great circle track (T) from A to B is 266.2°. The Approximate position of B is: A) B) C) D)

(31"00'5, 161°32'W) (31"00'5, 173°24'E) (31°00'5, 168°58'E) (31°00'5, 173°24'W)

1 4944 (A) 1230628 (e) 1230640 (e) 1230641 (e) 1

CPL

02 Magnetism and Compasses

MAGNETISM AND COMPASSES 02-01 Knowledge of the principles of direct reading (standby) compass 1127. Airplane ATPL CPL Hard iron magnetism in aircraft: A) B) C) D)

Heli

ATPL

CPL

is permanent of nature. will change with aircraft heading. is easily compensated for by degaussing (de-magnetization). will be the same in all aircraft of the same type.

(Refer to figures 061-E709 and 067-E770) Aircraft hard iron magnetism is permanent magnetism mostly induced by the Earth's magnetic field, especially at the aircraft assembly plant during construction. During the assembly process, the aircraft might be subject to a certain amount of hammering (beating) and riveting as part of the process, although this feature has been greatly reduced with modern construction techniques. Any vibration or banging on the aircraft structure while the aircraft is stationary will align the aircraft magnetic molecules with the Earth field in the same way that happens to an ordinary magnet. The result of the banging and vibrating is that any hard metal in the aircraft will be magnetized permanently to a certain extent. The polarity and strength of this permanent magnetism depends on the aircraft heading during manufacture. That part nearest the local magnetic North will have a red pole and that part nearest the local magnetic South will have a blue pole. If the aircraft is coincident with the local magnetic meridian, the nose and front halfwill be the north-seeking (red pole) and the rear will be the blue pole. Hard iron magnetism of an aircraft can be greatly affected by a permanent relocation of the aircraft to a place with significantly different Magnetic Latitude; being struck with a lightning; sustaining damage and subsequent structural repairs; transporting magnetized cargo, etc. ..

1153. Airplane ATPL CPL Heli ATPL CPL Magnetic compass calibration is carried out to reduce: A) B) C) D)

deviation. variation. parallax error. acceleration errors.

(Refer to figures 067-E709 and 067 -E7 70) The deviations in an aircraft can be minimised to a certain extent after a procedure known as a "Compass Swing'~ During this procedure, compass readings are taken on different headings and then compared to an accurate datum. Errors are noted and processed mathematically and are then compensated by means of corrector coils ("micro-adjusters" or "corrector magnets") which are part of the compass itself. The corrector coils neutralise much of the permanent magnetism and associated effects such as "soft iron" magnetism associated with it. They do so by applying a magnetic field in the opposite direction to the residual fields affecting the compass which can effectively neutralise most of the deviation errors. However, notall deviations can be removed from an aircraft and those remaining after completing the swing are noted and a "Compass Correction Card" (Deviation Card) is produced. The corrections noted on the card are applied by the pilot in flight to assist in steering the desired magnetic heading. The corrections on the compass deviation card basically represent the difference between the actual Magnetic North and the Compass North. The compass swing procedure should be performed on the following occasions:

I 1127 (A) I 1153 (A) I 1160 (8) I

• On delivery of a new aircraft; • When a new compass is fitted; • After major structural alterations to the aircraft; • After an electrical storm or lightning strike; • After fitting a different electrical equipment; • When the compass accuracy is in doubt; • When carrying a magnetic load; • When a new detector unit is fitted to a remote compass; • Ideally approximately every 3 months, but at least every 72 months; • Upon a large change of magnetic latitude of a permanent character (e.g. aircraft changes its permanent base from Spain to northern Scandinavia); • Major structural repairs involving significant hammering (hammering is basically hitting repeatedly a metal piece => this has the effect of slightly magnetizing the metal material, thus leading to the change in the aircraft's own magnetic field).

1160. Airplane ATPL CPL Heli ATPL CPL Turning right from 330 0 (C) to 040 0 (C) in the northern hemisphere. As the aircraft rolls out, does the compass overread or underread and will liquid swirl increase or decrease the error? A) B) C) D)

Underread / Decrease. Underread / Increase. Overread / Decrease. Overread / Increase.

(Refer to figure 067-E772) When the aircraft is turning right in the northern hemisphere, because of inertia, the magnet assembly will rotate clockwise. Thus, the aircraft and the magnet are rotating in the same direction and the compass is called sluggish. If the magnet rotates clockwise, then the compass underreads. This means that if the pilot stops the turn at 040° indicated, the actual heading will have a greater value such as 060° because the magnet rotated clockwise thus the angle between the direction indicated by the magnet and the heading direction will be smaller (i.e. 040°) then the angle between the real Magnetic North and the heading direction (i.e. 060°) While turning, the liquid is in contact with the interior side of the bowl and tends to be swept up with the bowl. This produces in the liquid small eddies which deviate the magnet assembly in the direction of turn. Therefore the liquid tends to swirl and rotate the magnet assembly with it in the same direction as the aircraft's turn, thus will increase the turning error.

Aviationexam Test Prep Edition 2012

1161. Airplane ATPL CPL Heli ATPL CPL When turning right from 330 0 (C) to 040 0 (C) in the northern hemisphere, the reading of a direct reading magnetic compass will: A) over-indicate the turn and liquid swirl will decrease the effect. B) under-indicate the turn and liquid swirl will increase the effect. C) under-indicate the turn and liquid swirl will decrease the effect. D) over-indicate the turn and liquid swirl will increase the effect.

For explanation refer to question #7760 on page 67. 1163. Airplane ATPL CPL Heli ATPL CPL Which of the following will probably not result in a deviation change on a ORe: A) a walk-man headset placed close to the compass. B) turning the ADF on in flight. C) relocating a steel iron construction in the cargo compartment close to the DRC. D) letting a passenger in the cockpit jump-seat put his mobile phone next to the DRC.

(Refer to figures 067-E709 and 067-E770) Be careful and read this question carefully. It says "not result" in a deviation change on a DRC (Direct Reading Compass). The DRC will typically be most affected by a change in deviation when a metal object or a piece of an electrical equipment is placed close to the ORe. The correct answer B} mentions switching an ADF on => typically this will have a much lower effect on the compass deviation as the unit is placed further away from the compass. 1164. Airplane ATPL CPL Heli ATPL CPL Which of the following statements is correct concerning the effect of turning errors on a direct reading compass? A) Turning errors are greatest on North and are least at high latitudes. B) Turning errors are greatest on East and are least at high latitudes. C) Turning errors are greatest on North and are greatest at high latitudes. D) Turning errors are greatest on East and are greatest at high latitudes.

/ South headings, / West headings, / South headings, / West headings,

(Refer to figure 067-E772) Turning errors: (UNGS => undershoot North. overshoot south) • When the aircraft is turning through a northerly heading, the pilot must undershoot, because the compass reading lags behind the actual heading; • When the aircraft is turning through a southerly heading, the pilot must overshoot, because the compass reading leads ahead the actual heading; • When turning through an easterly or westerly heading, the turning error is zero. While turning, the liquid is in contact with the interior side of the compass bowl and tends to be swept with the bowl. This produces small eddies in the liquid which deviate the magnet assembly in the direction of turn. Therefore the liquid tends to increase the turning error when passing through Northerly headings and decrease the error when passing through Southerly headings. Note 1: The info above is based on a Northern hemisphere. It is reversed in the Southern hemisphere. Note 2: The greater the latitude the greater the turning error. Note 3: A rate 7 turn is 3° per second. 1173. Airplane ATPL CPL Heli ATPL CPL Why are the detector units of slaved gyro compasses usually located in the aircraft wingtips or the helicopter tail-booms? A) With one detector unit in each wingtip, compass deviations are canceled out. B) To isolate the detector unit from the aircraft deviation sources.

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1161 (8)

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1163 (8)

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1164 (C)

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1173 (8)

I

C) To isolate the detector unit from the Earth's magnetic field. D) To reduce turning and acceleration errors.

Remote indicating compass principles are typically used in a so-called "slaved" gyro compass instrument. This type of instrument system contains a detector unit that detects the Earth's horizontal magnetic field and then transmits this information electronically to the instrument that uses this information to adjust the directional gyro on the flight deck (using a torque motor). Directional gyro indicator units are also typically equipped with an annunciator - the purpose of which is to indicate the need for manual re-synchronization (if the gyro precesses out of synch with the correct actual heading). The detector unit is typically located in the wing-tip of the aeroplane (or the tail boom of a helicopter) to minimize the effects of deviation = to place the unit as far as possible from the aircraft electrical equipment that can alter the magnetic field and also as far away from the airframe structure itself. The detector unit works on the principle of passively sensing the Earth's magnetic field rather than seeking the Magnetic North reference => precise magnetic heading is then calculated by the instrument based on this magnetic field measurement. Unlike the typical direct reading compass that uses a magnet assembly (suspended in liquid) which aligns itself with the horizontal component of the earth's magnetic field, i.e. it seeks the magnetic meridian, the detector unit of the remote indicating compass does not use any moving parts, even though it is a quite complex system. As such (no moving parts) it is more sensitive and it is not subject to the acceleration and turning errors of the direct reading compass. However, a disadvantage of a remote indicating compass is the fact that it requires electricity to operate. 1176. Airplane ATPL CPL Heli ATPL CPL A direct reading compass should be swung when: A) there is a large, and permanent, change in magnetic latitude. B) there is a large change in magnetic longitude. C) the aircraft is stored for a long period and is frequently

moved. D) the aircraft has made more than a stated number of landings.

For explanation refer to question #7753 on page 67. 1178. Airplane ATPL CPL In a remote indicating compass system the amount of deviation caused by aircraft magnetism and electrical circuits may be minimized by: A) B) C) D)

positioning the master unit in the center of the aircraft. the use of repeater cards. mounting the detector unit in the wingtip. using a vertically mounted gyroscope.

For explanation refer to question #7773 on this page. 1179. Airplane ATPL CPL Heli ATPL CPL A direct reading compass is used at a North magnetic latitude. Starting a right hand turn from heading 300 will result in: A) at first a compass indication of a left hand turn. B) the turn has to be broken off before the compass indicates ""~ the desired heading. C) the compass indication will lag during at least the first 90 0 of the turn. D) all answers are correct.

For explanation refer to question #7760 on page 67. 1181. Airplane ATPL CPL Heli ATPL CPL The forces acting upon the compass needle in a stand-by compass in an aircraft, are: A) the Earth's magnetic field, the coriolis effect and aircraft magnetism. B) the total magnetic field in the compass location. C) mechanic forces only. D) the Earth's magnetic field, the aircraft magnetic field and the effects of attitude and movement of the aircraft. A standby compass is a simple Direct Reading Compass (DRC). The needle

1176 (A)

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1178 (C)

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1179 (D)

c

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1181 (D)

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,

It·

•

02 Magnetism and Compasses

of the DRC aligns itself in parallel with the lines of Earth's magnetic field and thus indicates the direction towards the Magnetic pole. However, the needle of the DRC is also affected by any local magnetic forces, such as hard iron magnetism of the aircraft structure or the soft iron magnetism of the electrical equipment on board, etc. .. Thus it is affected by both the Earth magnetism and the aircraft magnetic field. The needle of the DRC is mounted on a pivot so that it is allowed to rotate freely around its attachment point => this can results in minor fluctuations of the needle as a result of aircraft manoeuvers or abrupt attitude changes and accelerations of the aircraft (remember the acceleration and turning compass errors).

2180. Airplane ATPL CPL Heli ATPL CPL The main reason for usually mounting the detector unit of a remote indicating compass in the wingtip of an aeroplane or in the helicopter tail-boom is to: A) facilitate easy maintenance of the unit and increase its exposure to the Earth's magnetic field. B) reduce the amount of deviation caused by aircraft magnetism and electrical circuits. C) place it in a position where there is no electrical wiring to cause deviation errors. D) place it where it will not be subjected to electrical or magnetic interference from the aircraft. For explanation refer to question #1173 on page 62.

2196. Airplane ATPL CPL A "Ianding compass" is:

Heli

ATPL

CPL

A) used to establish aircraft magnetic heading during a compass swing. B) painted on the ground at airfields to indicate the direction of the cardinal magnetic headings to observers on the ground or in the air. e) the compass used as reference during landing. D) a compass on which the runway direction for landing may be set as a "bug". The Landing Compass is a small portable bearing compass, having a high degree of accuracy, which can be used for ground calibration of aircraft compasses. It is designed for use on the ground and is fitted with a tripod and a bubble level. The horizontal compass card is read through a prism which forms part of the sighting head.

2197. Airplane ATPL CPL Heli ATPL CPL At the magnetic equator, when accelerating after takeoff on heading West, a direct reading compass: A) underreads the heading. B) overreads the heading. e) indicates the correct heading. D) indicates a turn to the South. (Refer to figure 061-E111) A typical compass system is made pendulous to counteract the effect of magnetic dip by displacing the CG and thus making the instrument effective over a greater latitude band. However, residual tilt always remains in the compass needle, except at the magnetic equator. The red end tilts down in the northern hemisphere and the blue end tilts down in the southern hemisphere. In the northern hemisphere the centre of gravity (CG) is therefore slightly displaced to the South of the pivot and in the southern hemisphere to the North of the pivot. It will be seen that the CG is always displaced towards the Equator. As a result we encounter the acceleration / deceleration errors when on East / West headings. Since the needle is not tilted at the Magnetic equator, these errors are not present at the Magnetic equator. Note: This is purely an.academical/ theoretical question.

2199. Airplane ATPL CPL Heli ATPL CPL You are in the northern hemisphere, heading West, .and the aircraft is accelerating. Will a direct reading magnetic compass over-read or under-read and is the compass indicating a turn to the North or to the south: A) over-reads North.

B) over-reads South. e) under-reads North.

D) under-reads South. (Refer to figures 061-E111, 061-E112 and 061-E57) On a Direct Reading Compass a magnet assembly is suspended in liquid. The magnet assembly which aligns itself with the horizontal component of the Earth's magnetic field, i.e. it seeks the magnetic meridian. It is subjected to turning and acceleration errors: Acceleration errors: (ANDS => accelerate North, decelerate south) ·If the aircraft accelerates on westerly or easterly heading, the compass will indicate an apparent turn towards north; ·If the aircraft decelerates on westerly or easterly heading, the compass will indicate an apparent turn towards south; • On a northerly or southerly heading the acceleration error is zero because the inertial force is in a north-south direction i.e. along the magnet and thus, it will displace it neither clockwise, nor anti-clockwise. Turning errors: (UNOS => undershoot North, overshoot south) • When the aircraft is turning through a northerly heading, the pilot must undershoot; , • When the aircraft is turning through a southerly heading, the pilot must overshoot; • When turning through an easterly or westerly heading, the turning error is zero. Note 1: All of the info above is based on in the Southern hemisphere.

a Northern hemisphere. It is reversed

Note 2: In some of the questions concerning compass errors we are concerned with which direction the compass points and in other questions we are concerned with the direction of the compass card movement - read each of the questions carefully. Realize that when reading the heading value from the direct reading compass you read it from the back of the compass card => for questions that ask about the direction of compass card rotation it helps to draw it on a piece of paper and see which way you have to rotate it to get the desired reading.

2200. Airplane ATPL CPL Heli ATPL CPL You are turning right from 330 0 (C) to 040 0 (C) in the northern hemisphere, using timing. You stop the turn at the correct time. Before the direct indicating magnetic compass settles down, does it over-read or under-read, and does the effect of liquid swirl increase or decrease? A) Under-read; increase. B) Over-read; decrease. e) Under-read; decrease. D) Over-read; increase. For explanation refer to question #1164 on page 62.

2208. Airplane ATPL CPL Heli ATPL CPL Concerning direct reading magnetic compasses, in the northern hemisphere, it can be said that: A) on an easterly heading, a longitudinal an apparent turn to the South. B) on an easterly heading, a longitudinal an apparent turn to the North. e) on a westerly heading, a longitudinal an apparent turn to the South. D) on a westerly heading, a longitudinal an apparent turn to the North.

acceleration causes acceleration causes acceleration causes deceleration causes

For explanation refer to question #2199 on this page.

2226. Airplane ATPL CPL Coefficient A is corrected for by:

Heli

ATPL

CPL

A) using the compensation magnets. B) moving the compass housing around vertical axis. e) removing disturbing magnetic material from the close vicinity of the compass. D) degaussing (demagnetizing) the compass. (Refer to figures 061-EI09 and 061-EII0) Apart from coefficients 8 and C a further coefficient can exist in the compass

I 2180 (8) I 2196 (A) I 2197 (C) I 2199 (A) I 2200 (A) I 2208 (8) I 2226 (8) I

Aviationexam Test Prep Edition 2012 - coefficient A. The error from this coefficient will be constant on all headings and can be resolved into two components:

to reduce oscillations. Another way to reduce the oscillations is to keep the CG of the needle assembly close to the compass point.

"Real A" (Magnetic) Real A is produced by horizontal soft iron indistinguishable from apparent A, but is seldom of any appreciable value. It is maximum where "H" is maximum => that is at the magnetic Equator. Real A is zero at the poles as "H" is zero therefore, the compass is unusable.

2258. Airplane ATPL CPL Heli ATPL CPL Which one of the following is an advantage of a remote reading compass as compared with a standby compass?

"Apparent A" (Mechanical) Apparent Acan be a significant error but can easily be removed.ltis a mechanical error caused by a mis-aligned lubber line (reference mark) which should be aligned exactly with the aircraft fore and aft axis. Apparent A gives constant deviations on all headings. Coefficient A is found by averaging the sum of the deviations on eight headings. It may also be calculated on 4, 6 or 8 headings but they must be reciprocal headings. In this case, the result would not be so accurate as that obtained from 8 headings. Coefficient A is caused by an incorrectly aligned lubber line and is corrected by rotating the compass back or forward against the lubber line by the required number of degrees.

2235. Airplane ATPL CPL Heli ATPL CPL Hard iron magnetism in aircraft may be caused by: A) B) C) D)

steel components, mainly in engines and undercarriage. magnetic qualities ofthe cargo or baggage. a strike of lightning. all answers are correct.

For explanation refer to question #1127 on page 67.

2247. Airplane ATPL CPL Heli ATPL CPL An aircraft in the northern hemisphere is making an accurate rate one turn to the right. If the initial heading was 135°, after 30 seconds the direct reading magnetic compass should read: A) 225°.

B) less than 225°. C) more or less than 225° depending on the pendulous suspension used. D) more than 225°.

A) It senses the magnetic meridian instead of seeking it, increasing compass sensitivity. B) It is lighter than a direct reading compass because it employs, apart from the detector unit, existing aircraft equipment. C) It eliminates the effect of turning and acceleration errors by pendulously suspending the detector unit. D) It is more reliable because it is operated electrically and power is always available from sources within the aircraft. For explanation refer to question #7773 on page 62.

2261. Airplane ATPL CPL Heli ATPL CPL What is the advantage of the remote indicating compass (slaved gyro compass) over the direct reading magnetic compass? A) It is lighter. B) It is connected to a source of electrical power and so is more accurate. C) It senses the Earth's magnetic field rather than seeks it, so is more sensitive. D) It is not affected by aircraft deviation. For explanation refer to question #1173 on page 62.

2285. Airplane ATPL CPL Soft iron magnetism In aircraft:

Heli

ATPL

CPL

A) will not cause any compass deviation. B) is non-permanent of nature, and cannot be reduced by degaussing (demagnetization). C) is easily compensated for in the compass swing procedure. D) will change at unknown times.

For explanation refer to question #7764 on page 62.

2248. Airplane ATPL CPL Heli ATPL CPL An aircraft in the northern hemisphere makes an accurate rate one turn to the right/starboard. If the initial heading was 330°, after 30 seconds of the turn the direct reading magnetic compass should read: A) 060°.

B) less than 060°. C) more than 060°. D) more or less than 060° depending on the pendulous suspension used. For explanation refer to question #1164 on page 62.

2255. Airplane ATPL CPL Heli ATPL CPL The direct reading magnetic compass is made aperiodic (dead beat) by: A) using the lowest acceptable viscosity compass liquid. B) keeping the magnetic assembly mass close to the compass point and by using damping wires. C) using long magnets. D) pendulous suspension of the magnetic assembly. Aperiodicity is the quality of moving slowly, damped or deadbeat motion. One of the requirements for a direct reading compass is to be aperiodic = the motion of the needle should be slow but responsive indication must be steady after any disturbances like the ones produced by heading changes, manoeuvres or turbulence. Particularly, there should be no oscillations. The whole compass system: needle, weights and compass card, is surrounded by a special liquid. This liquid dampens any motion and reduces or prevents oscillations. "Damping wires" which trail in the liquid and impede movement are attached to the system and these also help

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2247 (D)

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2248 (8)

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(Refer to figures 067-E709 and 067-El1O) Soft iron magnetism mainly occurs when soft iron type metal in the aircraft becomes magnetized by the earth's magnetic field. A secondary cause is from the aircraft radio and electrical circuits producing electromagnetic fields.

2307. Airplane ATPL CPL Heli ATPL CPL When an aircraft on a westerly heading on the northern hemisphere accelerates, the effect of the acceleration error causes the magnetic compass to: A) B) C) D)

lag behind the turning rate of the aircraft. indicate a turn towards the North. indicate a turn towards the South. to turn faster than the actual turning rate of the aircraft.

For explanation refer to question #2799 on page 63.

2321. Airplane ATPL CPL Heli An aircraft's compass must be swung:

ATPL

CPL

A) if the aircraft has been in the hangar for a long time and has been moved several times. B) if the aircraft has been subjected to hammering. C) every maintenance inspection. D) after a change of theatre of operations at the same magnetic latitude. For explanation refer to question #1153 on page 67.

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02 Magnetism and Compasses

2324. Airplane ATPL CPL Heli ATPL CPL In a typical remote reading compass, the gyro is kept aligned with the magnetic meridian by means of: A) B) C) D)

the annunciator. heading selector. a torque motor. the gyro rigidity controller.

compass North and Magnetic North. compass North and True North. True North and Magnetic North. compass North and the lubber line.

For explanation refer to question #1153 on page 61.

4147. Airplane ATPL CPL Heli ATPL CPL Deviation on MH 180 is -5 and on MH 000 it is +3. Calculate coefficient C:

For explanation refer to question #1173 on page 62.

2331. Airplane ATPL CPL Heli ATPL CPL Assume an aircraft has hard iron magnetism only and this hard iron magnetism is represented by a red pole in relative bearing 070 from the compass. On what heading will the westerly deviation be maximum? A) B) C) D)

A) B) C) D)

Heading 340. Heading 200. Heading 110. Heading 020.

A) B) C) D)

coefficient C =-1 coefficient C =+4 coefficient C = -2 coefficient C = +8

(Refer to figures 061-EI09 and 061-E110) Coefficient C is calculated from the deviations observed on North and South. The value of C is maximum on North/South headings while the value of Coefficient B is zero. Coefficient C = (deviation on North - deviation on South) + 2 Coefficient C = (N-S) + 2

(Refer to figures 061-EI09 and 061-EI1O) The deviating force of this hard iron will be strongest when the angle between this force and Magnetic North is 90", which occurs when the aircraft heading is 020 0 (westerly deviation) or 200 0 (easterly deviation).

In our case: CoeffC = ((+3) - (-5)) + 2 CoeffC=+B+2 CoeffC=+4

2907. Airplane ATPL CPL Heli ATPL CPL The main advantage of a remote indicating compass over a direct reading compass is that it:

4150. Airplane ATPL CPL Heli ATPL CPL Consider the following statements on coefficient A, as used to describe deviation:

A) is able to magnify the Earth's magnetic field in order to attain greater accuracy. B) has less moving parts. C) requires less maintenance. D) senses, rather than seeks, the magnetic meridian.

A) coefficient A is the average deviation on all headings. B) coefficient A will normally be calculated after coefficients B and C has been corrected for. C) coefficient A may be calculated at any stage during a compass swing. D) all answers are correct.

For explanation refer to question #1173 on page 62.

For explanation refer to question #2226 on page 63.

4033. Airplane ATPL CPL Heli ATPL CPL Permanent magnetism in aircraft arises chiefly from: A) exposure to the Earth's magnetic field during normal operation. B) hammering, and the effect of the Earth's magnetic field, whilst under construction. C) the combined effect of aircraft electrical equipment and the Earth's magnetic field. D) the effect of internal wiring and exposure to electrical storms ..

4152. Airplane ATPL CPL Heli ATPL CPL When accelerating on a westerly heading in the northern hemisphere, the compass card of a direct reading magnetic compass will turn: A) clockwise giving an apparent turn towards the North. B) clockwise giving an apparent turn towards the South. C) anti-clockwise giving an apparent turn towards the North. D) anti-clockwise giving an apparent turn towards the South.

For explanation refer to question #1127 on page 61.

For explanation refer to question #2199 on page 63.

4065. Airplane ATPL CPL Heli ATPL CPL When should a Direct Reading Compass (ORe) be "swung":

4153. Airplane ATPL CPL Heli ATPL CPL When accelerating on an easterly heading in the Northern Hemisphere, the compass card of a direct reading magnetic compass will turn:

A) every 6 months. B) following a permanent change of magnetic latitude. C) for night use. D) after flying in an area where lightning is visible.

A) B) C) D)

For explanation refer to question #1153 on page 61.

4139. Airplane ATPL CPL Heli ATPL CPL The annunciator of a remote indicating compass system is used when: A) B) C) D)

synchronizing the magnetic and gyro compass elements. compensating for deviation. setting local magnetic variation. setting the heading pointer.

For explanation refer to question #2199 on page 63.

4175. Airplane ATPL CPL Heli The purpose of compass check swing is to:

For explanation refer to question #1173 on page 62.

4143. Airplane ATPL CPL Heli ATPL CPL . One purpose of a compass calibration is to reduce the difference, if any, between:

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2331 (0) 4175 (C)

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4033 (8)

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anti-clockwise giving an apparent turn toward the South. clockwise giving an apparent turn toward the South. anti-clockwise giving an apparent turn toward the North. clockwise giving an apparent turn toward the North.

4139 (A)

ATPL

CPL

A) cancel out the horizontal component of the Earth's magnetic field. B) cancel out the vertical component of the Earth's magnetic field. C) measure the angle between Magnetic North and compass North. D) cancel out the effects of the magnetic fields found on board the aeroplane.

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4143 (A)

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4147 (8)

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Aviationexam Test Prep Edition 2012 For explanation refer to question #1153 on page 61.

4178. Airplane ATPL CPL Heli ATPL CPL The main reason for mounting the detector unit of a remote reading compass in the wingtip of an aeroplane or a helicopter tail-boom is: A) to ensure that the unit is in the most accessible position on the aircraft for ease of maintenance. B) by having detector units on both wingtips, to cancel out the deviation effects caused by the aircraft structure. C) to minimize the amount of deviation caused by aircraft magnetism and electrical circuits. D) to maximize the units exposure to the Earth's magnetic field. For explanation refer to question #7173 on page 62.

4186. Airplane ATPL CPL Heli ATPL CPL Which of the following is an occasion for carrying out a compass swing on a direct reading compass? A) After an aircraft has passed through a severe electrical storm or has been struck by lightning. B) Before an aircraft goes on any flight that involves a large change of magnetic latitude. C) After any of the aircraft radio equipment has been changed due to unserviceability. D) Whenever an aircraft carries a large freight load regardless of its content. For explanation refer to question #7153 on page 61.

4218. Airplane ATPL CPL Heli ATPL CPL When decelerating on a westerly heading in the northern hemisphere, the compass card of a direct reading magnetic compass will turn: A) B) C) D)

(Refer to figures 061-E109 and 061-E710) During a compass swing procedure, aircraft magnetism is separated into its horizontal components: • co-efficient B (longitudinal = fore-and-aft) • co-efficient C (lateral = athwartships). After calculation and analysis of deviation in the compass system various corrections are made to the compass system to compensate for the errors produced by co-efficients Band C. The corrections are made by means of microadjusters, which are small magnets attached to the compass itself and which exert a force in he opposite plane of the deviation which annuls it. Coefficient B is calculated from the deviations observed on East and West. The value of B is maximum on these headings while the value of coefficient C is zero. Coefficient B = (deviation on East - deviation on West) + 2 Coefficient B = (E-W) + 2 Coefficient C is calculated from the deviations observed on North and South. The value of C is maximum on North/South headings while the value of Coefficient B is zero. Coefficient C = (deviation on North - deviation on South) + 2 Coefficient C = (N-S) + 2

4253. Airplane ATPL CPL Heli ATPL CPL A direct reading compass is used. Accelerating an aircraft on heading 090 at South magnetic latitude will result in:

For explanation refer to question #2199 on page 63.

4228. Airplane ATPL CPL Heli ATPL CPL An aircraft is accelerating on a westerly heading in the northern hemisphere; the effect on a direct reading compass will result in: an apparent turn to the West. an indication of a turn to the North. a decrease in the indicated reading. an indication of a turn to the South.

For explanation refer to question #2199 on page 63.

4947. Airplane ATPL CPL Heli ATPL CPL In the calculation of deviation, the following headings are recorded: MH 358 0 091 0 1820 273 0

CH 356 0 087 0 1860 271 0

Coefficient B is:

4230. Airplane ATPL CPL Heli ATPL CPL In northern hemisphere, during an acceleration in an easterly direction, the magnetic compass will indicate: A) B) C) D)

A) the resultant deviation from magnetism along the aircraft lateral axis. B) a value representing the deviation registered on headings East and West. C) the resultant deviation from magnetism along the aircraft vertical (normal) axis. D) deviation values caused by hard iron magnetism only.

A) no change in compass indication. B) a stable oscillation ofthe compass indication around heading 090. C) an indication of a right turn on the compass. D) an indication of a left turn on the compass.

clockwise giving an apparent turn toward the South. anti-clockwise giving an apparent turn towards the South. clockwise giving an apparent turn towards the North. anti-clockwise giving an apparent turn towards the North.

For explanation refer to question #2199 on page 63.

A) B) C) D)

4251. Airplane ATPL CPL Heli ATPL CPL Coefficient B, as used in aircraft magnetism, presents:

a decrease in heading. an increase in heading. an apparent turn to the South. a heading of East.

A) +1

B) -3 C) -2 D) +2 (Refer to figures 061-E109 and 061-E710) Coefficient B is calculated from the deviations observed on East and West. The value of B is maximum on these headings while the value of coefficient C is zero. Coefficient B = (deviation on East- deviation on West) + 2 Coefficient B = (E-W) + 2

For explanation refer to question #2199 on page 63.

In our case: Deviation on East = +4° Deviation on West +2° Coeff B = ((+4) - (+2)) + 2 CoeffB=+2+2 CoeffB=+1

=

I

4178 (C)

I

4186 (A)

I

4218 (A)

I

4228 (8)

I

4230 (A)

I

4251 (8)

I

4253 (C)

I

4947 (A)

I

02 Magnetism and Compasses

4948. Airplane ATPL CPL Heli ATPL CPL In the calculation of deviation, the following headings are recorded:

MH

CH

3560 091 0 182 0 273 0

3560 087 0 1860 271 0

Coefficient A is: A) B) C) D)

+2 -2 -1 +1

(Refer to figures 061-E109 and 061-E110) Coefficient A is found by averaging the sum of the deviations on eight headings.lt may also be calculated on 4, 6 or 8 headings but they must be reciprocal headings. In this case, the result would not be so accurate as that obtained from 8 headings. In our case, the deviations for the given headings are: 0, +4, -4, +2 CoeffA = (4-4+2) +4 CoeffA =+2+4 CoeffA = +0,5 => round up to +1.

4958. Airplane ATPL CPL Heli ATPL CPL In the calculation of deviation, the following headings are recorded:

MH

CH

358 0 091 0 1820 273 0

3560 087 0 1860 271 0

Coefficient Cis: A) -2 B) +4 C) +3 D) -1 (Refer to figures 061-E109 and 061-E11O) Coefficient C is calculated from the deviations observed on North and South. The value of C is maximum on North/South headings while the value of Coefficient B is zero. Coefficient C = (deviation on North - deviation on South) + 2 Coefficient C= (N-S) + 2 In our case: Deviation on North = +2° Deviation on South =_4° Coeff C = ((+2) - (-4)) + 2 CoeffC = +6 +2 CoeffC=+3

230696. Airplane ATPL CPL Deviation on the standby compass is:

Heli

ATPL

CPL

A) independent of the latitude of the aircraft's position. B) dependent on the heading of the aircraft. C) positive if the Compass North is to the west of Magnetic North. D) zero on the magnetic equator.

1 4948 (0) 1 4958 (C) 1230696 (8) 1

Aviationexam Test Prep Edition 2012

03 Charts

CHARTS 03-01 General properties of miscellaneous types of projections 1132. Airplane ATPL CPL The Earth has been charted using:

Heli

ATPL

CPL

A) WGP84 B) WGS84 C) GD84 D) GPS84 The geographic term for the shape of the Earth is an ellipsoid (a three-dimensional ellipse) or an oblate spheroid as the Earth is not an exact sphere, as it is actually slightly compressed or flattened at the poles. This "squashed" shape is caused by the centrifugal effect of the rotation of the Earth. The navigational or geographic result of this "squashing" means that the length of a minute of latitude varies very slightly with the position on the Earth. One minute of latitude is equal to one nautical mile, the basic unit of distance in aviation navigation calculations. This compression is usually ignored when making maps, as a "reduced earth" or small scale model of the Earth is used, which treats the Earth as a perfect sphere for practical navigational purposes. The WGS 84 standard (World Geodetic System 1984) is used currently for very precise navigation and mapping purposes. This standard, amongst other things, defines the ellipsoid which is currently accepted as the best "fit" for the overall shape of the Earth. When an ellipsoid is fixed at a particular orientation and position with respect to the Earth, it forms a so-called "Geodetic Datum'~ WGS 84 is such a Geodetic Datum. Conventional surveying and projection techniques are then applied to the appropriate Geodetic Datum in order to produce a map. An ellipsoid itself is therefore insufficient to define a Geodetic Datum, the position and orientation of the ellipsoid to the Earth need to be defined also. All modern cartography and navigation systems are based on the WGS 84 system.

1151. Airplane ATPL CPL On a Mercator chart, the scale: A) B) C) D)

Heli

ATPL

CPL

varies as l/cosine of latitude (l/cosine=secant). varies as the sine of the latitude. is constant throughout the chart. varies as 1/2 cosine of the co-latitude.

(Refer to figures 061-E88, 06l-E89 and 06l-E90) The Direct Mercator projection is a cylindrical projection with the plane tangential to the equator of the Reduced Earth and the light source at the centre of the Reduced Earth. The parallel of origin is the Equator. Unlike the Transverse Mercator projection, the geographic poles cannot be projected (they are on the axis of the cylinder). It is clear that scale on a Mercator chart is not constant because of the increasing distance between parallels of latitude when moving North or South from the Equator. Scale is correct at the Equator and expands at a rate proportional to the secant of latitude (varies as lleosine of latitude). If a scale is given for a Mercator chart, it must always relate to a specific latitude - e.g. "scale at 35°S = 1:500000". As scale is constantly changing, care must be taken when measuring distances. The best way is to use the mean latitude covering the route portion. In practice, the measurement of distance is not difficult and speedily improves with time and practice.

1162. Airplane ATPL CPL Heli ATPL CPL On a direct Mercator projection a particular chart length is measured at 30 o N. What Earth distance will the same chart length be if measured at 60 0 N? A) B) C) D)

A larger distance. Twice the distance. The same distance. A smaller distance.

(Refer to figures 061-E88, 061-E89 and 061-E90) The scale on a Mercator chart is not constant because of the increasing distance between parallels of latitude when moving North or South from the Equator. Scale is correct at the Equator and expands as the secant (1leos) of the latitude. If a scale is given for a Mercator chart, it must always relate to a specific latitude - e.g. "scale at 35°S = 1:500 OOO'~ As scale is constantly changing, care must be taken when measuring distances. The best way is to use the mean latitude covering the route portion. In practice, the measurement of distance is not difficult and speedily improves with time and practice. Don't over-think this question - it is quite easy to solve. For example, let's take a distance of 500 NM at 30 0N => this distance will be represented by a change of Longitude of approx."577 minutes = approx. 9,6° (calculation = 500 NM .;cos 30°). Now if we measure the same chart length (577 minutes or 9,6°) along 60 0N the actual distance over the ground will be only 288,5 NM (577 minutes x cos 60°). As we can see, the actual distance over the ground for a chart Longitude change of 9,6° at 60 0N will be much smaller than 9,6° Longitude change at300N.

1191. Airplane ATPL CPL Heli ATPL CPL The term "oblique" in relation to map projections means that: A) the meridians and the parallels of latitude do not intersect at right angles on the whole map. B) the scale is different N/S from E/W. C) the axis of the cylinder or cone is neither parallel to or perpendicular to the Earth's axis of rotation. D) the projection may not be printed on a flat sheet of paper. The oblique Mercator projection is a cylindrical tangential projection with the origin at the Earth centre. The common (direct) Mercator is tangential to the Equator, while the Transverse Mercator is tangential to a chosen meridian. Any other tangential cylindrical projection is an "oblique" projection. The "oblique" tangent may be any great circle, which then appears as a straight line. Meridians are curves concave to the selected great circle datum, spaced along the datum in the same way as spacing on the reduced Earth. Parallels are curves along the selected datum great circle (Meridian) in the same way as the reduced Earth. Great circles intersecting the tangent at 90° are straight lines. Rhumb lines are complex curves. Meridians and parallels intersect at 90° Scale is correct along the great circle datum and remains accurate to +/- 1% within approximately 500 NM from the datum. It varies with distance from the datum GC as the secant of that distance, expressed in degrees and minutes. The projection is conformal convergency is assumed to be correct ± 600 NM each side of the datum great circle shapes are correct within small areas. The oblique Mercator is used for strip maps over long great circle routes.

I

1132 (8)

I

1151 (A)

I

1162 (D)

I

1191 (C)

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Aviationexam Test Prep Edition 2012

1192. Airplane ATPL CPL Heli ATPL CPL How does the scale vary in a direCt Mercator chart? A) The scale increases with increasing distance from the equator. B) The scale decreases with increasing distance from the equator. e) The scale is constant. D) The scale increases South ofthe equator and decreases North of the equator. \

For explanation refer to question #7151 on page 69. 1193. Airplane ATPL CPL Heli The scale quoted on a Lambert's chart is:

ATPL

1228. Airplane ATPL CPL Heli On an aeronautical chart it is common that:

CPL

(Refer to figures 061-E86 and 061-E87) Scale on the Lambert's projection is variable. It is constant along any parallel of latitude, however the scale varies with different parallels of latitude (with the exception of the standard parallels, where it is correct). Where the cone cuts the standard parallels, scale is the same as the Reduced Earth and is therefore correct along those parallels. The scale contracts between the standard parallels and expands outside them. By design, the rate of expansion and contraction is small and depends on the standard parallels chosen. The standard parallels are marked on the char~ and often match the chart limits. Note: The scale quoted on a Lambert's chart is the scale at the standard parallels. Heli

ATPL

CPL

(Refer to figures 061-E84 and 061-E85) Con formality (Orthomorphism) is a property of charts which means basically "same shape" = the directions on the Earth are accurately represented on the chart. The two basic conditions for a chart to be conformal are: • The graticule m~t be similar with the graticule on the Earth (parallels are perpendicular to the meridians) • The scale must be correct all over the chart or at least change at the same rate in all directions On the Lambert Conformal Conic Chart the scale is correct along the two standard parallels, reduces between them and expands outside them; i.e. it is constant along a parallel of latitude. The Polar Stereographic chart is orthomorphic (conformal) as well; the scale is correct atthe pole and expands away from it; i.e. it is constant along a parallel of latitude. On the Direct Mercator chart the scale is correct along the equator and expands away from it; i.e. it is constant along a parallel of latitude. On the Transverse Mercator chart the scale is correct along a meridian of longitude and its anti-meridian and expands away from them; i.e. it is constant along a parallel oflongitude. On the Oblique Mercator chart the scale is correct along a great circle and expands away from it. Note: The JAA defines the term "conformal" as: At any given point on the chart distortions (as a result of the projection) in east-west direction must be the same as in north-south direction. The meridians and parallels must cut each other at right angles. 1226. Airplane ATPL CPL Heli ATPL CPL On a transverse Mercator chart scale is correct at: the 1800 meridian. the false meridian. the great circle of tangency. the meridian of tangency.

I

1192 (A)

I

1193 (A)

I

1207 (8)

I

1226 (0)

CPL

Concerning the charts used in aviation, as the distance to the Latitude of Origin increases, the chart distance representing a constant Earth distance is increased, and consequently the chart scale becomes larger. On the Lambert Conformal Conic Cha;tthe scale is correct along the two standard parallels, reduces between them and expands outside them; i.e. it is constant along a parallel of latitude. The Polar Stereographic chart is orthomorphic (conformal) as well; the scale is correct atthe pole and expands away from it; i.e. it is constant along a parallel of latitude. On the Direct Mercator chart the scale is correct along the equator and expands away from it; i.e. it is constant along a parallel of latitude. On the Transverse Mercator chart the scale is correct along a meridian of longitude and its anti-meridian and expands away from them; i.e. itis constant along a parallel oflongitude. On the Oblique Mercator chart the scale is correct along a great circle and expands away

,

~m~

A) the variation information is printed on the map as isogonals. B) the meridians and the parallels of latitude intersect at right angles and when the scale from any selected point is the same in all directions. e) when it conforms to the specifications. D) the meridians are straight lines and the scale is constant.

A) B) e) D)

ATPL

A) the chart has best conformity only along one parallel of latitude. B) the exact scale vary within the chart. e) the exact meridians are curved lines. D) isoclinals are printed on the chart.

A) the scale at the standard parallels. B) the scale at the equator. e) the mean scale between the Pole and the equator. D) the mean scale at the parallel of the secant of the cone.

1207. Airplane ATPL CPL A map is conformal when:

The Transverse Mercator is a mathematical projection "constructed" with an imaginary cylinder tangential to a chosen meridian. This meridian is the central meridian of the projection and is a straight line on the chart. The origin of the chart is the centre of the Earth. The properties of the Transverse Mercator charts are: • The Transverse Mercator is a cylindrical orthomorphic projection with scale appro)(imately correct at the median of tangency; • Parallels are complex curves (ellipses) concave to the nearer pole; • Meridians are complex curves concave towards the central meridian; • This type of chart is ideal for maps with a large North/South area (stripcharts along selected North/South routes).

I

1266. Airplane ATPL Heli ATPL In which of the following projections does a plane surface touch the reduced Earth at one of the Poles? A) Direct Mercator. B) Stereographic. e) Lambert's. D) none of the above.

(Refer to figures 061-E91 and 061-E92) The Polar Stereographic projection is an azimuthal projection. The Earth graticule is projected onto a chart which is tangential to the selected pole. The point of projection is at the opposite pole. Meridians are drawn from the pole with angular spacing equal to the actual change of longitude. With the Pole as centre, circles are drawn with latitude radius 2R tan ~ co-Iat. (R = radius of the "Reduced Earth'~ co-Iat = 90° minus latitude). A complete hemisphere may be shown on one chart, although the accuracy suffers at latitudes lower than 70~ The chart is orthomorphic as scale varies at the same rate in all directions and meridians and parallels intersect at 90°. • Meridians are equally spaced straight lines radiating from the selected pole which is at the centre of the projection. • The angles shown are the true angles between meridians. • Parallels of latitude are a series of unequal concentric circles centred on the selected pole. The spacing betweenparallels increases with distance from the pole. The radius of the parallels is 2R tan 1/2co-lat. • Only one pole can be shown on the chart. • Meridians and para11els intersect at 90°. .........., • Except for meridians, Rhumb Lines are curves concave to the nearer pole which are difficult to measure accurately. • For practical purposes, all Great Circles within the area' covered are considered straight lines. The chart is constructed so that Great Circles are approximately straight lines, but in fact they are slightly concave to the pole. The closer a Great Circle is to the centre of the projection (the pole), the more it will appear as a straight line. This is completely true only for the lines crossing the central meridian at 90~ Great Circles can be measured at any convenient meridian. The Equator is a circle, furthest from thepole. • Convergency is constant and correct at the pole, incorrect elsewhere, with a value of 1. Chart convergency is equal to change of longitude.

1228 (8)

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1266 (8)

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03 Charts

1284. Airplane ATPL CPL Heli ATPL CPL A Lambert conformal conic projection, with two standard parallels: A) shows lines of longitude as parallel straight lines. B) shows all great circles as straight lines. e) the scale is only correct at parallel of origin. D) the scale is only correct along the standard parallels.

is also known as the Chart Convergence Factor (CCF). The value of chart convergence (CC) between any two meridians on the Lambert's projection may be calculated in either of the following ways, depending what data are known: • CC (chart convergence) = Ch long x sin parallel of origin • CC (chart convergence) = Ch long x constant of the cone • CC (chart convergence) = Ch long x "n" • CC (chart convergence) =Ch long x Chart Convergence Factor (CCF) It can be seen that each formula is basically the same as the constant of the cone = "n" = CCF = sin parallel of origin. Any of the values may be noted on the chart, or may be calculated from a knowledge of the standard parallels.

For explanation refer to question #1193 on page 70.

1304. Airplane ATPL CPL Heli ATPL CPL The nominal scale of a Lambert conformal conic chart is the: A) scale at the Equator. Bj -scale at the standard parallels. e) mean scale between the pole and the Equator. D) mean scale between the parallels of the secant cone.

2213. Airplane ATPL CPL Heli ATPL CPL If you want a chart where a particular great circle is an exact straight line, you should, look for a chart using the: A) Lambert conformal projection. B) oblique Mercator projection. e) polar stereographic projection. D) transverse Mercator projection.

For explanation refer to question #1193 on page 70.

For explanation refer to question #1191 on page 69.

1347. Airplane ATPL CPL Heli ATPL CPL The main use for an oblique Mercator chart would be: A) for countries with large changes in latitude but small changes in longitude. B) route charts for selected great circle routes. e) better topographical coverage of polar regions. D) topographical coverage of equatorial regions. For explanation refer to question #1191 on page 69.

2233. Airplane ATPL CPL Heli ATPL CPL Transverse Mercator projections are used for: A) maps of large north/south extent. B) maps of large east/west extent in equatorial areas. e) radio navigation charts in equatorial areas. D) plotting charts in equatorial areas. For explanation refer to question #1226 on page 70.

2182. Airplane ATPL CPL Heli ATPL CPL An oblique Mercator projection is used specifically to produce: A) plotting charts in equatorial regions. B) radio navigational charts in equatorial regions. e) topographical maps of large east/west extent. D) charts of the great circle route between two points.

2267. Airplane ATPL CPL Heli ATPL CPL How does the chart convergence change with latitude in a Lambert conformal projection? A) It changes with sine of latitude. B) It changes with cosine of latitude. e) It increases with increase of latitude. D) It is constant and.does not change with latitude.

For explanation refer to question #1191 on page 69.

2192. Airplane ATPL CPL Heli ATPL CPL In producing chart projections, the following projection surfaces may be used: A) plane, cylinder, cone. B) plane, sphere, cone. e) cylinder, sphere, plane. D) parabola, cone, plane, cylinder. (Refer to figures 061-E84 and 061-E85) Projections may be grouped under four main headings: • Mathematically constructed graticules not based on a geometric projection; • Conical projections and their modifications, i.e. a cone placed on either pole of the reduced earth; • Cylindrical projections and their modifications, i.e. a cylinder enclosing the reduced earth; • Azimuthal projections, i.e. a flat plane tangential to the reduced Earth's surface (for example in a Polar Stereographic projection the parallel of origin is the pole).

2195. Airplane ATPL CPL Heli ATPL CPL On a Lambert conformal conic chart Earth convergence is most accurately represented at the:

For explanation refer to question #2195 on this page.

2274. Airplane ATPL CPL Heli ATPL CPL On a Lambert conformal conic chart the convergence of the meridians: A) is the same as Earth convergency at the parallel of origin. B) is zero throughout the chart. e) varies as the secant of the latitude. D) equals Earth convergency at the standard parallels. For explanation refer to question #2195 on this page.

2294. Airplane ATPL CPL Heli ATPL CPL On a Lambert conformal conic chart the distance between parallels of latitude spaced the same number of degrees apart: A) is larger between the standard parallels and is smaller outside them. B) is constant between the standard parallels and is greater outside them. e) is smaller between the standard parallels than outside them. D) is constant throughout the chart. For explanation refer to question #1193 on page 70.

A) north and South limits of the chart. B) parallel of origin. e) standard parallels. D) equator. (Refer to figures 061-E86 and 061-E87) The chart convergence remains the same on a Lambert chart, Earth convergence does not. Chart convergence is that of the parallel of origin and is constant on the chart. The convergence between two positions is the Change in Longitude x sine of the Parallel of origin. This value of convergence

I I

1284 (D) 2294 (C)

1304 (8)

I

1347 (8)

I

2182 (D)

1

2192 (A)

I

2195 (8)

1 2213 (8) I

2233 (A)

I

2267 (D)

I

2274 (A)

I

Aviationexam Test Prep Edition 2012 2295. Airplane ATPL CPL Heli ATPL CPL The distance on a Lambert's chart, between two parallels of latitude the same number of degrees apart: A) is constant all over the chart. B) is constant between the standard parallels, and expands outside them. C) expands between the standard parallels, but reduces outside them. D) reduces between the standard parallels, but expands outside them. For explanation refer to question #1193 on page 70.

2296. Airplane ATPL CPL Heli ATPL CPL The scale on a Lambert conformal conic chart: A) B) C) D)

is constant along a meridian of longitude. is constant across the whole map. varies slightly as a function of latitude and longitude. is constant along a parallel of latitude.

For explanation refer to question #1193 on page 70.

2315. Airplane ATPL CPL Heli ATPL CPL On a Lambert conformal chart the scale is correct: A) at the standard parallels. B) in the middle of the chart. C) at the latitude where the convergence on the chart corresponds to the Earth convergence. D) at the selected parallel (parallel of origin). For explanation refer to question #1193 on page 70. .,_

4075. Airplane ATPL CPL Heli ATPL CPL How does scale change on a normal Mercator chart? A) Expands as the secant2 (1/2 co-latitude). B) Expands directly with the secant of the latitude. C) Correct on the standard parallels, expands outside them, contracts within them. D) Expands as the secant of the E/W great circle distance. For explanation refer to question #1151 on page 69.

4086. Airplane ATPL CPL Heli ATPL CPL On a transverse Mercator chart, the scale is exactly correct along the: A) prime meridian and the equator. B) equator and parallel of origin. C) meridian of tangency and the parallel of latitude perpendicular to it. D) meridians of tangency.

(Refer to figures 061-E88, 061-E89 and 061-E90) The Direct Mercator projection is a cylindrical projection with the plane tangential to the equator of the Reduced Earth and the light source at the centre of the Reduced Earth. The parallel of origin is the Equator. Unlike the Transverse Mercator projection, the geographic poles cannot be projected (they are on the axis of the cylinder). • The graticule is rectangular. • Meridians are vertical straight lines, equally spaced apart and parallel to each other. • Parallels of Latitude are horizontal straight lines, unequally spaced apart and parallel to each other. • Great Circles are complex curves, curved towards the nearest pole (convex towards the pole and concave towards the Equator). • Rhumb Lines are straight lines (Meridians are straight lines and since RL crosses all meridians at the same angle it is straight line as weI/). • Chart convergency (convergence) is the angle between meridians on the chart. As meridians are parallel, there is no angle between them. There is therefore no convergency on a Mercator chart. Alternatively, convergency is zero and constant everywhere on the chart. Since the value of convergency at the Equator on Earth is also zero, then convergency on a Mercator chart is only correct at the Equator Note: Concerning the plotting of Great Circles there are a few exceptions. Normally GCs are represented by curved lines on a Direct Mercator chart. However, the Equator and the Meridians are great circles by themselves - these and only these Great Circles (Equator + Meridians) are represented as straight lines.'

4110. Airplane ATPL Heli ATPL The chart that is generallyuseJ:! for navigation in polar areas is based on a: A) B) C) D)

stereographical projection. direct Mercator projection. gnomonic projection. Lambert conformal projection.

For explanation refer to question #1266 on page 70.

4135. Airplane ATPL CPL Heli Scale on a Lambert's conformal chart is: A) B) C) D)

ATPL

CPL

ATPL

CPL

constant along a parallel of latitude. constant along a meridian of longitude. constant over the whole chart. varies with latitude and longitude.

For explanation refer to question #1193 on page 70.

4136. Airplane ATPL CPL On a conformal chart, scale is:

Heli

A) constant. B) constant along a meridian of longitude. C) variable; it varies as a function of latitude and longitude. D) constant along a parallel of latitude. For explanation refer to question #1207 on page 70.

For explanation refer to question #1226 on page 70.

4098. Airplane ATPL CPL Heli ATPL CPL The angular difference, on a Lambert conformal conic chart, between the arrival and departure track is equal to: A) chart convergence. B) Earth convergence. C) conversion angle. D) difference in longitude. For explanation refer to question #2195 on page 71.

4106. Airplane ATPL CPL Heli ATPL CPL On which of the following chart projections is it not possible to represent the North or South Poles? A) B) C) D)

Lamberts conformal. Direct Mercator. Transverse Mercator. Polar stereographic.

4138. Airplane ATPL CPL Heli ATPL CPL A direct Mercator graticule is based on a projection that is: A) spherical B) concentric C) cylindrical D) conical For explanation refer to question #4106 on this page.

4164. Airplane ATPL CPL Heli ATPL CPL A straight line is drawn on a Lambert's conformal conic chart between two positions of different longitude. The angular difference between the initial true track and the final true track of the line is equal to: A) Earth convergence. B) chart convergence. C) conversion angle. D) difference in longitude.

I 2295 (D) I 2296 (D) I 2315 (A) I 4075 (8) I 4086 (D) I 4098 (A) I 4106 (8) I 4110 (A) I 4135 (A) I 4136 (D) I I 4138 (C) I 4164 (8) I

03 Charts

D) 40NM

For explanation refer to question #2795 on page 77.

4193. Airplane ATPL CPL Heli ATPL CPL Where on a direct Mercator projection is the chart convergence correct compared to the Earth convergence?

(Refer to figures 067-E84 and 067-E85) "Scale" is the relationship of the distance between two positions on a chart and the distance between the same positions on the Earth. • Scale Chart length.;- Earth distance

=

In our case, the distance on the chart is represented by 4 cm. The scale of the chartis 7: 7.850.000 (7 cm on the chart = 7.850.000 cm in reality). ·7 cmon the chart = 7.850.000cm in reality ·4 cm on the chart = Z400.000 cm in reality • Z400.000 cm = 74 km • 1 NM = 7,852 km ·74km=40NM

A) All over the chart. B) At the two parallels of tangency. C) At the poles. D) At the equator. For explanation refer to question #4706 on page 72.

4236. Airplane ATPL CPL Heli ATPL CPL On a Lambert conformal conic chart, with two standard parallels, the quoted scale is correct: A) along the prime meridian. B) along the two standard parallels. C) in the area between the standard parallels. D) along the parallel of origin.

The distance along the surface of the Earth will be approximately 40 NM.

4309. Airplane ATPL CPL Heli ATPL CPL On a Direct Mercator chart at latitude 45°N, a certain chart length along 45°N represents a distance of 90 NM on the surface of the earth. The same length on a chart along latitude 30 0 N will represent a distance on the earth of: A) 45 NM

For explanation refer to question #7793 on page 70.

4259. Airplane ATPL Heli The Polar Stereographic projection is:

B) 73,5 NM C) 78NM

ATPL

D) 110 NM

A) a cylinder projection. B) a plane projection. C) a variable cone projection. D) a conical projection. For explanation refer to question #2792 on page 77.

4294. Airplane ATPL CPL Heli ATPL CPL On a direct Mercator projection, at latitude 45° North, a certain length represents 70 NM. At latitude 30° North, the same length represents approximately: A) 57 NM B) 86NM C) 70 NM D) 81 NM (Refer to figures 067-E88, 067-E89 and 067-E90) The scale on a Mercator chart is not constant because of the increasing distance between parallels of latitude when moving North or South from the Equator. Scale is correct at the Equator and expands as the secant (7leos) , of the latitude. If a scale is given for a Mercator chart, it must always relate i to a specific latitude - e.g. "scale at 35°S = 7:500 000". As scale is constantly changing, care must be taken when measuring distances. The best way is to use the mean latitude covering the route portion. In practice, the measurement of distance is not difficult and speedily improves with time and practice. Since we know the actual distance over the surface of the Earth and the Latitude, we can determine the distance in terms of "Change in Longitude": ·70NM= ?xcos45° ·?=70NM.;-cos45° • ? = 99 minutes 0039') Now we will use the same formula again, but we will reverse the process for the second Latitude: • Distance =Change in Longitude (in minutes) x cos Lat • Distance = 99 minutes x cos 30° • Distance = 85,7 NM

At latitude 300N the same distance on the chart as measured at 45°N would represent a ground distance of approximately 86 NM.

4308. Given:

Airplane

ATPL

Heli

CPL

ATPL

CPL

Earth distance is approximately:

I

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Now we will use the same formula again, but we will reverse the process for the second Latitude: • Distance = Change in Longitude (in minutes) x cos Lat • Distance = 727,28 minutes x cos 30° • Distance = 770,2 NM At latitude 30 0N the same distance on the chart as measured at 45°N would represent a ground distance of approximately 770,2 NM.

4327. Airplane ATPL CPL Heli ATPL CPL An average true track of 120° is drawn between 'X' (61°30'N) and 'V' (58°30'N) on a Lambert Conformal conic chart with a scale of 1: 1000000 at 60 0 N. The chart distance between 'X' and 'V' is: A) 33,4cm

B) 66,7 cm C) 38,5 cm

D) 36,Ocm (Refer to figure 067-E42) "Scale" is the relationship of the distance between two positions on a chart and the distance between the same positions on the Earth. • Scale Chart length .;- Earth distance

=

In our case, the actual distance from X to Y = 360 NM (see attached illustration). • 7 NM = 7,852 km ·360 NM 666,72 km • 666,72 km = 66.672.000 cm • 7 cm on the chart = 7.000.000 cm in reality (scale) .? cm on the chart = 66.672.000 cm in reality ·?=66,67cm The distance X to Y will be represented by a distance of approx. 66,7 on the chart.

A) 4NM B) 74 NM C) 100 NM 4193 (0)

Since we know the actual distance over the surface of the Earth and the Latitude, we can determine the distance in terms of "Change ill Longitude": ·90NM=?xcos45° .? = 90 NM.;- cos 45° • ? = 727,28 minutes (2?')

=

Chart scale is 1:1.850.000. The chart distance between two points is 4 centimetres.

I

(Refer to figures 067-E88, 067-E89 and 067-E90) The scale on a Mercator chart is not constant because of the increasing distance between parallels of latitude when moving North or South from the Equator. Scale is correct at the Equator and expands as the secant (7leos) of the latitude. If a scale is given for a Mercator chart, it must always relate to a specific latitude - e.g. "scale at 35°S = 7:500 000". As scale is constantly changing, care must be taken when measuring distances. The best way is to use the mean latitude.(:overing the route portion. In practice, the measurement of distance is not difficult and speedily improves with time and practice.

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cm

Aviationexam Test Prep Edition 2012 4375. Airplane ATPL CPL Heli ATPL CPL The convergence factor of a Lambert conformal conic chart is quoted as 0,78535. At what latitude on the chart is Earth

"Scale" is the relationship of the distance between two positions on a chart and the distance between the same positions on the Earth. • Scale = Chart length + Earth distance

__

In our case, the distance on the chart is represented by 42 mm = 4,2 cm. The scale of the chart along is 7 : 7.600.000 (7 cm on the chart = 7.600.000 cm in reality). • 7 cm on the chart = 7.600.000 cm in reality ·4,2 cm on the chart = 6.720.000 cm in reality • 6 720 000 cm = 67,2 km ·7 NM= 7,852km • 67,2 km = 36,29 NM

-_ . --_ .... , .-.... ----..... --..

rnnVArnAnl"A rnrral"tlu rAnrAcAntArl7 M_~_";:J_----

A) 38°15' B) 51°45' C) 52°05' 0) 80°39' (Refer to figures 067-E86 and 067-E87) This question may appear difficult, but the fact is that it is one of the easiest ones in this subject and the solution is very simple. You just have to remember that the "Constant of the Cone" = sin of the Parallel of Origin. Parallel of Origin = the mean Latitude between the standard parallels. Therefore, to solve this question we have to find the inverted sin of 0,78535 => 57,75° => 57°45' (0,75° x 60 =45 minutes). The convergence is correctly represented along the parallel of origin => 57°45'.

4421. Airplane ATPL CPL Heli ATPL CPL On a Lambert conformal chart, the distance between two parallels of latitude (difference of latitude 2°), is measured to be 112 mm. The distance between two meridians, spaced 2° longitude, according to the chart is 70 NM. What is the latitude in the centre of the described square?

=

A) 49° B) 38° C) 54° 0) 42° The change in longitude = 2". At the Equator this would equal to 720 NM (2 x 60 NM). Since the actual distance stated by the question is 70 NM it means that the Latitude is quite far up North. • Distance = Change in Longitude (in minutes) x cos Latitude. ·70 NM = 720 minutes x cos 7 • 7 ~ 70 NM + 720 minutes ·7=0,5833 • Latitude = inv.cos 0,5833 • Latitude = 54,37 ° (54°79')

4422. Airplane ATPL CPL Heli ATPL CPL A chart has the scale 1:1.000.000. From A to B on the chart measures 3,8 cm. The distance from A to B in NM is: A) 70,4

B) 38,1 C) 20,5 0) 205 (Refer to figures 067-E84 and 067-E85) "Scale" is the relationship of the distance between two positions on a chart and the distance between the same positions on the Earth. • Scale = Chart length + Earth distance In our case, the distance on the chart is represented by 3,8 cm. The scale of the chartis 7 : 7.000.000 (7 cm on the chart 7.000.000cm in reality). • 7 cm on the chart = 7.000.000 cm in reality ·3,8 cm on the chart = 3.800.000 cm in reality ·3.800.000 cm = 38 km ·7 NM= 7,852km ·38 km =20,52 NM

=

The distance along the surface of the Earth will be approximately 20,5 NM.

4426. Airplane ATPL CPL Heli ATPL The distance measured between two points on tion map is 42 mm (millimetres). The scale of is 1:1.600.000. The actual distance between these is approximately: A) B) C) 0)

CPL

a navigathe chart two point

3,69 NM 370,00 NM 67,20NM 36,30NM

The distance along the surface of the Earth will be approximately 36,3 NM.

4427. Airplane ATPL CPL Heli ATPL CPL A Lambert conformal conic chart has a constant of the cone of 0,75. The initial course of a straight line track drawn on this chart from A (400N 0500W) to B is 043°(T} at A; course at B is 055°(T}. What is the longitude of B? A) 41°W B) 36°W C) 38°W 0) 34°W (Refer to figures 067-E86 and 067-E87) In this question the convergency = 055° - 043° = 72°. Convergency is calculated by multiplying the change in longitude with the sine of mean latitude of the two points. The constant of the cone, or otherwise called the convergence factor is 0,75 here. Constant of the cone = sin mean latitude. Convergency = Change in longitude x sine of the mean latitude. 72° = Change in long. x 0,75 Change in longitude = 76° Point A lies at 040 0 N 050 0 W and the heading is North-Easterly, therefore the longitude ofpoint B is 050 0 W - 76° = 034°W.

4470. Airplane ATPL CPL Heli ATPL CPL Assume a Mercator chart. The distance between positions A and B,located on the same parallel and 10° longitude apart, is 6 cm. The scale at the parallel is 1:9.260.000. What is the latitude of A and B? A) 45° N orS B) 30° N or S C) 0° 0) 60° N or S (Refer to figures 067-E88, 067-E89 and 067-E90) The scale on a Mercator chart is not constant because of the increasing distance between parallels of latitude when moving North or South from the Equator. Scale is correct at the Equator and expands as the secant (7leos) of the latitude. If a scale is given for a Mercator chart, it must always relate to a specific latitude - e.g. "scale at 35°S = 7:500 000': As scale is constantly changing, care must be taken when measuring distances. The best way is to use the mean latitude covering the route portion. In practice, the measurement of distance is not difficult and speedily improves with time and practice. Remember that Scale = "Chart length + Earth distance". In our case, the scale = 7 : 9.260.000: • 7 cm on the chart = 9.260.000 cm in reality ·6 cm on the chart = 55.560.000 cm in reality ·55.560.000 cm = 555,6 km ·7 NM= 7,852km '555,6 km = 300 NM Now we have to find the parallel, where 70° (600 minutes) of Longitude equals to 300 NM: • Distance = Change in Longitude (in minutes) x cos Latitude ·300 NM = 600 minutes x cos 7 • cos 7 = 300 NM + 600 minutes 'cos7=0,5 • ? = 60° (use the inverted cos function on your calculator to convert 0,5 tacos 60°) The latitude ofA and B is therefore going to be either 60 0 N or 60 0 S.

(Refer to figures 067-E84 and 067-E85)

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4375 (8)

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4421 (C)

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03 Charts

4481. Airplane ATPL CPL Heli ATPL CPL If the chart scale is 1:500.000, what Earth distance would be represented by 7 cm on the chart? A) 35 NM B) 3,5 km

C) 35.000 m D) 0,35 km (Refer to figures 061-E84 and 061-E85) "Scale" is the relationship of the distance between two positions on a chart and the distance between the same positions on the Earth. • Scale =Chart length -;- Earth distance In our case, the distance on the chart is represented by 7 cm. The scale of the chart is 1: 500.000 (1 cm on the chart = 500.000 cm in reality). • 1 cm on the chart = 500.000 cm in reality ·7 cm on the chart = 3.500.000 cm in reality ·3.500.000 cm =35.000 meters =35 km The distance along the surface of the Earth will be approximately 35 km (35.000 meters).

4559. Airplane ATPL CPL Heli ATPL CPL The constant of cone of a Lambert conformal conic chart is quoted as 0,3955. At what latitude on the chart is Earth convergence correctly represented? A) 68°25'

B) 21°35' C) 23°18'

D) 66°42'

(Refer to figures 061-E86 and 061-E87) This question may appear difficult, b!1t the fact is that it is one of the easiest ones in this subject and the solution is very simple. You just have to remember that the "Constant of the Cone" = sin of the Parallel of Origin. Parallel of Origin = the mean Latitude between the standard parallels. Therefore, to solve this question we have to find the inverted sin of 0,3955 => 23,29° => 23°18' (0,29° x 60 = 18 minutes). The convergency is correctly represented along the parallel of origin => 23°18'.

4599. Airplane ATPL CPL Heli ATPL CPL At 47° North the chart distance between meridians 10° apart is 12,7 cm. The scale of the chart at 47° North approximates: A) B) C) D)

D) 1:5.000.000 (Refer to figures 061-E84 and 061-E85) "Scale" is the relationship of the distance between two positions on a chart and the distance between the same positions on the Earth. • Scale =Chart length -;- Earth distance In our case, the distance on the chart is represented by 4,63 cm. The actual distance represented is 150 NM. • 1 NM = 1,852 km • 150NM=277,8km ·277,8 km =27.780.000 cm ·4,63 cm on the chart = 27.780.000 cm in reality • 1 cm on the chart =6.000.000 cm in reality

The scale is therefore 1: 6.000.000.

4601. Airplane ATPL CPL Heli ATPL CPL On a Lambert conformal chart the distance between meridians 5° apart along latitude 37° North is 9 cm. The scale ofthe chart at that parallel approximates: A) B) C) D)

1:3.750.000 1:5.000.000 1:2.000.000 1:6.000.000

(Refer to figures 061-E86 and 061-E87) Scale on the Lambert's projection is variable. Where the cone cuts the standard parallels, scale is the same as the reduced Earth and is therefore correct along those parallels. The scale contracts between the standard parallels and expands outside them. The actual distance along the surface of the Earth is: 1°along the Equator =60NM l°along 3rN= 60NM x co~ 3r 1° along 3rN = 47,92 NM 5° along 3rN =239,6 NM 1 NM = 1,852 km 239,6 NM = approx. 444 km 444 km =44 400 000 cm

9 cm on the chart = actual 44.400.000 cm 1 cm on the chart = actual 4.933.333 cm The scale is therefore 1: 4.933.333 => approximately 1 : 5.000.000.

Summary of calculation: 5° x 60 NM x cos 37x =239,6 NM 239,6 NM x 1,852 km = 443,7 km 9 cm = 443,7 km 1 cm = 49,3 km (4.930.000 cm)

1:2.500.000 1:8.000.000 1:3.000.000 1:6.000.000

(Refer to figures 061-E84 and 061-E85) "Scale" is the relationship of the distance between two positions on a chart and the distance between the same positions on the Earth. • Scale =Chart length -;- Earth distance

4602. Airplane ATPL CPL Heli ATPL CPL The chart distance between meridians 10° apart at latitude 65° North is 9,5 cm. The chart scale at this latitude approximates: A) 1:6.000.000 B) 1:5.000.000 C) 1:2.500.000 D) 1:3.000.000

The distance along the surface of the Earth: • along a Meridian =60 NM for each 1". • along the Equator = 60NM for each 1". • along a parallel = 60 x cos Latitude for each 1°. In our case, the distance on the chart is represented by 12,7 cm. This chart distance represents 10° along the parallel 4rN. In reality the distance equals: 70° x 60 x cos 4r = 409,2 NM. ·1 NM= 1,852km ·409,2 NM = 757,84 km ·757,84 km =75.784.000 cm • 12,7 cm on the chart = 75.784.000 cm in reality • 1 cm on the chart = 5.967.244 cm in reality

(Refer to figures 061-E84 and 061-E85) "Scale" is the relationship of the distance between two positions on a chart and the distance between the same positions on the Earth. • Scale = Chart length -;- Earth distance

The scale is therefore 1: 5.967.244 => approximately 1: 6.000.000.

In our case, the distance on the chart is represented by 9,5 cm. This chart distance represents 10° along the parallel 65"N. In reality the distance equals: 10° x 60 x cos 65° = 253,57 NM. ·1 NM= 1,852km ·253,57 NM =469,61 km ·469,61 km=46.961.000cm

4600. Airplane ATPL CPL Heli ATPL CPL On a chart, a straight line is drawn between two points and has a length of 4,63 cm. What is the chart scale if the line represents 150 NM? A) 1:1.000.000 B) 1:6.000.000 C) 1:3.000.000

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4484 (C)

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4559 (C)

The distance along the surface of the Earth: • along a Meridian = 60 NM for each 1°. • along the Equator = 60 NM for each 1o. • along a parallel = 60 x cos Latitude for each 1°.

·9,5 cm on the chart =46.961.000 cm in reality • 1 cm on the chart =4.943.263 cm in reality

The scale is therefore 1: 4.943.263 => approximately 1: 5.000.000.

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4599 (D)

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Aviationexam Test Prep Edition 2012

4603. Airplane ATPL CPL Heli ATPL CPL The total length of the 53°N parallel of latitude on a direct Mercator chart is 133 cm. What is the approximate scale of the chart at latitude 30 0 S? A) B) C) D)

1:26.000.000 1:30.000.000 1:18.000.000 1:21.000.000

(Refer to figures 061-E88, 061-E89 and 061-E90) The scale on a Mercator chart is not constant because of the increasing distance between parallels of latitude when moving North or South from the Equator. Scale is correct at the Equator and expands as the secant (1leos) of the latitude. If a scale is given for a Mercator chart, it must always relate to a specific latitude - e.g. "scale at 35°S = 1:500 OOO'~ As scale is constantly changing, care must be taken when measuring distances. The best way is to use the mean latitude covering the route portion. In practice, the measurement of distance is not difficult and speedily improves with time and practice. When the scale at one Latitude on a Mercator chart is known, the scale at any other Lat. may be calculated by a formula: • Scale Denominator '~" x cosine "8" =Scale denominator "8" x cosine '~" ... where '~" and "8" are the Latitudes. The formula is best remembered as '~88A"(A x8=8xA). In our case we have to first calculate the scale at the 53W parallel. Total length ofeach parallel =360~ Total length of the Equator =360 0 x 60 NM =21.600 NM. The total length of the 53 oN parallel = 21.600 NM x cos 53°= 12.999 NM (or 360° x 60x cos 53°= 12.999 NM). Next we need to calculate the scale at 53 oN: • 133 cm on the chart represents 12.999 NM in reality • 1 cm on the chart represents 97,73 NM in reality • 1 NM = 1.852 km ·97,73 NM = 181 km • 181 km = 18.100.000 cm • scale at 53 oN = 1: 18.100.000 Now we can apply the '~88A" formula: • 18.100.000 x cos 30° = 7x cos 53° • 7= (18.100.000 x cos 30°) 7 cos 53° • 7=26.046.309 The scale of the chart at 30 0 S is "1 : 26.046.309" = approximately 1 : 26.000.000.

4604. Airplane ATPL CPL Heli ATPL CPL A straight line on a chart 4,89 cm long represents 185 NM. The scale of this chart is approximately: A) B) C) D)

1:5.000.000 1:3.500.000 1:6.000.000 1:7.000.000

(Refer to figures 061-E84 and 061-E85) "Scale" is the relationship of the distance between two positions on a chart and the distance between the same positions on the Earth. • Scale = Chart length 7 Earth distance In our case, the distance on the chart is represented by 4,89 cm. This chart distance represents an actual distance of 185 NM. ·1 NM= 1,852km • 185 NM = 342,62 km • 342,62 km = 34.262.000 cm ·4,89 cm on the chart = 34.262.000 cm in reality • 1 cm on the chart = 7.006.544 cm in reality

The scale is therefore 1: 7.006.544 => approximately 1: 7.000.000.

4605. Airplane ATPL CPL Heli ATPL CPL At 60° N the scale of a direct Mercator chart is 1:3.000.000. What is the scale at the equator? A) B) C) D)

1:3.000.000 1:3.500.000 1:1.500.000 1:6.000.000

(Refer to figures 061-E88, 061-E89 and 061-E90) The scale on a Mercator chart is not constant because of the increasing distance between parallels of latitude when moving North or South from

the Equator. Scale is correct at the Equator and expands as the secant (1leos) of the latitude. If a scale is given for a Mercator chart, it must always relate to a specific latitude - e.g. "scale at 35°S = 1:500.000'~ As scale is constantly changing, care must be taken when measuring distances. The best way is to use the mean latitude covering the route portion. In practice, the measurement of distance is not difficult and speedily improves with time and practice. When the scale at one Latitude on a Mercator chart is known, the scale at any other Lat. may be calculated by a formula: • Scale Denominator '~" x cosine "8" =Scale denominator "8" x cosine '~" ... where '~" and "8" are the Latitudes. The formula is best remembered as '~88A" (A x 8 = 8 x A). We can apply the '~88A" formula directly: • Latitude of the Equator = 0° ·3.000.000 x cos 0°= 7x cos 60° ·7= (3.000.000 x cos 0°) 7 cos 60° ·7= 6.000.000 The scale of the chart at the Equator is 1 : 6.000.000.

4606. Airplane ATPL CPL Heli ATPL CPL On a Mercator chart, at latitude 60 0 N, the distance measured between 002°W and 008°E is 20 cm. The scale of this chart at latitude 60 0 N is approximately: A) B) C) D)

1:5.560.000 1:278.000 1:2.780.000 1:556.000

(Refer to figures 061-E88, 061-E89 and 061-E90) The scale on a Mercator chart is not constant because of the increasing distance between parallels of latitude when moving North or South from the Equator. Scale is correct at the Equator and expands as the secant (1leos) of the latitude. If a scale is given for a Mercator chart, it must always relate to a specific latitude - e.g. "scale at 35°S = 1:500 OOO'~ As scale is constantly 'changing, care must be taken when measuring distances. The best way is to use the mean latitude covering the route portion. In practice, the measurement of distance is not difficult and speedily improves with time and practice. First of all we need to find the actual distance over the surface of the Earth. The distance along the surface of the Earth between two meridians along a certain parallel of latitude can be calculated using a formula: • distance Change in Longitude (in minutes) x cos Lat

=

In our case the distance is measured between 002°W and 008°E = total of 10· of Longitude = 600 minutes. • Distance = 600 min x cos 60° • Distance = 300 NM Now we can easily calculate the scale. The distance on the chart is represented by 20 cm. The distance along the surface of the Earth is represented by 300 NM. • 1 NM = 1,852 km ·300 NM = 555,6 km ·555,6 km 55.560.000 cm ·20 cm on the chart = 55.560.000 cm in reality • 1 cm on the chart =2.778.000 cm in reality

=

The scale of the chart at 60 0 N is therefore 1 : 2.778.000.

4607. Airplane ATPL CPL Heli ATPL CPL On a chart, 49 NM is represented by 7cm. What is the scale? A) B) C) D)

1: 700 000 1: 7 000 000 1: 1 300 000 1: 130 000

(Refer to figures 061-E84 and 061-E85) "Scale" is the relationship of the distance between two positions on a chart and the distance between the same positions on the Earth. • Scale = Chart length 7 Earth distance In our case, the distance on the chart is represented by 7 cm. This chart distance represents an actual distance of 49 NM. ·1 NM= 1,852km ·49 NM = 90,75 km ·70,75 km = 9.075.000 cm

·7 cm on the chart = 9.075.000 cm in reality • 1 cm on the chart = 1.296.429 cm in reality The scale is therefore 1: 1.296.429 => approximately 1: 1 300000.

I 4603 (A) 14604(0) 14605(0) I 4606 (C) I 4607 (C) I

03 Charts 4610. Airplane ATPL CPL, Heli ATPL CPL On a chart, the distance along a meridian between latitudes 45°N and 46°N is 6 cm. The scale ofthe chart is approximately: A) 1:1.000.000 B) 1:1.850.000 C) 1:185.000 D) 1:18.500.000 (Refer to figures 061-E84 and 061-E85) "Scale" is the relationship of the distance between two positions and the distance between the same positions on the Earth. • Scale = Chart length.;. Earth distance The distance along the surface for each 1°.

on a chart

The scale is therefore 1: 1852000.

4652. Airplane ATPL CPL Heli ATPL CPL At latitude 60 0 N the scale of a Mercator projection is 1:5.000.000. The length on the chart between point C (60 0 N 008°E) and point 0 (60 0 N 008°W) is: A) 19,2 em B) 16,2 em C) 3S,6em

17,8em

(Refer to figures 061-E88, 061-E89 and 061-E90) The scale on a Mercator chart is not constant because of the increasing distance between parallels of latitude when moving North or South from the Equator. Scale is correct at the Equator and expands as the secant (1leos) of the latitude. If a scale is given for a Mercator chart, it must always relate to a specific latitude - e.g. "scale at 35°S = 1:500.000'~ As scale is constantly changing, care must be taken when measuring distances. The best way is to use the mean latitude covering the route portion. In practice, the measurement of distance is not difficult and speedily improves with time and practice. In our case we have to first calculate the actual distance over the surface of the Earth between 008°E and 008°W at 60 0 N. The total angular difference is 16°. The corresponding distance = 16° x 60 x cos 60° = 480 NM. Now let's review the question. The scale is 1 : 5.000.000 => 1 cm on the chart equals to 5 000 000 cm in reality: • 1 NM = 1,852 km ·480 NM =888,96 km '888,96 km =88.896.000 cm • 1 cm on the chart = 5.000.000 cm in reality .? cm on the chart = 88.896.000 cm in reality • ? =88.896.000 .;. 5.000.000

• ?= 17,78

17,78 cm on the chart equals to 88.896.000 cm (480 NM) in reality.

4665. Airplane ATPL CPL Heli ATPL CPL Approximately how many nautical miles correspond to 12 cm on a map with a scale of 1 : 2.000.000? A) 130 B) 150 C) 329 D) 43 (Refer to figures 061-E84 and 061-E85) "Scale" is the relationship of the distance between two positions and the distance between the same positions on the Earth. • Scale = Chart length.;. Earth distance

on a chart

In our case, the distance on the chart is represented by 12 cm. The scale of the chart is 1: 2.000.000 (1 cm on the chart =2.000.000 cm in reality). • 1 cm on the chart = 2.000.000 cm in reality • 12 cm on the chart =24.000.000 cm in reality ·24.000.000 cm = 240 km

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4610 (8)

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4652 (D)

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The distance along the surface of the Earth will be approximately 130 NM.

4675. Airplane ATPL CPL Heli ATPL CPL On a Lambert conformal chart the distance between two parallels of latitude (difference of latitude = 2°), is measured to be 112 mm. The distance between two meridians, spaced 2° longitude, according to the chart is 70 NM. The parallel of origin (selected parallel) runs through the middle of the described square. What is the convergence for a d-Iong of 15° on this map?

of the Earth along a Meridian = 60 NM

In our case, the distance on the chart is represented by 6 cm. This chart distance represents 1° along the Meridian in reality = 60 NM. • 1 NM = 1,852 km ·60NM= 111,12km ·111,12 km = 11112000 cm ·6cmonthechart= 11112000cminreality • 1 cm on the chart = 1852000 cm in reality

D)

• 1 NM = 1,852 km • 240 km = 129,6NM

4665 (A)

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4675 (C)

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4698 (8)

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A) 9,23° B) 14,56° C) 12,18° D)

7,50°

We will first find the Latitude of the Parallel of Origin. The change in longitude = 2°. At the Equator this would equal to 120 NM (2 x 60 NM). Since the actual distance stated by the question is 70 NM it means that the Latitude is quite far up North. • Distance =Change in Longitude (in minutes) x cos Latitude. ·70 NM = 120 minutes x cos? .? = 70 NM.;. 120 minutes ·?=0,5833 • Latitude = inv.cos 0,5833 • Latitude =54,31° (54°19') Now that we know the Parallel of Origin we can calculate the convergence: • Chart convergency (CC) = Ch.Long (in degrees) x the sin Latitude • Chart convergency (CC) = 15° x sin 54,31° • Chart convergency (CC) = 12,18°

4698. Airplane ATPL CPL Heli ATPL CPL What is the approximate distance from A to B, given: Direct Mercator chart with a scale of 1:200.000 at the Equator. Chart length from A to B, in the vicinity of the Equator =11 cm. A) 21NM

B) 12 NM C) 22 NM D) 14NM (Refer to figures 061-E88, 061-E89 and 061-£90) The scale on a Mercator chart is not constant because of the increasing distance between parallels of latitude when moving North or South from the Equator. Scale is correct at the Equator and expands as the secant (1leos) of the latitude. If a scale is given for a Mercator chart, it must always relate to a specific latitude - e.g. "scale at 35°S = 1:500.000'~ As scale is constantly changing, care must be taken when measuring distances. The best way is to use the mean latitude covering the route portion. In practice, the measurement of distance is not difficult and speedily improves with time and practice. In our case, the distance on the chart is represented by 11 cm. Remember that the Scale = "Chart length.;. Earth distance'~ The scale of the chart along the Equator is 1: 200.000 (1 cmon the chart =200.000 cm in reality) . • 1 cm on the chart = 200.000 cm in reality • 11 cm on the chart = 2.200.000 cm in reality ·2.200.000 cm =22 km • 1 NM = 1,852 km ·22km= 11,88NM The distance along the surface of the Earth at the Equator will be approximately12NM.

4708. Airplane ATPL CPL Heli ATPL CPL On a direct Mercator chart at latitude 15°5, a certain length represents a distance of 120 NM on the Earth. The same length on the chart will represent on the Earth, at latitude lOON, a distance of: A) 122,3 NM B) 117,7 NM

C) 124,2NM D) 118,2 NM (Refer to figures 061-E88, 061-E89 and 061-E90)

4708 (A)

I

Aviationexam Test Prep Edition 2012 The scale on a Mercator chart is not constant because of the increasing distance between parallels of latitude when moving North or South from the Equator. Scale is correct at the Equator and expands as the secant (7leos) of the latitude. If a scale is given for a Mercator chart, it must always relate to a specific latitude - e.g. "scale at 35°S = 1:500 OOO'~ As scale is constantly changing, care must be taken when measuring distances. The best way is to use the mean latitude covering the route portion. In practice, the measurement of distance is not difficult and speedily improves with time and practice. Since we know the actual distance over the surface of the Earth and the Latitude, we can determine the distance in terms of "Change in Longitude": • Distance = Change in Longitude (in minutes) x cos Lat • 120NM=?xcos 15° .?= 120NM.;.cos 15° • ? = 124,23 minutes (2°4') Now we will use the same formula again, but we will reverse the process for the second Latitude: • Distance = Change in Longitude (in minutes) x cos Lat • Distance = 124,23 minutes x cos 10° • Distance = 122,3 NM

At latitude lOON the same distance on the chart as measured at 15°S would represent a ground distance of approximately 122,3 NM.

4723. Airplane ATPL CPL Heli ATPL CPL What is the chart distance between longitudes 179°E and 175°W on a direct Mercator chart with a scale of 1:5.000.000 at the equator? 133 mm B) 106mm C) 167mm D) 72mm A)

(Refer to figures 061-E88, 061-E89 and 061-E90) The scale on a Mercator chart is not constant because of the increasing distance between parallels of latitude when moving North or South from the Equator. Scale is correct at the Equator and expands as the secant (7leos) of the latitude. If a scale is given for a Mercator chart, it must always relate to a specific latitude - e.g. "scale at 35°S = 1:500.000'~ As scale is constantly changing, care must be taken when measuring distances. The best way is to use the mean latitude covering the route portion. In practice, the measurement of distance is not difficult and speedily improves with time and practice. In our case, the scale of the chart is 1 : 5.000.000 => 1 cm on the chart equals 5.000.000 cm in reality. This is valid only at the Equator. At a different Latitude the scale expands. First of all we need to find the actual distance between 179°E and 175°Walong the Equator. For each 1° the distance = 60 NM. For 6° (775°W - 179°E) the distance will therefore be 360 NM. • 1 NM = 1,852 km ·360 NM = 666,72 km ·666,72 km = 66.672.000 cm • 1 cm on the chart = 5.000.000 cm in reality .? cm on the chart = 66.672.000 cm in reaiity .? = 66 672 000.;. 5.000.000 .?= 13,33cm • 13,33 cm = 133,3 mm The distance from 179°E to 175°Won this chart will be represented by 13,33 cm which equals to 133,3 mm.

4784. Airplane ATPL CPL Heli ATPL CPL A Mercator chart has a scale at the equator = 1 : 3 704 000. What is the scale at latitude 60 0 S? A) 1: 1.852.000 B) 1: 7.408.000 C) 1: 3.208.000 D) 1 : 185.200

When the scale at one Latitude on a Mercator chart is known, the scale at any other Lat. may be calculated by a formula: • Scale Denominator ';.'\" x cosine "8" =Scale denominator "8" x cosine ';.'\" ... where '!'I" and "8" are the Latitudes. The formula is best remembered as '!'I88A" (A x 8 = 8 x A). We can apply the '!'I88A" formula directly to find the solution: • The latitude of the Equator = 0° • 3.704.000 x cos 60° ? x cos 0° .? = (3.704.000 x cos 60°}.;. cos 0° • ? = 1.852.000 The scale of the chart at 60 0S is 1: 1.852.000.

=

4788. Airplane ATPL CPL Heli ATPL CPL On a direct Mercator projection, the distance measured between two meridians spaced 5° apart at latitude 60 0 N is 8 cm. The scale of this chart at latitude 60 0 N is approximately: A) B) C) D)

1: 4.750.000 1: 7.000.000 1: 6.000.000 1 : 3.500.000

(Refer to figures 061-E84 and 061-E85) "Scale" is the relationship of the distance between two positions on and the distance between the same positions on the Earth. • Scale = Chart length.;. Earth distance

a

chart

The distance along the surface of the Earth: • along a Meridian = 60 NM for each 1°. • along the Equator = 60 NM for each 1°. • along a parallel = 60 x cos Latitude for each 1~ In our case, the distance on the chart is represented by 8 cm. This chart distance represents 5° along the parallel600N. In reality the distance equals: 5° x 60 x cos 60° = 150 NM. • 1 NM = 1,852 km • 150NM=277,8km ·277,8 km = 27.780.000 cm

·8 cm on the chart =27.780.000 cm in reality • 1 cm on the chart =3.472.500 cm in reality The scale is therefore 1: 3.472.500 => approximately 1: 3.500.000.

4791. Airplane ATPL Heli ATPL On a polar stereographic chart the scale at the pole is 1 : 5.000.000. Calculate the scale of the chart at 65°N: A) B) C) D)

1 : 4.213.000 1: 5.250.000 1 : 5.000.000 1 : 4.766.000

(Refer to figures 061-E91 and 061-E92) On a Polar Stereographic chart the scale is correct at the pole only, expanding away from the pole as sec2 !h(co-Iat}. Distances can be measured with reasonable accuracy, utilising the latitude scale on a convenient meridian in the are where plotting is being done. shapes and areas are approximately correct, with some slight distortion away from the pole. The method of solving scale problems is similar to that used for the Mercator projection - the '!'I88A" formula is used, with the substitution of cos2 !h(co-Iat). (co-Iat = 90° -latitude). • The pole is taken as latitude ';.'\'~ latitude 60 0N is "8'~ • Scale denominator ';.'\" x cos2 !h(co-Iat "8"} =scale denominator "8" x cos2 !h(co-Iat ';.'\'7 ·5.000.000 x cos2 !h(900- 65°} = ? x cos2 !h(90° - 90°} ·5.000.000 X cos2 !h(25°} = ? x cos2 !h(OO} ·5.000.000 x cos 2 12,5° = ? x cos2 0° • ? = (5 000 000 x cos2 12,5°}.;. cos2 0°

• ?=4765 769

(Refer to figures 061-E88, 061-E89 and 061-E90) The scale on a Mercator chart is not constant because of the increasing distance between parallels of latitude when moving North or South from the Equator. Scale is correct at the Equator and expands as the secant (7leos) of the latitude. If a scale is given for a Mercator chart, it must always relate to a specific latitude - e.g. "scale at 35°S = 1:500.000'~ As scale is constantly changing, care must be taken when measuring distances. The best way is to use the mean latitude covering the route portion. In practice, the measurement of distance is not difficult and speedily improves with time and practice.

Scale at 65°N is therefore 1: 4j65.769 => approximately 1 : 4.766.000. Note:Tofindcos 2 12,5°,findcos 12,5°andsquaretheresult(cos 12,5° x cos 12,5°).

I 4723(A) I 4784 (A) 14788(0) 14791 (0) I

EI ------------ --------------------------

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03 Charts

4794. Airplane ATPL CPL Heli ATPL CPL In a navigation chart a distance of 69 NM is equal to 17 cm. The scale of the chart is approximately: A) 1 :40.000 B) 1: 400.000 C) 1: 750.000 D) 1: 1.300.000 (Refer to figures 067-E84 and 067-E85) "Scale" is the relationship of the distance between two positions on a chart and the distance between the same positions on the Earth. • Scale = Chart length + Earth distance In our case, the distance on the chart is represented by 77 cm. This chart distance represents an actual distance of 69 NM. • 7 NM = 7,852 km ·69 NM = 727,79 km • 727,79 km = 72.779.000 cm • 77 cm on the chart = 72.779.000 cm in reality • 7 cm on the chart = 757.705 cm in reality The scale is therefore 7 : 751.705 => approximately 7: 750.000.

4802. Airplane ATPL CPL Heli ATPL CPL On a Lambert conformal chart the distance between two parallels of latitude having a difference of 2°, is measured to be 112 mm. The distance between two meridians, spaced 2° longitude, is according to the chart 70 NM. What is the scale of the chart, in the middle ofthe square described? A) B) C) D)

1:756.000 1:1.056.000 1:1.233.000 1:1.984.000

(Refer to figures 061-E86 and 061-E87) Scale on the Lambert's projection is variable. Where the cone cuts the standard parallels, scale is the same as the reduced Earth and is therefore correct along those parallels. The scale contracts between the standard parallels and expands outside them. Scale is the relationship of the distance between two positions on a chart and the distance between the same pOSitions on the Earth. Scale = "Chart length + Earth distance". Let's start by finding the actual distance along the meridians. 1° along a meridian = 60 NM, therefore the 2° distance between the parallels = 120 NM."\ On the chart this distance is represented by 112 mm (11,2 cm). • 1 NM = 7,852 km • 120NM=222,24km ·222,24 km = 22.240.000 cm • 11,2 cm on the chart =22.240.000 cm in reality • 1 cm on the chart = 1.984.285 cm in reality The scale is therefore going to be somewhere in the range of 1 : 1.984.000.

4920. Airplane ATPL CPL Heli ATPL CPL A Lambert conformal conic chart has a constant of the cone of 0,80. A straight line course drawn on this chart from A (S3°N 004°W) to B is 080° at A; course at B is 092°(T). What is the longitudeofB? A) OWE B) 009°36'E C) 008°E D) 019°E (Refer to figures 061-E86 and 061-E87) In this question the convergency = 092° - 080° = 12". Convergency is calculated by multiplying the change in longitude with the sine of mean latitude of the two points. The constant of the cone, or otherwise called the convergence factor is 0,8 here. Constant of the cone =sin mean latitude. • Convergency = Ch.Long x sine of the mean latitude • 72"=Ch.Long x 0,8 ·Ch.Long= 15° Point A lies at 53 ON 004°W and the heading is easterly, therefore the longitude ofpoint B is OWE.

4929. Airplane ATPL CPL Heli ATPL CPL What is the constant of the cone for a Lambert conic projection whose standard parallels are at SOON and 700N? A) 0,500 B) 0,941 C) 0,866 D) 0,766 (Refer to figures 061-E86 and 061-E87) Finding the constant of the cone is one of the easier tasks involved in this subject. It is simply a fancy way of asking for the "sine of the mean latitude': The parallel of origin, by definition, lies midway between the standard parallels of SOON and 700N. The parallel of origin is therefore 60"N = halfway between the standard parallels. The value of "n" (Constant of the Cone) is equal to the sin of 60 0N = 0,866. The angular extent of the chart would therefore be 312° (0,866 x 360°). The CCF would also be equal to 0,866.

4930. Airplane ATPL CPL Heli ATPL CPL The two standard parallels of a conical Lambert projection are at 10040'N and 41°20'N. The cone constant of this chart is approximatively: A) 0,18 B) 0,90 C) 0,66

D) 0,44 (Refer to figures 061-E86 and 067-E87) Finding the constant of the cone is one of the easier tasks involved in this subject. It is simply a fancy way of asking for the "sine of the mean latitude': The parallel of origin, by definition, lies midway between the standard parallels of 10040'N and 41°20'N. The parallel of origin is therefore 26°N = halfway between the stondwd pwollels ((10,67" + 41,33°) + 2}. The value of "n" (Constant of the Cone) is equal to the sin of 26°N = 0,44. The angular extent of the chart would therefore be 158° (0,44 x 360°). The CCF would also be equal to 0,44.

4933. Airplane ATPL CPL Heli ATPL CPL A Lambert's conical conformal chart has standard parallels at 63°N and 41°N. What is the constant ofthe cone? A) 0,891 B) 0,788 C) 0,656 D) 0,707 (Refer to figures 067-E86 and 061-E87) Finding the constant of the cone is one of the easier tasks involved in this subject. It is simply a fancy way of asking for the "sine of the mean latitude': The parallel of origin, by definition, lies midway between the standard parallels of 63 oN and 41°N. The parallel of origin is therefore 52°N = halfway between the standard parallels ((63° + 41°) + 2}. The value of "n" (Constant of the Cone) is equal to the sin of 52"N = 0,788. The angular extent of the chart would therefore be approx. 284° (0,788 x 360°). The CCF would also be equal to 0,788.

4937. Airplane ATPL CPL Heli ATPL CPL Calculate the constant of the cone on a Lambert Chart given chart convergency between 0100E and 0300W as being 30°: A) 0,40

B) 0,75 C) 0,50

D) 0,64 (Refer to figures 061-E86 and 067-E87) We will begin by stating the formula for convergence which is the change in longitude x sine mean latitude. Remember that the "sine of the mean latitude" is another way of saying "Constant of the Cone': In our case the convergence on the chart = 30", while the Change in Longitude is 40° (30 0W - 100E). Chart convergence = Change in Longitude x Constant of the Cone 30 0=40 0x? ?=300+40° ?=0,75 The Constant of the Cone is 0,75.

I 4794 (C) 14802(0) I 4920 (A) I 4929 (C) 14930(0) 14933(8) 14937(8) I

Aviationexam Test Prep Edition 2012 4939. Airplane ATPL Heli ATPL What is the value of the convergence factor on a Polar Stereographic chart?

A) 0,866 B) 0,5 C) 0,0 0) 1,0

230763. Airplane ATPL CPL Heli An aeronautical chart is conformal when:

A) B) C) 0)

For explanation refer to question #1266 on page 70. 4943. Airplane ATPL CPL Heli ATPL CPL The standard parallels of a Lambert's conical orthomorphic projection are 07°40'N and 38°20'N. The constant of the cone for this chart is: A) 0,60 B) 0,39

C) 0,92 0) 0,42 (Refer to figures 061-E86 and 061-E87) Finding the constant of the cone is one of the easier tasks involved in this subject. It is simply a fancy way of asking for the "sine of the mean latitude". The parallel of origin, by definition, lies midway between the standard parallels of 7"40'N and 38°20'N. The parallel of origin is therefore 23 oN = halfway between the standard parallels ((7,67" + 38,33°) 72). The value of "n" (Constant of the Cone) is equal to the sin of 23 oN = 0,39. The angular extent of the chart would therefore be 140° (0,39 x 360°). The CCF would also be equal to 0,39.

ATPL

CPL

the meridians and parallels are perpendicular to each other. every great circle is represented by a straight line in the map. the map is an equidistant normal projection. At any point the scale over a short distance in the direction of the parallel is equal to the scale in the direction of the meridian and the meridians are perpendicular to the parallels.

Airplane ATPL CPL Heli ATPL CPL Which statement is true about the parallel of origin of a conformal chart? 230771.

A) The parallel of origin is the parallel at which the scale reaches its minimum value. B) The parallel of origin is the parallel at which the scale reaches its maximum value. C) The parallel of origin is the only parallel at which the chart is conformal. 0) The parallel of origin together with the standard parallel(s), are the only parallels at which the chart is conformal.

03-02 The representation of meridians, parallels, great circles and rhumb lines

For explanation refer to question #1226 on page 70.

the simple conical projection in order to achieve orthomorphism (preserving shape locally). The original standard parallel is now called "the Parallel of Origin". A cone is placed tangentialto a chosen latitude so that it lies on the North/ South axis of the Reduced Earth, with the cone apex above the pole. The cone is considered as if it cuts the Earth at two standard parallels instead of one. This is achieved mathematically. The Latitude halfway between the two standard parallels is termed the Parallel of Origin. It can be compared to the parallel of origin (standard parallel) of the si1l1ple conic projection. The origin is the centre of the reduced Earth. • Meridians are shown as straight lines radiating from the pole. • Parallels of latitude are arcs of concentric circles convex to the Equator (concave to the pole), spaced almost equally. • Rhumb Lines are curved on a Lambert's chart on the equatorial side of the Great Circle (concave to the GC line and convex to the Equator). ·An exact Great Circle results when two positions on the Parallel of Origin are connected by a straight line. Elsewhere, a straight line is only an approximate great circle (it is a curve concave to the parallel of origin). As the difference is so small and of no importance for practical purposes, any straight line on the chart may be taken as a Great Circle. • Great Circles are easily measured at any meridian, Rhumb Line tracks are measured at mid-meridian. This is useful to remember when plotting radio bearings. It is therefore much easier to plot radio bearings on a Lambert's chart than a Mercator chart.

1262. Airplane ATPL CPL Heli ATPL CPL The parallels on a Lambert conformal conic chart are represented by:

1263. Airplane ATPL CPL Heli ATPL CPL On a Lambert conformal conic chart great circles that are not meridians are:

Airplane ATPL CPL Heli ATPL CPL Which one of the following, concerning great circles on a direct Mercator chart, is correct? 1175.

A) They are all curves convex to the equator. B) They are all curves concave to the equator. C) They approximate to straight lines between the standard parallels. 0) With the exception of meridians and the equator, they are curves concave to the equator. For explanation refer to question #4106 on page 72. 1261. Airplane ATPL CPL Heli ATPL CPL On a transverse Mercator chart, with the exception of the equator, parallels of latitude appear as:

A) hyperbolic lines. B) straight lines. C) ellipses. 0) parabolas.

A) B) C) 0)

parabolic lines. straight lines. arcs of concentric circles. hyperbolic lines.

(Refer to figures 061-E86 and 061-E87)

--'

A) curves concave to the parallel of origin. B) straight lines. C) . curves concave to the pole of projection. 0) straight lines within the standard parallels. For explanation refer to question #1262 on this page.

A Lambert Conformal Conic chart is obtained by mathematically modifying

1 4939 (0) 1 4943 (8) 1230763 (D) 1230771 (A) 1 1175 (D)

1 1261 (e) 1 1262 (C)

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1 1263 (A)

1

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03 Charts at any convenient meridian. The Equator is a circle, furthest from the pole. • Convergency is constant and correct at the pole, incorrect elsewhere, with a value of I. Chart convergency is equal to change of/ongitude.

1264. Airplane ATPL CPL Heli ATPL CPL On a direct Mercator chart, great circles are shown as: A) curves convex to the nearer pole. B) straight lines. C) rhumb lines. D) curves concave to the nearer pole.

2303. Airplane ATPL CPL Heli ATPL CPL A straight line on a Lambert conformal projection chart for normal flight planning purposes:

For explanation refer to question #4106 on page 72.

A) B) C) D)

1276. Airplane ATPL CPL Heli ATPL CPL On a direct Mercator chart, a Rhumb Line appears as a: A) B) C) D)

straight line. small circle concave to the nearer pole. spiral curve. curve convex to the nearer pole.

For explanation refer to question #1262 on page 80.

4047. Airplane ATPL CPL Heli ATPL CPL On a Lambert chart (standard parallels 37°N and 65°N), with respect to the straight line drawn on the map the between A (49°N 030 0 W) and B (48°N 040 0 W), the:

For explanation refer to question #4106 on page 72.

2194. Airplane ATPL CPL Heli On a direct Mercator chart, meridians are:

ATPL

CPL

A) inclined, equally spaced, straight lines that meet at the nearer pole. B) parallel, equally spaced, vertical straight lines. C) parallel, unequally spaced, vertical straight lines. D) inclined, unequally spaced, curved lines that meet at the nearer pole. For explanation refer to question #4106 on page 72.

2253. Airplane ATPL Heli ATPL Which map projection Is described as follows:

Equatorial Mercator projection. Lambert conformal projection. Lambert conformal or a polar stereographic projection. Polar stereographic projection.

(Refer to figures 061-E86, 067-E87, 067-E91 and 061-E92) Properties of the Lambert Conformal Conic projection: • Meridians are shown as straight lines radiating from the pole. • Parallels of latitude are arcs of concentric circles convex to the Equator (concave to the pole), spaced almost equally. • Rhumb Lines are curved on a Lambert's chart on the equatorial side of the Great Circle (concave to the GC line and convex to the Equator). • An exact Great Circle results when two positions on the Parallel of Origin are connected bya straight line. Elsewhere, a straight line is only an approximate great circle (it is a curve concave to the parallel of origin). As the difference is so small and ofno importance for practical purposes, any straight line on the chart may be taken as a Great Circle. • Great Circles are easily measured at any meridian, Rhumb Line tracks are measured at mid-meridian This is useful to remember when plotting radio bearings. It is therefore much easier to plot radio bearings on a Lambert'schart than a Mercator chart. Properties of the Polar.Stereographic projection: • Meridians are equally spaced straight lines radiating from the selected pole which is at the centre of the projection. • The angles shown are the true angles between meridians. • Parallels of latitude are a series of unequal concentric circles centred on the selected pole. The spacing between parallels.increases with distance from the pole. The radius of the parallels is 2R tan 112co-lat. • Only one pole can be shown on the chart. • Meridians and parallels intersect at 90". • Except for meridians, Rhumb Lines are curves concave to the nearer pole which are difficult to measure accurately. • For practical purposes, all Great Circles within the area covered are considered straight lines. The chart is constructed so that Great Circles are approximately straight lines, but in fact they are slightly concave to the pole. The closer a Great Circle is to the centre of the projection (the pole), the more it will appear as a straight line. This is completely true only for the lines crossing the central meridian at 90°. Great Circles can be measured

1264 (A)

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1276 (A)

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2194 (8)

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2253 (C)

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2303 (0)

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A) B) C) D)

great circle is to the North, the rhumb line is to the South. great circle and rhumb line are to the North. great circle and rhumb line are to the South. rhumb line is to the North, the great circle is to the South.

(Refer to figures 067-E25, 061-E86 and 061-E87) On Lambert's projection chart the Rhumb Lines (RLs) are curved on the equatorial side of the great circle. An exact Great Circle (GC) results when two positions on the Parallel of Origin are connected by a straight line. Elsewhere, a straight line is only an approximate GC - the actual GC is a curve concave to the parallel of origin. However, in practice, as the difference is so small and of no importance for practical purposes, any straight line on the chart may be taken as a great circle. The Parallel of Origin is situated midway between the standard parallels => in our case between 3rN and 65°N = 51°N. Since the straight line quoted by the question is located a little bit South from the Parallel ofOrigin, the GC will be represented as a slightly curved line South of the straight line and the RL will be another curved line even further South of the straight line (on the southern side of the GC curve).

- Meridians are straight lines. - The scale vary with latitude. - Most rhumb lines are curved lines. A) B) C) D)

can only be a parallel of latitude. is a loxodromic line. is a rhumb line. is approximately a great circle.

4092. Airplane ATPL Heli ATPL Which one of the following describes the appearance of rhumb lines, except meridians, on a polar stereographic chart? A) B) C) D)

Straight lines. Ellipses around the pole. Curves convex to the pole. Curves concave to the pole.

For explanation refer to question #1266 on page 70.

4124. Airplane ATPL CPL Heli On a direct Mercator, Rhumb Lines are: A) B) C) D)

ATPL

CPL

straight lines. curves concave to the equator. ellipses. curves convex to the equator.

For explanation refer to question #4106 on page 72.

4125. Airplane ATPL CPL Heli ATPL CPL On a direct Mercator chart a great circle will be represented bya: A) B) C) D)

complex curve. curve concave to the equator. curve convex to the equator. straight line.

For explanation refer to question #4106 on page 72.

4047 (C)

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4092 (0)

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4124 (A)

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4125 (8)

I

Ell

Aviationexam Test Prep Edition 2012

4216. Airplane ATPL CPL Heli ATPL CPL Parallels of latitude on a direct Mercator chart are: A) parallel straight lines equally spaced. B) arcs of concentric circles equally spaced. e) straight lines converging above the pole. D) parallel straight lines unequally spaced. For explanation refer to question #4106 on page 72.

4217. Airplane ATPL Heli ATPL Which one of the following statements is correct concerning the appearance of great circles, with the exception of meridians, on a polar stereographic chart whose tangency is at the pole? A) The higher the latitude the closer they approximate to a straight line. B) Any straight line is a great circle. e) They are complex curves that can be convex and/or concave to the pole. D) They are curves convex to the pole. For explanation refer to question #1266 on page 70.

230764. Airplane ATPL CPL Heli ATPL CPL On a Mercator's projection the distance between (17°N, 035°E) and (17°N, 0400E) is 5 cm. The scale at 57°N is approximately: A) 1 :10 626 460 B) 1 :18658470 e) 1: 6 052 030 D) 1: 5 556 OOOA 230765. Airplane ATPL CPL Heli ATPL CPL From Rakovnik (50°05,9' N, 013°41,5' E) to Frankfurt FFM (50°05,9' N, 008°38,3' E) the True Track of departure along the straight line is 272,0°. The constant of the cone ofthis Lambert conformal projection is: A) 0,77 B) 0,79

C) 0,20

D) OAO 230766. Airplane ATPL CPL Heli ATPL CPL The positions A (30 000'N, 017°30'E) and B at longitude (30 000'N, 023°30'E) are plotted on a Lambert chart with a constant of the cone of 0,5. A and B are connected by a straight line. The True Track measured at A is 088,5°. What is the True Track measured at B? A) 085S

at 50 0S and 56°S. When passing 175°W, the True Track is: A) 078,0 0 B) 102,00 C) 282,0 0 D) 258,0 0 230770. Airplane ATPL CPL Heli ATPL CPL Which statement is correct about the scale of a Lambert projection? A) The scale reaches its minimum value at the standard parallels. B) The scale reaches its maximum value at the parallel of origin. e) The scale reaches its maximum value at the standard parallels. D) The scale reaches its minimum value at the parallel of origin. 230788. Airplane ATPL CPL Heli ATPL CPL The constant of the cone in a Lambert chart is 0,8666500. The angle between the north directions of the meridian in position A (65°00'N, 018°00'W) and the meridian of position B (75°00'N, 023°00'W) on the chart is: A) 5,0 0 B) 4,3 0 e) 10,00 D) 5,8 0

230789. Given:

Airplane

ATPL

CPL

Heli

ATPL

CPL

Lambert conformal conical projection, scale 1: 1 234000. Standard parallels 36°N and 60 0N. A (53°N, 01 OOW), B (53°N, 020 0W). The distance on the map between position A and position B measured along the rhumb line: A) is more than 57,13 cm. B) is 55,66 cm. e) is between 54,19 cm and 57,13 cm. D) is less than 54,19 cm. 230790. Airplane ATPL Heli ATPL A straight line from A (75°S, 120° E) to B (75°S, 1600E) is drawn on a Polar Stereographic chart. When passing the meridian 155°E, the True Track is: A) 1050 B) 095 0 e) 075 0 D) 255 0

230791. Airplane ATPL Heli ATPL A straight line from A (75°N, 1200E) to B (75°N, 1600E) is drawn on a Polar Stereographic chart. When passing the meridian 155°E, the True Track is:

B) 094S

e) 091S

D) 082,5° 230767. Airplane ATPL CPL Heli ATPL CPL A straight line from A (53°N, 155°W) to B (53°N, 1700E) is drawn on a Lambert Conformal conical chart with standard parallels at SooN and 56°N. When passing the meridian 175°E, the True Track is:

A) 255 0 B) 075 0 e) 1050 D) 285 0

230792. Airplane ATPL Heli ATPL Which statement is correct about the scale of a Polar Stereographic projection of the Northern polar area?

102S B) 100,0° C) 257S D) 260,0 A)

0

230768. Airplane ATPL CPL Heli ATPL CPL A straight line from A (53°S, 155°E) to B (53°S, 1700W) is drawn on a Lambert Conformal conical chart with standard parallels

A) The scale reaches its minimum value at the North pole. B) The scale reaches its minimum value at the equator. e) The scale reaches its maximum value at the North pole. D) The scale reaches its maximum value at the 45°N.

1 4216 (0) 1 4217 (A) 1230764 (C) 1230765 (8) 1230766 (C) 1230767 (0) 1230768 (A) 1230770 (0) 1230788 (8) 1230789 (0) 1 1230790 (C) 1230791 (C) 1230792 (A) 1 .,

03 Charts 230802. Airplane ATPL CPL Heli ATPL CPL On a Mercator's projection a straight line is drawn between (40 0 N, 050 0 W) and (SOON, 060 0 W). Calculate the angle between the straight line and the great circle in position A. A) 3,so B) 3,2°

230806. Airplane ATPL Heli ATPL Two places are situated on the same parallel in the Southern Hemisphere. The great circle, rhumb line and the straight line between these places are drawn on a Polar Stereographic Projection. Which statement is correct? A) The great circle is situated between the parallel and the straight line, because the concave side of the great circle is always pointed towards the equator. B) The rhumb line is situated between the great circle and the straight line because the shortest distance between to places on Earth is the great circle. C) The correct sequence from North to South is: Great circle, straight line, rhumb line. D) The great circle is situated between the parallel and the straight line, because the concave side of the great circle is always pointed towards the pole.

C) 1,80 D) 7,0°

03-03 The use of current aeronautical charts 1115. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-17) What feature is shown on the chart at position 52°12'N 006°12'W? A) B) C) D)

Tuskar Rock LT.H. NDB. WTD NDB. Kerry/Farranfore aerodrome. Clonbullogue aerodrome.

(Refer to figures 061-£69, 061-£70, 067-El1, 067-E72, 067-E73, 067-E74 and 067-E75) The key is to understand that each mark (tick) of Longitude and Latitude on the chart is equal to 70 minutes and there are 60 minutes in one degree. Refer to the chart and using the Lat/Long co-ordinates locate the position referenced by the question => identify the symbol and use your chart legend knowledge to determine it is an NOB station 'Tuskar Rock LT.H." with a frequencyof286 kHz and an identifier 'TW.lfyou are not sure about the meaning of chart symbols, refer to the attached figure (symbol #700) or Appendix 2 to ICAD Annex 4.

1116. Airplane ATPL CPL Heli ATPL CPL Contour lines on aeronautical maps and charts connect points: A) B) C) D)

of equal latitude. with the same variation. having the same longitude. having the same elevation above sea level.

(Refer to figure 067-E56) Contour lines are drawn connecting places at the same elevation above mean sea level (MSL). They can indicate gradient as well as height. The closer the lines are together on the chart, the steeper the terrain. Contour lines widely apart from each other indicate a shallow sloping terrain.

1139. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-17) What feature is shown on the chart at position 52°11'N 009°31'W? A) KERRY/Farranfore aerodrome. B) Waterford NDB. C) Connemara aerodrome. D) Punchestown aerodrome. (Refer to figures 067-E69, 067-E70, 067-El1, 067-E72, 067-E73, 061-£74 and 067-E75) The key is to understand that each mark (tick) of Longitude and Latitude on the chart is equal to 70 minutes and there are 60 minutes in one degree. Refer to the chart and using the Lat/Long co-ordinates locate the pOSition referenced

1230802 (A) 1230806 (0) 1 1115 (A)

1 1116 (0)

by the question => identify the symbol and use your chart legend knowledge to determine it is an aerodrome "KERRY / Farranfore". The adjacent symbols are an NDB station (Kerry NOB, frequency 334 kHz, identifier "KER") and an ILS to Kerry aerodrome. If you are not sure about the meaning of chart symbols, refer to the attached figure (symbol #93) or Appendix 2 to ICAD Annex 4.

1143. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-17) Which of the following lists all the aeronautical chart symbols shown at position 54°16,7'N 008°36,0'W? A) B) C) D)

VOR; DME; NDB; non-compulsory reporting point. VOR; DME; NDB; compulsory reporting point. Civil airport; VOR; DME; non-compulsory reporting point. Civil airport; NDB; DME; compulsory reporting point.

(Refer to figures 067-E69, 067-E70, 067-E77, 067-E72, 067-£73, 067-E74 and 067E75) The key is to understand that each mark (tick) of Longitude and Latitude on the chartis equal to 70 minutes and there are 60 minutes in one degree. Refer to the chart and using the Lat/Long co-ordinates locate the position referenced by the question => identify the symbols at this position (actually there are multiple symbols): • Sligo aerodrome with elevation of20 ft; • Sligo NOB (frequency 384 kHz, identifier "SLG"); • Sligo OME (frequency 709,0 MHz, identifier "SLG"); • Compulsory reporting point (solid triangle) - well "camouflaged" and blending-in together with the NDB symbol. If you are not sure about the meaning of chart symbols, refer to the attached figure or Appendix 2 to ICAD Annex 4. Note 7: It is very easy to distinguish military from civil aerodromes on the Jeppesen charts. Military aerodrome are circles only while the civil aerodromes are circles with slight protrusions on the circumference at 45° intervals. Note 2: It is easy to distinguish a compulsory reporting point from a non-compulsoryone => the compulsory reporting point is a solid triangle while the noncompulsory one is depicted as an outline of a triangle only.

1144. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-17) Which of the following lists all the aeronautical chart symbols shown at position 52°11'N 007°05'W? A) B) C) D)

NDB: Locator. VOR, NDB. Civil airport: Locator. Civil airport: NDB, Localiser.

(Refer to figures 067-E69, 067-E70, 067-El1, 067-E72, 067-E73, 067-E74

1 1139 (A) 1 1143 (0) 1 1144 (0) 1

Aviationexam Test Prep Edition 2012 and 061-E75) The key is to understand that each mark (tick) of Longitude and Latitude on the chart is equal to 10 minutes and there are 60 minutes in one degree. Refer to the chart and using the Lat/Long co-ordinates locate the position referenced by the question => identify the symbols at this position (actually there are multiple symbols): • Waterford aerodrome with elevation of 119 ft • Waterford NOB (frequency 368 kHz, identifier "WTD") ·ILS to Waterford aerodrome If you are not sure about the meaning of chart symbols, refer to the attached figure (symbols #93, #100 and #108) or Appendix 2 to ICAO Annex 4. Note: It is very easy to distinguish military from civil aerodromes on the Jeppesen charts. Military aerodrome are circles only while the civil aerodromes are circles with slight protrusions on the circumference at 45° intervals.

1145. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-12) What is the meaning of aeronautical chart symbol number 15? A) B) C) D)

Aeronautical ground light. Visual reference point. Hazard to aerial navigation. Lighthouse

(Refer to figures 061-E69, 061-E70, 061-E71, 061-E72, 061-E73, 061-E74, 061-E75 and 061-E76)

1265. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-16) Which of the following beacons is 185 NM from AKRABERG (61°24'N 006°40'W)? A) B) C) D)

(Refer to figure 061-E07) Remember that 1 minute of Latitude = 1 NM. Therefore a distance of 185 NM = 185 minutes = 3°05: Measure the length that is represented by 3°05' along any Meridian on the chart. Now use the drawing-compass and draw a circle with a center at AKRABERG and with a radius equal to the distance you have measured on the chart - the equivalent of 3°05' along a Meridian. This circle represents a distance of 185 NM from Akraberg. Notice that the closest symbol to this circle is KIRKWALL, although Stornoway and Sumburg are also pretty close. However, Kirkwall seems to be almost on the circle. The truth is that this chart is terrible for this kind ofprecise measurement.

Heli

ATPL

CPL

SHA VOR 52°43,3'N 008°53,1'W CRK VOR 51°50,4'N 008°29,7'W Aircraft position 52°30'N 008°20'W Which of the following lists two radials that are applicable to the aircraft position? A) B) C) D)

SHA 131° CRK 017°. SHA 304° CRK 189°. SHA 312° CRK 197°. SHA 124° CRK 009°.

Heli

ATPL

CPL

SHA VOR 52°43,3'N 008°53,1'W CRK VOR 51°50,4'N 008°29,7'W Aircraft position 52°20'N 009°10'W Which of the following lists two radials that are applicable to the aircraft position? A) B) C) D)

SHA 025° CRK 141°. SHA 212° CRK 328°. SHA 205° CRK 321°. SHA 033° CRK 149°.

Use the Lat/Long information stoted by the question to locate the position of the aircraft on the chart and mark it with a pen. Then locate the SHA and CRK VORs on the chart. Draw a line from each of the two VOR symbols to the mark representing your position. The easiest way to find the radials from the VORs to your position is to measure the angle between an airway (= known radial) and your line using a protractor. In our case it is an airway G4 => the angle between G4 airway and the line originating from CRK VOR is approx. 025°, therefore the radial to our position is going to be 353° (radial of the airway) minus 025° = 328~ The angle between G4 airway and the line originating from SHA VOR is approx. 039", therefore the radial to our position is going to be 039° + 173° (radial of the airway) =212°. There is no need for any variation adjustments - remember that VOR radials are magnetic bearings.

1293. Airplane ATPL CPL (Refer to figure 061-17) Given:

Heli

ATPL

CPL

SHA VOR 52°43,3'N 008°53,1'W CON VOR 53°54,8'N 008°49,1'W Aircraft position 53°20'N 009°50'W

KIRKWALL (58°58'N 002°54'W). STORNOWAY (58°15'N 006°17'W). SUMBURGH (59°55'N 001°15'W). SAXAVORD (60 050'N 000050'W).

1291. Airplane ATPL CPL (Refer to figure 061-17) Given:

1292. Airplane ATPL CPL (Refer to figure 061-17) Given:

Which of the following lists two radials that are applicable to the aircraft position? A) B) C) D)

SHA 325° CON 235°. SHA 137° CON 046°. SHA 317° CON 226°. SHA 145° CON 055°.

Use the Lat/Long information stated by the question to locate the position of the aircraft on the chart and mark it with a pen. Then locate theSHAandCON VORs on the chart. Draw a line from each of the two VOR symbols to the mark representing your position. The easiest way to find the radials from the VORs to your position is to measure the angle between an airway (= known radial) and your line using a protractor. In our case it is an airway Bl for CON VOR => the angle between Bl airway and the line originating from CON VOR is approx. 051°, therefore the radial to our position is going to be 285° (radial of the airway) minus 051° =234°. For SHA VOR we will use the airway W13 => the angle between W13 airway and the line originating from SHA VOR is approx. 086°, therefore the radial to our position is going to be 052° (radial of the airway) minus 086° = 326°. There is no need for any variation adjustments - remember that VOR radials are magnetic bearings.

1298. Airplane ATPL CPL (Refer to figure 061-17) Given: SHA VOR 52°43,3'N 008°53,1'W CRK VOR 51°50,4'N 008°29,7'W Aircraft position 52°30'N 009°30'W

Heli

ATPL

CPL

Use the Lat/Long information stated by the question to locate the position of the aircraft on the chart and mark it with a pen. Then locate the SHA and CRK Which of the following lists two radials that are applicable VORs on the chart. Draw a line from each of the two VOR symbols to the mark to the aircraft position? representing your position. The easiest way to find the radials from the VORs A) SHA 068° CRK 145°. to your position is to measure the angle between an airway'(= known radial) and your line using a protractor. In our case it is an airwayGl for the radial from B) SHA 060° CRK 138°. SHA => the angle between Gl airway and the line originating from SHA VOR C) SHA 240° CRK 13]0. is approx. 016", therefore the radial to our position is going to be 115° (radial D) SHA 248° CRK 325°. of the airway) plus 016° = 131~ For CRK VOR the relevant airway could be for example W10 - the angle between Wl0 airway and the line originating from . Use the Lat/Long information stated by the question to locate the position of the aircraft on the chart and mark it with a pen. Then locate the SHA and CRK CRK VOR is approx. 036", therefore the radial to our position is going to be 053° VORs on the chart. Draw a line from each of the two VOR symbols to the mark (radial of the airway) minus 036° = 01r. There is no need for any variation adrepresenting your position. The easiest way to find the radials from the VORs justments - remember that VOR radials are magnetic bearings.

I 1145 (A) I 1265 (A) I 1291 (A) I 1292 (8) I 1293 (A) I 1298 (D) I

03 Charts

to your position is to measure the angle between an airway (= known radial) and your line using a protractor. In our case it is an airway G4 => the angle between G4 airway and the line originating from CRK VOR is approx. 028~ therefore the radial to our position is going to be 353° (radial of the airway) minus 028° = 325°. The angle between G4 airway and the line originating from SHA VOR is approx. 075~ therefore the radial to our position is going to be 075° + 173° (radial of the airway) = 248". There is no need for any variation adjustments - remember that VOR radials are magnetic bearings.

1301. Airplane ATPL CPL (Refer to figure 061-17) Given:

Heli

ATPL

A) B) C) D)

Which of the following lists two radials that are applicable to the aircraft position? SHA 213° CON 310°. SHA 042° CON 138°. SHA 033° CON 130°. SHA 221° CON 318°.

Use the Lat/Long information stated by the question to locate the position of the aircraft on the chart and mark it with a pen. Then locate the SHA and CON VORs on the chart. Draw a line from each of the two VOR symbols to the mark representing your position. Now use your protractor to measure the angles between these lines and the VOR's Magnetic Noth reference line (the Mag.North reference is indicated for each VOR and NOB on the chart by a thin line protruding from the navaid symbol upwards and this line looks like a flag). In the case of CON VOR you will measure an angle of approx. 138° while for the SHA VOR you will measure an angle of approx. 042°. These are your magnetic bearings from the respective VORs to the aircraft pOSition => magnetic bearings are the radials. Another (easier and probably more precise) method would be to measure the angle between another VOR radial and the line from the VOR to our position.ln case of CON VOR we can use the line representing the airway Bl (radial 113°) and measure the angle between this airway and out line => result will be approx. 25° => we simply add the 25° to the 113° to get the result of 138°. Same procedure can be applied for SHA VOR and airway W13 => in this case the angle will beapprox. -10°=> 052°-10°= 042°.

2210. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-12) What is the meaning of aeronautical chart symbol number 16? A) B) C) D)

Shipwreck showing above the surface at low tide. Off-shore lighthouse. Lightship. Off-shore helicopter landing platform.

(Refer to figures 061-E69, 061-E70, 061-E71, 061-E72, 061-E73, 061-E74, 061-E75 and 061-E76)

4099. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-17) Which of the following lists all the aeronautical chart symbols shown at position 52°43,3'N 008°53,1'W? A) B) C) D)

4126. Airplane ATPL CPL (Refer to figure 061-17) The airport at 52°11'N 009°32'W is:

CPL

SHA VOR 52°43,3'N 008°53,1'W CON VOR 53°54,8'N 008°49,1'W Aircraft position 53°30'N 008°00'W

A) B) C) D)

Note: It is easy to distinguish a compulsory reporting point from a non-compulsory one => the compulsory reporting point is a solid triangle while the noncompulsory one is depicted as an outline of a triangle only.

VOR; DME; non-compulsory reporting point. DME; non-compulsory reporting point. VOR; DME; compulsory reporting point. VOR; compulsory reporting point.

(Refer to figures 061-E69, 061-E70, 061-E71, 061-E72, 061-E73, 061-E74 and 061-E75) The key is to understand that each mark (tick) of Longitude and Latitude on the chart is equal to 10 minutes and there are 60 minutes in one degree. Refer to the chart and using the Lat/Long co-ordinates locate the position referenced by the question => identify the symbols at this position (actually there are multiple symbols): • Shannon VOR / OME (frequency 113,3 MHz, identifier "SHA"); • Compulsory reporting point (open triangle).

1301 (8)

I

2210 (C)

I

4099 (C)

I

4126 (A)

I

4137 (0)

I

ATPL

CPL

Kerry. Cork. Shannon. Waterford.

(Refer to figures 061-E69, 061-E70, 061-E71, 061-E72, 061-E73, 061-E74 and 061-E75) The key is to understand that each mark (tick) of Longitude and Latitude on the chart is equal to 10 minutes and there are 60 minutes in one degree. Refer to the chart and using the Lat/Long co-ordinates locate the position referenced by the question => identify the symbol and use your chart legend knowledge to determine it is an aerodrome "KERRY / Farranfore" with an elevation of 113 ft. The adjacent symbols are an NOB station (Kerry NOB, frequency 334 kHz, identifier "KER") and an ILS to Kerry aerodrome. Ifyou are not sure about the meaning of chart symbols, refer to the attached figure or Appendix 2 to ICAO Annex 4.

4137. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-17) What feature is shown on the chart at position 53°11'N 006°37'W? A) Clonbullogue aerodrome. B) Connemara aerodrome. C) Kerry/Farranfore aerodrome. D) Pllnr.hestown aerodrome. (Refer to figures 061-E69, 061-E70, 061-E71, 061-E72, 061-E73, 061-E74 and 061-E75) The key is to understand that each mark (tick) of Longitude and Latitude on the chart is equal to 10 minutes and there are 60 minutes in one degree. Refer to the chart and using the Lat/Long co-ordinates locate the position referenced by the question => identify the symbol and use your chart legend knowledge to determine it is an aerodrome "Punchestown" with an elevation of 420 ft. If you are not sure about the meaning of chart symbols, refer to the attached figure or Appendix 2 to ICAO Annex 4.

4154. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-17) At position 52°11'N 009°31'W, which of the following denotes all the symbols? A) B) C) D)

Military airport; ILS; NDB. Civil airport; VOR; ILS. Military airport; VOR; ILS. Civil airport; ILS; NDB.

(Refer to figures 061-E69, 061-E70, 061-E71, 061-E72, 061-E73, 061-E74 and 061-E75) The key is to understand that each mark (tick) of Longitude and Latitude on the chart is equal to 10 minutes and there are 60 minutes in one degree. Refer to the chart and using the Lat/Long co-ordinates locate the position referenced by the question => identify the symbol and use your chart legend knowledge to determine it is an aerodrome "KERRY / Farranfore" with an elevation of 113 ft. The adjacent symbols are an NOB station (Kerry NOB, frequency 334 kHz, identifier "KER") and an ILS to Kerry aerodrome. Ifyou are not sure about the meaning of chart symbols, refer to the attached figure or Appendix 2 to ICAO Annex 4. Note: It is very easy to distinguish military from civil aerodromes on the Jeppesen charts. Military aerodrome are circles only while the civil aerodromes are circles with slight protrusions on the circumference at 45° intervals.

If you are not sure about the meaning of chart symbols, refer to the attached figure or Appendix 2 to ICAO Annex 4.

I

Heli

4154 (D)

I

Aviationexam Test Prep Edition 2012 4157. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-17) Which of the following lists all the aeronautical chart symbols shown at position 51°50,4'N 008°29,7'W? A) B) C) D)

Civil airport; VOR; non-compulsory reporting point. Civil airport; VOR; DME; compulsory reporting point. VOR; DME; NDB; compulsory reporting point. VOR; DME; NDB; ILS.

If you are not sure about the meaning of chart symbols, refer to the attached figure or Appendix 2 to ICAD Annex 4. Note 7: It is very easy to distinguish military from civil aerodromes on the Jeppesen charts. Military aerodrome are circles only while the civil aerodromes are circles with slight protrusions on the circumference at 45° intervals. Note 2: It is easy to distinguish a compulsory reporting point from a non-compulsory one => the compulsory reporting point is a solid triangle while the noncompulsory one is depicted as an outline of a triangle only.

(Refer to figures 067-E69, 067-E70, 067-E77, 067-E72, 067-E73, 067-E74 and 067 -E75) The key is to understand that each mark (tick) of Longitude and Latitude on the chart is equal to 70 minutes and there are 60 minutes in one degree. Refer to the chart and using the Lat/Long co-ordinates locate the position referenced by the question => identify the symbols at this position (actually there are multiple symbols): • Cork aerodrome with elevation of502 ft; • Cork VDR /OME (frequency 774,6 MHz, identifier UCRK"); • Compulsory reporting point (solid triangle).

4174. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-17) What feature is shown on the chart at position 53°51'N 009°17'W? A) B) C) D)

Connaught aerodrome. Castlebar aerodrome. Connemara aerodrome. Brittas Bay aerodrome.

Note 2: It is easy to distinguish a compulsory reporting point from a non-compulsory one => the compulsory reporting point is a solid triangle while the noncompulsory one is depicted as an outline of a triangle only.

(Refer to figures 067-E69, 067-E70, 067-E77, 067-El2, 067-E73, 067-E74 and 067-E75) The key is to understand that each mark (tick) of Longitude and Latitude on the chart is equal to 70 minutes and there are 60 minutes in one degree. Refer to the chart and using the Lat/Long co-ordinates locate the position referenced by the question => identify the symbol and use your chart legend knowledge to determine it is an aerodrome UCastlebar" with an elevation of 750 ft. If you are not sure about the meaning of chart symbols, refer to the attached figure or Appendix 2 to ICAD Annex 4.

4158. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-17) Which of the following lists all the aeronautical chart symbols shown at position 53°18,0'N 006°26,9'W?

4185. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-17) What feature is shown on the chart at position 54°17'N 010 0 05'W?

If you are not sure about the meaning of chart symbols, refer to the attached figure or Appendix 2 to ICAD Annex 4. Note 7: It is very easy to distinguish military from civil aerodromes on the Jeppesen charts. Military aerodrome are circles only while the civil aerodromes are circles with slight protrusions on the circumference at 45° intervals.

A) B) C) D)

A) B) C) D)

VOR; DME; danger area. Civil airport; VOR; DME. Military airport; VOR; NDB. Military airport; VOR; DME.

(Refer to figures 067-E69, 067-E70, 067-E77, 067-E72, 067-E73, 067-E74 and 067-E75) The key is to understand that each mark (tick) of Longitude and Latitude on the chart is equal to 70 minutes and there are 60 minutes in one degree. Refer to the chart and using the Lat/Long co-ordinates locate the position referenced by the question => identify the symbols at this position (actually there are multiple symbols): • Baldonnel Casement AB (Airforce Base) with elevation of379 ft; • Baldonnel VDR /OME (frequency 775,8 MHz, identifier "BLN'J; If you are not sure about the meaning of chart symbols, refer to the attached figure or Appendix 2 to ICAD Annex 4. Note: It is very easy to distinguish military from civil aerodromes on the Jeppesen charts. Military aerodrome are circles only while the civil aerodromes are circles with slight protrusions on the circumference at 45° intervals.

4161. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-17) What are the symbols at Galway Carnmore (53°18,1'N 008°56,5'W)? A) B) C) D)

(Refer to figures 067-E69, 067-E70, 067-E77, 067-El2, 067-E73, 067-E74 and 067-E75) The key is to understand that each mark (tick) of Longitude and Latitude on the chart is equal to 70 minutes and there are 60 minutes in one degree. Refer to the chart and using the Lat/Long co-ordinates locate the position referenced by the question => identify the symbols at this position (actually there are multiple symbols): • GALWAY Cam more aerodrome with elevation of87 ft; • Cam more NOB (frequency 327 kHz, identifier UCRN'J; • Cammore OME (frequency 770,0 MHz, identifier UCRN'J; • Non-compulsory reporting point (open triangle).

4157 (8)

I

4158 (D)

I

4161 (8)

I

4174 (8)

(Refer to figures 067-E69, 067-E70, 067-E77, 067-E72, 067-E73, 067-E74 and 067-E75) The key is to understand that each mark (tick) of Longitude and Latitude on the chart is equal to 70 minutes and there are 60 minutes in one degree. Refer to the chart and using the Lat/Long co-ordinates locate the position referenced by the question => identify the symbol and use your chart legend knowledge to determine it is an NOB station "Eagle Island LT.H." with a frequency of 307 kHz and an identifier "GJ.:'. Do not confuse the co-ordinates with the Belmullet aerodrome - the Lat/Long of the aerodrome are different than the coordinates specified by the question. If you are not sure about the meaning of chart symbols, refer to the attached figure (symbol #700) or Appendix 2 to ICAD Annex 4.

4207.

Airplane

ATPL

CPL

Heli

ATPL

CPL

How is a non controlled route marked on a map/chart? A) B) C) D)

As solid line. As a dashed line. As an alternate dotted/dashed line. As a dotted line.

(Refer to figures 067-E69, 067-E70, 067-E77, 067-E72, 067-E73, 067-E74, 067-E75 and 067-E76)

VOR; NDB; DME; compulsory reporting point. Civil airport; NDB; DME; non-compulsory reporting point. Civil airport; VOR; DME; non-compulsory reporting point. VOR; NDB; DME; non-compulsory reporting point.

I

Carnmore aerodrome. Belmullet aerodrome. Eagle Island Lt. H. NDB. Clonbullogue aerodrome.

I

4316. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-13) Which of the aeronautical chart symbols indicates a VORTAC? A) 6 B) 7

C) 3 D) 5 (Refer to figures 067-E69, 067-E70, 067-E77, 067-E72, 067-E73, 067-E74, 067-E75 and 067-E76)

4185 (C)

I

4207 (C)

I

4316 (8)

I

03 Charts 4317.

Airplane

ATPL

CPL

Heli

ATPL

CPL

(Refer to figure 061-13) Which of the aeronautical chart symbols indicates a TAeAN?

C) 3

4346.

D) 1 (Refer to figures 061-£69, 061-E70, 061-E71, 061-E72, 061-E73, 061-E74, 061-E75 and 061-E76)

Airplane

ATPL

CPL

Heli

ATPL

CPL

(Refer to figure 061-16) An aircraft on. a magnetic bearing 315° at a range of 150 NM from MYGGENES NDB (62°06'N 007°32'W) is at position? A) B) C) D)

(Refer to figure 061-E06) Use the Lat/Long information mentioned by the question to locate the MYGENNES NOB on the chart. The question states that we are on the 315° magnetic bearing from the NOB. In order to plat this bearing on the chart we will first need to convert it into a True bearing using the value of the magnetic variation. You can find the variation using the Isogonic lines running through the chart => for Mygennes NOB the variation would be approximately 13°W. True bearing will be 315° - 13°W variation => 302°. Using a protractor, plot an angle of 58° (360° - 302°) from the Meridian (lines of Longitude running North <=> South) passing through Mygennes NOB and draw a straight line representing a bearing of 302° from Mygennes NOB. Now the second part - the distance. Remember that 1 minute of Latitude = 1 NM. Therefore a distance of 150 NM = 150 minutes = 2°30'. Measure the length that is represented by 2°30' along any Meridian on the chart. Now apply this measured distance to the line that represents your bearing from Mygennes NOB and mark the point that is located at this distance from Mygennes NOB => this point represents your location and it will be more or less around the co-ordinates 63°20'N 01rw.

Airplane

ATPL

CPL

Heli

ATPL

CPL

(Refer to figure 061-16) An aircraft on radial 110° at a range of 120 NM from SAXAVORD VOR (60 050'N 000050'W) is at position: A) B) C) D)

61°27'N 004°43'W 60010'N 002°55'E 61°09'N 002°55'E 60 027'N 003°07'E

(Refer to figure 061-E02) Use the Lat/Long information mentioned by the question to locate the SAX,1IVORO VOR on the chart. The question states that we are on the 110° radial (magnetic bearing) from the VOR. In order to plot this radial on the chart we will first need to convert it into a True bearing using the value of the magnetic variation. You can find the variation using the Isogonic lines running through the chart => for Saxavord VOR the variation would be approximately 9°W. True bearing will be 110°- 9°Wvariation => 701~ Using a protractor, plot an angle of 101° from the Meridian (lines of Longitude running North <=> South) passing through Saxavord VOR and draw a straight line representing a radial of 701° from Saxavord. Now the second part - the distance. Remember that 1 minute of Latitude = 1 NM. Therefore a distance of 120 NM = 120 minutes = r. Measure the length that is represented by r along any Meridian on the chart. Now apply this measured distance to the line that represents your bearing from Saxavord VOR and mark the point that is located at this distance from Saxavord VOR => this point represents your location and it will be more or less around the coordinates 60 025'N 003°30'E.

4345. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-12) Which aeronautical chart symbol indicates an exceptionally high unlighted obstacle? A) 13

Airplane

ATPL

CPL

Heli

ATPL

CPL

(Refer to figure 061-12) Which aeronautical chart symbol indicates a f1yover waypoint? A) 15

B) 6

C) 7 D) 8 (Refer to figures 061-E69, 061-E70, 061-E71, 061-E72, 061-E73, 061-E74, 061-£75 and 061-E76)

63°20'N 012°05'W 60 020'N 004°05'W 63°45'N 011°25'W 60 040'N 003°20'W

4334.

D) 12 (Refer to figures 061-E69, 061-£70, 061-E71, 061-E72, 061-E73, 061-E74, 061-E75 and 061-E76)

A) 6 B) 7

4332.

B) 6

C) 9

4347. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-12) Which aeronautical chart symbol indicates a non-compulsory reporting point? A) 15

B) 6

C) 7 D) 8 (Refer to figures 061-E69, 061-£70, 061-E71, 061-E72, 061-E73, 061-E74, 061-£75 and 061-E76)

4352.

Airplane

AWL

CPL

Heli

ATPL

CPL

(Refer to figure 061-17) Given: CON VOR (53°54.8'N 008°49.1'W) DME 30 NM eRN DME (53°18.1'N 008°56.5'W) DME 25 NM Aircraft heading is 2700(M) and both DME distances are decreasing. What is the aircraft position? A) B) C) D)

53°30'N 008°20'W 53°43'N 009°25'W 53°35'N 009°25'W 53°37'N 008°20'W

Use the Lat/Long information stated by the question to locate the CON VOR and CRN OME on the chart. Now you will need to use the drawing-compass. Refer to the scale at the left edge of the chart and identify the distance of 30 NM => adjust your drawing-compass to this distance and draw a 30 NM circle around CON VOR. Repeat the same step with a distance of 25 NM for CRN OME. The circles you have drawn on the chart intersect at two locations - now you need to read the question again carefully and realize it is stating that the OME distances are decreasing while on a west-bound heading => this is only possible if the aircraft is in the position to the North-East of CRN OME (at the position to the NW of CRN the OME values would increase with a westbound heading) => the coordinates of the correct position are approximately 53°30'N / 008°21'W.

4353.

Airplane

ATPL

CPL

Heli

ATPL

CPL

(Refer to figure 061-17) Given: SHA VOR/DME (52°43,3'N 008°53,1'W) Radial 025°/49 NM What is the aircraft position? A) B) C) D)

53°28'N 008°20'W. 53°30'N 008°30'W. 51°55'N 009°15'W. 52°00'N 009°25'W.

Use the Lat/Long information stated by the question to locate the SHA VOR on the chart. Probably the easiest way to draw the specific radial requested by the question is to use the existing radials (airways) as a reference and plot the requested radial as an angular reference to one of these airways.

I 4317 (A) I 4332 (A) I 4334 (D) I 4345 (A) I 4346 (D) 14347(8) I 4352 (A) 14353(8) I

Ell

Aviationexam Test Prep Edition 2012 We can choose an airway WI3 which represents a radial 05r from SHA VOR. The angular difference between the requested radial 025° and the airway radial 052° is 021" => using a protractor we plot an angle of -021" to the airway WI3 and obtain a radial 025° from SHA => we draw a straight line from the VOR symbol representing this radial. Now we need to plot the distance information onto the chart. Refer to the scale on the left edge of the chart and measure the distance of 49 NM => now apply this distance to the line representing your radial from the VOR to find your position => approximately 53°30'N / 00B030W

4354. Airplane ATPL CPL (Refer to figure 061-17) Given:

Heli

ATPL

CPL

4357. Airplane ATPL CPL (Refer to figure 061-17) Given:

CRN DME (53°18,l'N 008°56,5'W) DME 18 NM SHA VOR (5r43,3'N 008°53,1'W) DME 30 NM Aircraft heading is 270 0 (M) and both DME distances are decreasing. What is the aircraft position? A) B) C) D)

52°52'N 009°23'W 53°10'N 008°30'W 53°0TN 009°23'W 53°55'N 008°25'W

Use the Lat/Long information stated by the question to locate the SHA VOR and eRN OME on the chart. Now you will need to use the drawing-compass. Refer to the scale at the left edge of the chart and identify the distance of 30 NM => adjust your drawing-compass to this distance and draw a 30 NM circle around SHA VOR. Repeat the same step with a distance of IB NM for eRN OME. The circles you have drawn on the chart intersect at two locations - now you need to read the question again carefully and realize it is stating that the OME distances are decreasing while on a west-bound heading => this is only possible if the aircraft is in the position to the North-East ofSHA VOR (at the position to the NW of SHA the OME values would increase with a westbound heading) => the coordinates of the correct position are approximately 53°09'N / 00B030'W.

4355. Airplane ATPL CPL (Refer to figure 061-17) Given:

Heli

ATPL

CPL

Use the Lat/Long information stated by the question to locate the SHA VOR on the chart. Probably the easiest way to draw the specific radial requested by the question is to use the existing radials (airways) as a reference and plot the requested radial as an angular reference to one of these airways. We can choose an airwayG4 which represents a radial 173 °from SHA VOR. The angular difference between the requested radial 222° and the airway radial 173° is 049° => using a protractor we plot an angle of +049° to the airway G4 and obtain a radial 222° from SHA => we draw a straight line from the VOR symbol representing this radial. Now we need to plot the distance information onto the chart. Refer to the scale on the left edge of the chart and measure the distance of 40 NM => now apply this distance to the line representing your radial from the VOR to find your position => approximately 5r10'N / 009°30W

4356. Airplane ATPL CPL (Refer to figure 061-17) Given:

Heli

ATPL

CPL

What is the aircraft position? A) B) C) D)

52°28'N 009°20'W 53°00'N 008°30'W 52°58'N 008°25'W 52°25'N 009°17'W

Use the Lat/Long information stated by the question to locate the SHA VOR on the chart. Probably the easiest way to draw the specific radial requested by the question is to use the existing radials (airways) as a reference and plot the requested radial as an angular reference to one of these airways. We can choose an airway WI3 which represents a radial 052° from SHA VOR. The angular difference between the requested radial 04Bo and the airway radial 052° is 004° => using a protractor we plot an angle of -004° to the airway WI3 and obtain a radial 04Bo from SHA => we draw a straight line from the VOR symbol representing this radial. Now we need to plot the distance information onto the chart. Refer to the scale on the left edge of the chart and measure the distance of 22 NM => now apply this distance to the line representing your radial from the VOR to find your position => approximately 53°00'N / 00B030'W.

Heli

ATPL

CPL

A) B) C) D)

52°55'N 008°15'W 52°50'N 0Q3°00'W 53°05'N 009°30'W 53°10'N 008°20'W

Use the Lat/Long information stated by the question to locate the SHA VOR and eRN OME on the chart. Now you will need to use the drawing-compass. Refer to the scale at the left edge of the chart and identify the distance of 26 NM => adjust your drawing-compass to this distance and draw a 26 NM circle around SHA VOR. Repeat the same step with a distance of 34 NM for eRN OME. The circles you have drawn on the chart intersect at two locations - now you need to read the question again carefully and realize it is stating that the OME distances are increasing while on an east-bound heading => this is only possible if the aircraft is in the position to the North-East of SHA VOR (at the position to the NWofSHA the OME values would decrease with an eastbound heading) => the coordinates of the correct position are approximately 52°55'N /00BoI7'W.

4360.

Airplane

ATPL

CPL

Heli

ATPL

CPL

(Refer to figure 061-17)

SHA VOR/DME (52°43,3'N 008°53,l'W) Radial 120°/35 NM

Given: SHA VOR (52°43,3'N 008°53,1'W) radial 223° CRKVOR (51°50,4'N 008°29,7'W) radial 322°

What is the aircraft position?

What is the aircraft position?

52°30'N 008°00'W 53°00'N 009°45'W 52°25'N 008°05'W 52°50'N 009°50'W

Use the Lat/Long information stated by the question to locate the SHA VOR

4354 (8)

CPL

Aircraft heading is 090 0 (M) and both DME distances are increasing. What is the aircraft position?

53°03'N 008°10'W 53°05'N 008°15'W 52°28'N 009°35'W 52°10'N 009°30'W

I

ATPL

CRN DME (53°18,l'N 008°56,5'W) DME 34 NM SHA VOR (52°43,3'N 008°53,1'W) DME 26 NM

What is the aircraft position?

A) B) C) D)

Heli

SHA VOR/DME (52°43,3'N 008°53,l'W) Radial 048°/22 NM

4359. Airplane ATPL CPL (Refer to figure 061-17) Given:

SHA VOR/DME (52°43,3'N 008°53,1'W) Radial 222°/40 NM

A) B) C) D)

on the chart. Probably the easiest way to draw the specific radial requested by the question is to use the existing radials (airways) as a reference and plot the requested radial as an angular reference to one of these airways. We can choose an airway GI which represents a radial 115° from SHA VOR. The angular difference between the requested radial 120° and the airway radial 115° is 005° => using a protractor we plot an angle of +005° to the airway GI and obtain a radial 120° from SHA => we draw a straight line from the VOR symbol representing this radial. Now we need to plot the distance information onto the chart. Refer to the scale on the left edge of the chart and measure the distance of 35 NM => now apply this distance to the line representing your radial from the VOR to find your pOSition => approximately 52°30W / OOBoOOW

I

4355 (D)

I

4356 (A)

I

4357 (8)

I

A) B) C) D)

4359 (A)

52°20'N 009°20'W 52°30'N 009°1O'W 52°1O'N 009°1O'W 52°10'N 009°30'W

I

4360 (A)

I

Ir.r.. ~

- - - - - - - - - -..

----~

03 Charts Use the Lat/Long information stated by the question to locate the SHA and CRK VORs on the chart. Probably the easiest way to draw the specific radials requested by the question is to use the existing radials (airways) as a reference and plot the requested radials as an angular reference to these airways. Let's start with the radial 223° from SHA VOR. We can choose an airway G4 which represents a radial 173°from SHA. The angular difference between 223° and 173° is 050°=> we plot an angle of +050° to the airway G4 and obtain a radial 223° from SHA => we draw a straight line from the VOR symbol representing this radial. For the CRK VOR we can also choose the airway G4 (radial 353°) => the requested radial from CRK VOR is 322° => the angular difference between 322" and 353° is 031° => we plot an angle of-031° to an airway G4 and obtain a radial 322° from CRK VOR => draw a straight line from the VOR symbol representing this radial. Now find the point where these two lines intersect and this is your position => approximately 52°20'N /009°20'w.

4361. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-17) An aircraft is inbound to KERRY aerodrome (52°10,9'N 009°32,0'W) on an extended runway centerline for a straightin ILS approach. The ILS DME is indicating 41 NM. The pilot performs a position check using a CRN DME and obtains a DME distance of 50 NM from CRN (53°18,1'N 008°56,5'W). What is the position of the aircraft? A) B) C) 0)

52°42'N 008°27'W 52°30'N 008°35'W 52°55'N 008°19'W 52°19'N 008°09'W

Use the Lat/Long information stated by the question to locate the KERRYaerodrome and CRN OME on the chart. Now you will need to use the drawing-compass. Refer to the scale at the left edge of the chart and identify the distance of 41 NM => adjust your drawing-compass to this distance and draw a 41 NM circle around the Kerry aerodrome symbol. Repeat the same step with a distance of50 NM for CRN OME. The circles you have drawn on the chart intersect at two locations - now you need to read the question again and realize it is stating that the aircraft is positioned on the extended runway centerline inbound for an ILS approach => this is only possible if the aircraft is in the position to the North-East of the aerodrome => the coordinates of this position are approximately 52°30W / 008°35'w.

4362. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-17) You are on a compass heading of 105°, deviation 3°E. WTD NDB (52°11,3'N 007°05,0'W) bears 013°R, CRK VOR (51°50,4'N 008°29,7'W) QDM is 211°. What is your position? A) B) C) 0)

52°45'N 007"57'W 52°28'N 008°02'W 54°12'N 006°39'W 52°17'N 007°45'W

QOM = Magnetic bearing TO a station. If we are on a QOM of 211" to CRK VOR it means that we are on a 031° radial (QOR - Magnetic bearing FROM a station) from CRK VOR. The best way to start with the solution of this question is to draw this radia/. The best way to draw a specific radial from a VOR is to use the existing radials (airways) that are already drawn on the chart=> in our case we can for example select the airway W70 (radial 053 0 from CRK VORl - the difference between 053 0 and our radial of 031° is 022°=> using a protractor we will measure an angle of -022° from the W70 airway and draw a straight line from the CRK VOR symbol representing a radial 031~ Magnetic heading of the aircraft = Compass heading plus Easterly deviation => 105° + 3 0 = 108° MH. Question states that the AOF indicator is showing a relative bearing (RB) of+13~ To obtain the QOM (Magnetic bearing) TO the WTO NOB we have to recall that MH (108°) + RB (13") =QOM (121°). To get a QOR (Magnetic bearing FROM a station) simply add 180°=> 121°+ 1800=301~ We are on a 301° Magnetic bearing from WTO NOB. Since the readouts are taken from the magnetic compass and the RBI indicator, the bearing is magnetic. To plot it on the chart we need a True value => to convert Magnetic to True we need to know the value of magnetic variation - this can be read just above the lower edge of the chart => approx. JOIN. True bearing = Magnetic bearing (301°) minus Westerly variation (JOW) = 294°. Now we are ready to plot the line that will represent the True Bearing of 294° from WTO NOB. After plotting the line find the place where it intersects with the 031° radial from CRK VOR and that is your position => approx. 52°28'N 008°02'IN.

I

4361 (8)

I

4362 (8)

I

4363 (A)

I

4364 (8)

I

4365 (C)

I

4363. Airplane ATPL CPL (Refer to figure 061-17) Given:

Heli

ATPL

CPL

SHA VOR (52°43,3'N 008°53,1'W) radial 120° CRK VOR (51°50,4'N 008°29,7'W) radial 033° What is the aircraft position? A) 52°30'N 008°00'W B) 52°25'N 008°05'W C) 52°20'N 007°50'W 0) 52°40'N 007"50'W Use the Lat/Long information stated by the question to locate the SHA and CRK VORs on the chart. Probably the easiest way to draw the specific radials requested by the question is to use the existing radials (airways) as a reference and plot the requested radials as an angular reference to these airways. Let's start with the radial 120° from SHA VOR. We can choose an airway Gl which represents a radial 115° from SHA. The angular difference between 120° and 115° is 005°=> we plot an angle of +005° to the airway Gl and obtain a radial 120° from SHA => we draw a straight line from the VOR symbol representing this radia/. For the CRK VOR we can choose the airway W70 (radial 053°) => the requested radial from CRK VOR is 033° => the angular difference between 033° and 053° is 020° => we plot an angle of -020° to an airway Wl0 and obtain a radial 033° from CRK VOR => draw a straight line from the VOR symbol representing . this radial. Now find the point where these two lines intersect and this is your position => approximately 52°30W / 008°00'IN.

4364. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-17) What are the lat/long coordinates of the point on SHA VOR (52°43'N 008°53'W) radial 239°M at a range of 36 NM? A) B) C) 0)

52 u 15'N 009"30'W 52°20'N 009°37'W 52°12'N 009°30'W 52°12'N 009°15'W

Use the Lat/Long information stated by the question to locate the SHA VOR on the chort. Probably the easiest way to draw the specific radial requested by the question is to use the existing radials (airways) as a reference and plot the requested radial as an angular reference to one of these airways. We can choose an airway G4 which represents a radial 173 0 from SHA VOR. The angular difference between the requested radial 239° and the airway radial 173° is 066° => using a protractor we plot an angle of +066° to the airway G4 and obtain a radial 239° from SHA => we draw a straight line from the VOR symbol representing this radial. Now we need to plot the distance information onto the chart. Refer to the scale on the left edge of the chart and measure the distance of 36 NM => now apply this distance to the line representing your radial from the VOR to find your position => approximately 52°20'N / 009°37'IN.

4365. Airplane ATPL CPL (Refer to figure 061-17) Given:

Heli

ATPL

CPL

SHA VOR (52°43,3'N 008°53,1'W) DME 50 NM CRK VOR (51°50,4'N 008°29,7'W) DME 41 NM Aircraft heading is 270° (M) and both DME distances are increasing. What is the aircraft position? A) B) C) 0)

52°15'N 007"45'W 52°15'N 009°40'W 52°00'N 009°35'W 52°35'N 007"50'W

Use the Lat/Long information stated by the question to locate the SHA and CRK VORs on the chart. Now you will need to use the drawing-compass. Refer to the scale at the left edge of the chart and identify the distance of 50 NM => adjust your drawing-compass to this distance and draw a 50 NM circle around SHA VOR. Repeat the same step with a distance of 41 NM for CRK VOR. The circles you have drawn on the chart intersect at two locations - now you need to read the question again carefully and realize it is stating that the OME distances are increasing while on a west-bound heading => this is only possible if the aircraft is in the position to the West of CRK VOR (at the position to the East of CRK the OME values would decrease with a west-bound heading) => the coordinates of the correct position are approximately 51 °59W / 009°31'IN.

Aviationexam Test Prep Edition 2012

4366. Airplane ATPL CPL (Refer to figure 061-17) Given:

Heli

ATPL

(at the position to the NW ofCRKthe DME readouts would increase with a westbound heading) => the coordinates of the correct position are approximately 52"15'N / 00Bo06'w.

CPL

SHA VOR (52°43,3'N 008°53,1'W) radial 129° CRKVOR (51°50,4'N 008°29,7'W) radial 047°

4374. Airplane ATPL CPL (Refer to figure 061-17) Given:

What is the aircraft position?

4367. Airplane ATPL CPL (Refer to figure 061-17) Given:

Heli

ATPL

CPL

SHA VOR/DME (52°43,3'N 008°53,1'W) Radial 165"/36 NM

A) B) C) D)

52°05'N 008°05'W 51°55'N 008°10'W 52°10'N 008°00'W 52°00'N 008°00'W

Use the Lat/Long information stated by the question to locate the SHA and CRK VORs on the chart. Probably the easiest way to draw the specific radials requested by the question is to use the existing radials (airways) as a reference and plot the requested radials as an angular reference to these airways. Let's start with the radial 143° from SHA VOR. We can choose an airway Gl which represents a radial 115° from SHA. The angular difference between 143° and 115° is 02Bo => we plot an angle of +02BO to the airway Gl and obtain a radial 143° from SHA => we draw a straight line from the VOR symbol representing this radial. For the CRK VOR we can choose the airway WlO (radial 053°) => the requested radial from CRK VOR is 050° => the angular difference between 050° and 053° is 003° => we plot an angle of -003° to an airway Wl0 and obtain a radial 050° from CRK VOR => draw a straight line from the VOR symbol representing this radial. Now find the point where these two lines intersect and this is your position => approximately 52°1O'N / OOBoOO'w.

4376. Airplane ATPL CPL (Refer to figure 061-17) Given:

What is the aircraft position?

Heli

ATPL

CPL

SHA VOR (52°43,3'N 008°53,1'W) radial 205° CRK VOR (51°50,4'N008°29,7'W) radial 317°

52"10'N 008°30'W 52"08'N 008°40'W 53°15'N 009°15'W 53°17'N 009°08'W

What is the aircraft position?

Use the Lat/Long information stated by the question to locate the SHA VOR on the chart. Probably the easiest way to draw the specific radial requested by the question is to use the existing radials (airways) as a reference and plot the requested radial as an angular reference to one of these airways. We can choose an airway Gl which represents a radial 115°from SHA VOR. The angular difference between the requested radial 165° and the airway radial 115° is 050° => using a protractor we plot an angle of +050° to the airway Gl and obtain a radial 165° from SHA => we draw a straight line from the VOR symbol representing this radial. Now we need to plot the distance information onto the chart. Refer to the scale on the left edge of the chart and measure the distance of 36 NM => now apply this distance to the line representing your radial from the VOR to find your position => approximately 52°10'N / 00B030'W.

4368. Airplane ATPL CPL (Refer to figure 061-17) Given:

Heli

ATPL

CPL

SHA VOR (52"43'N 008°53'W) DME 41 NM CRK VOR (51°50'N 008°29'W) DME 30 NM Aircraft is heading westbound and both DME values are decreasing. What is the position of the aircraft? A) 52°15'N 008°05'W B) 52°05'N 009°15'W C) 52°15'N 009°15'W D) 52°25'N 008°1O'W Use the Lat/Long information stated by the question to locate the SHA and CRK VORs on the chart. Now you will need to use the drawing-compass. Refer to the scale at the left edge of the chart and identify the distance of 41 NM => adjust your drawing-compass to this distance and draw a 41 NM circle around SHA VOR. Repeat the same step with a distance of 30 NM for CRK VOR. The circles you have drawn on the chart intersect at two locations - now you need to read the question again carefully and realize it is stating that the DME distances are decreasing while on a west-bound heading => this is only possible if the aircraft is in the position to the North-East ofCRK VOR

4366 (0)

CPL

What is the aircraft position?

Use the Lat/Long information stated by the question to locate the SHA and CRK VORs on the chart. Probably the easiest way to draw the specific radials requested by the question is to use the existing radials (airways) as a reference and plot the requested radials as an angular reference to these airways. Let's start with the radial 129° from SHA VOR. We can choose an airway Gl which represents a radial 115° from SHA. The angular difference between 115° and 129° is 014° => we plot an angle of +014° to the airway Gl and obtain a radial 129° from SHA => we draw a straight line from the VOR symbol representing this radial. For the CRK VOR we can choose an airway Wl0 (radial 053°) => the requested radial from CRK VOR is 047" => the angular difference between 047" and 053° is 006°=> we again plot an angle of-006°to an airway Wl0 and obtain a radial 047" from CRK VOR => draw a straight line from the VOR symbol representing this radial. Now find the point where these two lines intersect and this is your . position => approximately 52°20'N / 007"50'w.

I

ATPL

SHA VOR (52°43,3'N 008°53,1'W) radial 143° CRKVOR (51°50,4'N 008°29,7'W) radial 050°

A) 52°05'N 007°55'W B) 52°15'N 007"55'W C) 52°10'N 007°50'W D) 52°20'N 007°50'W

A) B) C) D)

Heli

I

4367 (A)

I

4368 (A)

I

4374 (C)

I

A) B) C) D)

52°10'N 009°lO'W 51°18'N 009°13'W 52°05'N 009°15'W 52°15'N 009°17'W

Use the Lat/Long information stated by the question to locate the SHA and CRK VORs on the chart. Probably the easiest way to draw the specific radials requested by the question is to use the existing radials (airways) as a reference and plot the requested radials as an angular reference to these airways. Let's start with the radial 205° from SHA VOR. We can choose an airway G4 which represents a radial 173° from SHA. The angular difference between 205° and 173° is 032° => we plot an angle of +032" to the airway G4 and obtain a radial 205° from SHA => we draw a straight line from the VOR symbol representing this radial. For the CRK VOR we can also choose the airway G4 (radial 353°) => the requested radial from CRK VOR is 317" => the angular difference between 317" and 353° is 036° => we plot an angle of -036° to an airway G4 and obtain a radial 317" from CRK VOR => draw a straight line from the VOR symbol representing this radial. Now find the point where these two lines intersect and this is your position => approximately 52"1O'N / 009°10'W.

4408. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-15) At 1215 UTC LAJES VORTAC (38°46'N 027°05'W) RMI reads 178°, range 135 NM. Calculate the aircraft position at 1215 UTC: A) B) C) D)

41°00'N 028°30'W 41°15'N 027°30'W 40 0 55'N 028°00'W 40 0 30'N 027°30'W

(Refer to figure 061-E01) Use the Lat/Long information mentioned by the question to locate LAJES VOR on the chart. The question states that the RMI is indicating 17Bo = bearing to the station = radial of 35Bo (17Bo + lBOO). Remember that radials are magnetic. In order to plot radial 35Bo on the chart we will first need to convert it into a True bearing using the value of the magnetic variation. You can find the variation using the Isogonic lines running through the chart => for Lajes

4376 (A)

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- - - - _..._ - - - - _.._ - - - - - - - - - - - - - - - - - - _...

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03 Charts the variation would be approximately 15°W. True bearing will be 35BO - 15°W variation => 343°. Using a protractor, plot an angle of 343° from the Meridian (lines of Longitude running North <=> South) passing through Lajes and draw a straight line representing a radial of343° from Lajes. Now the second part - the distance. Remember that 1 minute of Latitude = 1 NM. Therefore a distance of 135 NM = 135 minutes = 2°15' . Measure the length that is represented by 2°15' along any Meridian on the chart. Now apply this measured distance to the line that represents your bearing from Lajes VOR and mark the point that is located at this distance from Lajes VOR => this point represents your location and it will be more or less around the coordinates 40 055'N 02Bo05'£.

4415. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-13) Which of the aeronautical chart symbols indicates an NOB? A) 5 B) 4

C) 2 0) 3 (Refer to figures 061-E69, 061-E70, 061-E71, 061-E72, 061-E73, 061-E74, 061-E75 and 061-E76)

4416. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-13) Which of the aeronautical chart symbols indicates a basic, non-specified, navigation aid? A) 5

(51°50,4'N 008°29,7'W) and CRN NOB (53°18,1'N 008°56,5'W)? A) B) C) 0)

Use the Lat/Long information mentioned by the question to locate the CRK VOR and CRN NOB on the chart => draw a straight line between them (use the mid-points / centers of the navaid symbols for the line start and end points). Measure the length of this line and compare it to the scale at the left edge of the chart to determine the distance of approximately 90 NM. Now the second part of the solution - to find the average True course between the two points. You can obtain the True course directly from the map by measuring the angle between the lines of Longitude (Meridians running North <=> South) and the line you have drawn representing your track. Since the question asks for the average course we will have to measure the angles at multiple positions along the intended route of flight. The angle at B040'W is approx. 11° => the average True course is therefore going to be approx. 360° - 011° => 349°. Since the question is asking about the average Magnetic course we have to convert the average True course into Magnetic using the value of Magnetic variation. You can find the variation information at the very bottom of the chart - just above the lower edge. Variation will be BOW. With an avg. TC of349° and variation ofBowe will obtain an avg. Magnetic Course ofapprox.

357" (349° + BOW).

4441. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-17) What is the radial and OME distance from SHA VOR/OME (52°43,3'N 008°53,1'W) to position 52°10'N 009°20'W?

B) 4 C) 2

A) 3460 - 34 NM. B) 3540 - 34 NM.

0) 6

C) 1980

(Refer to figures 061-E69, 061-E70, 061-E71, 061-E72, 061-E73, 061-E74, 061-E75 and 061-E76)

4417. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-12) Which aeronautical chart symbol indicates an uncontrolled route? A) 3 B) 4 C) 5

0) 2 (Refer to figures 061-E69, 061-E70, 061-E71, 061-E72, 061-E73, 061-E74, 061-E75 and 061-E76)

4438. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-17) What is the average track (0T) and distance between SHA VOR (52°43,3'N 008°53,1 'W) and CON VOR (53°54,8'N 008°49,1 'W)? A) B) C) 0)

oor

4439. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-17) What is the average track (OM) and distance between CRK VOR 4415 (8)

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37 NM.

Use the Lat/Long information mentioned by the question to locate the SHA VOR and the defined position on the chart => place a mark on the defined Lat/Long position and draw a straight line between this mark and the SHA VOR. Measure the length of this line and compare it to the scale at the left edge of the chart to determine the distance of approximately 37 NM. Now the second part of the solution - to find the radial between the VOR and the defined position. Probably the easiest way is to use the existing radials (airways) as a reference. In this case we can choose for example the airway G4 which represents a radial 173° from SHA VOR. Now measure the angle between the line representing the airway and the line you have drawn => you will come up with approximately 042°. To find out the radial (magnetic bearing from the VORl that is represented by the line you have drawn simply calculate => airway radial 173°plus 042°=> radial 215°. The radial and distance from SHA VOR to the specified position is approximately 215°/37 NM.

4457. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-17) What is the average track (OT) and distance between BAL VOR (53°18,0'N 006°26,9'W) and CFN NOB (55°02,6'N 008°20,4'W)? A) B) C) 0)

oor

-

0) 214"-37 NM.

010 0 -71 NM. 358 0 - 72 NM. 0060 - 71 NM. 002 0 - 72 NM.

Use the Lat/Long information mentioned by the question to locate the SHA and CON VORs on the chart => draw a straight line between them (use the mid-points of the navaid symbols for the line start and end points). Measure the length of this line and compare it to the scale at the left edge of the chart to determine the distance ofapproximately 72 NM. Now the second part of the solution - to find the average True course between the two points. For this task there is no need to work with the magnetic variation (it would only be the case if the questions asked about a magnetic course) => you can obtain the True course directly from the map by measuring the angle between the lines ofLongitude (Meridians running North <=> South) and the line you have drawn representing your track. The angle at B050'W is approx. => the average True course is therefore going to be approx. and the distance 72 NM.

I

177 0 - 92 NM. 357 0 - 90 NM. 1690 - 91 NM. 349 0 - 89 NM.

335 0 - 128 NM. 327" - 124 NM. 3360 - 126 NM. 320 0 - 127 NM.

Use the Lat/Long information mentioned by the question to locate the BAL VOR and CFN NOB on the chart => draw a straight line between them (use the midpoints / centers ofthe navaid symbols for the line start and end points). Measure the length of this line and compare it to the scale at the left edge of the chart to determine the distance of approximately 124 - 124,5 NM. Now the second part of the solution - to find the average True course between the two points. For this task there is no need to work with the magnetic variation (it would only be the case if the questions asked about a magnetic course) => you can obtain the True course directly from the map by measuring the angle between the lines ofLongitude (Meridians running North <=> South) and the line you have drawn representing your track. Since the question asks for the average True course we will have to measure the angles at multiple positions along the intended route of flight. The angle at 7"W is approx. 032° and at BOW approx. 033° => the average True course is therefore going to be approx. 327,5° (360° - 32,5°) and the distance 124 - 124,5 NM.

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Aviationexam Test Prep Edition 2012

4459. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-17) What is the average track (OM) and distance between CRN NOB (53°18,1'N 008°56,5'W) and WTO NOB (52°11,3'N 007°05,0'W)? A) B) C) D)

135° - 96 NM. 322° - 95 NM. 142° - 95 NM. 3W-94NM.

Use the Lat/Long information mentioned by the question to locate the CRN and WTD NDBs on the chart => draw a straight line between them (use the mid-points / centers of the navaid symbols for the line start and end points). Measure the length of this line and compare it to the scale at the left edge of the chart to determine the distance of approximately 95 NM. Now the second part of the solution - to find the average True course between the two points. You can obtain the True course directly from the map by measuring the angle between the lines of Longitude (Meridians running North <=> South) and the line you have drawn representing your track. Since the question asks for the average course we will have to measure the angles at mUltiple is approx. 44,5°, positions along the intended route of flight. The angle at at 8°W approx. 045° and at 9°W approx. 046° => the average True course is therefore going to be approx. 180° - 045°=> 135~ Since the question is asking about the average Magnetic course we have to convert the average True course into Magnetic using the value ofMagnetic variation. You can find the variation information at the very bottom of the chart - just above the lower edge. Notice that for our route of flight the variation varies between 8°W and => average variation will be 7,5°W. With an avg. TC of 135° and variation of7,5°we will obtain an avg. Magnetic Course of approx. 142,5° (135° + 7,5°W).

rw

rw

4471. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-17) What is the radial and OME distance from SHA VOR/OME (52°43,3'N 008°53,1'W) to position 53°00'N 009°40'W? A) 057° - 27 NM. B) 309° - 33 NM. C) 293 0 - 33 NM.

D) 3240

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17 NM.

4474. Airplane ATPL CPL (Refer to figure 061-17) Given:

Heli

ATPL

CPL

CON VOR/OME (53°54,8'N 008°49,1'W) Abbey Shrule aerodrome (53°35'N 007°39'W) What is the CON radial and OME distance when overhead Abbey Shrule aerodrome? 2960 - 46 NM. 3040 - 47 NM. 1240 - 46 NM. 1160 -47 NM.

Use the Lat/Long information mentioned by the question to locate the CON VOR and the Abbey Shrule airport on the chart => draw a straight line between the airport symbol and the CON VOR. Measure the length of this line and compare it to the scale at the left edge of the chart to determine the distance of approximately 46 NM. Now the second part of the solution - to find the radial between the VOR and the airport. Probably the easiest way is to use the existing radials (airways) as a reference. In this case we can choose

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4487. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-12) Which aeronautical chart symbol indicates a Flight Information Region (FIR) boundary? A) 1 B) 3

C) 4 D) 5 (Refer to figures 061-E69, 061-E70, 061-E71, 061-E72, 061-E73, 061-E74, 061-E75 and 061-E76)

4488. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-12) Which aeronautical chart symbol indicates a Control Zone boundary?

A) 2 B) 3

C) 4 D) 5 (Refer to figures 061-E69, 061-E70, 061-E71, 061-E72, 061-E73, 061-E74, 061-E75 and 061-E76)

4489. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-12) Which aeronautical chart symbol indicates the boundary of advisory airspace?

A) 2

Use the Lat/Long information mentioned by the question to locate the SHA VOR and the defined position on the chart => place a mark on the defined Lat/Long position and draw a straight line between this mark and the SHA VOR. Measure the length of this line and compare it to the scale at the left edge of the chart to determine the distance of approximately 33 NM. Now the second part of the solution - to find the radial between the VOR and the defined position. Probably the easiest way is to use the existing radials (airways) as a reference. In this case we can choose for example the airway W13 which represents a radial 052" from SHA VOR. Now measure the angle between the line representing the airway and the line you have prawn => you will come up with approximately 102°. To find out the radial (magnetic bearing from the VORl that is represented by the line you have drawn simply calculate => airway radial 052° minus 102° => radial 310°. The radial and distance from SHA VOR to the specified position is approximately 310°/33 NM.

A) B) C) D)

for example the airway Bl which represents a radial 113° from CON VOR. Now measure the angle between the line representing the airway and the line you have drawn => you will come up with approximately 01O~ To find out the radial (magnetic bearing from the VORl that is represented by the line you have drawn simply calculate => airway radial 113° plus 010° => radial 123°. The radial and distance from CON VOR to the Abbey Shrule airport is approximately 123°/46 NM.

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B) 3

C) 4 D) 5 (Refer to figures 061-E69, 061-E70, 061-E71, 061-E72, 061-E73, 061-E74, 061-E75 and 061-E76)

4491. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-17) What is the radial and OME distance from BEL VORIDME (54°39,7'N 006°13,8'W) to position 55°00'N 007°00'W? A) B) C) D)

1260 - 33 NM. 2960 - 65 NM. 3W-34NM. 222 0 - 48 NM.

Use the Lat/Long information mentioned by the question to locate the BEL VOR and the defined position on the chart => place a mark on the defined Lat/Long position and draw a straight line between this mark and the BEL VOR. Measure the length of this line and compare It to the scale at the left edge of the chart to determine the distance of approximately 34 NM. Now the second part of the solution - to find the radial between the VOR and the defined position. Remember that the radials are magnetic bearings from the VOR. Notice the thin line that looks a little bit like a flag that protrudes from the VOR symbol upwards => this is your magnetic North reference. What we need to do is to measure the angle between this Magnetic North reference and the line we have drawn => using a protractor you will measure an angle of approximately 046° => now simply deduct 046° from 360° to get a result of 316°. The radial and distance from BEL VOR to the specified position is approximately 316°/34NM.

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03 Charts 4493. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-17) What is the radial and OME distance from CRK VOR/OME (Sl°S0,4'N 008°29,7'W) to position S2°10'N 009°20'W? A) 350 0 - 22 NM. B) 295 0 - 38 NM. C) 1700 - 22 NM. 0 D) 311 - 38 NM. Use the Lat/Long information mentioned by the question to locate the CRK VOR and the defined position on the chart => place a mark on the defined Lat/Long position and draw a straight line between this mark and the CRK VOR. Measure the length of this line and compare it to the scale at the left edge of the chart to determine the distance of approximately 38 NM. Now the second part of the solution - to find the radial between the VOR and the defined position. Probably the easiest way is to use the existing radials (airways) as a reference. In this case we can choose for example the airway G4 which represents a radial 353° from CRK VOR. Using a protractor measure the angle between the line representing the airway and the line you have drawn => you will come up with approximately 041". To find out the radial (magnetic bearing from the VORl that is represented by the line you have drawn simply calculate => airway radial 353° minus 041° => radial 312". The radial and distance from CRK VOR to the specified position is approximately 312° / 38 NM.

4494. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-17) What is the average track (0T) and distance between WTO NOB (S2°11,3'N 007°0S,O'W) and FOY NOB (S2°34,0'N 009°11,7'W)?

4512. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-17) What is the radial and OME distance from BEL VOR/OME (S4°39,7'N 006°13,8'W) to position S4°40'N 007°30'W? 0

- 46 NM. B) 278° - 44 NM. C) 278° -10 NM. D) 098 0 - 45 NM.

Use the Lat/Long information mentioned by the question to locate the 8EL VOR and the defined position on the chart => place a mark on the defined Lat/Long position and draw a straight line between this mark and the BEL VOR. Measure the length of this line and compare it to the scale at the left edge of the chart to determine the distance of approximately 44 NM. Now the second part of the solution - to find the radial between the VOR and the defined position. Remember that the radials are magnetic bearings from the VOR. Notice the thin line that looks a little bit like a flag that protrudes from the VOR symbol upwards => this is your magnetic North reference. What we need to do is to measure the angle between this Magnetic North reference and the line we have drawn => using a protractor you will measure an angle of approximately 083° => now simply deduct 083° from 360° to get a result of 27r. The radial and distance from BEL VOR to the specified position is approximately 27r / 44 NM.

Airplane

ATPL

Heli

CPL

ATPL

CPL

(Refer to figure 061-17)

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2780 270 0 262 0 262 0

89 NM. 91 NM. 91 NM. 89 NM.

-

Use the Lat/Long information mentioned by the question to locate the BAL VOR and CRN NOB on the chart => draw a straight line between them (use the mid-points of the navaid symbols for the line start and end points). Measure the length of this line and compare it to the scale at the left edge of the chart to determine the distance of approximately 91 NM. Now the second part of the solution - to find the average True course between the two points. For this task there is no need to work with the magnetic variation (it would only be the case if the questions asked about a magnetic course) => you can obtain the True course directly from the map by measuring the angle between the lines ofLongitude (Meridians running North <=> South) and the line you have drawn representing your track. Since the question asks for the average True course we will have to measure the angles at multiple positions along the intended route of flight. The angle at rw is approx. 090 0 and at 8°W also approx. 090 0 => the average True course is therefore going to be approx. 270° (360° - 090°) and the distance 91 NM.

4536. Airplane ATPL CPL (Refer to figure 061-17) Given:

Heli

ATPL

CPL

SHA VOR/OME (S2°43,3'N 008°S3,1'W) Birr aerodrome (S3°04'N 007°S4'W)

A) 068°-41 NM.

Use the Lat/Long information mentioned by the question to locate the WTO and FOY N08s on the chart => draw a straight line between them (use the mid-points of the navaid symbols for the line start and end points). Measure the length of this line and compare it to the scale at the left edge of the chart to determine the distance of approximately 82 NM. Now the second part of the solution - to find the average True course between the two points. For this task there is no need to work with the magnetic variation (it would only be the case if the questions asked about a magnetic course) => you can obtain the True course directly from the map by measuring the angle between the lines ofLongitude (Meridians running North <=> South) and the line you have drawn representing your track. The angle at rw is approx. 074~ at 8°W approx. 074° and at 9°W approx. 075° => the average True course is therefore going to be approx. 286° (360° - 074°) and the distance 82 NM.

4516.

A) B) C) D)

What is the SHA radial and OME distance when overhead Birr aerodrome?

A) 075 0 - 81 NM. B) 2940 - 80 NM. C) 2860 - 82 NM. D) 27]0 - 83 NM.

A) 090

What is the average True track and distance between BAL VOR (S3°18,O'N 006°26,9'W) and CRN NOB (S3°18,1'N 008°S6,S'W)?

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B) 248 0 - 42 NM. C) 060 0 - 42 NM. D) 240 0 - 41 NM. Use the Lat/Long information mentioned by the question to locate the SHA VOR and the BIRR aerodrome on the chart => draw a straight line between the aerodrome symbol and the VOR. Measure the length of this line and compare it to the scale at the left edge of the chart to determine the distance of approximately 41 NM. Now the second part of the solution - to find the radial between the VOR and the aerodrome. Probably the easiest way is to use the existing radials (airways) as a reference. In this case we can choose for example the airway W13 which represents a radial 052° from SHA VOR. Now measure the angle between the line representing the airway and the line you have drawn => you will come up with approximately 016°. To find out the radial (magnetic bearing from the VORl that is represented by 0 the line you have drawn simply calculate => airway radial 05r plus 016 => radial 068°. The radial and distance from SHA VOR to the Birr aerodrome is approximately 068°141 NM.

4543. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-17) What is the radial and" OME distance from BEL VOR/OME (S4°39,7'N 006°13,8'W) to position S4°10'N 007°10'W? A) 223 0 B) 2360 C) 320 0 D) 333 0

-

-

36 NM. 44 NM. 44 NM. 36 NM.

Use the Lat/Long information mentioned by the question to locate the BEL VOR and the defined position on the chart => place a mark on the defined Lat/Long position and draw a straight line between this mark and the BEL VOR. Measure the length of this line and compare it to the scale at the left edge of the chart to determine the distance of approximately 44 NM. Now the second part of the solution - to find the radial between the VOR and the defined position. Remember that the radials are magnetic bearings from the VOR. Notice the thin line that looks a little bit like a flag that protrudes from the VOR symbol upwards => this is your magnetic North reference. What we need to do is to measure the angle between this Magnetic North reference and the Iinewe have drawn => using a protractor you will measure an angle of approximately 125° => now simply deduct 125° from 360° to get a result of 235°. The radial and distance from BEL VOR to the specified position is approximately 235° / 44 NM.

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Aviationexam Test Prep Edition 2012 4545. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-17) What is the radial and OME distance from CON VOR/OME (53°54,8'N 008°49,1 'W) to position 53°30'N 009°30'W? A) B) C) 0)

165 - 27 NM. 233 0 - 35 NM. 335 0 - 43 NM. 025 0 - 38 NM.

4562. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-17) What is the radial and OME distance from CON VOR/OME (53°54,8'N 008°49,1'W) to position 54°30'N 009°00'W? 049 0 - 45 NM. 2140 - 26 NM. 358 0 - 36 NM. 1690 - 35 NM.

Use the Lat/Long information mentioned by the question to locate the CON VOR and the defined position on the chart => place a mark on the defined Lat/Long position and draw a straight fine between this mark and the CON VOR. Measure the length of this fine and compare it to the scale at the left edge of the chart to determine the distance of approximately 36 NM. Now the second part of the solution - to find the radial between the VOR and the defined position. Probably the easiest way is to use the existing radials (airways) as a reference. In this case we can choose for example the airway 81 (west-bound) which represents a radial 285° from CON VOR. Now measure the angle between the fine representing the airway and the fine you have drawn => you will come up with approximately 072". To find out the radial (magnetic bearing from the VORl that is represented by the fine you have drawn simply calculate => airway radial 285° plus 072°=> radial 357". The radial and distance from CON VOR to the specified position is approximately 357"/36 NM.

4567. Airplane ATPL CPL Heli ATPL CPL An aircraft is over position HO (55°30'N 060 0 15'W), where YYR VOR (53°30'N 060 0 15'W) can be received. The magnetic variation is 31°W at HO and 28°W at YYR. What is the radial from YYR? A) 031 0 B) 208 0 C) 028 0

0) 332 0 (Refer to figures 061-£53 and 061-E97) Notice that the aircraft and the VOR are on the same meridian (060 0 15'W) => therefore the aircraft is positioned on the 360° True bearing from the YYR VOR. Remember that the radials are magnetic bearings from the station. Magnetic variation = the angle between the True North and the Magnetic North. It is expressed as how many degrees East or West is the Magnetic North situated from the True North. Therefore, knowing the value of the variation as well as the true bearing we can easily calculate the magnetic bearing (radia/): If the variation is "West" it means that the magnetic bearing is greater than the true bearing. Thus the magnetic radial will be greater than the true radial by 28° which is the westerly variation at the YYR VOR => radial 028°.

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4612. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-17) What is the average track (OM) and distance between KER NOB (52°10,9'N 009°31,5'W) and eRN NOB (53°18,1'N 008°56,5'W)? A) 025 0 - 70 NM. B) 197 0 - 71 NM. C) 205 0 - 71 NM. 0) 017 0 - 70 NM. Use the Lat/Long information mentioned by the question to locate the KER and CRN NOBs on the chart => draw a straight fine between them (use the mid-points / centers of the navaid symbols for the fine start and end points). Measure the length of this fine and compare it to the scale at the left edge of the chart to determine the distance of approximately 70 NM. Now the second part of the solution - to find the average True course between the two points. You can obtain the True course directly from the map by measuring the angle between the fines of Longitude (Meridians running North <=> South) and the fine you have drawn representing your track. Since the question asks for the average course we will have to measure the angles at mUltiple positions along the intended route of flight. The angle at 9°30'W is approx. 017" and at 9°W approx. 017" => the average True course is therefore going to be approx. 017". Since the question is asking about the average Magnetic course we have to convert the average True course into Magnetic using the value of Magnetic variation. You can find the variation information at the very bottom of the chart - Just above the lower edge. Notice that for our route of fljght the variation equals to approximately 8°W. With an avg. TC of 017" and variation of 8° we will obtain an avg. Magnetic Course of approx. 25° (017"+ 8°W).

4646. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-17) What is the average course (OT) and distance from WTO NOB (52°12'N 007°05'W) to SLG NOB (54°17'N 008°36'W)? A) B) C) 0)

344° (T); 139 NM. 1640 (T); 138 NM. 1560 (T); 136 NM. 3360 (T); 137 NM.

Use the Lat/Long information mentioned by the question to locate the WTO and SLG NOBs on the chart => draw a straight fine between them (use the mid-points of the navaid symbols for the fine start and end points). Measure the length of this fine and compare it to the scale at the left edge of the chart to determine the distance of approximately 137 NM. Now the second part of the solution - to find the average True course between the two points. For this task there is no need to work with the magnetic variation (it would only be the case if the questions asked about a magnetic course) => you can obtain the True course directly from the map by measuring the angle between the fines of Longitude (Meridians running North <=> South) and the fine you have drawn representing your track. Since the question asks for the average True course we will have to measure the angles at multiple positions along the intended route of flight. The angle at 7"20'W is approx. 023° and at 8°20'W also approx. 023 0 => the average True course is therefore going to be approx. 337" (360° - 023°) and the distance 137 NM.

4653. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-12) Which aeronautical chart symbol indicates an aeronautical ground light? A) B) C) 0)

14 16 10

15

(Refer to figures 061-E69, 061-E70, 061-E71, 061-E72, 061-E73, 061-E74, 061-E75 and 061-E76)

4593. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-13) Which of the aeronautical chart symbols indicates a VOR?

I

0) 7 (Refer to figures 061-E69, 061-E70, 061-E71, 061-E72, 061-E73, 061-E74, 061-E75 and 061-E76)

0

Use the Lat/Long information mentioned by the question to locate the CON VOR and the defined position on the chart => place a mark on the defined Lat/Long position and draw a straight line between this mark and the CON VOR. Measure the length of this line and compare it to the scale at the left edge of the chart to determine the distance of approximately 36 NM. Now the second part of the solution - to find the radial between the VOR and the defined position. Probably the easiest way is to use the existing radials (airways) as a reference. In this case we can choose for example the airway 81 (west-bound) which represents a radial 285° from CON VOR. Now measure the angle between the fine representing the airway and the fine you have drawn => you will come up with approximately 052°. To find out the radial (magnetic bearing from the VORl that is represented by the fine you have drawn simply calculate => airway radiall85° minus 052° => radial 233". The radial and distance from CON VOR to the specified position is approximately 233°/36 NM.

A) B) C) 0)

A) 1 B) 2 C) 3

I

4612 (A)

I

4646 (D)

I

4653 (D)

I

03 Charts 4656. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-17) What is the average track (0T) and distance from SLG NOB (54°16,7'N 008°36,0'W) to WTO NOB (52°11,3'N 007°05,0'W)? A) 343° - 139 NM.

B) 156° - 137 NM. C) 336° - 136 NM. 0) 163° - 138 NM. Use the Lat/Long information mentioned by the question to locate the WTD and SLG NOBs on the chart => draw a straight line between them (use the mid-points of the navaid symbols for the line start and end points). Measure the length of this line and compare it to the scale at the left edge of the chart to determine the distance of approximately 137 NM. Now the second part of the solution - to find the average True course between the two points. For this task there is no need to work with the magnetic variation (it would only be the case if the questions asked about a magnetic course) => you can obtain the True course directly from the map by measuring the angle between the lines of Longitude (Meridians running North <=> South) and the line you have drawn representing your track. Since the question asks for the average True course we will have to measure the angles at multiple positions along the intended route of flight. The angle at r20'W is approx. 023° and at 8°20'W also approx. 023° => the average True course is therefore going to be approx. 15r (180° - 023°) and the distance 137 NM.

4659. Airplane ATPL CPL (Refer to figure 061-17) Given:

Heli

ATPL

What is the SHA radial and OME distance when overhead Connemara aerodrome? 333° - 37 NM. 154° - 38 NM. 326° - 37 NM. 146° - 38 NM.

Use the Lat/Long information mentioned by the question to locate the SHA VOR and the CONNEMARA aerodrome on the chart => draw a straight line between the aerodrome symbol and the VOR. Measure the length of this line and compare it to the scale at the left edge of the chart to determine the distance of approximately 37 NM. Now the second part of the solution - to find the radial between the VOR and the aerodrome. Probably the easiest way is to use the existing radials (airways) as a reference. In this case we can choose for example the airway W13 which represents a radial 052° from SHA VOR. Now measure the angle between the line representing the airway and the line you have drawn => you will come up with approximately 079°. To find out the radial (magnetic bearing from the VOR) that is represented by the line you have drawn simply calculate => airway radial 052" minus 079° => radial 333~ The radial and distance from SHA VOR to the Connemara aerodrome is approximately 333°/37 NM.

4661. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-17) What is the radial and OME distance from CRK VOR (51°51'N 008°30'W) to position 52°20'N 009°10'W? A) 322° - 39 NM. B) 150°-41 NM. C) 329° - 40 NM. 0) 321° - 41 NM. Use the Lat/Long information mentioned by the question to locate the CRK VOR and the defined position on the chart => place a mark on the defined Lat/Long position and draw a straight line between this mark and the CRK VOR. Measure the length of this line and compare it to the scale at the left edge of the chart to determine the distance of approximately 40 NM. Now the second part of the solution - to find the radial between the VOR and the defined position. Probably the easiest way is to use the existing radials (airways) as a reference. In this case we can choose for example the airway G4 which represents a radial 353° from CRK VOR. Using a protractor measure the angle between the line representing the airway and the line you have drawn => you will come up with approximately 024°. To find out the radial (magnetic bearing from

I

4656 (8)

I

4659 (A)

I

4661 (C)

I

4668 (D)

I

4668. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-12) Which aeronautical chart symbol indicates a compulsory reporting point? A) 8 B) 15

C) 6 0) 7 (Refer to figures 061-E69, 061-E70, 061-E71, 061-E72, 061-E73, 061-E74, 061-E75 and 061-E76)

4669. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-17) What is the mean true track and distance from the BAL VOR (53°18'N 006°27'W) to CFN NOB (55°02'N 008°20'W)? A) 328° (T); 125 NM. B) 148° (T); 125 NM. C) 328° (T); 134 NM. 0) 148° (T); 134 NM.

CPL

SHA VOR/OME (52°43,3'N 008°53,1'W) Connemara aerodrome (53°14'N 009°28'W)

A) B) C) 0)

the VOR) that is represented by the line you have drawn simply calculate => airway radial 353° minus 024° => radial329~ The radial and distance from CRK VOR to the specified position is approximately 329°/40 NM.

4669 (A)

I

Use the Lat/Long information mentioned by the question to locate the BAL VOR and CFN NOB on the chart => draw a straight line between them (use the mid-points of the navaid symbols for the line start and end points). Measure the length of this line and compare it to the scale at the left edge of the chart to determine the distance of approximately 125 NM. Now the second part of the solution - to find the average True course between the two points. For this task there is no need to work with the magnetic variation (it would only be the case if the questions asked about a magnetic course) => you can obtain the True course directly from the map by measuring the angle between the lines of Longitude (Meridians running North <=> South) and the line you have drawn representing your track. Since the question asks for the average True course we will have to measure the angles at mUltiple positions along the intended route of flight. The angle at rw is approx. 032" and at 8°W approx. 033° => the average True course is therefore going to be approx. 327,5° (360° - 032,5°) and the distance 125 NM.

4678.

Airplane

ATPL

CPL

Heli

ATPL

CPL

(Refer to figure 061-15) 1300 UTC DR position 37°30'N 021°30'W alter heading PORTO SANTO NOB (33°03'N 016°23'W) TAS 450 kts, Forecast W/V 360°/30 kts. Calculate the ETA at PORTO SANTO NOB: A) B) C) D)

1335 1341 1348 1354

(Refer to figure 061-E03) Use the Lat/Long information mentioned by the question to locate the referenced position and the Porto Santo NOB on the chart => draw a straight line between these two points. We have to find the average True course for this route => you can obtain the True courses directly from the map by measuring the angle between the lines ofLongitude (Meridians running North <=> South) and the line you have drawn representing your track. Since the question asks for the average course we will have to measure the angles at the beginning of our route and at the end of our route, then calculate an average value. The angle at 21°30'W is apptox. 134° and at lrWapprox. 137,5° => find the average of these two values = Average True course ofapprox. 136°. Second step is finding the distance of the route. Remember that 1 minute ofLatitude = 1 NM. Measure the length that is represented by the line ofyour route => draw a line of the same distance along any Meridian on the chart => see how many degrees and minutes of Latitude your line represents => you will come up with approximately 6°10' => 370 minutes => 370 NM. Now we have the true course and distance and the last step is to calculate the Groundspeed (GS) and Time. Use your flight computer to calculate the GS => approximately 471 kts and then use this GS to calculate the time => you will come-up with approximately 47 minutes. It will take approx. 47 minutes to fly this sector => ETA at Porto Santo is 13:4Z

4678 (C)

Aviationexam Test Prep Edition 2012

4685. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-12) Which aeronautical chart symbol indicates an exceptionally high lighted obstacle? A) 14 B) 13 C) 12 D) 16 (Refer to figures 061-E69, 061-E70, 061-E71, 061-E72, 061-E73, 061-E74, 061-E75 ond 061-E76)

4690. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-17) What is the radial and OME distance from SHA VOR/OME (52°43,3'N 00so53,1'W) to position 52°20'N 00S010'W? A) 139° - 35 NM. B) 129° - 46 NM.

C) 132° - 36 NM. D) 212° - 26 NM. Use the Lat/Long information mentioned by the question to locate the SHA VOR and the defined position on the chart => place a mark on the defined Lat/Long position and draw a straight line between this mark and the SHA VOR. Measure the length of this line and compare it to the scole at the left edge of the chart to determine the distance ofapproximately 35 NM. Now the second part of the solution - to find the radial between the VOR and the defined position. Probably the easiest way is to use the existing radials (airways) as a reference. In this case we can choose for example the airway Gl which represents a radial 115° from SHA VOR. Now measure the angle between the line representing the airway and the line you have drawn => you will come up with approximately U24". To find out the radial (magnetic bearing from the VON) that is represented by the line you have drawn simply calculate => airway radial 115° plus 024° => radial 139". The radial and distance from SHA VOR to the specified position is approximately 139°/35 NM.

4692. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-17) What is the average track (OM) and distance between BAL VOR (53°1S,O'N 006°26,9'W) and SLG NOB (54°16,7'N 00So36,O'W)? A) 262° - 86 NM. B) 128° - 99 NM. C) 308° - 96 NM. D) 3W-98NM. Use the Lat/Long information mentioned by the question to locate the BAL VOR and SLG NOB on the chart => draw a straight line between them (use the mid-points / centers of the navaid symbols for the line start and end points). Measure the length of this line and compare it to the scale at the left edge of the chart to determine the distance of approximately 98 NM. Now the second part of the solution - to find the average True course between the two points. You can obtain the True course directly from the map by measuring the angle between the lines of Longitude (Meridians running North <=> South) and the line you have drawn representing your track. Since the question asks for the average course we will have to measure the angles at multiple is approx. positions along the intended route of flight. The angle at and at 8°W approx. 52,5° => the average True course is therefore going to be approx. 360° - 052° => 308°. Since the question is asking about the average Magnetic course we have to convert the average True course into Magnetic using the value of Magnetic variation. You can find the variation information at the very bottom of the char( - just above the lower edge. Variation will be between and 8°W. With an avg. TC of 349° and avg. variation of 7,5° we will obtain an avg. Magnetic Course of approx. 315,5° (308° + 7,5°W).

r'w

5r

rw

4705. Airplane ATPL CPL (Refer to figure 061-17) Given:

Heli

ATPL

CPL

--~---

I

4690 (A)

I

4692 (0)

-----~-------~.

4706. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-17) What is the radial and OME distance from CON VOR/OME (53°54,SiN 00so49,1'W) to position 53°40'N 00S020'W)? A) 3W-22 NM.

B) 119°-42 NM. C) 140° - 23 NM. D) 240° - 24 NM. Use the Lat/Long information mentioned by the question to locate the CON VOR and the defined position on the chart => place a mark on the defined Lat/Long position and draw a straight line between this mark and the CON VOR. Measure the length of this line and compare it to the scale at the left edge of the chart to determine the distance of approximately 23 NM. Now the second part of the solution - to find the radial between the VOR and the defined pOSition. Probably the easiest way is to use the existing radials (airways) as a reference. In this case we can choose for example the airway Bi (est-bound) which represents a radial 113° from CON VOR. Now measure the angle between the line representing the airway and the line you have drawn => you will come up with approximately 026". To find out the radial (magnetic bearing from the VORl that is represented by the line you have drawn simply calculate => airway radial 113° plus 026° => radial 139". The radial and distance from CON VOR to the specified position is approximately 139° / 23 NM.

4707. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-16) What are the average magnetic course and distance between INGO VOR (63°50'N 016°40'W) and SUMBURG VOR (59°55'N 001°15'W)? A) 131 ° - 494 NM. B) 118° - 440 NM.

C) 117°-494 NM. D) 105° - 440 NM. (Refer to figures 061-E05 and 061-E100) Use the Lat/Long information mentioned by the question to locate the positions on the chart => draw a straight line between them. You can obtain the True courses directly from the map by measuring the angle between the lines of Longitude (Meridians running North <=> South) and the line you have drawn representing your track. Since the question asks for the average course we will have to measure the angles at the beginning of our route and at the end ofourroute, then calculate an average value. The angle at 16°W is approx. 114° and at approx. 124° => find the average of these two values = Average True course of 119". Since the question is asking about the average Magnetic course we have to convert the True course into Magnetic using the value of Magnetic variation. You can find the variation using the Isogonic lines running through the chart => for our route of flight the variation values of 20°1N, 15°1N, 12°W and 9°W are relevant => now average these out to get an avg. variation of 14°W. Avg. true course (119°) + Westerly variation (14°W) average magnetic course of 133°.

rw

For the approximate distance calculation we can use the following formula:

What is the CRK radial and OME distance when overhead Kerry aerodrome? 4685 (A)

Use the Lat/Long information mentioned by the question to locate the CRK VOR and the KERRY aerodrome on the chart => draw a straight line between the aerodrome symbol and the VOR. Measure the length of this line and compare it to the scale at the left edge of the chart to determine the distance of approximately 44 NM. Now the second part of the solution - to find the radial between the VOR and the aerodrome. Probably the easiest way is to use the existing radials (airways) as a reference. In this case we can choose for example the airway G4 which represents a radial 353° from CRK VOR. Using a protractor measure the angle between the line representing the airway and the line you have drawn => you will come up with approximately 045°. To find out the radial (magnetic bearing from the VORl that is represented by the line you have drawn simply calculate => airway radial 353° minus 045° => radial 308". The radial and distance from CRK VOR to the Kerry aerodrome is approximately 308°/44 NM.

=

CRKVOR/OME (51°50,4'N 00so29,7'W) Kerry aerodrome (52°10,9'N 009°31,4'W)

I

A) 307" - 43 NM. B) 119° - 44 NM. C) 127°-45NM. D) 299° - 42 NM.

I

4705 (A)

I

o = ,; (change in Latitude in minutesY + (change in Longitude in minutes x Cosine Mean LatitudeY o =,; (3°55'Y + (W25' x Cosine 61 °52'307

4706 (C)

I

4707.(A)

I

---------------_. - - -

03 Charts

o = ,; (2351 + (925' x Cosine 67,875°)1 o =,; (235')1 + (436)1

(Refer to figures 067-£69, 067-E70, 067-E77, 067-E72, 067-E73, 067-E74, 067-£75 and 067-E76)

0='; (55.225 + 790.096) 0=';245.327 o = 495NM

4724. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-17) What is the radial and DME distance from CRK VOR/DME (51°50,4'N 008°29,7'W) to position 51°40'N 007°30'W? A) 106° - 40 NM. B) 104° - 76 NM. C) 293° - 39 NM. D) 113° - 39 NM.

11

12

symbols

CPL

indicates

A) 3 B) 1

C) 4 D) 7 (Refer to figures 067-E69, 067-E70, 067-E17, 067-E72, 067-E73, 067-E74, 067-E75 and 067-£76)

4813. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-13) Which ofthe aeronautical chart symbols indicates a DME? A) 2 B) 1

C) 6

4826. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-16) What are the average magnetic course and distance between position 60 0 00'N 020 0 00'W and Sumburg VOR (59°55'N 001°15'W)? A) 105° - 562 NM. B) 091° - 480 NM. C) 091 ° - 562 NM. D) 105° - 480 NM.

9

10 11

12

4740. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-12) Which aeronautical chart symbol indicates a lightship?

(Refer to figures 067-E04 and 067-E700) Use the Lat/Long information mentioned by the question to locate the positions on the chart => draw a straight line between them. You can obtain the True courses directly from the map by measuring the angle between the lines of Longitude (Meridians running North <=> South) and the line you have drawn representing your track. Since the question asks for the average course we will have to measure the angles at the beginning of our route and at the end of our route, then calculate an average value. The angle at 200W is approx. 084° and at approx. 098° => find the average of these two values = Average True course of 097~ Since the question is asking about the average Magnetic course we have to convert the True course into Magnetic using the value of Magnetic variation. You can find the variation using the isogonic lines running through the chart => for our route of flight the variation values of 20°W, 75°W, 72°W and 9°W are relevant => now average these out to get an avg. variation of 74°W. Avg. true course (097°) + Westerly variation (74°W) = average magnetic course of 705°.

rw

12

10 15 16

(Refer to figures 067-E69, 067-E70, 067-E17, 067-E72, 067-E73, 067-E74, 067-E75 and 067 -E76)

4741. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-12) Which aeronautical chart symbol indicates a lighted obstacle? 9

For the approximate distance calculation we can use the following formula:

o = ,; (change in Latitude in minutes)1 + (change in Longitude in minutes x Cosine Mean Latitude)1 0='; (5')1 + (78°45' x Cosine 59°57'30"1 0='; (5')1 + (7.725' x Cosine 59,9583°)2

o =,; (5')1 + (563,2)1

10 15 16

4724 (0)

chart

ATPL

(Refer to figures 067-E69, 067-E70, 067-E17, 067-E72, 067-E73, 067-E74, 067-E75 and 067-E76)

(Refer to figures 067-£69, 067-E70, 067-E17, 067-E72, 067-E73, 067-E74, 067-E75 and 067-E76)

I

Heli

D) 3

10

4739. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-12) Which aeronautical chart symbol indicates a group of lighted obstacles?

A) B) C) D)

4806. Airplane ATPL CPL (Refer to figure 061-13) Which of the aeronautical aVOR/DME?

9

(Refer to figures 067-E69, 067-E70, 067-E77, 067-E72, 067-E73, 067-E74, 067-E75 and 067-E76)

A) B) C) D)

B) 10

C) 8 (Refer to figures 067-E69, 067-E70, 067-E17, 067-E72, 067-£73, 067-E74, 067-E75 and067-E76)

4738. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-12) Which aeronautical chart symbol indicates a group of unlighted obstacles?

A) B) C) D)

A) 9

D) 15

Use the Lat/Long information mentioned by the question to locate the CRK VOR and the defined position on the chart => place a mark on the defined Lat/Long position and draw a straight line between this mark and the CRK VOR. Measure the length of this line and compare it to the scale at the left edge of the chart to determine the distance of approximately 39 NM. Now the second part of the solution - to find the radial between the VOR and the defined position. Probably the easiest way is to use the existing radials (airways) as a reference. In this case we can choose for example the airway 870 which represents a radial 094° from CRK VOR. Using a protractor measure the angle between the line representing the airway and the line you have drawn => you will come-up with approximately 020~ To find-out the radial (magnetic bearing from the VORl that is represented by the line you have drawn simply calculate => airway radial 094°plus 020°=> radial 774°. The radial and.distance from CRKVOR to the specified position is approximately 774°/39 NM.

A) B) C) D)

4742. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-12) Which aeronautical chart symbol indicates an unlighted obstacle?

0='; (25 + 37Z794,2) 0='; 37Z279,2 0=563NM

I

4738 (C)

I

4739 (0)

I

4740 (0)

I

4741 (8)

I

4742 (A)

I

4806 (8)

I

4813 (A)

I

4826 (A)

I

Aviationexam Test Prep Edition 2012 4832. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-17) What is the average track (OM) and distance between WTO NOB (52°11,3'N 007°05,0'W) and KER NOB (52°10,9'N 009°31,5'W)? A) 270 0 B) 090 0 0 C) 278 0) 098 0 -

89 NM. 91 NM. 90 NM. 90 NM.

Use the Lat/Long information mentioned by the question to locate the WTD NOB and KER NOB on the chart => draw a straight line between them (use the mid-points / centers of the navaid symbols for the line start and end points). Measure the length of this line and compare it to the scale at the left edge of the chart to determine the distance of approximately 90 NM. Now the second part of the solution - to find the average True course between the two points. You can obtain the True course directly from the map by measuring the angle between the lines of Longitude (Meridians running North <=> South) and the line you have drawn representing your track. Since the question asks for the average course we will have to measure the angles at mUltiple positions along the intended route of flight. The angle at r is approx. 089~ at 8°W approx. 090° and at 9°W approx. 091° => the average True course from WTO to KER is therefore going to be approx. 270° (360° - 90°). Since the question is asking about the average Magnetic course we have to convert the average True course into Magnetic using the value of Magnetic variation. You can find the variation information at the very bottom of the chart -just above the lower edge. Notice that for our route of flight the variation is between rw and 8°W => avg. variation = 7,5°W. With an avg. TC of 270° and avg. variation of 7,5° we will obtain an avg. Magnetic Course of approx. 277,5° (270° + 7,5°W).

4839. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-17) What is the average track (OM) and distance between CRN NOB (53°18,1'N 008°56,5'W) and BEL VOR (54°39,7'N 006°13,8'W)? A) 229 0 - 125 NM. B) 089 0 - 95 NM. C) 057 0 -128 NM. 0) 23r - 130 NM. Use the Lat/Long information mentioned by the question to locate the CRN NOB and BEL VOR on the chart => draw a straight line between them (use the mid-points / centers of the navaid symbols for the line start and end points). Measure the length of this line and compare it to the scale at the left edge of the chart to determine the distance of approximately 128 NM. Now the second part of the solution - to find the average True course between the two points. You can obtain the True course directly from the map by measuring the angle between the lines of Longitude (Meridians running North <=> South) and the line you have drawn representing your track. Since the question asks for the average course we will have to measure the angles at mUltiple positions along the intended route offlight. The angle at 9°W is approx. 048°, at 8°W approx. 049° and at rw approx. 050° => the average True course is therefore going to be approx. 049°. Since the question is asking about the average Magnetic course we have to convert the average True course into Magnetic using the value of Magnetic variation. You can find the variation information at the very bottom of the chart - just above the lower edge. Notice that for our route of flight the variation varies between 8°Wand rw => average variation will be 7,5°W. With an avg. TC of 049° and variation of7,5°we will obtain an avg. Magnetic Course of approx. 56,5° (049° + 7,5°W).

4841. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-17) What is the radial and OME distance from CON VOR/OME (53°54,8'N 008°49,1'W) to position 54°00'N 008°00'W? A) 320 0 B) 088 0 0 C) 094 0) 260 0 -

8 NM. 29 NM. 64 NM. 30 NM.

Use the Lat/Long information mentioned by the question to locate the CON VOR and the defined position on the chart => place a mark on the defined Lat/Long position and draw a straight line between this mark and the CON VOR. Measure the length of this line and compare it to the scale at the left edge of the chart to determine the distance of approximately 30 NM. Now the second part of the solution - to find the radial between the VOR

and the defined position. Probably the easiest way is to use the existing radials (airways) as a reference. In this case we can choose for example the airway Bl which represents a radial 113° from CON VOR. Now measure the angle between the line representing the airway and the line you have drawn => you will comeup with approximately 025~ To find out the radial (magnetic bearing from the VORl that is represented by the line you have drawn simply calculate => airway radial 113° minus 025°=> radial 088°. The radial and distance from CON VOR to the specified position is approximately 088°/30 NM.

4849. Airplane ATPL CPL (Refer to figure 061-17) Given:

Heli

ATPL

CPL

CON VOR/OME (5r54,8'N 008°49,1'W) Castlebar aerodrome (53°51'N 009°17'W) What is the CON radial and OME distance when overhead Castlebar aerodrome? A) 265 0 - 17 NM. B) 07r - 18 NM. C) 25r - 17 NM. 0) 0860 -18 NM. Use the Lat/Long information mentioned by the question to locate the CON VOR and the Castlebar airport on the chart => draw a straight line between the airport symbol and the CON VOR. Measure the length of this line and compare it to the scale at the left edge of the chart to determine the distance of approximately 18 NM. Now the second part of the solution - to find the radial between the VOR and the airport. Probably the easiest way is to use the existing radials (airways) as a reference. In this case we can choose for example the airway Bl (west-bound) which represents a radial 285° from CON VOR. Now measure the angle between the line representing the airway and the line you have drawn => you will come up with approximately 019°. To find out the radial (magnetic bearing from the VORl that is represented by the line you have drawn simply calculate => airway radial 285° minus 019° => radial 266°. The radial and distance from CON VOR to the Castlebar airport is approximately 266°/18 NM.

4855. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-17) What is the radial and OME distance from SHA VOR/OME (52°43,3'N 008°53,1'W) to position 53°10'N 008°30'W? A) B) C) 0)

019 0 -31 NM. 070 0 - 58 NM. 20r - 31 NM. 035 0 - 30 NM.

Use the Lat/Long information mentioned by the question to locate the SHA VOR and the defined position on the chart => place a mark on the defined Lat/Long position and draw a straight line between this mark and the SHA VOR. Measure the length of this line and compare it to the scale at the left edge of the chart to determine the distance of approximately 30 NM. Now the second part of the solution - to find the radial between the VOR and the defined position. Probably the easiest way is to use the existing radials (airways) as a reference. In this case we can choose for example the airway W13 which represents a radial 052° from SHA VOR. Now measure the angle between the line representing the airway and the line you have drawn => you will come-up with approximately 017~ To find-out the radial (magnetic bearing from the VORl that is represented by the line you have drawn simply calculate => airway radial 052° minus 01r => radial 035~ The radial and distance from SHA VOR to the specified position is approximately 035°/30 NM.

4856. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-17) What is the average track (0T) and distance between CON VOR (53°54,8'N 008°49,1 'W) and BEL VOR (54°39,7'N 006°13,8'W)? A) 293 0 - 98 NM. B) 070 0 - 100 NM. C) 113 0 - 96 NM. 0) 063 0 -102 NM. Use the Lat/Long information mentioned by the question to locate the CON and BEL VORs on the chart => draw a straight line between them (use the mid-points of the navaid symbols for the line start and end points).

I 4832 (C) I 4839 (C) 14841 (8) I 4849 (A) I 4855 (D) I 4856 (D) I

----.----~----.-------------~-

03 Charts Measure the length of this line and compare it to the scale at the left edge of the chart to determine the distance of approximately 102 NM. Now the second part of the solution - to find the average True course between the two points. For this task there is no need to work with the magnetic variation (it would only be the case if the questions asked about a magnetic course) => you can obtain the True course directly from the map by measuring the angle between the lines of Longitude (Meridians running North <=> South) and the line you have drawn representing your track. Since the question asks for the average True course we will have to measure the angles at mUltiple positions along the intended route of flight. The angle at 8°W is approx. 062,5° and at 7"W approx. 064° => the average True course is therefore going to be approx. 063 0 and the distance 702 NM.

4874. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-17) What is the average track (0T) and distance between CRN NOB (53°18,1'N 008°56,5'W) and EKN NOB (54°23,6'N 007°38,7'W)?

Probably the easiest way is to use the existing radials (airways) as a reference. In this case we can choose for example the airway W70 which represents a radial 053° from CRK VOR. Using a protractor measure the angle between the line representing the airway and the line you have drawn => you will come up with approximately 023°. To find out the radial (magnetic bearing from the VORl that is represented by the line you have drawn simply calculate => airway radial 053° minus 023°=> radial 030~ The radial and distance from CRK VOR to the specified position is approximately 030°/33 NM.

4887. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-16) What are the initial true course and distance between positions 58°00'N 013°00'W and 66°00'N 002°00'E? A) 032° - 470 NM. B) 029° - 570 NM. C) 042° - 635 NM. D) 036° - 638 NM.

A) 0440 -82 NM.

B) 042 0 - 83 NM. C) 031° - 81 NM. D) 035° - 82 NM. Use the Lat/Long information mentioned by the question to locate the CRN and EKN NDBs on the chart => draw a straight line between them (use the mid-points of the navaid symbols for the line start and end points). Measure the length of this line and compare it to the scale at the left edge of the chart to determine the distance ofapproximately 82 NM. Now the second part of the solution - to find the average True course between the two points. For this task there is no need to work with the magnetic variation (it would only be the case if the questions asked about a magnetic course) => you can obtain the True course directly from the map by measuring the angle between the lines ofLongitude (Meridians running North <=> South) and the line you have drawn representing your track. Since the question asks for the average True course we will have to measure the angles at multiple positions along the intended route of flight. The angle at 9°W is approx. 034° and at 8°W approx. 035° => the average True course is therefore going to be approx. 34,5° and the distance 82 NM.

4877. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-17) You are at position 53°40'N 008°00'W. What is the QOR from the SHA VOR (52°43'N 008°53'W)? A) 217 B) 037

(Refer to figure 067-E08) Use the Lat/Long information mentioned by the question to locate the positions on the chart => draw a straight line between them. You can obtain the True courses directly from the map by using the protractor and measuring the angle between the lines ofLongitude (Meridians running North <=> South) and the line you have drawn representing your track. Since the question asks for the initial course we will have to measure the angles at the beginning of our route => the result will be approx. 036~ For the approximate distance calculation we can use the following formula: D = ,; (change in Latitude in minutes1 + (change in Longitude in minutes x Cosine Mean Latitude)2 Ch. Lat = 66° - 58° = 8° = 480 min Ch.Long = 73°W + 2°E = 75°= 900 min Mean Lat = (58° + 66°) .;. 2 = 62° D = ,; (480'Y + (900' x Cosine 62"Y D = ,; (480'1 + (422,52'1 D (230.400+ 778.523,75) D = ,; 408.923,75 D=639,4NM

=,;

4891. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-17) What is the radial and OME distance from CRK VOR/OME (51°50,4'N 008°29,7'W) to position 52°30'N 007°50'W? A) 039 0 - 48 NM. B) 024° - 43 NM.

C) 209

D) 029 QDR: Magnetic heading FROM a station. Use the Lat/Long information mentioned by the question to locate the SHA VOR and the defined position on the chart => place a mark on the defined Lat/Long position and draw a straight line between this mark and the SHA VOR. In order to find the radial between the VOR and the defined position probably the easiest way would be to use the existing radials (airways) as a reference. In this case we can choose for example the airway W73 which represents a radial 052° from SHA VOR. Now measure the angle between the line representing the airway and the line you have drawn => you will come up with approximately075~ To find out the radial (QDR - Magnetic Bearing from the VORl that is represented by the line you have drawn simply calculate => airway radial 052° minus 075° => radial 037".

4879. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-17) What is the radial and OME distance from CRK VOR/OME (51°50,4'N 008°29,7'W) to position 52°20'N 008°10'W? A) B) C) D)

048° - 40 NM. 030° - 33 NM. 014"-33 NM. 220° - 40 NM.

Use the Lat/Long information mentioned by the question to locate the CRK VOR and the defined position on the chart => place a mark on the defined Lat/Long position and draw a straight line between this mark and the CRK VOR. Measure the length of this line and compare it to the scale at the left edge of the chart to determine the distance of approximately 33 NM. Now the second part of the solution - to find the radial between the VOR and the defined position.

I

4874 (D)

I

4877 (8)

I

4879 (8)

I

4887 (D)

I

4891 (A)

I

C) 023° - 48 NM. D) 017" - 43 NM. Use the Lat/Long information mentioned by the question to locate the CRK VOR and the defined position on the chart => place a mark on the defined Lat/Long position and draw a straight line between this mark and the CRK VOR. Measure the length of this line and compare it to the scale at the left edge of the chart o determine the distance of approximately 48 NM. Now the second part of the solution - to find the radial between the VOR and the defined position. Probably the easiest way is to use the existing radials (airways) as a reference. In this case we can choose for example the airway W70 which represents a radial 053° from CRK VOR. Using a protractor measure the angle between the line representing the airway and the line you have drawn => you will come up with approximately 074~ To find out the radial (magnetic bearing from the VORl that is represented by the line you have drawn simply calculate => airway radial 053° minus 074°=> radial 039°. The radial and distance from CRK VOR to the specified position is approximately 039°/48 NM.

4902. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-17) What is the average track (OT) and distance between SLG NOB (54°16,7'N 008°36,0'W) and CFN NOB (55°02,6'N 008°20,4'W)? A) B) C) D)

191° - 45 NM. 020° - 46 NM. 348° - 46 NM. 011°-47 NM.

Use the Lat/Long information mentioned by the question to locate the SLG and CFN NDBs on the chart => draw straight line between them

4902 (D)

a

I

Aviationexam Test Prep Edition 2012

(use the mid-points of the navaid symbols for the line start and end points). Measure the length of this line and compare it to the scale at the left edge of the chart to determine the distance ofapproximately 47 NM. Now the second part of the solution - to find the average True course between the two points. For this task there is no need to work with the magnetic variation (it would only be the case if the questions asked about a magnetic course) => you can obtain the True course directly from the map by measuring the angle between the lines of Longitude (Meridians running North <=> South) and the line you have drawn representing your track. The angle at 8°30'W is approx. 071° => the average True course is therefore going to be approx. 071° and the distance

230889. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-12) Which aeronautical chart symbol indicates an exceptionally high unlighted obstacle? A) 13

B) 9 C) 11 D) 14

47NM.

4912. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-17) What is the average track (OM) and distance between WTO NOB (52°11,3'N 007°05,0'W) and BAL VOR (53°1S,0'N 006°26,9'W)? A) B) C) D)

206° - 71 NM. 018° -153 NM. 026° - 71 NM. 198° - 72 NM.

Use the Lat/Long information mentioned by the question to locate the WTO NOB and BAL VOR on the chart => draw a straight line between them (use the mid-points / centers of the navaid symbols for the line start and end points). Measure the length of this line and compare it to the scale at the left edge of the chart to determine the distance of approximately 71 NM. Now the second part of the solution - to find the average True course· between the two points. You can obtain the True course directly from the map by measuring the angle between the lines of Longitude (Meridians running North <=> South) and the line you have drawn representing yourtrack. Since the question asks for the average course we will have to measure the angles at mUltiple positions along the intended route of flight. The angle at 6°50'W Is approx. 019 => the average True course from WTO to BAL is therefore going to be approx. 79°. Since the question is asking about the average Magnetic course we have to convert the average True course into Magnetic using the value of Magnetic variation. You can find the variation information at the very bottom of the chart -just above the lower edge. Notice that for our route of flight the variation is approx. rw. With an avg. TC of 079° and variation ofr we will obtain an avg. Magnetic Course of approx. 26° (079° + rW). 0

230841. Airplane ATPL CPL (Refer to figure 061-17) Given:

Heli

ATPL

CPL

SHA VOR/OME (N52°43.3 WOOS053.1) radial 232°/32 NM. What is the aircraft position? A) B) C) D)

N52°20 W009°30 N53°05 W008°15 N52°28 W009°35 N53°03 W008°1 0

230844. Airplane ATPL CPL (Refer to figure 061-17) Given:

Heli

ATPL

CPL

SHA VOR N52°43.3 WOOS053.1 CRK VOR N51°50.4 WOOS029.7 Aircraft position N52°20 W009°1 0 Which of the following lists two radials that are applicable to the aircraft position? A) B) C) D)

SHA 033° CRK 149° SHA 214° CRK 330° SHA 205° CRK 321° SHA 025° CRK 141°

230895. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-0S) The fix of the aircraft position is determined by radials from three VOR stations. The measurements contain small random errors, known systematic errors and unknown systematic errors. The measured radials are corrected for known systematic errors and are plotted on a navigation chart. The result is shown in the annex. What is the most probable position of the aircraft? A) 1 B) 2

C) 3 D) 4 230909. Given:

Airplane

ATPL

CPL

Heli

ATPL

CPL

Position NOB (55°10'N, 012°55'E) DR Position (54°53'N, 009°58'E) NOB on the RMI reads 090° Magnetic variation 100W The position line has to be plotted on a Lamberts conformal chart with standard parallels at 400N and 4soN. Calculate the direction (T) of the bearing to be plotted from the NOB. A) 262°

B) 272° C) 258° D) 265° Airplane ATPL CPL Heli ATPL CPL 230910. A VOR is situated at position (N55°26', W005°42'). The variation at the VOR is 9°W. The position of the aircraft is (N6°00'N, W010000'). The variation at the aircraft-position is 11°W. The initial TT-angle of the great circle from the aircraft position to the VOR is 101,5°. Which radial is the aircraft on? A) B) C) D)

278° 296° 294° 276°

230911.

Airplane

ATPL

CPL

Heli

ATPL

CPL

An NOB is located at position (N55°26', W005°42'). The variation at the NOB is 9°W. The position of the aircraft is (56°00'N, 010 000'W). The variation at the aircraft-position is 11 oW. The initial TT of the great circle from the aircraft position to the NOB position, is 101,5°. What is the Magnetic Bearing of the NOB from the aircraft? A) B) C) D)

108,so 110,so 112,so 114,so

1 4912 (C) 1230841 (A) 1230844 (8) 1230889 (A) 1230895 (A) 1230909 (A) 1230910 (C) 1230911 (C) 1

04 Dead Reckoning Navigation (DR)

DEAD RECKON I NG

NAVIGATION (DR) 04-01 Basis of dead reckoning Airplane ATPL CPL Heli ATPL CPL Ground speed is 540 knots. 72 NM to go. What is time to go? 4287.

A) B) C) 0)

8mins 9 mins 18 mins 12 mins

4407.

Airplane

ATPL

Heli

ATPL

A) B) C) 0)

CPL

122 kts 985NM

7 hrs 48 min 8 hrs 04 min 7 hrs 49 min 8 hrs 10 min

Airplane

Airplane

ATPL

Heli

CPL

CPL

ATPL

ATPL

CPL

510 kts 43NM

GS: Distance A to B: What is the time (min) from A to B?

Heli

CPL

ATPL

A) 6 B) 4 C) 5

CPL

0) 7 105 kts 103NM

GS: Distance from A to B:

Distance =Rate x Time. 43 NM =570kts x? hrs ? hrs =43 NM +570kts ? = 0,0843 hrs = 5 minutes (0,0843

What is the time from A to B? 1 hr01 min 57 min 58min 59min

4424.

Airplane

Distance A to B: PlannedGS: ATD:

ATPL

Heli

CPL

ATPL

CPL

A) B) C) 0)

95 kts 480NM

GS: Distance from A to B:

CPL

Heli 325NM 315 kts 1130 UTe

335 kts 375 kts 395 kts 355 kts

Let's summarize the information we have. Distance to cover is 325 NM and our planned speed is 375 kts. If we departed at 11:30 then our planned flight time will be 1,0317 hrs (325 NM + 375 kts) = 61,9 minutes = estimated arrival at approx. 72:32 UTC.

What is the time from A to B? A) 4 hrs 59 min B) 5 hrs 03 min C) 5 hrs 00 min 0) 5 hrs 08 min

The flight progress check at 12:05 indicated a distance covered of 765 NM = distance remaining to go of 760 NM. If our ETA is 12:32 urc we have 27 minutes (0,45 hrs) to cover the distance of 760 NM. Our speed for the remainder

Distance = Rate x Time.

4299 (8)

ATPL

1205 UTe - fix obtained 165 NM along track. What GS must be maintained from the fix in order to achieve planned ETA at B?

Given:

I

Airplane

x 60)

Given:

Distance =Rate x Time. 703 NM = 705 kts x? hrs ? hrs = 703 NM + 105 kts ? =0,98 hrs =58,8 minutes (0,98 x 60)

4287 (A)

ATPL

120 kts 84NM

42 min 43min 44min 45 min

4414.

Given:

I

CPL

Given:

Distance = Rate x Time. 985 NM = 722 kts x ? hrs ? hrs = 985 NM + 722 kts ? =8,0737 hrs =484 minutes (8,0837 x 60) 484 minutes = 8 hrs 04 minutes

4388.

ATPL

Distance = Rate x Time. 84NM= 720kts x ?hrs ?hrs=84NM+ 720kts ? = OJ hrs '" 42 minutes (0,7 x 60)

What is the time from A to B?

A) B) C) 0)

Heli

CPL

What is the time from A to B?

CPL

GS: Distance from A to B:

4351.

ATPL

GS: Distance from A to B:

Given:

A) B) C) 0)

Airplane

Given:

Distance = Rate x Time. 72 NM = 540kts x ?hrs ?hrs = 72 NM + 540kts ? =0,1333 hrs =8 minutes (0,7333 x 60)

4299.

480NM = 95 kts x? hrs ? hrs=480NM + 95 kts ? =5,05 hrs =303 minutes (5,05 x 60) 303 minutes =5 hrs 03 min

I

4351 (D)

I

4388 (8)

I

4407 (A)

I

4414 (C)

I

4424 (D)

I

Aviationexam Test Prep Edition 2012 of the flight would have to be 355,5 kts (160 NM + 0,45 hrs).

4482. Given:

Airplane

ATPL

Heli

CPL

GS: Distance from A to B:

C) 235° (T) 0) 240° (T)

ATPL

CPL

Maximum drift angle will be experienced during a direct crosswind (130° or 310°). As the angle between the wind and the track decreases the drift also decreases. Greatest drift will therefore be experienced when the wind angle is the greatest. With a track of 040° we will have the greatest wind angle if the wind is from 240° (20° = 240° - 040° - 180°). With all other answers the value of wind angle is lower than 20°. It helps to take a look at the wind side of your navigation computer - place the track 040° under the TC pointer and look at the directions from which the wind is blowing in the respective answers A to D.

135 kts 433NM

What is the time from A to B? A) 3 hrs 20 min B) 3 hrs 25 min C) 3 hrs 19 min 0) 3 hrs 12 min

4587. Airplane ATPL CPL Heli ATPL CPL How many NM would an aircraft travel in 1 min 45 sec if GS is 135 kts?

Distance = Rate x Time. 433 NM = 135 kts x ? hrs ? hrs = 433 NM + 135 kts ? =3,2 hrs = 192 minutes (3,2 x 60) 192 min =3 hrs 12 min

4483. Given:

Airplane

ATPL

Heli

CPL

GS: Distance from A to B:

ATPL

A) 39,0 NM B) 2,36 NM C) 3,25 NM 0) 3,94 NM

CPL

1 min 45 sec = 1,15 minutes (1 min + (45 + 60) 1,75 minutes = 0,02916 hrs (1,75 + 60)

435 kts. 1.920 NM.

Distance = Rate x Time: ? = 135 kts x 0,02916 hrs ?=3,94NM

What is the time from A to B? A) 4 hrs 10 min B) 3 hrs 25 min C) 3 hrs 26 min 0) 4 hrs 25 min

4648. Airplane ATPL CPL Heli ATPL CPL An aircraft travels 2,4 statute miles in 47 seconds. What is its ground speed?

Distance = Rate x Time. 1.920 NM =435 kts x ? hrs ? hrs = 1.920 NM + 435 kts ? =4,41 hrs =265 minutes (4,41 x 60) 265 minutes =4 hrs 25 minutes

A) 183 kts B) 160 kts C) 209 kts 0) 131 kts

4498. Airplane ATPL CPL Heli ATPL CPL If it takes 132,4 min to travel 840 NM, what is your speed in km/h? A) B) C) 0)

705 km/h 290 km/h 120 km/h 966 km/h

4715. Given:

132,4 min = 2,2066 hours (132,4 + 60). If the distance traveled in this time = 860 NM then the speed is 380,67 kts (840 NM + 2,2066). 1 NM (kts) = 1,852 km 380,67 kts = 705 kmlhr

4500. Given:

Airplane

ATPL

GS: Distance to go:

Heli

CPL

ATPL

CPL

20min 29min 2 hrs 05 min 2 hrs 12 min

Heli

CPL

GS: Distance from A to B:

ATPL

CPL

480 kts 5.360NM

What is the time from A to B? A) B) C) 0)

A) B) C) 0)

4547. Airplane ATPL CPL Heli ATPL CPL Course 0400T, TAS 120 kts, wind speed 30 kts. From which direction will the wind give the greatest drift: A) 2W(T) B) 230° (T) 4482 (0)

ATPL

11 11 11 11

hrs 07 min hrs 06 min hrs 10 min hrs 15 min

4729. Airplane ATPL CPL Heli ATPL CPL An aircraft travels 100 statute miles in 20 min, how long does it take to travel 215 NM?

Distance = Rate x Time. 500 NM = 240 kts x ? hrs ? hrs =500 NM + 240 kts ? =2,0833 hrs = 125 minutes (2,0833 x 60) 125 minutes =2 hrs 05 min

I

Airplane

Distance = Rate x Time. 5.360 NM =480 kts x ? hrs ? hrs =5.360 NM + 480 kts ? = 11,1666 hrs = 670 minutes (11,1666 x 60) 670 minutes = 11 hrs 10 minutes

240 knots 500NM

What is time to go? A) B) C) 0)

47 seconds = 0,0130555 hrs (47 + 60 + 60). If the distance traveled in this time = 2,4 SM then the speed is 183,83 MPH . (2,4SM + 2,2066). 1 SM = 0,86897 NM 183,83 MPH (SM) = 159,7 kts (NM)

I

4483 (0)

I

4498 (A)

I

4500 (C)

I

50min 100 min 90 min 80 min

100 Statute Miles (SM) in 20 minutes = 300 MPH (100 x 3) 1 SM = 0,8689 NM 300 SM =260,7 kts (NM) Distance = Rate x Time 215 NM =260,7 kts x ?hrs ? hrs =215 NM + 260,7 kts ? = 0,82 hrs =49,2 min (0,82

4547 (0)

I

4587 (0)

I

x 60)'

4648 (8)

I

4715 (C)

I

4729 (A)

I

04 Dead Reckoning Navigation (DR)

4737. Given:

Airplane

ATPL

Heli

CPL

ATPL

CPL

Distance = Rate x Time. 354 NM = 236 kts x ? hrs

? hrs = 354 NM + 236 kts ? = 7,5 hrs = 90 minutes (7,5 x 60)

345 kts 3560NM

GS: Distance from A to B:

230912. Airplane ATPL CPL Heli ATPL CPL What may cause a difference between a DR-position and a Fix?

What is the time from A to B? A) 10 hrs 19 min B) 10 hrs 05 min C) 11 hrs 00 min D) 11 hrs 02 min

A) The difference between no-wind and the actual wind. B) The difference between no-wind and the forecasted wind. C) The difference between the actual wind and the forecasted wind. D) The difference between the magnetic and the true wind direction.

Distance = Rate x Time. 3.560 NM = 345 kts x ? hrs ? hrs = 3.560 NM + 345 kts ? = 10,3788 hrs = 679 minutes (70,3788 x 60) 679 minutes = 10 hrs 79 min

4765. Airplane ATPL CPL Heli ATPL CPL How long will it take to travel 284 NM at a speed of 526 km/h?

230913. Airplane ATPL CPL Heli ATPL CPL The accuracy of the, manually calculated, DR-position of an aircraft is, among other things, affected by: A) the accuracy of the forecasted wind.

A) 1,6 hrs B) 1,9 hrs C) 45 min D) 1 hour

B) the accuracy of the actual wind. C) the accuracy of the adjustment of the position lines for

the motion of the aircraft between the last fix and the DRposition. D) the accuracy of the adjustment of the position lines for the motion of the aircraft between the last and the new DRposition.

7NM = 7,852 km 526 km = 284 NM (kts)

Distance = Rate x Time. 284 NM = 284 kts x ? hrs

? hrs = 284 NM + 284 kts ?= 7 hr

4772. Airplane ATPI (PI Heli ATPI ,PI How long will it take to fly 5 NM at a ground speed of 269 kts? A) 1 min 07 sec B) 1 min 55 sec C) 2 min 30 sec D) 0 min 34 sec Distance = Rate x Time. 5 NM =269 kts x ?hrs ?hrs =5 NM +269 kts ? = 0,078587 hrs = 7,72 minutes (0,078587 x 60) 7,72 min = 7 minute 07 seconds

4777. Given:

Airplane

ATPL

CPL

Heli

ATPL

CPL

A) the accuracy of the adjustment of the position lines for the motion of the aircraft between the last fix and the DRposition. B) the accuracy of the actual wind. C) the flight time since the last position update. D) the accuracy of the adjustment of the position lines for the motion of the aircraft between the last and the new DRposition. 230915. Airplane ATPL CPL Heli ATPL CPL Consider the following factors that determine the accuracy of a DR position: 1) The flight time since the last position update. 2) The accuracy of the forecasted wind. 3) The accuracy of the TAS. 4) The accuracy of the steered heading.

236 kts 354NM

GS: Distance from A to B:

230914. Airplane ATPL CPL Heli ATPL CPL The accuracy of the,· manually calculated, DR-position of an aircraft is, among other things, affected by:

What is the time from A to B? A) 1 hr09 min B) 1 hr 30 min C) 1 hr10min D) 1 hr40 min

Using the list above which of the following contains the most complete answer? A) 1,2

B) 1,2,3 C) 1,2,4 D) 1,2,3,4

04-02 Use of the navigational computer 1258. Given:

Airplane

ATPL

CPL

Heli

CPL

Calculate the values for TAS and local speed of sound (LSS). A) B) C) D)

FL350 Mach 0,80 OAT-55°C 1 4737 (A)

ATPL

1 4765 (0) 1 4772 (A)

1 4777 (8)

TAS: 460 kts; LSS: 296 kts. TAS: 237 kts; LSS: 296 kts. TAS: 490 kts; LSS: 460 kts. TAS: 460 kts; LSS: 575 kts.

1230912 (C) 1230913 (A) 1230914 (C) 1230915 (0) 1 1258 (0)

1

Aviationexam Test Prep Edition 2012 Mach number is the ratio of the aircraft's speed (TA5) to the local speed of sound (L55). Mach No. TA5 ... L55. The 5peed of 50und varies only with the temperature. As the temperature increases so does the local speed ofsound. . Because the temperature reduces with altitude, the speed of sound reduces bspltitude increases. A formula for calculating the L55 for a given temperature is: L55 38,95 x ';Absolute temperature (in °Kelvin) ... Absolute temp in oK = °C + 273. Temperature in °C = oK - 273. We can also use an alternative formula, using 0c, but this does not yield a precise result - only an estimation: L55 = 644 + (1,2 x Temp °C).

=

=

In our case: oK = 278° (-55° + 273) L55 =575 kts (38,95 x ';278°) Mach Nr. = TA5 -;- L55 TA5 = Mach Nr. x L55 TA5 = 0,8 x 575 TA5 =460kts

4428. Given:

Airplane

ATPL

CPL

Heli

ATPL

CPL

310° 200 kts 176 kts 7° right

Calculate the WIV. ATPL

CPL

5pecific gravity (5G) = the ratio of the density of the fluid to the density of water. The 5G of water is 7, which means that 7kg of water is equal to 7litre of water.

CPL

Heli

B) 360°/33 kts.

(Refer to figures 067-E775, 067-E776, 067-E777, 067-E55, 067-E67 and 067-E62) For an example of solution of a wind direction and velocity problem using a navigation computer refer to the attached illustration. Please note that the illustration to this type of question only explains the solution in general and does not necessarily need to depict the solution of this specific question. This type of question can also be easily solved by drawing the triangle ofvelocities - see the attached illustration for information on the triangle of velocities along with examples ofproblem solutions.

7 U5 Gallon =3,785 litres 265 U5 Gallons = 70031itres. 7003litres x 0,8 =802,4 kg =approximately 803 kg

ATPL

A) 090°/33 kts. C) 270°/33 kts. D) 180°/33 kts.

A) 862 kg B) 803 kg C) 895 kg D) 940 kg

Airplane

(Refer to figures 067-E58 and 067-£59) Refer to the attached illustration for an example ofsolution of this type ofproblem. Please note that the solution only demonstrates the method in general and does not necessarily have to illustrate solution of the problem with these specific values.

True heading: TAS: GS: Drift angle:

4297. Airplane ATPL CPL Heli 265 US-GAL equals? (Specific gravity 0,80)

4381.

D) 444kts

ATPL

CPL

A fuel amount of 160 US Gal allows a endurance of 3 hrs 10 min with a light twin engine piston aircraft. What is the corresponding fuel flow per engine?

A) 25,3 US Gallhr B) 50,5 US Gal/hr C) 51,6 US Gallhr D) 25,8 US Gal/hr

4431. Airplane ATPL CPL Heli ATPL CPL The tank capacity of an aircraft is 310 US GAL. Fuel specific gravity is 0,78 kg/litre. The tanks are now 3/4 full. You want to refuel so that total fuel will be 850 kg. How much fuel will you have to refuel? / A) 1641bs

B) 360lbs

C) 320lbs D) 410lbs

3 hrs 70 minutes =3, 7666 hrs (Refer to figure 067-E68) If 760 U5 Gallons provides an endurance of3,7666hrs for a two-engine aircraft, 5ee the attached illustration for solving unit conversion problems using then the hourly fuel flow is 760 -;- 3,7666 = 50,53 U5 Gallons per hour of flight. a navigation computer. Please note that the illustration demonstrates Note that it is a 2-engine aircraft, therefore we divide this figure by 2 to get ", the methods in general and does not necessarily have to demonstrate the soluthe hourly fuel flow per each engine => 50,53 -;- 2 =25,27 U5 Gallons per engine tion (jf the problem using the values stated in this specific question. per hour. 5pecific gravity (5G) = the ratio of the density of the fluid to the density of water. The 5G ofw(/ter is 7, which means that 7 kg of water is equal to 7 litre of water. 4387. Airplane ATPL CPL Heli ATPL CPL

730 ft/min equals: A) B) C) D)

7 U5 Gallon =3,185 litres 370 U5 Gallons'; 7, 773,51itres (total capacity of the tanks) The tanks are -% full =75% of 7. 773,51itres =880,7 litres in the tanks.

3,7 m/sec 5,2 m/sec 1,6 m/sec 2,2 m/sec

If we want to refuel and have 850 kg of fuel available, this would be 7.089,7litres (850 kg -;- 0,78 5G).lt means that we need to refuel a total amount

(Refer to figure 067-E68) 5ee the attached illustration for solving unit conversion problems using a navigation computer. Please note that the illustration demonstrates the methods in general and does not necessarily have to demonstrate the solution of the problem using the values stated in this specific question.

7 m =3,2808 ft (1 ft =0,3048 m) 730 ft per minute =222,5 meters per minute 222,5 mlmin =3,7 mlsec (222,5 -;- 60) 4399. Given:

Airplane

CAS: FL290 OAT:

CPL

324kts -46°C

What is the TAS? A) 487 kts B) 473 kts C) 458 kts

ATPL

of209,61itres (7.089,7- 880,7). 209,61itres = 763,49 kg (209,6Iitres x 0,785G) 7 kg=2,2046Ibs 763,49 kg = 360,4lbs

4440.

Airplane

ATPL

CPL

Heli

ATPL

CPL

Given:

Heli

ATPL

CPL

Magnetic track: HOG: VAR: TAS:

315° 301°(M) SoW 225 kts

The aircraft flies 50 NM in 12 min. Calculate the WIV (OT). A) B) C) D)

195° 1 63 kts. 355° 1 15 kts. 195° 1 61 kts. 190° 1 63 kts.

(Refer to figures 067-E775, 067-E776, 067-E777, 067-E55, 067-E67, 067-E62 and 067-E96) First of all we need to convert the magnetic value of the heading to true value:

14297(8) 14381 (A) I 4387 (A) I 4399 (A) I 4428 (C) 14431 (8) I 4440 (D) I

04 Dead Reckoning Navigation (DR) TH = MH - Westerly variation (or + Easterly variation) TH=301°-5°W TH=296°

4449. Given:

Now we need to do the same with the track - convert it from magnetic to true, using the same procedure of deducting 5° of Westerly variation: True Track (TT) =315° - 5°W TT=310° The last thing that we need in order to use the navigation computer is the Groundspeed (GS). If the aircraft covers a distance of 50 NM in 12 minutes it will cover a distance of 250 NM in 60 minutes => GS is 250 kts (50 NM + 0,2 hrs). For an example of solution of a wind direction and velocity problem using a navigation computer refer to the attached illustration. Please note that the illustration to this type of question only explains the solution in general and does not necessarily need to depict the solution of this specific question.

Airplane

TrueHDG: TAS: Track: GS:

ATPL

Heli

CPL

ATPL

CPL

035° 245 kts 046° (T) 220 kts

Calculate the W/V.

A) 335/55 kts B) 335/45 kts C) 340/50 kts 0) 340/45 kts

This type of question can also be easily solved by drawing the triangle of velocities - see the attached illustration for information on the triangle of velocities along with examples ofproblem solutions.

(Refer to figures 067-£115, 067-£116,067-£117, 061-E55, 061-E61 and 067-£62) For an example of solution of a wind direction and velocity problem using a navigation computer refer to the attached illustration. Please· note that the illustration to this type of question only explains the solution in general and does not necessarily need to depict the solution of this specific question.

4448. Given:

This type of question can also be easily solved by drawing the triangle ofvelocities - see the attached illustration for information on the triangle of velocities along with examples ofproblem solutions.

Airplane

ATPL

Compass heading: Deviation: Variation: TAS:

Heli

CPL

ATPL

CPL

090° 2°W 12°E 160 kts

4461. Airplane ATPL CPL Heli ATPL CPL The following information is displayed on an Inertial Navigation System:

Whilst maintaining a radial 070° from a VOR station, the aircraft flies a ground distance of 14 NM in 6 min. What is the W/V

0(T}? A) 155°/25 kts. B) 340°/75 kts. C) 340°/98 kts. 0) 160°/50 kts.

GS: TrueHDG: Drift angle: TAS: SAT (static air temperature):

520 kts 090° 5° right 480kts -51°C

The W/V being experienced is:

(Refer to figures 061-E115, 067-£116, 061-E117, 061-E55, 067-£61, 067-£62, 067-£95 and 067-£96) First of all we need to convert the compass heading into magnetic heading using the deviation: MH CH - Westerly deviation (or + Easterly deviation) MH 090° - 2°W MH 088°

= = =

Next we need to convert the magnetic heading into a true heading using the variation: TH = MH - Westerly variation (or + Easterly variation) TH=088°+ 12°E TH= 100° Next step is to find the track. The question states that the aircraft is maintaining a VOR radial of 070°. We know that radials are magnetic bearings - therefore, our magnetic track is 070 0• To convert it from into a true track we will use the same procedure as for the heading =adding 12" of Easterly variation: True Track (TT) 070 0 + 12°E TT=082"

=

The last thing that we need in order to use the navigation computer Is-the Groundspeed (GS). If the aircraft covers a distance of 14 NM in 6 minutes it will cover a distance of 140NM in 60 minutes => GSis 140kts (14NM + 0,1 hrs). Now we have all of the information required to proceed with the navigation computer. For an example ofsolution ofa wind direction and velocity problem using a navigation computer refer to the attached illustration. Please note that the illustration to this type of question only explains the solution in general and does not necessarily need to depict the solution of this specific question. This type of question can also be easily solved by drawing the triangle of velocities - see the attached illustration for information on the triangle of velocities along with examples ofproblem solutions.

A) B) C) 0)

225°/60 kts. 320°/60 kts. 220°/60 kts. 325°/60 kts.

(Refer to figures 067-£115, 067-£116, 061-E117, 061-E55, 061-E61, 061-E62 and 061-E96) To obtain the track we need to realize that we are experiencing a drift of 5° to the right. It means that our wind is from the left and pushing us to the right => our track will be greater than our heading => track will be a True Heading of 090° + 5° of drift => 095 0• Now we have all of the information required to use the navigation computer to determine the wind. For an example of solution of a wind direction and velocity problem using a navigation computer refer to the attached illustration. Please note that the illustration to this type of question only explains the solution in general and does not necessarily need to depict the solution of this specific question. Note: You will not be able to use the E68 navigation computer for this question - use the CR3 one. This type of question can also be easily solved by drawing the triangleofvelocities - see the attached illustration for information on the triangle of velocities along with examples ofproblem solutions.

4463. Airplane ATPL CPL Heli ATPL CPL A fuel amount of 146 Imp Gal allows a endurance of 4 hrs 26 min. What is the corresponding fuel flow? A) 34,3 Imp Gal/hr

B) 32,9 US Gall hr C) 39,5 US Gal/hr 0) 39,5 Imp Gal/hr (Refer to figure 061-E68) See the attached illustration for solving unit conversion problems using a navigation computer. Please note that the illustration demonstrates the methods in general and does not necessarily have to demonstrate the solution of the problem using the values stated in this specific question. liMP Gallon = 1,2 US Gallons =4,546 litres 146 IMP Gallons = 175,2 US Gallons 4 hrs 26 min =4,43 hrs 175,2 + 4,43 =39,55 US Gallons per hour.

I

4448 (D)

I

4449 (C)

I

4461 (8)

I

4463 (C)

I

-

Aviationexam Test Prep Edition 2012

4468. Given:

Airplane

ATPL

True track: True heading: TAS:

Heli

CPL

ATPL

liMP Gallon = 1;2 US Gallons 83 IMP GAL = 99,6 US GAL With a fuel flowof22 US GALlhrand total fuel amountof99,6 US GAL the flight time available (endurance) is 4,53 hrs (99,6 + 22) => 4 hrs 32 minutes.

CPL

239° 229° 555 kts 577 kts

GIS:

4514. Given:

Calculate the wind velocity.

Heli

ATPL

CPL

A) 280 kts

Note: You will not be able. to use the E68 navigation computer for this type of question - use the CR3 one. This type of question can also be easily solved by drawing the triangle of velocities - see the attached illustration for information on the triangle of velocities along with examples ofproblem solutions. ATPL

Magnetic track: HDG: VAR: TAS:

Heli

CPL

ATPL

075° 066°(M)

274 kts 292 kts 287 kts

(Refer to figures 061-E58 and 061-E59) You can solve this question in two steps: 1) Determine the TAS from the given Mach number and FL. If no temperature is given, use ISA temp at the given FL. 2) Convert the TAS determined in step 1) above to CAS.

4515. Given:

Airplane

ATPL

CPL

Heli

ATPL

CPL

ATPL

CPL

230 kts

What is the TAS?

11°E

A) 266 kts B) 273 kts C) 280 kts D) 287 kts

275 kts

A) 335°/45 kts. B) 300°/50 kts. C) 210°/15 kts. D) 180°/45 kts.

For explanation refer to question #4399 on page 104.

(Refer to figures 061-E115, 061-E116, 061-E117, 061-E55, 061-E61, 061-E62 and 061-E96) First of all we need to convert the magnetic value of the heading to true value: TH =MH - Westerly variation (or + Easterly variation) TH=066°+ WE TH=07r Now we need to do the same with the track - convert it from magnetic to true, using the same procedure of adding 11 0 of Easterly variation: True Track (TT) = 075 0 + WE TT=086° The last thing that we need in order to use the navigation computer is the Groundspeed (GS). If the aircraft covers a distance of 48 NM in 10 minutes it will cover a distance of 288 NM in 60 minutes => GS is 288 kts (48 NM + 0,16666 hrs). For an example of solution of a wind direction and velocity problem using a navigation computer refer to the attached illustration. Please note that the illustration to this type of question only explains the solution in general and does not necessarily need to depict the solution of this specific question. This type of question can also be easily solved by drawing the triangle ofvelocities - see the attached illustration for information on the triangle of velocities along with examples ofproblem solutions.

4484. Airplane ATPL CPL Heli ATPL CPL Fuel flow per HR is 22 US-GAL, total fuel on board is 83 IMP GAL. What is the endurance? A) 4 hrs 32 min

B) 3 hrs 12 min C) 3 hrs 53 min D) 2 hrs 15 min

4533. Given:

Airplane

ATPL

CPL

Heli

307° 230 kts 313°(T) 210 kts

TrueHDG: TAS: Track: GS: Calculate the WIV. A) 255°/25 kts B) 257°/35 kts C) 260°/30 kts D) 265°/30 kts

(Refer to figures 061-E115, 061-E116, 061-El17, 061-E55, 061-E61 and 061-E62) For an example of solution of a wind direction and velocity problem using a navigation computer refer to the attached illustration. Please note that the illustration to this type of question only explains the solution in general and does not necessarily need to depict the solution of this specific question. This type of question can also be easily solved by drawing the triangle of velocities - see the attached illustration for information on the triangle of velocities along with examples ofproblem solutions.

4542. Given:

Airplane

ATPL

True track: TAS: True heading: GS:

CPL

Heli

ATPL

CPL

095° 160 kts 087° 130 kts

Calculate WIV.

(Refer to figure 061-E68) See the attached illustration for solving unit conversion problems using a navigation computer. Please note that the illustration demonstrates the methods in general and does not necessarily have to demonstrate the solution of the problem using the values stated in this specific question.

4468 (C)

B) C) D)

CAS: FL120 OAT:

CPL

Aircraft flies 48 NM in 10 min. Calculate the true WIV.

I

CPL

What is the CAS?

(Refer to figures 061-E115, 061-E116, 061-E117, 061-E55, 061-E61 and 061-E62) For an example af solution of a wind direction and velocity problem using a navigation computer refer to the attached illustration. Please note that the illustration to this type of question only explains the solution in general and does not necessarily need to depict the solution of this specific question.

Airplane

ATPL

FL310 MO,76

A) 300°/100 kts B) 310°/100 kts C) 130°/100 kts D) 165°/100 kts

4476. Given:

Airplane

I

4476 (A)

I

4484 (A)

I

4514 (A)

I

A) B) C) D)

124°/36 kts 23r136kts 30rl36kts 057°/36 kts

(Refer to figures 061-E115, 061-E116, 061-El17, 061-E55, 061-E61 and 061-E62) For an example of solution of a wind direction and velocity problem using

4515 (8)

I

4533 (C)

I

4542 (0)

I

.---------------

04 Dead Reckoning Navigation (DR)

a navigation computer refer to the attached illustration. Please note that the illustration to this type of question only explains the solution in general and does not necessarily need to depict the solution of this specific question. This type of question can also be easily solved by drawing the triangle ofvelocities - see the attached illustration for information on the triangle of velocities along with examples ofproblem solutions.

A) 265°/50 kts B) 195°/50 kts C) 235°/50 kts 0) 300°/30 kts (Refer to figures 061-E115, 061-E116, 061-El17, 061-E55, 061-E61, 061-E62 and 061-E96) First of all we need to convert the magnetic value of the heading to true value: TH = MH - Westerlx variation (or + Easterly variation)

4563. Airplane ATPL CPL Heli ATPL CPL You are flying at FL80 and air temperature is ISA +15. What CAS. --TF/=2150+YSOf- ----is required to make TAS 240 kts? TH =230° A) 208 kts B) 214 kts C) 220kts

Now we need to do the same with the track - convert it from magnetic to true, using the same procedure of adding 15° of Easterly variation: True Track(TT) =210°+ WE

0) 226 kts

IT =225° The last thing that we need in order to use the navigation computer is the Groundspeed (GS). If the aircraft covers a distance of 64 NM in 12 min-

For explanation refer to question #4399 on page 104.

4573. Given:

Airplane

ATPL

True altitude: OAT: CAS:

Heli

CPL

ATPL

utes it will cover a distance of 320 NM in 60 minutes => GS is 320 kts (64 NM + 0,2 hrs).

CPL

For an example of solution of a wind direction and velocity problem using a navigation computer refer to the attached illustration. Please note that the illustration to this type of question only explains the solution in general and does not necessarily need to depict the solution of this specific question. This type of question can also be easily solved by drawing the triangle ofvelocities - see the attached illustration for information on the triangle of velocities along with examples ofproblem solutions.

9.000ft -32°C 200 kts

What is the TAS? A) 215 kts

B) 200 kts C) 210 kts 0) 220 kts (Refer to figures 061-E58 and 061-E59) Refer to the attached illustration for an example ofsolution of this type ofproblem. Please note that the solution only demonstrates the method in general and does not necessarily have to illustrate solution of the problem with these specific values. Before you start with the TAS calculation you have to convert the True Altitude to Pressure altitude and enter the navigation computer with the correct Pressure altitude for the TAS determination: ISA at 9.000 ft is -3°C. Our actual temp is -32°C = ISA -29~ For every 10° of ISA deviation we apply a 4% correction of the altitude: 4% x 2,9 = 11,6% 11,6% of9.000 ft = 1.044 ft Pressure altitude = approx. 10.044 ft Then with the navigation computer yeiuwill find a TAS of approx.220 knots.

4590. Airplane ATPL CPL Heli ATPL CPL Fuel flow per HR is 31 US-GAL, total fuel on board is 260 I. What is the endurance?

A) 8 hrs 39 min B) 2 hrs 22 min C) 2 hrs 13 min 0) 8 hrs 23 min (Refer to figure 061-E68) See the attached illustration for solving unit conversion problems using a navigation computer. Please note that the illustration demonstrates the methods in general and does not necessarily have to demonstrate the solution of the problem using the values stated in this specific question. 1 US Gallon =3,785 Iitres 31 US Gallons ~ 117,34Iitres With 260 Iitres of fuel on board an an hourly fuel flow of 117,34 we can remain airborne for 2,21 hrs (260 + 117,34) => 2 hrs 13 min.

4615. Given:

Airplane

ATPL

Heli

CPL

ATPL

4618. Airplane ATPL CPL Heli ATPL CPL An aircraft is flying at FL180 and the outside air temperature is -30°C. If the CAS is 150 kts, what is the TAS? A) 115 kts

B) 195 kts

C) 180 kts 0) 145 kts For explanation refer to question #4399 on page 104.

4620.

I

4573 (0)

I

4590 (C)

I

4618 (8)

I

ATPL

CPL

090° 180 kts 180 kts 5° right

(Refer to figures 061-E115, 061-E116, 061-E117, 061-E55, 061-E61 and 061-E62) For an example of solution of a wind direction and velocity problem using a navigation computer refer to the attached illustration. Please note that the illustration to this type of question only explains the solution in general and does not necessarily need to depict the solution of this specific question. This type of question can also be easily solved by drawing the triangle ofvelocities - see the attached illustration for information on the triangle of velocities along with examples ofproblem solutions.

4626.

Airplane

ATPL

CPL

Given:

A) B) C) 0)

I

Heli

A) 360°/15 kts B) 190°/15 kts C) 010°115 kts 0) 180°/15 kts

145° 240 kts 150° 210 kts

Calculate the W/V.

4615 (A)

CPL

Calculate the W/V.

Calculate the true W/V.

4563 (A)

ATPL

True Heading: TAS: GS: Drift:

TrueHDG: TAS: Track (OT): GS:

CPL

210° Magnetic track: 215° Magnetic HOG: VAR: 15°E 360 kts TAS: Aircraft flies 64 NM in 12 min

I

Airplane

Given:

4620 (A)

I

360/35kts 180/35kts 295/35kts 115/35kts

4626 (0)

I

Heli

ATPL

CPL

Aviationexam Test Prep Edition 2012 (Refer to figures 061-E115, 061-E116, 061-El17, 061-E55, 061-E61 and 061-E62) For an example of solution of a wind direction and velocity problem using a navigation computer refer to the attached illustration. Please note that the illustration to this type of question only explains the solution in general and does not necessarily need to depict the solution of this specific question.

4651. Given:

This type of question can also be easily solved by drawing the triangle ofvelocities - see the attached illustration for information on the triangle of velocities along with examples ofproblem solutions.

OAT:

4627. Given:

Airplane

ATPL

TrueHDG: TAS: Track: GS:

Heli

CPL

ATPL

CPL

074° 230 kts 066°(T) 242 kts

Heli

ATPL

CPL

ATPL

CPL

140 kts +20°C

What is the TAS? A) B) C) D)

156 kts 160 kts 164 kts 168 kts

Airplane

ATPL

TAS:

CPL

Heli

168 kts

FL85 OAT: What is the CAS?

(Refer to figures 061-£115, 061-E116, 061-El17, 061-E55, 061-E61 and 061-£62) For an example of solution of a wind direction and velocity problem using a navigation computer refer to the attached illustration. Please note that the illustration to this type of question only explains the solution in general and does not necessarily need to depict the solution of this specific question. This type of question can also be easily solved by drawing the triangle ofvelocities - see the attached illustration for information on the triangle of velocities along with examples ofproblem solutions.

Airplane

ATPI

Heli

('PI

Fuel flow: Specific gravity: Mach number: OAT:

ATPI

A) B) C) D)

145 kts 150 kts 195 kts 188 kts

For explanation refer to question #4399 on page 104.

4676. Airplane ATPL CPL Heli ATPL CPL The equivalent of 70 mlsec is approximately: A) 145 kts B) 136 kts C) 210kts D) 35 kts

('PI

8,4t/hr

0,80

(Refer to figure 061-E68) 5ee the attached illustration for solving unit conversion problems using a navigation computer. Please note that the illustration demonstrates the methods in general and does not necessarily have to demonstrate the solution of the problem using the values stated in this specific question.

0,76 -36°C

What is the specific fuel consumption? A) 14,7 kg/NM air distance. B) 18,4 kg/NM air distance. C) 19,5 kg/NM air distance. D) 15,6 kg/NM air distance.

1 NM = 1,852 km

(Refer to figures 061-E58 and 061-E59) 5pecific gravity (5G) =the ratio of the density of the fluid to the density of water. The 5G of water is 1, which means that 1 kg of water is equal to 1 litre of water.

70 meters per second x 60 =4.200 meters per minute. 4.200 meters per minute x 60 =252.000 meters per hour 252.000 meters per hour + 1.000 = 252 kilometers per hour. 1 km =0,54 NM. 252 km x 0,54 = 136 NM = 136 knots.

Note: We do not need to apply the specific gravity in this solution at all - we would have to apply it if we were asked about the fuel consumption in terms of liters (volume) per hour, but not in terms of kg (mass) per hour.

4691. Given:

1st step will be to determine the TA5 based on the Mach number and temperature. Mach number is the ratio of the aircraft's speed (TA5) to the local speed of sound (L55). Mach No. = TA5 + L55. The 5peed of 50und varies only with the temperature. As the temperature increases so does the local speed ofsound. Because the temperature reduces with altitude, the speed of sound reduces as altitude increases. A formula for calculating the L55 for a given temperature is: L55 = 38,95 x {Absolute temperature (in °Kelvin) ... Absolute temp in oK =°C + 273. Temperature in °C =oK - 273. We can also use an alternative formula, using 0c, but this does not yield a precise result- only an estimation: L55 = 644 + (1,2 x Temp °C).

TAS:

In our case: oK =23r (-36° + 273) L55 = 599,6 kts (38,95 x {231") Mach Nr. =TA5 + L55 TAS =Mach Nr. x L55 TA5 =0,76 x 599,6 TA5 =456 kts

Airplane

4627 (8)

I

4630 (8)

I

4651 (C)

I

4664 (8)

I

ATPL

CPL

Heli

140 kts

FL80 OAT:

+20°C

What is the CAS? A) 120 kts B) 129 kts C) 151 kts D) 163 kts For explanation refer to question #4399 on page 104.

2 nd step will be to determine the specific fuel consumption (5FC). If the engines burn 8.400 kg of fuel (8,4 tons) to cover the air distance of 456 NM (TA5 of 456 kts), then the fuel required to cover 1 NM of air distance will be 8.400 + 456 = approx. 18,42 kg => 18,42 kg of fuel per 1 NM ofAIR distance.

I

CPL

FL80

4664. Given:

180/30 kts 180/35 kts 185/35 kts 180/40 kts

4630. Given:

ATPL

For explanation refer to question #4399 on page 104.

Calculate the WIV A) B) C) D)

CAS:

Airplane

4676 (8)

I

4691 (A)

I

ATPL

CPL

04 Dead Reckoning Navigation (DR) 4713. Given:

Airplane

ATPL

Heli

CPL

ATPL

4775. Airplane ATPL CPL Heli ATPL CPL What is the ratio between the litre and the US gallon?

CPL

233° 480 kts 2400 (T) 523 kts

TrueHDG: TAS: Track: GS:

A) 1 US-GAL equals 4,55 lit res. B) 1 litre equals 4,55 US-GAL. e) 1 US-GAL equals 3,78Iitres. 0) 1 litre equals 3,78 US-GAL.

Calculate the W/V.

(Refer to figure 061-E68) See the attached illustration for solving unit conversion problems using a navigation computer. Please note that the illustration demonstrates the methods in general and does not necessarily have to demonstrate the solution of the problem using the values stated in this specific question.

A) 115/70 kts B) 11 0/75 kts e) 110/80 kts 0) 105/75 kts (Refer to figures 061-E175, 061-E176, 061-E177, 061-E55, 061-E61 and 061-E62) For an example of solution of a wind direction and velocity problem using a navigation computer refer to the attached illustration. Please note that the illustration to this type of question only explains the solution in general and does not necessarily need to depict the solution of this specific question.

Summary of conversion factors for calculations without navigation computer: 1 US Gallon =3,785 litres liMP Gallon = 1,2 US Gallons =4,546 litres 1 litre = 0,2641 US Gallons 1 NM = 1,852 km""'= 1,1507 SM 1 SM = 1,609 km =0,8689 NM

Note: You will not be able to use the E68 navigation computer for this question - use the CR3 one.

1 m = 3,2808 ft = 39,37 inches 1 ft =0,3048 m = 12 inches

This type of question can also be easily solved by drawing the triangleofvelocities - see the attached illustration for information on the triangle of velocities along with examples ofproblem solutions.

1 kg = 2,20461bs lib = 0,4535 kg

4727. Given:

Airplane

ATPL

Heli

CPL

ATPL

4824. Given:

CPL

+20°C

What is the TAS?

Airplane

Fuel flow: Specific gravity: TAS:

ATPL

Heli

CPL

ATPL

ATPL

CPL

353° 130 kts 132 kts

(Refer to figures 061-E175, 061-El16, 061-E177, 061-E55, 061-E61 and 061-E62) For an example of solution of a wind direction and velocity problem using a navigation computer refer to the attached illustration. Please note that the illustration to this type of question only explains the solution in general and does not necessarily need to depict the solution of this specific question.

CPL

28 Imp Gal/hr 0,72 154MPH

This type of question can also be easily solved by drawing the triangle ofvelocities - see the attached illustration for information on the triangle of velocities along with examples ofproblem solutions.

What is the specific range?

4852. Given:

A) 1,46 NM air distance / kg. B) 1,67 NM air distance / kg. e) 0,68 NM air distance / kg. 0) 2,0 NM air distance / kg. (Refer to figure 061-E68) Range is the distance an aircraft can fly on the amount of fuel available. The term "Specific Range (SR)" is used to define the number of miles per 1 unit of fuel (eg.? NM per 1 kg offuel) => for example 10NMI 1 kg offuel.lfwehave a total fuel on board (excluding reserves) of 100 kg, then our maximum range can be determined by dividing the total fuel by the SR => 100 kg x 10 NM (per 1 kg) => 1.000 NM range. Solution in our case will involve the following steps: liMP Gallon = 1,2 US Gallons = 4,546 litres 28 IMP Gallons = 127,29 litres Specific Gravity of 0,72 means that 1 liter of fuel equals to 0,72 kg. 127,29 litres with SG of 0,72 = 91,65 kg Note that the abbreviation MPH stands for STATUTE miles per hour. We need the TAS in NAUTICAL miles per hr (kts). 1 SM = 1,609 km =0,8689 NM 154 SM = 133,82 kts

I

4727 (0)

I

4766 (A)

I

4775 (C)

I

4824 (8)

I

Airplane

TrueHDG: TAS: Track: GS:

ATPL

CPL

Heli

ATPL

CPL

133° 225 kts 144° (T) 206 kts

Calculate the WIV. A) 070/40 kts B) 075/45 kts e) 070/45 kts 0) 075/50 kts (Refer to figures 061-E115, 061-E176, 061-E177, 061-E55, 061-E61 and 061-E62) For an example of solution of a wind direction and velocity problem using a navigation computer refer to the attached illustration. Please note that the illustration to this type of question only explains the solution in general and does not necessarily need to depict the solution of this specific question. This type of question can also be easily solved by drawing the triangle ofvelocities - see the attached illustration for information on the triangle of velocities along with examples ofproblem solutions.

Now we have the information that our aircraft burns 91,65 kg of fuel to cover the air distance of 133,82 NM. From this information we can easily determine the distance covered on 1 kg of fuel => 133,82 kts .,. 91,65 kg = 1,46 NM per 1 kg of fuel.

4713 (8)

Heli

A) 088/15 kts B) 092 / 20 kts e) 270 / 20 kts 0) 095 / 25 kts

For explanation refer to question #4399 on page 104.

I

CPL

Calculate the WIV:

A) 132 kts B) 102 kts e) 120 kts 0) 141 kts

4766. Given:

ATPL

True HOG: True Track: TAS: GS:

120 kts

lAS: FL80 OAT:

Airplane

4852 (8)

I

Aviationexam Test Prep Edition 2012

4875. Given:

Airplane

ATPL

FL120 OAT:

CPL

Heli

ATPL

CPL

ISA standard

CAS:

200kts 222°(M) 215°(M) 15°W. 21 min.

Track: Heading: Variation: Time to fly 105 NM:

To obtain the track we need to realize that we are experiencing a drift of 11° to the right. It means that our wind is from the left and pushing us to the right => our track wi/l be greater than our heading => track will be a True Heading of7200+ 11°ofdrift=> 737°. Now we have all of the information required to use the navigation computer to determine the wind.

What is the WIV? A) B) C) D)

TA5 =466 kts Now we need to convert the magnetic value of the heading to true value: TH =MH - Westerly variation (or + Easterly variation) TH= 7400-200W TH= 720°

For an example of solution of a wind direction and velocity problem using a navigation computer refer to the attached illustration. Please note that the illustration to this type of question only explains the solution in general and does not necessarily need to depict the solution of this specific question.

0

050 (T)/70 kts 040 0 (T)/l05 kts 055°(T)/l05 kts 065°(T)/70 kts

(Refer to figures 067-E175, 067-E116, 067-E117, 067-E55, 067-E58, 067-E59, 067-E67, 067-E62 and 067-E96) First of all we need to convert the magnetic value of the heading to true value: TH = MH - Westerly variation (or + Easterly variation) TH=275°-75°W TH=200°

Note: You will not be able to use the E6B navigation computer for this question - use the eR3 one. This type ofquestion can also be easily solved by drawing the triangle ofvelocities - see the attached i/Iustration for information on the triangle of velocities along with examples ofproblem solutions.

4922. Given:

Airplane

Now we need to do the same with the track - convert it from magnetic to true, using the same procedure of deducting 75° of Westerly variation: TrueTrack(TT)=222°-75°W TT=20r

TAS:

The next thing is the Groundspeed (G5) calculation. If the aircraft covers a distance of 705 NM in 27 minutes it will cover a distance of 300 NM in 60 minutes => G5 is 300 kts (50 NM -;- 0,35 hrs).

Calculate the WIV.

Now you are ready for the use of the navigation computer, however not yet for the wind calculation. First of all you need to use the navigation computer to convert the CAS into TA.'>. You wi/l get a TAS value of approximately 239 kts (use a temperature of _9° = ISA at FL720). Now you are ready to proceed with the actual wind calculation. For an example of solution of a wind direction and velocity problem using a navigation computer refer to the attached i/Iustration. Please note that the illustration to this type of question only explains the solution in general and does not necessarily need to depict the solution of this specific question.

4890. Given:

Airplane

ATPL

M:

CPL

Heli

ATPL

CPL

0,80 -50°C

OAT: FL330 GS: VAR: Magnetic heading: Drift:

490 kts

A) B) C) D)

TrueHDG: Track: GS:

Heli

ATPL

CPL

A) 010/55 kts B) 005/50 kts C) 010/50 kts D) 010/45 kts (Refer to figures 067-E175, 067-E116, 067-E117, 067-E55, 067-E67 and 067-E62) For an example of solution of a wind direction and velocity problem using a navigation computer refer to the attached illustration. Please note that the illustration to this type of question only explains the solution in general and does not necessarily need to depict the solution of this specific question. Note: You will not be able to use the E6B navigation computer for this question - use the CR3 one. This type of question can also be easily solved by drawing the triangle ofvelocities - see the attached illustration for information on the triangle of velocities along with examples ofproblem solutions.

4928. Given:

Airplane

TrueHDG:

200°/95 kts. 025°/47 kts. 020°/95 kts. 025°/45 kts.

CPL

054° 450kts 059°(T) 416 kts

TAS:

Calculate the true W/V.

ATPL

Track: GS:

ATPL

CPL

Heli

ATPL

CPL

206° 140 kts 207°(T) 135 kts

Calculate the WIV.

(Refer to figures 067-E115, 067-E116, 067-E777, 067-E55, 067-E67, 067-E62 and 067-E96) Mach number is the ratio of the aircraft's speed (TA5) to the local speed of sound (L55). Mach No. TA5 -;- L55. The Speed of 50und varies only with the temperature. As the temperature increases so does the local speed ofsound. Because the temperature reduces with altitude, the speed of sound reduces as altitude increases. A formula for calculating the L55 for a given temperature is: L55 = 38,95 x ,jAbsolute temperature (in °Kelvin) ... Absolute temp in OK = °C + 273. Temperature in °e = OK - 273. We can also use an alternative formula, using 0(, but this does not yield a precise result - only an estimation: L55 =644 + (1,2 x Temp °e).

=

In our case: OK = 223° (-50° + 273) L55 = 582 kts (38,95 x ';223°) Mach Nr. = TA5 -;- L55 TA5 = Mach Nr. x L55 TA5 = 0,8 x 582

I 4875 (A) I 4890 (C) I 4922 (C) 14928(0) I

A) B) C) D)

180/10 kts 000105 kts 000/10 kts 180105 kts

(Refer to figures 067-E115, 067-E116, 067-E117, 067-E55, 067-E67 and 067-E62) For an example of solution of a wind direction and velocity problem using a navigation computer refer to the attached illustration. Please note that the illustration to this type of question only explains the solution in general and does not necessarily need to depict the solution of this specific question. This type ofquestion can also be easily solved by drawing the triangle ofvelocities - see the attached illustration for information on the triangle of velocities along with examples ofproblem solutions.

04 Dead Reckoning Navigation (DR)

4931. Given: TAS: OAT: FL410

Airplane

ATPL

CPL

Heli

ATPL

CPL

Mach Nr. = TAS .,. LSS Mach Nr. = 545 kts .,. 571 kts Mach Nr. = 0,95

485 kts ISA+10°C

4935.

Calculate the Mach number.

Fuel flow: Specific gravity: TAS:

(Refer to figures 061-E58 and 061-E59) Refer to the attached illustration for an example ofsolution ofthis type ofproblem. Please note that the solution only demonstrates the method in general and does not necessarily have to illustrate solution of the problem with these specific values. If you do not have a navigation computer, you can also solve this question mathematically. Remember that Mach number is the ratio of the aircraft's speed (TAS) to the local speed of sound (LSS). Mach No. TAS + LSS. The Speed of Sound varies only with the temperature. As the temperature increases so does the local speed of sound. Because the temperature reduces with altitude, the speed of sound reduces as altitude increases. A formula for calculating the LSS for a given temperature is: LSS = 38,95 x ,fAbsolute temperature (in °Kelvin) ... Absolute temp in OK = °C + 273. Temperature in °C = OK - 273. We can also use an alternative formula, using °C, but this does not yield a precise result - only an estimation: LSS = 644 + (1,2 x Temp °C).

=

To obtain the Mach Number from TAS and OAT: 1) Find OAT in OK (= OAT in °C + 273) 2) Find LSS (=38,95 x ,fOAT in OK) 3) Find Mach nr. (= TAS + LSS)

ATPL

CPL

Heli

ATPL

CPL

Airplane

ATPL

CPL

42 US Gal/hr 0,72 210 kts

What is the specific fuel consumption? A) 1,052 kg/NMair distance. B) 0,757 kg/NM air distance. C) 0,144 kg/NM air distance. D) 0,545 kg/NM air distance. (Refer to figure 061-E68) Solution in our case will involve the following steps: 1 US Gallon =3,785litres 42 US Gallons = 159litres Specific Gravity of 0,72 means that 1 liter of fuel has a mass of 0,72 kg. 159litres with SG of 0,72 = 114,5 kg Now we have the information that our aircraft burns 114,5 kg of fuel to cover the air distance of210 NM (TAS of 210 kts). From this information we can easily determine the specific fuel consumption = fuel required to cover the distance of 1 NM => 114,5 kg.,. 210 NM = 0,545 kg of fuel per 1 NM of air distance.

4938.

Airplane

ATPL

CPL

Heli

ATPL

CPL

Given:

Note: Watch out for the questions where the JAA asks for a temperature calcu lation at high flight levels. Above FL 360 the temp is fairly constant - the ISA = -56.5°C.

Heli

ATPL

CPL

Given:

Fuel flow: Specific gravity: TAS:

281mp Gal/hr 0,72 154mph

What is the specific fuel consumption? A) 0,60 kg/NM air distance.

487 kts TAS: FL330 Temperature: ISA +15 Calculate the Mach number. A) B) C) D)

Airplane

Given:

A) 0,85 B) 0,90 C) 0,825 D) 0,87

4932.

Actual temp (OK) = 2W (-58° + 273) LSS = 571 kts (38,95 x ,f2W)

0,81 0,84 0,76 0,78

B) 0,68 kg/NM air distance. C) 1,46 kg/NM air distance. D) 0,50 kg/NM air distance.

(Refer to figure 061-E68) Solution in our case will involve the following steps: liMP Gallon = 1,2 US Gallons =4,546litres 28 IMP Gallons = 121,291itres Specific Gravity of 0,72 means that 1 liter of fuel equals to 0,72 kg. 121,29litres with SG of 0,72 = 91,65 kg

For explanation refer to question #4931 on this page.

4934. Airplane ATPL CPL Heli ATPL CPL If the headwind component is 50 kts, the FL is 330, temperature ISA -7°C and the ground speed is 495 kts, what is the Mach number? A) 0,98 B) 0,78 C) 0,95

Note that the abbreviation MPH stands for STATUTE miles per hour. We need the TAS in NAUTICAL miles per hr (kts). 1SM = 1,609 km =0,8689 NM 154SM= 133,82kts Now we have the information that our aircraft burns 91,65 kg of fuel to caver .the air distance of 133,82 NM. From this information we can easily determine the specific fuel consumption = fuel required to cover the distance of 1 NM => 91,65 kg.,. 133,82 kts =0,68 kg offuel per 1 NM of air distance.

4942.

D) 0,75

Airplane

ATPL

CPL

Heli

ATPL

CPL

Given:

Mach number is the ratio of the aircraft's speed (TAS) to the local speed of sound (LSS). Mach No. TAS + LSS. The Speed of Sound varies only with the temperature. As the temperature increases so does the local speed ofsound. Because the temperature reduces with altitude, the speed of sound reduces as altitude increases. A formula for calculating the LSS for a given temperature is: LSS = 38,95 x ,fAbsolute temperature (in °Kelvin) ... Absolute temp in OK = °C + 273. Temperature in °C = OK - 273. We can also use an alternative formula, using 0(, but this does not yield a precise result - only an estimation: LSS = 644 + (1,2 x Temp °C).

=

In our case: TAS = 495 kts (GS) + 50 kts headwind TAS =545kts

FL250 OAT: TAS:

-15°C 250 kts

Calculate the Mach number. A) 0,44 B) 0,40 C) 0,39 D) 0,42 For explanation refer to question #4931 on this page.

Standard temp (OC) at FL330 = -51° Actual temp (OC) at FL330 = -58° (-51 0_7")

14931 (C) I 4932 (A) I 4934 (C) 14935(0) 14938(8) 14942(8) I

III

Aviationexam Test Prep Edition 2012 4956. Given:

Airplane

ATPL

Heli

CPL

ATPL

(Refer to figure 061-E52) On your flight computer, find the value of for example 120 kts TAS on the outer ring scale and align it with the value of 100 kts CAS on the inner ring scale (these values meet the requirement that TAS is 20% more than CAS). Then in the True Airspeed window, read the temperature opposite the value of 10.000 ft. You will find approx. +13°C.

CPL

140 kts

CAS: FL130 TAS: What is the OAT?

174 kts

4959. Given:

A) -11°C B) +6°C C) O°C D) -6°C

CAS:

ATPL

CPL

Heli

ATPL

CPL

ATPL

CPL

ATPL

CPL

ATPL

CPL

130 kts 1.000ft 127 kts

PA: TAS:

For explanation refer to question #4399 on page 104.

Airplane

What is the OAT?

4957. Airplane ATPL CPL Heli ATPL CPL If the TAS exceeds the CAS by 20% at FL100, the OAT should be: A) -5°C. B) +5 dc. C) +15°C. D) is not defined.

A) +20°C B) +10°C C) O°C D) -8°C For explanation refer to question #4399 on page 104.

04-03 The triangle of velocities 4279. Airplane ATPL CPL Given: TAS: 485 kts True HOG: 226° 11 00(T)/95 kts WIV:

Heli

ATPL

4300. Airplane Given: TAS: True HOG:

CPL

W/V:

Calculate the drift angle and GS. A) B) C) D)

ATPL

CPL

Heli

472 kts 005° 1100 (T)/50 kts

Calculate the drift angle and GS.

7°R - 531 kts. 9°R - 533 kts. 9°R - 433 kts. SOL - 435 kts.

A) B) C) D)

(Referto figures 061-E115, 061-E116, 061-E117, 061-E55 and 061-E65) For an example of the solution of a Drift and Ground speed problem refer to the attached illustration. Please note that the illustration to this type of question only explains the solution in general and does not necessarily need to depict the solution of this specific question. This type of question can also be easily solved by drawing the triangle ofvelocities - see the attached illustration for information on the triangle of velocities along with examples ofproblem solutions.

6°L - 487 kts. 7°R - 491 kts. 7°L - 491 kts. 7°R - 487 kts.

For explanation refer to question #4279 on this page.

4301. Given:

Airplane

TAS: HOG: WIV:

ATPL

CPL

Heli

235 kts 076° (T) 040/40 kts

Note: Some questions give the heading value as Magnetic. Remember that the winds aloft (cruise winds) are given in True directions (0T), so a conversion from Magnetic to True might be required for the heading. Surface winds received by ATiS or given by ATC over the radio are in ° Magnetic. Winds listed in METARs are in ° True.

Calculate the drift angle and GS.

4280. Given:

For explanation refer to question #4279 on this page.

Airplane

ATPL

Heli

CPL

ATPL

CPL

A) 5°R - 207 kts. B) 7°L - 269 kts. C) SOL - 255 kts. D) 7°R - 204 kts.

4302. Given:

TAS: HOG: WIV:

140 kts 005°(T) 265/25 kts Calculate the drift and GS.

Airplane

ATPL

Heli

CPL

TAS: 190 kts True HOG: 085° 110 0 (T)/50 kts W/V: Calculate the drift angle and GS.

A) noR - 140 kts. B) 9°R -140 kts. C) noR -142 kts. D) 1QoR - 146 kts.

A) 8°L - 146 kts. B) rL -156 kts. C) 4°L -168 kts. D) 4°R - 145 kts.

For explanation refer to question #4279 on this page.

For explanation refer to question #4279 on this page.

I

4956 (C)

I

4957 (C)

I

4959 (0)

I

4279 (8)

I

4280 (0)

I

4300 (A)

I

4301 (0)

I

4302 (A)

I

04 Dead Reckoning Navigation (DR)

4303. Airplane ATPL CPL Heli Given: Magnetic heading: 255° 400W VAR: GS: 375 kts 235°(T)1120 kts WIV: Calculate the drift angle.

A) B)

ATPL

4339. Given:

CPL

Airplane

True Heading: TAS: WIV:

ATPL

Heli

CPL

ATPL

CPL

ATPL

CPL

ATPL

CPL

ATPL

CPL

ATPL

CPL

180° 500 kts 225°/100 kts

Calculate the GS. A) 450 kts B) 600 kts C) 535 kts 0) 435 kts

r left. r right.

C) 9° left. 0) 16° right.

For explanation refer to question #4279 on page 112.

For explanation refer to question #4279 on page 112.

4318. Given:

Airplane

TAS: True HOG: Actual wind:

ATPL

Heli

CPL

ATPL

4348. Given:

CPL

Airplane

ATPL

Heli

CPL

470 kts TAS: 317° True HOG: 045°(T)/45 kts WIV: Calculate the drift angle and GS.

270 kts 145° 205°(T)/30 kts

Calculate the drift angle and GS.

A) 3°R - 470 kts. B) SOL - 470 kts. C) SOL - 475 kts. 0) 5°R - 475 kts.

A) gOR - 261 kts. B) 6°R - 251 kts. C) 6°L - 256 kts. 0) 6°R - 259 kts.

For explanation refer to question #4279 on page 772.

For explanation refer to question #4279 on page 112.

4319. Given:

Airplane

TAS: True HOG: Actual wind:

ATPL

Heli

CPL

ATPL

4391. Given:

CPL

Airplane

Heli

CPL

132 kts 257° 095°(T)/35 kts

TAS: True HOG: W/V:

270 kts 270° 205° (T)/30 kts

ATPL

Calculate the drift angle and GS.

Calculate the drift angle and GS.

A) 2°R - 166 kts. B) 4°R - 165 kts. C) 4°L -167 kts. 0) 3°L -166 kts.

A) 6°R - 259 kts. B) 6°L - 256 kts. C) 6°R - 251 kts. 0) gOR - 259 kts.

For explanation refer to question #4279 on page 112.

For explanation refer to question #4279 on page 772.

4320. Given:

Airplane

ATPL

Heli

CPL

ATPL

4392. Given:

CPL

ATPL

Heli

CPL

290 kts 171° 310°(T)/30 kts

TAS: True HOG: WIV:

TAS: 205 kts 180° (T) HOG: 240/25 kts WIV: Calculate the drift and GS. A) B) C) 0)

Airplane

Calculate the drift angle and GS. A) B) C) 0)

7°L - 192 kts. 6°L - 194 kts. 3°L -190 kts. 4°L -195 kts.

4°R - 310 kts. 4°L - 314 kts. 4°R - 314 kts. 4°L - 310 kts.

For explanation refer to question #4279 on page 112.

For explanation refer to question #4279 on page 772.

4322. Given:

Airplane

ATPL

TAS: Actual HOG: Wind:

Heli

CPL

ATPL

4393. Given:

CPL

ATPL

Heli

CPL

150 kts 270° 245/12 kts

TAS: Actual HOG: Wind:

120 kts 150° 245112 kts

What is the wind correction angle?

What is the wind correction angle? A) B) C) 0)

Airplane

A) B) C) 0)

6° to the right. 6° to the left. 12° to the right. 12° to the left.

4° to the right. 4° to the left. 2° to the right. 2° to the left.

For explanation refer to question #4279 on page 112.

For explanation refer to question #4279 on page 772.

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4303 (A)

I

4318 (C)

I

4319 (A)

I

4320 (8)

I

4322 (A)

I

4339 (0)

I

4348 (8)

I

4391 (8)

I

4392 (8)

I

4393 (0)

I

Aviationexam Test Prep Edition 2012

4419. Given:

Airplane

ATPL

TAS: HOG: WIV:

Heli

CPL

ATPL

should be:

CPL

A) 322 0 (M) B) 328 0 (M) C) 3160 (M) 0) 3260 (M)

465 kt$ 124° (T) 170/80 kts

Calculate the drift and GS.

(Refer to figures 067-E775, 067-E116, 067-E117, 067-E55, 067-E63 and 067-E64) Let's summarize the information we have: Track =322" TAS=95kts WN=350120

A) 8°L - 415 kts. B) 3°L - 415 kts. C) 4°L - 400 kts. 0) 6°L - 400 kts.

Remember that the surface winds received by ATC over the radio or given in ATIS are in ° Magnetic. Winds listed in METARs and TAFs are in ° True. Therefore, in this case the wind reported by the tower is in °Magnetic. Runway headings are also in ° Magnetic.

For explanation refer to question #4279 on page 112.

4437. Given:

Airplane

Runway direction: Surface WIV:

ATPL

Heli

CPL

ATPL

CPL

For an example of solution of a wind correction and Ground speed problem refer to the attached illustration. Please note that the illustration to this type of question only explains the solution in general and does not necessarily need to depict the solution of this specific question.

230°(T) 280°(T)/40 kts

Calculate the effective cross-wind component. A) B) C) 0)

/fyou do not have the navigation computer available you can use a calculation method:

21 kts 36 kts 31 kts 26 kts

Wind angle = 350° - 322° = 28° XWC = sin (wind angle) x wind speed XWC = sin 28° x 20 kts = 9,4 kts

(Refer to figure 067-E66) For an example of solution of a wind component problem using a navigation computer or the calculation method refer to the attached illustration. Please note that the illustration to this type of question only explains the solution in general and does not necessarily need to depict the solution of this specific question. • Headwind =cos (wind angle) x wind speed • Crosswind =sin (wind angle) x wind speed In our case the wind angle =280° - 230° =50° cos 50° x 40 kts => approx. 25,7 kts headwind sin 50° x 40 kts => approx. 30,6 kts crosswind

4444. Given: TAS: HOG: WIV:

Airplane

ATPL

Heli

CPL

ATPL

CPL

This type of question can also be easily solved by drawing the triangle ofvelocities - see the attached illustration for information on the triangle of velocities along with examples ofproblem solutions.

4467. Given:

Airplane

ATPL

CPL

Heli

ATPL

CPL

227 kts 316° (T) 205/15 kts

Calculate the HOG and GS. 90kts 355° (T) 120/20 kts

A) 313 0 (T) -

(Refer to figures 067-E115, 067-E116, 067-E117, 067-E55, 067-E63 and 067-E64) For an example of solution of a wind correction and Ground speed problem refer to the attached illustration. Please note that the illustration to this type of question only explains the solution in general and does not necessarily need to depict the solution of this specific question.

95 kts. 102 kts. 101 kts. 0) 359° (T) - 102 kts.

For explanation refer to question #4279 on page 112.

4446. Given:

Airplane

TAS: True Heading: WIV(T):

ATPL

Heli

CPL

/fyou do not have the navigation computer available you can use a calculation method: ATPL

CPL

Windangle=376°-205°= 117° XWC = sin (wind angle) x wind speed XWC=sin mox 75kts= 74kts

125 kts 355° 320°/30 kts

Drift = XWC x 60 + TAS Drift= 74 x 60+227= approx. 4° Heading = 376° - 4° = 372"

Calculate the True track and GS.

HWC = cos (wind angle) x wind speed HWC= cos mox 75 kts =5,4kts tailwind

002 - 98 kts. B) 345° -100 kts. C) 348° - 102 kts. 0) 005 0 - 102 kts. 0

Effective TAS = TAS x cos WCA Eff.TAS =227x cos 4°= 226,4kts GS = 226,4 kts + 5,4 kts tailwind = 237,8 kts

For explanation refer to question #4279 on page 112.

Airplane

235 kts.

B) 311" (T) - 230 kts. C) 312° (T) - 232 kts. 0) 310° (T) - 233 kts.

A) 0060 (T) B) 3460 (T) C) 358 0 (T) -

4455. Given:

Heading = 322° + 5,9° = 327,9° (328°)

TAS: Track: WIV:

Calculate the track and GS.

A)

Drift=XWCx 60+ TAS Drift = 9,4 x 60 + 95 = 5,9°

ATPL

Heli

CPL

ATPL

This type of question can also be easily solved by drawing the triangle ofvelocities - see the attached illustration for information on the triangle of velocities along with examples ofproblem solutions.

CPL

An aircraft is on final approach to runway 32R (322°). The wind velocity reported by the tower is 350°/20 kts. TAS on approach is 95 kts. In order to maintain the center line, the aircraft's heading

I

4419 (A)

I

4437 (C)

I

4444 (8)

I

4446 (D)

I

4455 (8)

I

4467 (C)

I

04 Dead Reckoning Navigation (DR)

4469. Given:

Airplane

ATPL

Heli

CPL

ATPL

For explanation refer to question #4279 on page 772.

CPL

4490. 20 kts 06 063°(M) 1000 (M)

Maximum allowable crosswind component: Runway: RWYQDM: Wind direction:

(Refer to figure 067 -E66) For an example of solution of a wind component problem using a navigation computer or the calculation method refer to the attached illustration. Please note that the illustration to this type of question only explains the solution in general and does not necessarily need to depict the solution of this specific question. In our case the wind angle = 700° - 063° =3r Crosswind = sin (wind angle) x wind speed Max. 20 kts =sin 3r x ? kts ? kts = 20 kts + sin 3r ? kts = 20 kts + 0,6 ?=33,3kts

Airplane

ATPL

ATPL

4492. Airplane ATPL CPL Heli ATPL CPL Determine the WIV by using the multi-drift method (multiple drift WIV) when the following observations have been made while TAS was 187 kts:

Wind angle =305°- 760°= 745° XWC = sin (wind angle) x wind speed XWC=sin 745°x 78kts= 70,3kts

What is the track in this period of time? A) 322° B) 290° C) 310 0 D) 316°

4497.

TAS: Magnetic course: WIV(OM):

HWC = cos (wind angle) x wind speed HWC = cos 745° x 78 kts = 74,7 kts tailwind Effective TAS =TAS x cos WCA Eff.TAS = 755 x cos 4° = 754,6 kts

A) B) C) D)

This type of question can also be easily solved by drawing the triangle of velocities - see the attached illustration for information on the triangle of velocities along with examples ofproblem solutions.

Heli

CPL

ATPL

290° 250 kts 135/75 kts

CPL

186 kts 290 kts 246 kts 250 kts

Drift =XWC x 60 + TAS

I

ATPL

220 kts 212° 160°/50 kts

Windangle=272°-7600=52° XWC =sin (wind angle) x wind speed XWC =sin 52° x 50 kts =39,4 kts

320 kts 300 kts 175 kts 200 kts 4472 (A)

Heli

/fyou do not have the navigation computer available you can use a calculation method:

What is the ground speed?

I

CPL

(Refer to figures 067-E775, 067-E776, 067-E777, 067-E55, 067-E63 and 067-£64) For an example of solution of a wind correction and Ground speed problem refer to the attached illustration. Please note that the illustration to this type of question only explains the solution in general and does not necessarily need to depict the solution of this specific question.

CPL

Given:

4469 (C)

ATPL

Calculate the GS.

GS = 754,6 kts+ 74,7 kts tailwind = 769,3 kts

I

Airplane

Given:

Heading =305° - 4° =307 °

A) B) C) D)

305° 135 kts 230°/40 from 11:30to 11:45

For explanation refer to question #4279 on page 772.

Drift = XWC x 60 + TAS Drift = 70,3 x 60 + 755 = approx. 4°

Actual HOG: TAS: Wind:

310° (M)/41 296° (M)/36 320° (M)/18 328° (M)/29

True Heading: TAS: WIV: Period of time:

/fyoudo not have the navigation computer available you can use a calculation method:

ATPL

Drift 7°R Drift8°R Drift 3°L

4496. Airplane ATPL CPL Heli ATPL CPL Construct the triangle of velocities on a piece of paper, showing the following data:

(Refer to figures 067-E775, 067-£776, 067-E777, 067-E55, 067-E63 and 067-E64) For an example of solution of a wind correction and Ground speed problem refer to the attached illustration. Please note that the illustration to this type of question only explains the solution in general and does not necessarily need to depict the solution of this specific question.

Airplane

CPL

(Refer to figure 067-E67) Refer to the attached figure for explanation ofthe multi-drift method of finding the Will.

301 0 (T) - 169 kts. 305 0 (T) - 169 kts. 309° (T) - 170 kts. 309° (T) - 141 kts.

4478.

ATPL

For explanation refer to question #4279 on page 772.

CPL

Calculate the HOG and GS. A) B) C) D)

Heli

4°L-415 kts. 2°L - 420 kts. 6°L - 395 kts. SOL - 385 kts.

A) B) C) D)

155 kts 305° (T) 160/18 kts

TAS: Track: WIV:

A) B) C) D)

MH 015: MH075: MH 177:

Heli

CPL

CPL

Calculate the drift and GS.

26 kts 31 kts 33 kts 25 kts

4472. Given:

ATPL

440 kts 349°(T) 040/40kts

TAS: HOG: WIV:

Calculate the maximum allowable windspeed? A) B) C) D)

Airplane

Given:

4478 (A)

I

4490 (A)

I

4492 (0)

I

4496 (A)

I

4497 (A)

I

Aviationexam Test Prep Edition 2012 Drift = 39,4 x 60 -;- 220 = 10,1° (approx. W)

Drift = 7,8 x 60 -;- 270 = approx. 2°

Heading =212"- W=201°

Heading = 260° + 2" = 262°

HWC = cos (wind angle) x wind speed HWC = cos 52° x 50 kts = 30,8 kts headwind

HWC =cos (wind angle) x wind speed HWC = cos 15° x 30 kts = 29 kts headwind

Effective TAS = TAS x cos WCA Eff.TAS = 220 x cos 70,7" = 216,2 kts

Effective TAS = TAS x cos WCA Eff.TAS = 270 x cos 2° = 269,8 kts

GS =216,2 kts - 30,8 kts headwind = 185,4 kts

GS = 269,8 kts - 29 kts headwind = 240,8 kts

This type of question can also be easily solved by drawing the triangle ofvelocities - see the attached illustration for information on the triangle of velocities along with examples ofproblem solutions.

This type of question can also be easily solved by drawing the triangle ofvelocities - see the attached illustration for informatioll on the triangle of velocities along with examples ofproblem solutions.

4507. Given:

4532. Given:

Airplane

ATPL

CPL

Heli

ATPL

CPL

135 kts 278° 140/20 kts

TAS: True Heading:

W/V:

True heading: TAS: WIV:

Calculate the True track and GS.

A) 279 B) 283 C) 272 0) 275

0 0 0 0 -

152 kts. 150 kts. 121 kts. 150 kts.

Airplane

Runway direction: Surface WIV:

CPL

Heli

ATPL

CPL

ATPL

CPL

265° 290kts 210°/35 kts

0

A) 259 and 305 kts B) 259 and 272 kts C) 260 0 and 315 kts 0) 271 0 and 272 kts 0

For explanation refer to question #4279 on page 112.

ATPL

CPL

Heli

ATPL

CPL

083°(M) 035/35 kts

4540. Given:

Airplane

Runway direction: Surface W/V:

Calculate the effective headwind component.

ATPL

CPL

Heli

305°(M) 260 0 (M)/30 kts

Calculate the cross-wind component.

A) 24 kts B) 27 kts C) 31 kts 0) 34 kts

A) 18 kts B) 24kts C) 27 kts 0) 21 kts

(Refer to figure 061-E66) For an example of solution of a wind component problem using a navigation computer or the calculation method refer to the attached illustration. Please note that the illustration to this type of question only explains the solution in general and does not necessarily need to depict the solution of this specific question.

=

0

In our case the wind angle = 083° - 035° = 48° cos 48°x 35 kts => approx. 23,5 kts headwind sin 48° x 35 kts => approx. 26 kts crosswind

Airplane

TAS: Track:

ATPL

CPL

In our case the wind angle = 305° - 260 = 45° cos 45° x 30 kts => approx. 21 kts headwind sin 45° x 30 kts => approx. 21 kts crosswind

Heli

ATPL

CPL

270 kts 260° (T) 275/30 kts

W/V:

Calculate the HDG and GS. 0

A) 264 (T) - 237 kts. B) 262 0 (T) - 237 kts. C) 264 (T) - 241 kts. 0) 262 (T) - 241 kts. 0

0

(Refer to figures 061-E115, 061-E116, 06H117, 061-E55, 06H63 and 061-E64) For an example of solution of a wind correction and Ground speed problem . refer to the attached illustration. Please note that the illustration to this type ofquestion only explains the solution in general and does not necessarily need to depict the solution of this specific question.

Ifyou do not have the navigation computer available you can use a calculation method: Wind angle =275°-260°= 15° XWC = sin (wind angle) x wind speed XWC = sin W x 30 kts = 7,8 kts

(Refer to figure 061-E66) For an example of solution of a wind component problem using a navigation computer or the calculation method refer to the attached illustration. Please note that the illustration to this type of question only explains the solution in general and does not necessarily need to depict the solution of this specific question . • Headwind = cos (wind angle) x wind speed • Crosswind = sin (wind angle) x wind speed

• Headwind cos (wind angle) x wind speed • Crosswind = sin (wind angle) x wind speed

4527. Given:

ATPL

What is the true track and GS?

For explanation refer to question #4279 on page 112.

4522. Given:

Airplane

4541. Airplane ATPL CPL Heli ATPL CPL The reported surface wind from the control tower is 240°/35 kts. Runway 30 (300°). What is cross-wind component?

A) 30 kts B) 24 kts C) 27 kts 0) 21 kts (Refer to figure 061-E66) For an example of solution of a wind component problem using a navigation computer or the calculation method refer to the attached illustration. Please note that the illustration to this type of question only explains the solution in general and does not necessarily need to depict the solution of this specific question. • Headwind = cos (wind angle) x wind speed • Crosswind = sin (wind angle) x wind speed In our case the wind angle =300° - 240° = 60° cos 60 0 x 35 kts => approx. 17,5 kts headwind sin 60° x 35 kts => approx. 30 kts crosswind

Drift=XWCx60-;- TAS

I 4507 (8) I 4522 (A) I 4527 (D) I 4532 (D) I 4540 (D) I 4541 (A) I

-----------~.-----~-----~----

04 Dead Reckoning Navigation (DR)

4546. Airplane ATPL CPL Given: TAS: 155 kts TrueHDG: 216° W/V: 090/60kts Calculate the True track and GS. A) B) C) D)

Heli

ATPL

CPL

0

Note: Remember that the surfacewinds are reported as magnetic directions you use the magnetic runway heading to ,find the wind angle - no need to use the variation to convert the runway heading from magnetic to true!

2240 - 175 kts. 231 0 - 196 kts. 2220 -181 kts. 2260 - 186 kts.

4572. Airplane ATPL CPL Heli ATPL CPL An aircraft is following a true track of 048° at a constant TAS of 210 kts. The wind velocity is 350°/30 kts. The GS and drift angle are:

For explanation refer to question #4279 on page 772.

4548. Airplane Given: True heading: TAS: W/V: Calculate the GS. A) B) C) D)

ATPL

CPL

Heli

ATPL

CPL

090° 200 kts 220/30 kts

180 kts 230 kts 220 kts 200 kts

For explanation refer to question #4279 on page 772.

4555. Airplane ATPL CPL Heli ATPL CPL An aircraft is landing on runway 23 (QFU 227°), surface wind 180°/30 kts from ATIS; variation is 13°E. The cross wind component on landing is: A) B) C) D)

• Crosswind = sin (wind angle) x wind speed In our case the wind angle =22r - 180 =4r cos 4r x 30 kts => approx. 20,5 kts headwind sin 4r x 30 kts => approx. 22 kts crosswind

26 kts 22 kts 20 kts 15 kts

(Refer to figure 061-E66) For an example of solution of a wind component problem using a navigation computer or the calculation method refer to the attached illustration. Please note that the illustration to this type of question only explains the solution in general and does not necessarily need to depict the solution of this specific question. In our case we first have to find out the angle between the runway and the wind => 22r - 180 0 = 04r Note that you can disregard the Variation because the wind direction stated in the ATIS or provided by the ATe over the radio is given in degrees Magnetic. Winds aloft (cruise), in METARs or TAFs are given in True directions. • Headwind =cos (wind angle) x wind speed • Crosswind =sin (wind angle) x wind speed In our case the wind angle = 04r cos 4r x 30 kts => approx. 20,5 kts headwind sin 4r x 30 kts => approx. 22 kts crosswind Note: QFU = Magnetic bearing of the runway in use.

4557. Airplane ATPL CPL Heli ATPL CPL For a landing on runway 23 (227° magnetic) surface W/V reported by the ATIS is 180/30 kts. VAR is 13°E. Calculate the cross wind component. A) 20 kts B) 22 kts C) 26 kts D) 15 kts (Refer to figure 061-E66) For an example of solution of a wind component problem using a navigation computer or the calculation method refer to the attached illustration. Please note that the illustration to this type of question only explains the solution in general and does not necessarily need to depict the solution of this specific question.

A) 192 kts, 7° left B) 200 kts, 3,5" right C) 192 kts, 7 0 right D) 225 kts, 7" left (Refer to figures 061-E775, 061-E776, 061-E777, 061-E55, 061-E63 and 061-E64) Let's summarize the information that we have: Track = 048° TAS = 270kts WN= 350"130 kts For an example of solution of a wind correction and Ground speed problem refer to the attached illustration. Please note that the illustration to this type of question only explains the solution in general and does not necessarily need to depict the solution of this specific question. Ifyou do not have the navigation computer available you can use a calculation method: Wind angle =048° - 350° =58° XWC = sin (wind angle) x wind speed XWC = sin 58° x 30 kts = 25,4 kts Drift=XWCx 60.,. TAS Drift =25,4 x 60.,. 270 =7,2° (approx. r) Heading = 048°- r= 041°, HWC =cos (wind angle) x wind speed HWC =cos 58° x 30 kts = 15,9 kts headwind Effective TAS =TAS x cos WCA Eff.TAS =270x cos 7,2°=208,3 kts GS =208,3 kts - 15,9 kts headwind = 192,4 kts This type of question can also be easily solved by drawing the triangle ofvelocities - see the attached illustration for information on the triangle of velocities along with examples ofproblem solutions.

4575. Airplane ATPL CPL Heli ATPL CPL Given: For takeoff an aircraft requires a headwind component of at least 10 kts and has a cross-wind limitation of 35 kts. The angle between the wind direction and the runway is 60°. Calculate the minimum and maximum allowable wind speeds. A) B) C) D)

12 kts; 38 kts 20 kts; 40 kts 15 kts; 43 kts 18 kts; 50 kts

(Refer to figure 061-E66) For an example of solution of a wind component problem using a navigation computer or the calculation method refer to the attached illustration. Please note that the illustration to this type of question only explains the solution in general and does not necessarily need to depict the solution of this specific question. • Headwind = cos (wind angle) x wind speed • Crosswind = sin (wind angle) x wind speed In our case the wind angle = 060 0 sin 60° =maximum crosswind component.,. maximum wind intensity: max wind intensity = 35 knots.,. sin 60° = 40 knots cos 60° = minimum headwind component.,. minimum wind intensity: min wind intensity = 70 knots.,. cos 60 0 = 20 knots

• Headwind =cos (wind angle) x wind speed

14546(8) I 4548 (C) 14555(8) 14557(8) I 4572 (C) 14575(8) I

Aviationexam Test Prep Edition 2012

4594. Given:

Airplane

ATPL

CPL

Heli

ATPL

CPL

250 kts 029° (T) 035/45 kts

TAS: HOG: WIV:

4635. Airplane ATPL CPL Heli ATPL CPL The wind velocity is 359/25. An aircraft is heading 180° at a TAS of 198 knots. (All directions are true). What is its track and ground speed?

Calculate the drift and GS. A) B) C) D)

1°L - 205 kts. lOR - 205 kts. 1°L - 265 kts. lOR - 295 kts.

A) 180 0 (T)/223

B) 179°(T)/220 C) 180 0 (T)/220

D) 179°(T)/223

For explanation refer to question #4279 on page 112.

4597. Given:

Airplane

TAS: HOG (OT): W/V(OT):

ATPL

CPL

Heli

ATPL

CPL

230 kts 250° 205/10 kts

lL - 225 kts. lR - 221 kts. 2R - 223 kts. 2L - 224 kts.

779,8r.

4636. Given:

For explanation refer to question #4279 on page 112.

4598. Given:

Airplane

TAS: HOG: WIV:

ATPL

CPL

Heli

ATPL

CPL

A) 1R - 165 kts. B) 1L - 225 kts. C) 1R - 175 kts. D) 1L - 215 kts.

CPL

Heli

ATPL

CPL

ATPL

CPL

ATPL

CPL

198 kts 180° (T) 359/25 180° (T) - 223 kts. 179° (T) - 220 kts.

181°(T)-180kts. 180° (T) - 183 kts.

For explanation refer to question #4279 on page 772.

4637. Given:

For explanation refer to question #4279 on page 112.

ATPL

CPL

Maximum allowable tailwind component for landing: Planned runway: The direction of the surface wind reported by ATIS: Variation:

Heli

ATPL

CPL

Airplane

ATPL

CPL

Heli

370 kts 181° 095°/35 kts

TAS: HOG: W/V:

Calculate the true track and GS. 10 kts 05 (047° magnetic)

Calculate the maximum allowable wind speed that can be accepted without exceeding the tailwind limit. A) B) C) D)

ATPL

TAS: HOG: WIV: A) B) C) D)

Calculate the drift and GS.

Airplane

Airplane

Calculate the track and GS.

190 kts 355° (T) 165/25 kts

4632. Given:

With a heading of 780° and the wind coming from 359° at 25 kts we can say we have almost a direct tailwind (ok, 7° off, which is negligible). With a direct tailwind we will not have a wind correction angle => our track will remain 780°. The Ground speed will be 298 kts (TAS) + 25 kts (tailwind) = 223 kts. Note: Yes, we will have a drift angle because the wind is not 700% tailwind, but coming at an angle of 7°. This will result in a crosswind component of 0,43 kts and with a TAS of 798 kts the drift angle will be 0,73°. Therefore, even with a calculation method (or using the nav computer) the answer "780°" remains the closest one to the correct result. Actual track would be roughly

Calculate the drift and GS. A) B) C) D)

70 kts max. tailwind = cos 763° x wind speed (1) Wind speed (1) = 70 kts + cos 763° 1= 70,5 kts

15 kts 18 kts 8 kts 10kts

A) B) C) D)

186° (T) - 370 kts. 176° (T) - 370 kts. 192° (T) - 370 kts. 189° (T) - 370 kts.

For explanation refer to question #4279 on page 112.

4638. Given:

Airplane

TAS: True HOG: WIV:

(Refer to figure 067-£66) For an example of solution of a wind component problem using a navigation computer or the calculation method refer to the attached illustration. Please note that the illustration to this type of question only explains the solution in general and does not necessarily need to depict the solution of this specific question. In our case we first have to find out the angle between the runway and the wind => 210° - 04r = 763° (tailwind 7r from the right). Note that you can disregard the Variation because the wind direction stated in the ATIS or provided by the ATC over the radio is given in degrees Magnetic. Winds aloft (cruise), in METARs or TAFs are given in True directions.

ATPL

CPL

Heli

485 kts 168° 130°175 kts

Calculate the True track and GS. A) 175° - 432 kts. B) 173° - 424 kts. C) 175° - 420 kts. D) 174° - 428 kts. For explanation refer to question #4279 on page 112.

• Headwind = cos (wind angle) x wind speed • Crosswind =sin (wind angle) x wind speed In our case the wind angle = 763°. Now apply the formula:

I 4594 (A) I 4597(C) 14598(0) 14632(0) I 4635 (A) I 4636 (A) I 4637 (A) 14638(0) I

04 Dead Reckoning Navigation (DR)

4641. Given:

Airplane

ATPL

Heli

CPL

ATPL

4673. Given:

CPL

140 kts TAS: 302° True HOG: 045°(T)/45 kts W/V: Calculate the drift angle and GS:

Heli

Airplane

ATPL

"-

Heli

CPL

ATPL

040 0 (T) 120 kts 30 kts

4677. Given:

27.000 feet -35°C 0,45

Airplane

ATPL

CPL

Heli

4693. Given:

Airplane

ATPL

CPL

Heli

CPL

ATPL

CPL

ATPL

CPL

305° (T) 135 kts 230/40

Heading: TAS: WV:

What is the ground speed? A) B) C) D)

130 kts 125 kts 145 kts 97 kts

For explanation refer to question #4279 on page 112.

Wind angle = 270° - 200° = 70° XWC = sin (wind angle) x wind speed XWC = sin 70° x 85 kts = 79,9 kts

4694. Given:

Heading = 200° + 17,8° = approx. 218° HWC = cos (wind angle) x wind speed HWC = cos 70° x 85 kts = 29 kts headwind

G5 = 257 kts - 29 kts headwind = approx. 228 kts

This type of question can also be easily solved by drawing the triangle ofvelocities - see the attached illustration for information on the triangle of velocities along with examples ofproblem solutions.

Heading: TAS: W/V: What is your true track? A) 160 0 B) 152 0 C) 1040 D) 2220

Heli

CPL

ATPL

CPL

156°T 320 kts 130°/45

4695. Given:

Heli

I

Airplane

ATPL

CPL

Heli

200 kts 110° (T) 015/40 kts

Calculate the HOG and GS. A) B) C) D)

4660 (A)

CPL

For explanation refer to question #4279 on page 112.

TAS: Track: WIV:

4673 (D)

097° (T) - 201 kts. 121° (T) - 207 kts. 121° (T) -199 kts. 099° (T) -199 kts.

(Refer to figures 061-E115, 061-E116, 061-E117, 061-E55, 061-E63 and 061-E64) For an example of solution of a wind correction and Ground speed problem refer to the attached illustration. Please note that the illustration to this type

For explanation refer to question #4279 on page 112.

I

ATPL

A) 125° (T) - 322 kts. B) 123° (T) - 320 kts. C) 126° (T) - 320 kts. D) 125° (T) - 318 kts.

Effective TA5 = TA5 x cos WCA Eff.TA5 =270 x cos 17,75°=257kts

ATPL

Airplane

375 kts TAS: 124° True HOG: 1300 (T)/55 kts WIV: Calculate the true track and GS.

Drift=XWCx 60.;. TA5 Drift =79,9 x 60.;. 270 =approx. 17,8°

4655 (C)

ATPL

For explanation refer to question #4279 on page 112.

Ifyou do not have the navigation computer available you can use a calculation method:

I

CPL

B) 134° - 188 kts. C) 120° -190 kts. D) 123° - 180 kts.

To obtain the TA5 from Mach number and OAT: I} Find OAT in OK (= OAT in °c + 273) = 238 OK 2} Find L55 (=38,95 x ..jOATin OK) = 601 kts 3} Find TA5 (= Mach x L55) = 270 kts

4641 (8)

ATPL

225 kts, TAS: HOG (OT): 123° 090/60 kts WIV: Calculate the track (OT) and GS. A) 1340-178 kts.

Note: Remember that the winds aloft (cruise winds) are given in True directions (0T). Surface winds received by ATl5 or given by ATC over the radio are in °Magnetic. Therefore, the wind stated by this question is in ° True.

I

= 040° +

CPL

(Refer to figures 061-E115, 061-E116, 061-E117, 061-E55, 061-E63 and 061-E64) For an exampie of solution of a wind correction and Ground speed problem refer to the attached illustration. Please note that the illustration to this type of question only explains the solution in general and does not necessarily Med to depict the solution of this specific question.

Airplane

CPL

Maximum drift angle will be experienced during a direct crosswind 90° = 130° (or 040° - 90° =310°).

18l / 252 knots 15R / 310 knots 17l / 228 knots 17R / 287 knots

4660. Given:

ATPL

B) 145° C) 115° D) 1300

270°/85 WIV: 200°(T) Track: What is drift and ground speed?

A) B) C) D)

CPL

Maximum drift angle will be obtained for a wind direction of: A) 1200

For explanation refer to question #4279 on page 112.

Pressure Altitude: OAT: Mach number:

ATPL

Course: TAS: Wind speed:

A) 9°R - 143 kts. B) 16°l -156 kts. C) 9°l - 146 kts. D) 18°R - 146 kts.

4655. Given:

Airplane

I

4677 (A)

I

4693 (A)

I

4694 (8)

I

4695 (D)

I

Aviationexam Test Prep Edition 2012 ofquestion only explains the solution in general and does not necessarily need to depict the solution of this specific question. Ifyou do not have the navigation computer available you can use a calculation method: Wind angle = 110° - 015° = 95° XWC = sin (wind angle) x wind speed XWC = sin 95° x 40 kts = 39,8 kts

HWC = cos (wind angle) x wind speed HWC = cos 95° x 40 kts = 3,5 kts tailwind Effective TAS = TAS x cos WCA Eff.TAS =200 x cos 1]0= 195,6kts GS =195,6 kts + 3,5 kts tailwind = approx. 199,1 kts This type of question can also be easily solved by drawing the triangle ofvelocities - see the attached illustration for information on the triangle of velocities along with examples ofproblem solutions. ATPL

TAS: HOG: WIV:

Heli

CPL

ATPL

077° (T) - 214 kts. 079° (T) - 211 kts. 075° (T) - 213 kts. 077° (T) - 210 kts.

(Refer to figures 061-E115, 061-E116, 061-El17, 061-E55, 061-E63 and 061-E64) For an example of solution of a wind correction and Ground speed problem refer to the attached illustration. Please note that the illustration to this type ofquestion only explains the solution in general and does not necessarily need to depict the solution of this specific question. Ifyou do not have the navigation computer available you can use a calculation method:

GS = 199,5 kts + 14,6 kts tailwind =approx. 214 kts

ATPL

TAS: HOG (0T): WIV:

This type of questloll can also be easily solved by drawlllg the triangle ofvelocities - see the attached illustration for information on the triangle of velocities along with examples ofproblem solutions.

Heli

CPL

ATPL

CPL

4861. Given:

170 kts 100° 350/30 kts

Airplane

Required course: WIV: FL55 ISA Variation: CAS:

Calculate the track (OT) and GS. 098 - 178 kts. 109 -182 kts. 091 - 183 kts. 103 -178 kts.

ATPL

CPL

Heli

ATPL

CPL

045°(T) 190/30

What is the magnetic heading and GS?

A) 052°(M) 154 kts B) 067°(M) 154 kts C) 03r(M) 154 kts D) 03r(M) 113 kts

For explanation refer to question #4279 on page 112.

Airplane

ATPL

Heli

CPL

ATPL

CPL

(Refer to figures 061-E115, 061-E116, 061-EII7, 061-E55, 061-E63 and 061-E64) First of all we need to convert the CAS into TAS using the navigation computer. Refer to the attached illustration for a general explanation of this procedure. You should get a result TAS = approx. 130 kts. Note that the ISA temp at FL55 is 4°C.

210 0 (M) 230 (M)/30 kts 0

Calculate the cross-wind component.

A) 19 kts B) 10 kts C) 16 kts D) 13 kts (Refer to figure 061-E66) For an example of solution of a wind component problem using a navigation computer or the calculation method refer to the attached illustration. Please note that the illustration to this type of question only explains the solution in general and does not necessarily need to depict the solution of this specific question. • Headwind = cos (wind angle) x wind speed • Crosswind =sin (wind angle) x wind speed

4718 (A)

I

4721 (8)

With a TAS of 130 kts, wind of 190"/35 kts and the information that our track must be 045°T we can calculate the True Heading and Groundspeed using the navigation computer. Again, for an explanation of the general procedure, refer to the attached illustration. You will get a result of approx. 053° (TH) and 153 kts (GS). Note that this is the True Heading (TH). Co convert it to Magnetic Heading (MH) we have to apply the value ofvariation: MH = TH + Westerly variation (or - Easterly variation) MH=053°- WE MH=038° For an example of solution of a wind correction and Ground speed problem refer to the attached illustration. Please note that the illustration to this type of question only explains the solution in general and does not necessarily need to depict the solution of this specific question.

In our case the wind angle =230° - 210°= 20° cos 20° x 30 kts => approx. 28 kts headwind sin 20° x 30 kts => approx. 10 kts crosswind

I

200kts 073° (T) 210/20 kts

Effective TAS = TAS x cos WCA Eff.TAS =200x cos 4 = 199,5 kts

Airplane

Runway direction: Surface W/V:

CPL

HWC =cos (wind angle) x wind speed HWC =cos 13r x 20 kts = 14,6 ktstailwind

9°R - 108 kts. 100L -104 kts. 9°L - 105 kts. 8°R - 104 kts.

4735. Given:

ATPL

Heading =073° + 4° =07r

For explanation refer to question #4279 on page 112.

A) B) C) D)

Heli

Drift = XWC x 60 + TAS Drift = 13,6 x 60 + 200 =approx. 4°

95 kts 075°(T) 310/20 kts

4721. Given:

CPL

Windangle=2100-073°= 13r XWC =sin (wind angle) x wind speed XWC=sin 13rx20 kts = 13,6kts

CPL

Calculate the drift and GS.

A) B) C) D)

ATPL

TAS: Track: WIV:

A) B) C) D)

Heading = 1100-12°=approx. 098°

Airplane

Airplane

Calculate the HOG and GS.

Drift=XWCx 60+ TAS Drift =39,8 x 60 + 200 =approx. 1]0

4718. Given:

4848. Given:

Ifyou do not have the navigation computer available you can use a calculation method: Wind angle = 190°-045°= 145°

I

4735 (8)

I

4848 (A)

I

4861 (C)

I

04 Dead Reckoning Navigation (DR) XWC =sin (wind angle) x wind speed XWC =sin 745° x 35 kts =20 kts

Eff.TAS =465 x cos 9° =459 kts

Drift=XWCx60+ TAS Drift =20 x 60 + 730 =approx. 9,2°

This type of question can also be easily solved by drawing the triangle ofvelocities - see the attached illustration for information on the triangle of velocities along with examples ofproblem solutions.

GS = 459 kts - 37 kts headwind = approx. 428 kts

Heading (True) = 045° + 9,2° = approx. 054° Magnetic Heading = 054° - 75°E = 039°

4924.

HWC = cos (wind angle) x wind speed HWC = cos 745° x 35 kts = 28,7 kts tailwind

Airplane

ATPL

Effective TAS =TAS x cos WCA Eff.TAS = 730 x cos 9,2 = 728,3 kts GS = 728,3 kts + 28,7 kts tailwind =approx. 757 kts

TAS: Track: WIV:

This type of question can also be easily solved by drawing the triangle of velocities - see the attached illustration for information on the triangle of velocities along with examples ofproblem solutions.

Calculate the HOG and GS. A) 002 0 (T) - 173 kts.

4869.

Airplane

ATPL

CPL

Heli

ATPL

CPL

Given: TAS: 132 kts HOG: 053° (T) WIV: 205/15 kts Calculate the track (0T) and GS. A) 057 0 (T) - 144 kts.

ATPL

CPL

CPL

130 kts 003°(T)

190/40 kts

B) 001 0 (T) - 170 kts. C) 3590 (T) -166 kts. 0) 357 0 (T) - 168 kts.

Wind angle = 790° - 003° = 78r XWC = sin (wind angle) x wind speed XWC=sin 78rx40kts=4,9kts

Heli

ATPL

CPL

Drift=XWCx60+ TAS Drift =4,9 x 60 + 730 =approx. 2° Heading = 003° - 2° = 007°

Given: TAS: HOG: WIV:

HWC = cos (wind angle) x wind speed HWC =cos 78r x 40 kts =39,7 kts tailwind

480 kts 040° (T)

Effective TAS = TAS x cos WCA Eff.TAS = 730 x cos]O= 729,9 kts

090/60 kts

Calculate the track and GS. A) 032 0 (T) - 425 kts. B) 028 0 (T) - 415 kts. C) 0340 (T) - 445 kts. 0) 0360 (T) - 435 kts.

GS = 729,9 kts + 39,7 kts tailwind = 769,6 kts This type of question can also be easily solved by drawing the triangle ofvelocities - see the attached illustration for information on the triangle of velocities along with examples ofproblem solutions.

4946. Airplane ATPL CPL For a given heading the:

For explanation refer to question #4279 on page 772.

4923.

ATPL

/fyou do not have the navigation computer available you can use a calculation method:

For explanation refer to question #4279 on page 772.

Airplane

Heli

(Refer to figures 067-E775, 067-E776, 06H771, 06H55, 067-E63 and 06H64) For an example of solution of a wind correction and Ground speed problem refer to the attached illustration. Please note that the illustration to this type of question only explains the solution in general and does not necessarily need to depict the solution of this specific question.

B) 050° (T) - 145 kts. C) 052° (T) - 143 kts. 0) 0510 (T) -144 kts.

4889.

CPL

Given:

Airplane

ATPL

CPL

Heli

ATPL

CPL

Given: TAS: 465 kts 007° (T) Track: 300/80 kts WIV: Calculate the HOG and GS. A) 000 0 (T) - 430 kts. B) 001 0 (T) -432 kts. C) 3580 (T) - 428 kts. 0) 357 0 (T) - 430 kts. (Refer to figures 067-E775, 06H776, 067-E771, 06H55, 067-E63 and 067-E64) For an example of solution of a wind correction and Ground speed problem refer to the attached illustration. Please note that the illustration to this type of question only explains the solution in general and does not necessarily need to depict the solution of this specific question. /fyou do not have the navigation computer available you can use a calculation method: XWC = sin (wind angle) x wind speed XWC = sin 6r x 80 kts = 73,6 kts Drift = XWC x 60+ TAS Drift =73,6 x 60 + 465 =approx. 9° Heading = oor - 9° =358° HWC = cos (wind angle) x wind speed HWC=cos6rx80kts=37 ktsheadwind Effective TAS = TAS x cos WCA

14869(8) I 4889 (C) I 4923 (C) I 4924(8) 14946(8) I

Heli

ATPL

CPL

Wind component: +45 kts 15° left Orift angle: 240 kts TAS: What is the wind component on the reverse track? A) B) C) 0)

-55 kts -65 kts -45 kts -35 kts

(Refer to figures 067-E775, 06H776, 067-E771, 06H55, 067-E67 and 06H62) For an example of solution of a wind direction and velocity problem using a navigation computer refer to the attached illustration. Please note that the illustration to this type of question only explains the solution in general and does not necessarily need to depict the solution of this specific question. Let's choose a heading offor example 000°. With a drift angle of/soL our track will be 360° - 75° = 345°. Now calculate the W/V that will give us a tailwind component of 45 kts. Remember that wind component with a positive preequals a tailwind and with a negative prefix "-" equals a headwind => fix think of it what kind of wind component you would like to see when you are flying from point A to point B => a "positive" thing to have would be a tailwind, right?

'+"

/fyou have a tailwind component of 45 kts, then your GS will be 285 kts. Knowing the Heading, Track and TAS you can derive the wind direction and speed using your navigation computer=> approx. 776°/82 kts. Now assume you reverse the track: Track = 345°- 780°= 765° TAS =240 kts WN= 7W/82kts

Aviationexam Test Prep Edition 2012 Again use your navigation computer and determine the head/tailwind component for this case => you should get approximately 65 kts headwind = -65 kts.

231105. Airplane ATPL CPL Heli ATPL CPL An aircraft tracks radial 200 inbound to a VOR station with a Magnetic Heading (MI:I) of 010°. After being overhead the VOR station the aircrafuracks radial 090 outbound with a MH of 080°. The TAS is 240 kts and the magnetic variation in the area is SoW. What is the wind vector (T)? A) B) e) D)

231106. Airplane ATPL CPL Heli ATPL CPL At 10:15 the reading from a VOR/DME station is 211°/90 NM, at 10:20 the reading from the same VOR/DME station is 211°/120 NM. Compass Heading: Variation in the area: Deviation: TAS:

200° 31°W +1° 390 kts

The wind vector (T) is approximately:

300°/50 kts 330°/50 kts 320°/50 kts 310°/60 kts

A) 100°/60 kts B) 110°/70 kts C) 110°/40 kts D) 120°/50 kts 231107. Airplane ATPL CPL Heli ATPL CPL On a True Heading of 090° the aircraft experiences drift of 5 0 S. On a True Heading of 180° the aircraft experiences no drift. On both headings the'TAS is 200 ktsand it is assumed that the wind is the same. What is the experienced wind speed and direction? A) B) e) D)

360° / 34 kts 360° /17 kts 180° / 34 kts 180° /17 kts

04-04 Determination of DR position 1180. Airplane ATPL CPL Heli ATPL CPL The evaluation of your plotting work shows a WCA +3° and a drift 3° left: A) B) e) D)

your actual position is on the intended track. the track error is 6°. the expected W/V and the actual W/V coincide. the GS was exactly calculated.

(Refer to figure 061-£55) If you calculated a WCA (Wind Correction Angle) of +3° (heading to the right of the intended track) and you sustained a drift 3° left (pushing the aircraft to the left), your actual position should be exactly on the intended track.

1221. Airplane ATPL CPL Heli ATPL CPL You should follow a track due North taking account of a North westerly wind. The line connecting your last known position with the DR position represents: A) the estimated track. B) the heading line and the TAS. C) the TAS only. D) the no wind line and the GS.

DR Air Position is the position of the aircraft along the air vector in terms of the true heading and true airspeed. It is not the position with regard to the ground, but where the aircraft would be if there were no wind. The distance traveled along the air vector is found by mUltiplying the TAS by time. DR Ground position is the position underneath the aircraft - it takes the wind into account.

1229. Airplane ATPL CPL The DR position represents:

"DR" stands for Dead Reckoning or Deduced Reckoning. It means estimated or calculated positions and/or speeds. The under/ying principle of DR navigation is that the progress of the aircraft is plotted on the chart according to predicted or expected conditions and relevant data is recorded in the navigation log. As new information is acquired during the flight, it is applied to the current situation and the heading and ETA are continually amended and revised in the light of up-to-date information. The accuracy of the DR pOSition will depend on the quality and the time of the latest information obtained and used in modifying the DR position or directions etc, as opposed to actual established or measured data. A "DR Position" is a position of the aircraft which has been calculated or estimated, taking into account the actual heading and actual lAS + the estimated TAS and WN to determine the estimated actual track of the aircraft (desired track). It is the best position at the time for example if no radio navigation

1231105 (C) 1231106 (8) 1231107 (8) 1 1180 (A)

or a precise visual checkpoint is available. A DR position may be many miles in error or it may be very accurate. A DR position is not a fix, which is a definite established position. Thus, a line joining the last known position (fix) and the current DR position is an estimated track, which may differ from the actual track, depending on the accuracy of the DR position. The accuracy of a DR position depends on many factors: • Time elapsed since the last fix, • Accuracy of the information obtained to construct the fix, • Accuracy of the wind velocity used, • Accuracy of headings and airspeeds flown • Skill and knowledge of the pilot navigating.

Heli

ATPL

CPL

A) the estimated position taking account of the estimated TAS and wind condition. B) the estimated position in no wind condition. e) the actual position corrected by the track error. D) the air position corrected by the track error. For explanation refer to question #1221 on this page.

1231. Airplane ATPL CPL Heli ATPL CPL You should follow a track due North taking account of a northwesterly wind. You calculated a WCA _8°, A) B) e) D)

1 1221 (A)

The drift will be 8° righi:. The drift will be 8° left. A track error of _2° (left) shows a weA of only _6°. A track error of 2° (right) shows a drift of 10° right. 1 1229 (A)

1 1231 (A)

1

04 Dead Reckoning Navigation (DR) (Refer to figure 061-E55) Drift is the angular difference between heading and track measured from heading to track.

XWC =sin 90 0 x 100kts= 100kts

Wind Correction Angle (WCA) is similar to drift but is in the opposite direction. It is the amount of degrees offset to give the required track and is measured from track to heading. It is equal in value to the drift but is in the opposite sense.

True Heading = 270° + 13°= 283°

Therefore, if you wish to follow a Northerly track while experiencing a NorthWesterly wind, you will have to slightly alter your heading towards the West (crab into the wind) to compensate for the wind and not to get blown-offyour intended Northerly track. This "crabbing into the wind" is the WCA. Regardless of the fact whether you apply the correct WCA or not, you will be subject to the drift = the wind will blow you towards the East anyway. However, if you apply the correct WCA then you offset this drift and you will be able to maintain your desired Northerly track. In terms ofvalues the WCA =Drift in order to offset the drift. .

1277. Airplane The air position:

ATPL

Heli

CPL

ATPL

For explanation refer to question #1221 on page 122.

1286. Airplane ATPL CPL Heli ATPL CPL You should follow a track due North taking account of a northwesterly wind. The line connecting your last known position with the air position A) Shows a North easterly direction. B) Shows a North westerly direction. e) Represents the track line. D) Represents the wind v@locity.

Heli

ATPL

CPL

4506. Airplane ATPL CPL Heli ATPL CPL If the true track from A to B is 270°, TAS is 460 kts, wind velocity is 360°/100 kts, variation is 100W and deviation is +2°. Calculate the compass heading and ground speed. 291° and 448 kts. 283 0 and 451 kts. 25r and 446 kts. 275 0 and 453 kts.

For an example of solution of a wind correction and Ground speed problem refer to the attached illustration. Please note that the illustration to this type of question only explains the solution in general and does not necessarily need to depict the solution of this specific question. /fyou do not have the navigation computer available you can use a calculation method: Wind angle = 360° - 270° = 90° XWC = sin (wind angle) x wind speed 1286 (8)

When applying the values of Deviation and Variation using the mnemonic "CDMVT" we usually start off with the True heading => we work from the right to the left. When applying the Deviation and Variation values we have to apply it in the correct sense - either with a "plus" or a "minus" sign. Remember the mnemonic "West is Best, East is Least" = in other words West = plus sign, East = minus sign. This mnemonic works when starting with True heading and working your way towards the left though the Magnetic heading to the Compass heading. When working the other way around = starting with the Compass heading and wishing to obtain the Magnetic heading or the True heading, you have to reverse this logic => West = minus sign and East = plus sign. C D M v T ? <= 2°E - r <= 100W + 283° Magnetic Heading (?) = 293° Compass Heading (1?) = 291°

4851. Given:

Airplane

ATPL

Course required: Forecast WIV: TAS: Distance:

CPL

Heli

ATPL

CPL

085° (T) 030/100 kts 470 kts 265NM

0960 - 29 min. 076° - 34 min. 075 0 - 39 min. 095 0 - 31 min.

(Refer to figures 061-E115, 061-E116, 061-E117, 061-E55, 061-E63 and 067-£64) Let's start by the determination of the True Heading (TH) + Ground Speed (GS) based on the track required and the wind: For an example of solution of a wind correction and Ground speed problem refer to the attached illustration. Please note that the illustration to this type of question only explains the solution in general and does not necessarily need to depict the solution of this specific question. /fyou do not have the navigation computer available you can use a calculation method:

(Refer to figures 061-E115, 061-E116, 061-E117, 061-E55, 061-E63, 067-£64 and061-E97) Let's start by the determination of the True Heading (TH) + Ground Speed (GS) based on the track required and the wind:

I

We have determined that our TH = 283° and our GS = 448 kts. Now we need to convert the TH to MH using the variation and subsequently the MH to CH using the deviation. A very useful mnemonic for the actual calculation exists: CDMVT = Can Dead Men Vote Twice? It stands for Compass Heading; Deviation; Magnetic Heading; Variation; True Heading.

A) B) C) D)

For explanation refer to question #1221 on page 122.

1277 (8)

GS = 448,2 kts - 0 kts headwind = 448,2 kts

Calculate the true HDG and flight time.

A) heading line corrected by the actual drift. B) actual track corrected by the track error. e) desired track. D) desired track corrected by the track error.

I

Effective TAS =TAS x cos WCA Eff.TAS = 460 x cos 13° = 448,2 kts

Note: Remember that you have to work your way from the right (from TH) to the left (to CH).

With the wind coming from the left your heading must be less than the track (left of the track to account for the drift). The air position will then be West of the track => in this case in the North-Westerly direction. /fyou make a quick sketch it will help you visualize the situation.

A) B) C) D)

HWC =cos (wind angle) x wind speed HWC = cos 90° x 100 kts = 0 kts

Easterly Deviation as well as Easterly Variation is a positive value (+ °E). Westerly Deviation as well as Westerly Variation is a negative value (- OW).

CPL

A) is along the intended track. B) shows where the aircraft would be as a result if its lAS and true heading were not affected by wind. e) can never coincide with the DR position. D) shows where the aircraft will be, taking account of the lAS and actual wind conditions.

4242. Airplane ATPL CPL A DR position is to be found on the:

Drift=XWCx60..,. TAS Drift = 100 x 60 ..,. 460 = approx. 13°

I

4242 (C)

I

4506 (A)

I

4851 (C)

I

Wind angle = 085° - 030° = 55° XWC = sin (wind angle) x wind speed XWC = sin 55° x 100 kts = 81,9 kts Drift=XWCx60..,. TAS Drift = 81,9 x 60..,. 470 = 10,4° Heading = 085° - 10° = 075° HWC = cos (wind angle) ~ wind speed HWC = cos 55° x 100 kts = 57,3 kts headwind Effective TAS = TAS x cos WCA Eff.TAS = 470 x cos 10,4°= 462,3 kts GS = 462,3 kts - 57,3 kts headwind = 405 kts We have determined that our heading will be 075 and the Ground Speed 405 kts. Now the last step is to calculate the time it will take us to fly a distance of265NM: 0

Aviationexam Test Prep Edition 2012 Distance = Rate x Time. 265 NM = 405 kts x ? hrs ? hrs = 265 NM -;- 405 kts ? =0,6543 hrs =39 minutes (0,6543 x 60)

4853. Given:

Airplane

ATPL

True course A to B: Distance A to B: TAS:

CPL

method: Wind angle =090° - 000° =90° XWC = sin (wind angle) x wind speed XWC=sin900x 100kts= 700kts

Heli

ATPL

CPL

True Heading = 090° - 73° = 077"

250° 315NM 450 kts 200°/60 kts 0650UTC

WIV: ETDA:

Drift=XWCx60-;- TAS Drift = 700 x 60 -;- 460 =approx. 73° HWC = cos (wind angle) x wind speed HWC= cos 90 0 x 700kts = Okts Effective TAS = TAS x cos WCA Eff.TAS = 460 x cos 73" = 448,2 kts GS = 448,2 kts - 0 kts headwind = 448,2 kts

What is the ETA at B?

We have determined that our TH = 077" and our GS = 448 kts. Now we need to convert the TH to MH using the variation and subsequently the MH to CH using the deviation. A very useful mnemonic for the actual calculation exists: CDMVT = Can Dead Men Vote Twice? It stands for Compass Heading; Deviation; Magnetic Heading; Variation; True Heading.

A) 0730UTe B) 0736 UTe C) 0810 UTe D) 0716 UTe (Refer to figures 067-E175, 067-E176, 067-E777, 067-E55, 067-E63 and 067-E64) For the solution of this problem we need to determine the Ground Speed (GS) and then use it to calculate the flight time. For an example of solution of a wind correction and Ground speed problem refer to the attached illustration. Please note that the illustration to this type of question only explains the solution in general and does not necessarily need to depict the solution of this specific question. Ifyou do not have the navigation computer available you can use a calculation method: Wind angle =250° - 200° =50° XWC =sin (wind angle) x wind speed XWC =sin 50° x 60 kts = 46 kts

=

Easterly Deviation as well as Easterly Variation is a positive value (+ °E). Westerly Deviation as well as Westerly Variation is a negative value (- OW). When applying the values of Deviation and Variation using the mnemonic "CDMVT" we usually start off with the True heading => we work from the right to the left. When applying the Deviation and Variation values we have to apply it in the correct sense - either with a "plus" or a "minus" sign. Remember the mnemonic "West is Best, East is Least" = in other words West = plus sign, East = minus sign. This mnemonic works when starting with True heading and working your way towards the left though the Magnetic heading to the Compass heading. When working the other way around = starting with the Compass heading and wishing to obtain the Magnetic heading or the True heading, you have to reverse this logic => West = minus sign and East = plus sign.

C

D ? <= 2°W

Drift XWC x 60 -;- TAS Drift = 46 x 60 -;- 450 = 6,7°

M

+

V

r

T

0 10 E - 077°

<=

Magnetic Heading (?) = 067" Compass Heading (??) 069°

HWC = cos (wind angle) x wind speed HWC = cos 50° x 60 kts = 38,6 kts headwind

=

Effective TAS =TAS x cos WCA Eff.TAS = 450 x cos 6,7°= 447,4 kts

Note: Remember that you have to work your way from the right (from TH) to the left (to CH).

GS = 447,4 kts - 38,6 kts headwind = 409 kts

4866. Given:

We have determined that our Ground Speed will be approx. 409 kts. Now the last step is to calculate the time it will take us to fly a distance of375NM: Distance = Rate x Time. 375 NM =409 kts x ?hrs ?hrs =375 NM -;-409kts ? =0,77 hrs =46 minutes (0,77 x 60)

Airplane

ATPL

True course from A to B: TAS:

WIV: Average variation: Deviation:

CPL

Heli

ATPL

CPL

090° 460 kts 360/100 kts 100E _2°

Calculate the compass heading and GS.

A) B) C) D)

ATPL

CPL

Heli

ATPL

CPL

Required course 045°(M) Variation is 15°E WIVis 190°(T)/30 kts CAS is 120 kts at FL55 in standard atmosphere What are the heading (OM) and GS?

If we departed at 06:50 UTC and the flight time is 46 minutes, the estimated time of arrival will be at 07:36 UTe.

4860. Given:

Airplane

078 0 - 450 kts. 068 0 - 460 kts. 069 0 - 448 kts. 070 0 - 453 kts.

(Refer to figures 067-E775, 067-E776, 067-E177, 067-E55, 067-E63, 067-E64 and 067 -E97) Let's start by the determination of the True Heading (TH) + Ground Speed (GS) based on the track required and the wind: For an example of solution of a wind correction and Ground speed problem refer to the attached illustration. Please note that the illustration to this type of question only explains the solution in general and does not necessarily need to depict the solution of this specific question. Ifyou do not have the navigation computer available you can use a calculation

14853(8) I 4860 (C) 14866(8) I

A) 0360 and 151 kts. B) 055 0 and 147 kts. C) 049 0 and 154 kts. D) 0560 and 137 kts. (Refer to figures 067-E175, 067-E176, 067-E177, 067-E55, 067-E58, 067-E59, 067-E63, 067-E64 and 067-E97) Let's start by the determination of the TAS from CAS and FLiTemp information. Use your navigation computer - remember that the ISA temp at FL55 is 4°e. You should obtain a TAS of 730 kts. Next step will be the determination of the Heading + Ground Speed (GS) based on the track required and the wind. Notice that the track required is given in °Magnetic and the Wind is given in °True. Let's convert the wind into °Magnetic - in this way we will obtain the Magnetic Heading after determining the drift angle. Remember the mnemonic "West is Best, East is Least" =in other words West = plus sign, East = minus sign. This m?iemoilic works when starting with True value and working your way to obtain the Magnetic value, which is our case here. Therefore, 790° True wind direction - 75° Easterly variation will result in 775° Magnetic wind direction. Let's continue with a value of WN 775"130 kts (M). For an example of solution of a wind correction and Ground speed problem refer to the attached illustration. Please note that the illustration to this type of question only explains the solution in general and does not necessarily need to depict the solution of this specific question. Ifyou do not have the navigation computer available you can use a calculation method:

04 Dead Reckoning Navigation (DR) Wind angle = 175°-045°= 130° XWC = sin (wind angle) x wind speed XWC = sin 130° x 30 kts = 23 kts

• NAM = NGM x TAS + GS • NGM =NAM x GS + TAS

230896. Airplane ATPL CPL Heli ATPL CPL An aircraft is flying from SALCO to BERRY HEAD on Magnetic Track 007°, TAS 445 kts. The wind is 050 0 (T)/40 kts. Variation SoW, deviation +2° At 1000 UTC the RB of locator PY is 311°. At 1003 UTC the RB of locator PY is 266°. Calculate the True bearing of locator PY at 1003 UTC from the aircraft.

Drift=XWCx 60+ TAS Drift = 23 x 60 + 130 = 10,6° True Heading = 045° + 10,6° = 055,6° HWC =cos (wind angle) x wind speed HWC =cos 130° x 30 kts = 19,3 kts tailwind Effective TAS =TAS x cos WCA Eff.TAS= 130 x cos 10,6°= 127,8kts GS = 127,8 kts + 19,3 kts tailwind = 147, 1 kts

A) 275° (T)

We have determined that our MH =055,6° and ourGS = 147 kts.

4960. Given: AD: GO: TAS: GS:

Airplane

ATPL

CPL

Heli

ATPL

CPL

B) 27]o(T) e) 268° (T) D) 272°(T)

Air distance Ground distance True airspeed Ground speed

Which of the following is the correct formula to calculate ground distance (GO) gone? A) GD = (AD x GS) + TAS B) GD = (AD - TAS) + TAS C) GD = AD x (GS -TAS) + GS D) GD =TAS+ (GS x AD) NAM =Nautical Air Miles =Air Distance NGM =Nautical Ground Miles =Ground Distance TAS =True Air Speed GS = Ground Speed

230897. Airplane ATPL CPL Heli ATPL CPL An aircraft is flying from SALCO to BERRY HEAD on Magnetic Track 007°, TAS 445 kts. The wind is 050 0 (T)/40 kts. Variation SoW, deviation +2° At 1000 UTC the RB of locator PY is 311°. At 1003 UTC the RB of locator PY is 266°. Calculate the distance of the aircraft from locator PY at 1003 UTC. A) 22 NM B) 21 NM e) 23 NM D) 24NM

I

04-05 Measurement of DR elements 1257. Airplane ATPL CPL Heli ATPL CPL Calibrated airspeed (CAS) is indicated airspeed (lAS) corrected for: A) density. B) temperature and pressure error. e) compressibility error. D) instrument error and position error. Indicated Airspeed (lAS) = the direct instrument reading obtained from the ASI, uncorrected for variations in atmospheric density, installation error, or instrument error. lAS drops as you climb, because as the density of the air decreases with altitude, fewer air molecules hit the pitot tube. This effect is most noticeable in high-performance aircraft that operate at high altitude. For example, at cruise altitude, the airspeed indicator on the B737 may indicate about 280 kts when the actual speed through the air is more than 400 knots. Pilots use indicated airspeed to get the proper performance from their aircraft (takeoff, climb, approach, and landing speeds are based on lAS). Calibrated Airspeed (CAS) = lAS corrected for position (pressure) and instrument errors (at certain airspeeds and with certain flap settings, the position and instrument errors may total several knots). This error is generally greatest at low airspeeds. In the cruising and higher airspeed ranges, indicated airspeed and calibrated airspeed are approximately the same. Refer to the airspeed calibration chart to correct for possible airspeed errors. Equivalent airspeed (EAS) = the speed at sea level that would produce the same incompressible dynamic pressure as the true airspeed at the altitude at which the vehicle is flying. An aircraft in forward flight is subject to the effects of compressibility. Likewise, the calibrated airspeed is a function of the compressible impact pressure. EAS, on the other hand, is a measure of airspeed that is a function of incompressible dynamic pressure. Structural analysis is often in terms of incompressible dynamic pressure, so that equivalent airspeed is a useful speed for structural testing. At sea level, standard day, CAS and EAS are equal (or equivalent), but only under those conditions.

1 4960 (A) 1230896 (D) 1230897 (8) 1 1257 (D) 1 2245 (e) 1

True Airspeed (TAS) = CAS corrected for altitude and nonstandard temperature. It is the actual speed of an aircraft through the air. Because air density decreases with an increase in altitude, an airplane has to be flown faster at higher altitudes to cause the same pressure difference between pitot impact pressure and static pressure. Therefore, for a given CAS, the TAS increases as altitude increases; or fora given TAS, the CAS decreases as altitude increases. The airspeed indicator displays TAS only at sea level under standard conditions, so you must calculate TAS based on lAS, the current pressure altitude, and air temperature. As a rule of thumb, you can estimate TAS by adding 2% to lAS for each 1.000 ft ofaltitude. Pilots use TAS in navigation calculations and when filing flight plans. Groundspeed (GS) = the actual speed of the airplane over the ground. It is true airspeed adjusted for wind. Groundspeed decreases with a headwind, and increases with a tailwind. Summary: CAS = lAS + Pressure Error correction EAS = CAS + Compressibility Error correction TAS = EAS + Density Error correction TAS = CAS + Compressibility Error correction + Density Error correction

2245. Airplane ATPL CPL Heli ATPL CPL Using mental navigation, the local speed of sound may be found using the following equation: A) LSS = TAS + Mach number x TATc B) LSS = 333 x TAS + Mach number C) LSS = 644 + 1,2 TATc D) LSS = 735 - 1,05 TATc Remember that Mach number is the ratio of the aircraft's speed (TAS) to the local speed of sound (LSS). Mach No. = TAS + LSS. The Speed of Sound varies only with the temperature. As the temperature increases so does the local speed of sound. Because the temperature reduces with altitude, the speed

Aviationexam Test Prep Edition 2012 of sound reduces as altitude increases. A formula for calculating the L55 for a given temperature is: LSS = 38,95 x ,jAbsolute temperature (in °Kel_ vin) ... Absolute temp in oK =°C + 273. Temperature in °C = oK - 273.

We can also use an alternative formula, using TAT in 0(, but this does not yield a precise result - only an estimation: L55 = 644 + (1,2 x Temp °C). To obtain the Mach Number from TA5 and OAT: 7) Find OAT in OK (= OAT in °c + 273) 2) Find L55 (=38,95 x ,fOAT in OK) 3) Find Mach nr. (= TA5 .;- L55)

4311. Airplane ATPL CPL Heli ATPL CPL Your pressure altitude is FL55, the QNH is 998, and the OAT is +30 °C. What is density altitude? A) B) C) D)

6.980 feet. 7.750 feet. 8.620 feet. 10.020 feet.

(Refer to figure 067-E60) Refer to the attached illustration for an example of general method of Density Altitude problem solution. Please note that the illustration only explains the procedure in general and does not necessarily demonstrate the solution of the problem with the values of this specific question. Definitions: • Density Altitude (DA) =a Pressure Altitude corrected for non-standard temperature. Aircraft performance is greatly affected by Density altitude. • Pressure Altitude (PA) = reading of the altimeter with 7073 mb (29.92 inHg) set in the reference window. When maintaining Flight Levels your altimeter is indicating a Pressure Altitude. Density altitude problems can also be solved using calculation method. The key information to remember is that Density Altitude correction = 720 ft for eacll 7" of temperalwe devicJlioll (rolll ISA. If llle lemperature is colder, then the Density altitude will be lower than Pressure Altitude. If the temperature is warmer than ISA, the Density Altitude will be higher than Pressure Altitude. In this case: • PA =5.500 ft ·/5A at 5.500 ft =4°C • Actual temp at 5.500 ft = +30°C • Temp deviation =15A+26° .26° of deviation = 720 ft x 26° =3. 720 ft • Temp is warmer tllan 15A => Density Alt will be approximately 3. 720 ft higher => 8.620 ft (5.500 ft PA + 3. 720 ft DA correction).

4341. Airplane ATPL CPL Heli ATPL CPL An aircraft takes off from the aerodrome of BRIOUDE (elevation 1.483 ft, QFE 963 hPa, temperature 32°C). Five minutes later, passing 5.000 ft on QFE, the second altimeter is reset to 1013 hPa. After the reset it will indicate approximately:

=

=

=

A) 6.900ft B) 6.400 ft C)3.500 ft D) 4.000ft The question is quite easy to solve, although it might appear a bit complicated. When the altimeter is set to QFE it will indicate the height above aerodrome level. If the value of 7073 hPa is set in the altimeter window a Pressure Altitude will be indicated. If we change the pressure reference value in the altimeter setting window, the altimeter reading will also change, even if we maintain our actual vertical position constant. During this change when we are changing to a lower pressure value, the altitude indication decreases. When changing to a higher pressure value, the altitude indication increases. For each 7 hPa of change the altimeter reading will change by 27 ft up or do"wn (at lower altitudes). Therefore, if we change the reading from 963 hPa to 7073 hPa we actually increase the pressure reference value by 50 hPa. Total altimeter indication change will be 50 hPa x 27 ft = 7.350 ft. Since the pressure reference is now higher the altimeter will indicate a higher altitude => 5.000 ft (original) + 7.350 ft (change) = 6.350 ft (new reading) = approximately 6.400 ft.

4382. Given:

Airplane

OAT: Pressure alt:

ATPL

Heli

CPL

ATPL

4311 (C)

I

4341 (8)

A) B) C) D)

4.550 feet. 5.550 feet. 4.290 feet. 5.320 feet.

(Refer to figure 067-E60) Refer to the attached illustration for an example of general method of True Altitude problem solution. Please note that the illustration only explains the procedure in general and does not necessarily demonstrate the solution of the problem with the values of this specific question. True altitude problems can also be solved using calculation method. The key information to remember is that altitude correction for non-standard temperature = 0,4% of altitude for each 7° of temperature deviation from 15A. If the temperature is colder, then the True altitude will be lower than Indicated Altitude. If the temperature is warmer than 15A, the True Altitude will be higher than Indicated Altitude. In this case: ·/5A at 5.000 ft = 5°C • Actual temp at 5.000 ft =+35°C • Temp deviation = 15A+30° ·0,4% of5.000 ft (per 7° deviation) =20 ft .30° of deviation = 20 ft x 30° = 600 ft • Temp is warmer than 15A => True Alt will be approximately 600 ft higher => 5.600 ft (5.000 ft + 600 ft TA correction).

4458. Given:

Airfield elevation: QNH:

4382 (8)

I

4458 (C)

ATPL

CPL

Heli

ATPL

CPL

1000ft 988 hPa

What is the pressure altitude? (Note: assume 1 hPa = 27 ft) A) B) C) D)

320ft 680ft 1675 ft -320ft

(Refer to figure 067-E60) Remember that by definition the Pressure Altitude (PA) = reading of the altimeter with 7073 mb (29.92 inHg) set in the reference window. When using the Flight Levels as the altitude reference we are actually flying a constant PA, because to maintain a Flight Level we have 7073 hPa set in the altimeter window. When we fly with an actual QNH set in the altimeter window the altimeter is showing an "Indicated Altitude': Ifwe change the QNH value in the altimeter setting window, the altimeter reading will also change, even if we maintain our actual vertical position constant. During this QNH change when we are changing to a lower QNH, the altitude indication decreases. When changing to a higher QNH, the altitude indication increases. In the case of this question we have an altimeter reading of 7.000 ftwith a QNH of 988 hPa. If we change the setting to 7073 hPa (definition of PAl the altitude indication will increase by 27 ft for each 7 hPa. Total change will be 25 hPa (1073 hPa - 988 hPa) x 27 ft = 675 ft. Add this to the previous indication of 7.000 ft to obtain the PA of 7.675 ft.

4510. Given:

Airplane

Pressure Altitude: OAT:

ATPL

CPL

Heli

ATPL

CPL

29.000ft -50°

Calculate the density altitude. A) B) C) D)

32.500ft 28.000ft 31.000ft 27.500ft

Definitions:

I

Airplane

(Refer to figure 067-E60) Refer to the attached illustration for an example of general method of Density Altitude problem solution. Please note that the illustration only explains the procedure in general and does not necessarily demonstrate the solution of the problem with the values of this specific question.

CPL

+35°C 5.000 feet

I

What is true alt?

I

4510 (8)

I

04 Dead Reckoning Navigation (DR) • Density Altitude (DA) = a Pressure Altitude corrected for non-standard temperature. Aircraft performance is greatly affected by Density Altitude. • Pressure Altitude (PA) = reading of the altimeter with 7073 mb (29.92 inHg) set in the reference window. When maintaining Flight Levels your altimeter is indicating a Pressure Altitude. Density altitude problems can also be solved using calculation method. The key information to remember is that Density Altitude correction = 720 ft for each 7° of temperature deviation from ISA.1f the temperature is colder, then the Density altitude will be lower than Pressure Altitude. If the temperature is warmer than ISA, the Density Altitude will be higher than Pressure Altitude. In this case: • PA = 29.000 ft ·ISA at 29.000 ft = -43°C • Actual temp at 29.000 ft =-50°C • Temp deviation = ISA-r • rofdeviation = 720 ft x r= 840 ft • Temp is colder than ISA => Density Alt will be approximately 840 ft lower => 28.760 ft (29.000 ft PA - 840 ft DA correction).

Airplane ATPL CPL Heli ATPL CPL The pressure alt is 29.000 feet and the SAT is -55°C. What is density altitude? 4529.

A) 27.500 feet. B) 26.000 feet. C) 30.000 feet. 0) 31.000 feet. (Refer to figure 067-£60) Refer to the attached illustration for an example of general method of Density Altitude problem solution. Please note that the illustration only explains the procedure in general and does not necessarily demonstrate the solution of the problem with the values of this specific question. Definitions: • Density Altitude (DA) = a Pressure Altitude corrected for non-standard temperature. Aircraft performance is greatly affected by Density altitude. • Pressure Altitude = reading of the altimeter with 1073 mb (29.92 inHg) set in the reference window. When maintaining Flight Levels your altimeter is indicating a Pressure Altitude. Density altitude problems can also be solved using calculation method. The key information to remember is that Density Altitude correction = 720 ft for each 7°oftemperature deviation from ISA.lfthe temperature is colder, then the Density altitude will be lower than Pressure Altitude. If the temperature is warmer than ISA, the Density Altitude will be higher than Pressure Altitude. In this case: • Pressure altitude =29.000 ft ·ISA at 29.000 ft =-43°C • Actual temp at 29.000 ft = -55°C • Temp deviation = ISA-72° • 72°ofdeviation= 720ftx 7r= 7.440ft • Temp is colder than ISA => Density Alt will be approximately 7.440 ft lower => 2Z560 ft (29.000 ft PA - 7.440 ft DA correction).

4640.

Airplane

ATPL

CPL

Heli

ATPL

CPL

Given: Pressure altitude: OAT:

15.000ft -35°C

What is the density altitude:

A) 12.500ft B) 17.500ft C) 11.000 ft 0) 18.000ft (Refer to figure 067 -£60) Refer to the attached illustration for an example of general method of Density Altitude problem solution. Please note that the illustration only explains the procedure in general and does not necessarily demonstrate the solution of the problem with the values of this specific question. Definitions: • Density Altitude (DA) a Pressure Altitude corrected for non-standard temperature. Aircraft performance is greatly affected by Density altitude. • Pressure Altitude (PA) = reading of the altimeter with 1073 mb (29.92 inHg) set in the reference window. When maintaining Flight Levels your altimeter

=

I 4529 (A) I 4640 (A) 14666(8) I

is indicating a Pressure Altitude. Density altitude problems can also be solved using calculation method. The key information to remember is that Density Altitude correction = 720 ft for each 7° of temperature deviation from ISA.1f the temperature is colder, then the Density altitude will be lower than Pressure Altitude. If the temperature is warmer than ISA, the Density Altitude will be higher than Pressure Altitude. In this case: • PA = 75.000 ft ·ISA at 75.000 ft = - WC • Actual temp at 75.000 ft = -35°C • Temp deviation = ISA-20° • 20° of deviation = 720ftx200=2.400ft • Temp is colder than ISA => Density Alt will be approximately 2.400 ft lower => 72.600 ft (15.000 ft PA - 2.400 ft DA correction).

4666. Airplane ATPL CPL Heli ATPL CPL An aircraft at FL150 overhead an airport. Given:

Elevation of airport: QNH: OAT at FL150:

720 ft 1003 hPa -5°C

What is the true altitude of the aircraft? (Note: assume 1 hPa = 27 tt)

A) 15.840ft B) 15.300 ft C) 14.160 ft 0) 14.720ft (Refer to figure 067-£60) Refer to the attached illustration for an example of general method of True Altitude problem solution. Please note that the illustration only explains the procedure in general and does not necessarily demonstrate the solution of the problem with the values of this specific question. • Indicated Altitude = the altitude reading of the altimeter, assuming it is correctly set (correct QNH in the reference window). It indicates the approximate height of the aircraft above the Mean Sea Level (MSL). • Calibrated Altitude = indicated altitude corrected for the instrument, position and installation errors. • True Altitude = calibrated altitude corrected for non-standard atmospheric conditions. It is the actual height of the aircraft above the sea level. True altitude problems can also be solved using calculation method. The key information to remember is that altitude correction for non-standard temperature = 0,4% of altitude for each 7° of temperature deviation from ISA. If the temperature is colder, then the True altitude will be lower than Indicated Altitude. If the temperature is warmer than ISA, the True Altitude will be higher than Indicated Altitude. In this case: 7) First of all let's obtain the Calibrated Altitude (CA). From the definition above we can see that the CA is an indicated altitude corrected for various instrument errors. Let's disregard the instrument errors and assume that Calibrated =Indicated. Indicated Alt. =reading of the altimeter with QNH set. Pressure Alt. = reading of the altimeter with 7073 set as the reference. With QNH of 7003 hPa the altimeter would read a lower altitude than with 1073 => ifeach 7 hPaequals27ftthen 70hPa (7073 -1003) will represent 270ft. OurCA = 75.000ft-270 ft= 74.730 ft. 2) ISA Deviation: ·ISA temp at 74.730ft=-74,soC • Our actual temp = -5°C • Temp deviation = ISA +9,5°C 3) True Alt correction ·If ground elevation is given, we only apply the temp correction to the Dcolumn of airD above the ground level • Calibrated Altitude above ground = 74.730 ft- 720 ft airport elevation = 74.01Oft ·0,4% of altitude fof each 7° of temperature deviation from ISA ·0,4% of 74.070 =56 ft (per 7° ISA deviation) • Total correction = 9,5° x 56 ft =532 ft 4) Final True Alt calculation • Ground elevation =720 ft . • Calibrated Altitude above ground = 74.010 ft • True Alt correction = 532 ft (positive correction, because temp is warmer thanlSA) • True Altitude = 720 ft + 74.070ft+532ft= 75.262ft

Aviationexam Test Prep Edition 2012 4697. Airplane ATPL CPL Heli ATPL CPL You want to fly 12.000 ft above a frozen lake at elevation 930 ft AMSL. You have obtained QNH from an airfield in the area. Climbing, you observe that the air temperature at FL80 is -20°C. What should your indicated altitude be when you are 12.000 ft above the frozen lake? Use the mechanical computer for the calculations. A) 11.950 ft B) 12.000ft () 12.560 ft D) 13.950ft (Refer to figure 067-E60) The question states that the pilot wishes to maintain a height of 72.000 ft directly above a frozen lake. We are therefore interested in finding the true altitude above the ground. At the pressure altitude of 8.000 ft the outside air temp is -20°C. Using your navigation computer, align the value -20°C with the value of 8.000 ft in the 'TRUE ALT" window, then find the value of 72.000 ft on the outer ring scale (the required True Altitude above the ground) and read the corresponding Indicated altitude of approx. 72.970 ft on the inner ring scale. Now remember that this is our True Altitude above the lake = above the ground. The question asks about the True Altitude above MSL = we have to add the elevation of the lake = 72.970 ft + 930 ft = 73.900 ft. True altitude problems can also be solved using calculation method. The key information to remember is that altitude correction for non-standard temperature = 0,4% of altitude for each 7° of temperature deviation from ISA. If the temperature is colder, then the True altitude will be lower than Indicated Altitude. If the temperature is warmer than ISA, the True Altitude will be higher than Indicated Altitude. In this case: ·ISA atB.OOO ft= -7°C • Actual temp at 8.000 ft = -20'C • Temp deviation = ISA-79° ·0,4% of 72.000 ft (per 7° deviation) = 4B ft • 79° of deviation =4B ftx 79°= 972 ft • Temp is colder than ISA => True Alt will be approximately 972 ft lower than Indicated Altitude. ·If we wish to maintain 72.000 ft TA AGL then our indicated altitude AGL = 72.000 ft + 972 ft = 72.972 ft • Since the question asks for the Indicated altitude MSL we have to add the lake elevation => 72.972 ft + 930 ft = 73.B42 ft.

4734. Airplane ATPL CPL Given: Pressure Altitude: 10.000 ft OAT: +15°C What is true altitude: A) B) () D)

Heli

ATPL

CPL

4754. Airplane ATPL CPL Heli ATPL CPL Airfield elevation is 1.000 feet. The QNH is 1035. Use 27 feet per millibar. What is the pressure altitude? 1.675 ft 1.594 ft () 406ft D) -594ft A) B)

(Refer to figure 067-E60) Remember that by definition the Pressure Altitude (PA) = reading of the altimeter with 1073 mb (29.92 inHg) set in the reference window. When using the Flight Levels as the altitude reference we are actually flying a constant PA, because to maintain a Flight Level we have 1073 hPa set in the altimeter window. When we fly with an actual QNH set in the altimeter window the altimeter is showing an "Indicated Altitude'~ If we change the QNH value in the altimeter setting window, the altimeter reading will also change, even if we maintain our actual vertical position constant. During this QNH change when we are changing to a lower QNH, the altitude indication decreases. When changing to a higher QNH, the altitude indication increases. In the case of this question we have an altimeterreading of 7.000 ftwith a QNH of 7035 hPa.lf we change the setting to 1073 hPa (definition of PAl the altitude indication will decrease by 27 ft for each 7 hPa. Total change will be 22 hPa (7035 hPa - 7073 hPa) x 27 ft =594 ft. Subtract this from the previous indication of 7.000 ft to obtain the PA of 406 ft.

4945. Airplane ATPL CPL Heli ATPL CPL What is the ISA temperature value at FL330? -56°( -66°( () -81 o( D) -51°(

A) B)

This is the quickest and easiest way to find ISA at a particular altitude or FL. Simply take the altitude or FL and double it. If we do this we get 66. Now, subtract this value from 75 => you get -57°C. Note: Remember that the standard temperature at. sea level is 75°C and the lapse rate is 2° per 7.000 ft of altitude.

231028. Airplane ATPL CPL Heli ATPL CPL An aircraft is flying at FL150, with an outside air temperature of -30°, above an airport where the elevation is 1.660 ft and the QNH is 993 hPa. Calculate the true altitude. A) B) () D)

9.260ft 10.750 ft 8.480ft 11.830ft

(Refer to figure 067 -E60) Refer to the attached illustration for an example of general method of True Altitude problem solution. Please note that the illustration only explains the procedure in general and does not necessarily demonstrate the solution of the problem with the values of this specific question. True altitude problems can also be solved using calculation method. The key information to remember is that altitude correction for non-standard temperature = 0,4% of altitude for each 7° of temperature deviation from ISA. If the temperature is colder, then the True altitude will be lower than Indicated Altitude. If the temperature is warmer than ISA, the True Altitude will be higher than Indicated Altitude. In this case: ·ISA at 10.000 ft =-5°C • Actual temp at 70.000 ft=+75°C • Temp deviation =ISA+20° ·0,4% of 10.000 ft (per 7°deviation) = 40 ft .20° of deviation =40 ft x 20° =BOO ft • Temp is warmer than ISA => True Alt will be approximately BOO ft higher => 1O.BOO ft (70.000 ft + BOO ft TA correction).

13.660ft 15.210 ft 17.160 ft 14.120ft

231029. Airplane ATPL CPL Heli ATPL CPL An aircraft is flying at FL200. The QNH, given by a meteorological station at an elevation of 1.300 ft is 998.2 hPa. OAT 40°C. The elevation of the highest obstacle along the route is 8.000 ft. Calculate the aircraft's approximate clearance above the highest obstacle on this route. A) 11.800ft B) 9.200ft C) 10.500ft D) 20.200ft

Given: An aircraft is flying at FL100, OAT ISA -15°C. The QNH, given by a meteorological station with an elevation of 100 ft below MSL is 1032 hPa. 1 hPa = 27 ft Calculate the approximate True Altitude of this aircraft. A) 10.600ft B) 9.900ft C) 11.200 ft D) 9.400ft

1 4697 (D) 14734(8) 1 4754(C) 1 4945 (D) 1231028 (A) 1231029 (C) 1231030(8) 1

.-~---.-~---.-~-~-~-------~----~--~---------

04 Dead Reckoning Navigation (DR)

231031.

Airplane

ATPL

CPL

Heli

ATPL

CPL

An aircraft is flying at FL250, OAT -45°C. The QNH, given by a station at MSL, is 993,2 hPa. Calculate the approximate True Altitude. A) B) C) 0)

25.500ft 26.100ft 23.400ft 24.000ft

=

Airplane ATPL CPL Heli ATPL CPL 231033. An aircraft flies at FL250. OAT -45°C. The QNH, given by a meteorological station with an elevation of 2.830 ft, is 1033 hPa~. Calculate the clearance above a mountain ridge with an elevation of 20.410 ft. A) B) C) 0)

231088. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-18) An aircraft is flying from Inverness VOR/DME (N57°32,6', W 004°02,5W) to Aberdeen VORDME (N57°18,6', W002°16,0'W). At 1000 UTC the fix of the aircraft is determined by VOR/DME Inverness: radial = 114; DME-distance = 20,5 NM. At 1006 UTC the fix of the aircraft is determined by VOR/OME Aberdeen: radial 294; DME-distance 10,5 NM. What is the average GS of the aircraft between 1000 UTC and 1006 UTC? A) B) C) 0)

3.000ft 4.200ft 3.500ft 4.600ft

231034. Airplane ATPL CPL Heli ATPL CPL An aircraft is flying at FL200. OAT 0 °C. When the actual air pressure on an airfield at MSL is placed in the subscale of the altimeter the indicated altitude is 19.300 ft. Calculate the aircraft's true altitude. A) B) C) 0)

19.300ft 21.200ft 20.200ft 20.700ft

231036.

Airplane

ATPL

CPL

Heli

ATPL

CPL

An aircraft has to fly over a mountain ridge. The highest obstacle, indicated in the navigation chart, has an elevation of 9.800 ft. The QNH, given by a meteorological station at an elevation of 6.200 ft, is 1022 hPa. The OAT ISA +5 °C. Calculate the approximate indicated altitude to obtain a clearance of 2.000 ft. A) B) C) 0)

lf800 ft and 12.000 ft 10.900 ft and 11.100 ft 11.900 ft and 11.200 ft 11.500 ft and 11.700 ft

231052. Airplane ATPL CPL Heli ATPL CPL An aircraft is at position (53°N, 006°W) and has a landmark at position (52°47'N, 004°45'W), with a relative bearing of 060°. Given: Compass Heading: Variation: Deviation:

051° 16°W rE

What is the true bearing of the position line to be plotted from the landmark to the aircraft on a Lambert chart with standard parallels at 37°N and 65°N? A) B) C) 0)

2760 278° 250 0 277 0

1231031 (C) 1231033 (8) 1231034 (8) 1231036 (D) 1231052 (8) 1231088 (A) 1

280 kts 485 kts 385 kts 180 kts

=

Aviationexam Test Prep Edition 2012

05 In-Flight Navigation

IN-FLIGHT NAVIGATION 05-01 Use of visual observations and application to in-flight navigation

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1129. Airplane ATPL CPL Heli ATPL What do you understand by the term "white-out"?

1214. Airplane ATPL Transferring position lines:

CPL

A) Flight conditions when you suddenly enter a cloud and everything "gets white" outside the cockpit windows. B) When the terrain is covered with snow and the horizon blend with the sky, visualdetermination of height becoming difficult. C) Flying in heavy snow. D) laking off from a snow-covered lake. In areas where snow and/or ice cover the terrain from horizon to horizon and where the sky is covered with a uniform layer of clouds so that no shadows are cast, the horizon disappears causing the ground and the sky to blend and making a pilot disorientation a real danger. In a "white-out" there is a complete lack of contrast => distance and height above the ground are virtually impossible to estimate.

1169. Airplane ATPL CPL Heli ATPL Transferring range position lines, you should:

(Refer to figures 061-E713 and 061-E114) It is unusual to be able to obtain simultaneous position lines (P/Ls). It is more normal to use two or three successive P/Ls. Three lines give greater accuracy. These may be from a beacon when flying near to or abeam it or from other sources. When P/Ls are obtained at different times, a technique known as "transferring the position line" is used. All P/Ls are transferred along track line at the Ground Speed to the time of the last PlL. If all is well, they will intersect at one position with no error. If there is error, a small triangle is formed called a "cocked hat" because of the shape. The intersection, or center of the cocked hat is the position of the fix. An alternative method to transferring the P/L along track is to transfer the actual radio station itself. Whether or not to use this method is a matter of personal preference. The radio station is moved along a track parallel to the planned aircraft track. Each position line is then plotted normally from the transferred station or "base':

A) B) C) D)

Heli

CPL

ATPL

CPL

1217. Airplane ATPL CPL To establish a track plot you need: A) B) C) D)

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1169 (8)

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1213 (A)

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1214 (A)

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1217 (C)

Heli

1281. Airplane The track plot: A) B) C) D)

ATPL

CPL

Heli

CPL

ATPL

CPL

is used to determine the actual WIV. shows the path of the aircraft relative to the ground. is used to determine the drift angle. leads from the air position to the new fix.

For explanation refer to question #1217 on this page.

2183. Airplane ATPL CPL Heli Transferring position lines can be done with: A) B) C) D)

radials and QDM/QDR only. QDM/QDR and QlE. radials and DME only. radials, DME, QDM/QDR.

For explanation refer to question #1169 on this page.

I

ATPL

(Refer to figures 061-E113 and 061-E114) The "track plot" on your chart represents the path of the aircraft relative to the ground (track over the ground). When using DR (Dead Reckoning or Deduced Reckoning) navigation technique you are able to plot the line representing your actual track only based on at least 2 fixes that you have obtained either from mUltiple (2 or more) position lines or from a pinpoint of a prominent ground feature (overflying a significant landmark). It makes sense => ifyou wish to plot the actual track line on your chart, you need to have an actual starting point and an actual "along the track" point ofyour aircraft.

the lines of position are transferred at ground speed. the lines of position are transferred at lAS effective. the lines of position are transferred at lAS. the lines of position are transferred lAS effective plus wind speed.

1129 (8)

CPL

the last position, air position and DR position. the last position and the DR position. at least two pinpoints or fixes. the air position and a pinpoint or fix.

For explanation refer to question #1169 on this page.

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ATPL

For explanation refer to question #1169 on this page.

When a DME position line (range position line) has to be transferred, the DME station must be transferred - there is no other way to do it.

1213. Airplane ATPL Transferring position lines:

Heli

A) the lines of position are transferred along the track line. B) the lines of position are transferred along the heading line. C) the lines of position are transferred along the track line corrected by the track error. D) the lines of position are transferred along the heading line corrected by the track error.

CPL

A) always plot the original position line as well as the transferred range position line. B) transfer the origin and plot the range position lines from the transferred origin. C) never transfer the origin (DME station). D) include at least 1 straight position line.

CPL

1281 (8)

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2183 (0)

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ATPL

CPL

Aviationexam Test Prep Edition 2012

2190. Airplane ATPL CPL Heli ATPL CPL During a low level flight, 2 parallel roads that are crossed at right angles by an aircraft. The time between these roads can be used to check the aircraft: A) groundspeed B) position e) track D) drift

position using other landmarks, radio navigation fixes, ATC, etc...) does not yield any positive results then the best solution may be to turn towards a good visual line feature such as a coastline, motorway, river or railway => there is a good chance that after a while you will be able to identify a suitable visual checkpoint from which navigation can be resumed.

4266. Airplane ATPL CPL Heli ATPL CPL An aircraft follows a coastline during a particular time. This coastline is:

(Refer to figures 061-E113 and 061-E114) If the distance between the two parallel roads is known from the map then the best use of the knowledge of the time it takes the aircraft to fly the distance between these two roads is to determine the Ground Speed. Note that the question asks for the use of the time, hence the position is not the correct answer.

2257. Airplane ATPL CPL Transferring position lines:

Heli

ATPL

A) a speedline. B) a line of position. e) a VFR line. D) an unknown line of position. (Refer to figures 061-E113 and 061-E114) A position line is a line along which the aircraft was at a particular time.

CPL

A) it is necessary to plot the lines of position in its original position before transferring them. B) it is unnecessary to plot the lines of position in its original position before transferring them. e) in practice it is usual the position lines to intersect in one point. D) the fix obtained by transferring position lines is called off track correction. For explanation refer to question #1169 on page 731.

4067. Airplane ATPL CPL Heli ATPL CPL A useful method of a pilot resolving, during a visual flight, any uncertainty in the aircraft's position Is to maintain visual contact with the ground and: A) set heading towards a line feature such as a coastline, motorway, river or railway. B) fly the reverse of the heading being flown prior to becoming uncertain until a pinpoint is obtained. e) fly expanding circles until a pinpoint is obtained. D) fly reverse headings and associated timings until the point of departure is regained. A situation may arise from time to time when serious uncertainty of position arises. This situation frequently resolves itself when a "missing" feature is found in an unexpected position perhaps, or at a different time to that calculated, or possibly when a large navigational error is discovered. More rarely, it may become necessary after a period of uncertainty to declare that the pilot is officially "lost': The worst response is to circle purposelessly trying to decide what to do. While events will dictate immediate actions, two factors must be considered as soon as possible and, in the vast majority of cases, certain steps should be taken. There may be a risk of controlled flight into terrain (CFIT) and of fuel exhaustion. These dangers must be minimized or removed as soon as possible. If the troubleshooting of your position (checking compass, speed, fixing your

It can be straight, curved or even an irregular line if it is a fiver or a road. It is not a definite position itself but it marks a line of position where the aircraft is or was at a particular time. For example, crossing a rai/way line, a canal or a highway or the coast would give a line ofposition which can then be plotted on the chart. What can be said is that the aircraft was somewhere along the position line at the time of the observation. This is true even though it is not exactly certain whereabouts on the position line it actually was. Although a P/L is not a fix, it is very useful for navigation. When a P/L is plotted on the chart, it is usually marked with a single arrow at each end indicating it is a position line.

4773. Airplane ATPL CPL Heli ATPL CPL Transferring position lines (LOP): An aircraft should follow a true course 120°, given TAS 100 kts and W/V 360/50. It obtains position lines at 14:00, 14:03, 14:06 hours. A) 1 LOP is transferred by 8,0 NM, 1 LOP by 1i,0 NM, the third one is not transferred. B) 1 LOP is transferred by 10,0 NM, 1 LOP by 5,0 NM, the third one is not transferred. e) 1 LOP is transferred by 12,5 NM, 1 LOP by 6,25 NM, the third one is not transferred. D) 1 LOP is transferred by 11,6 NM, 1 LOP by 5,8 NM, the third one is not transferred. (Refer to figures 061-E113, 061-EI14, 067-£63 and 061-E64) The aircraft is following a True Course of 120~ With a TAS of 100 kts and a WN of 360"150 kts the Ground Speed (GS) will be approx. 115 kts (use your navigation computer). With a GS of 115 kts each 1 minute will equal to 1,92 NM (115 kts + 60). When transferring the position lines (P/Ls), you transfer all of the previous P/Ls to the time of the last one (last one is not being transferred). Therefore, from the time of the 1st P/L to the 3rd P/L is 6 minutes (1" P/L taken at 14:00 and the 3'" P/L taken at 14:06) => 6 minutes x 1,92 NM = 11,52 NM (1" P/L will be transferred by 11,52 NM). From the time of the 2nd P/L to the 3,d PlL is 3 minutes (2nd P/L taken at 14:03 and the 3,d P/L taken at 14:06) => 3 minutes x 1,92 NM = 5,76 NM (2nd P/L will be transferred by 5,76 NM). Third P/L is not being transferred.

05-02 Navigation in climb and descent 2241. Airplane ATPL CPL Heli ATPL CPL What is the effect on the Mach number and TASin an aircraft that is climbing with constant CAS? A) Mach number decreases; lAS decreases. B) Mach number remains constant; lAS increases. e) Mach number increases; lAS increases. D) Mach number increases; lAS remains constant.

The speeds quoted for take-off and landing are normally quoted in terms of calibrated airspeed. 1) A constant lAS/CAS climb will result in an increasing TAS and Mach number as the climb progresses. 2} A constant. TAS climb will cause the Mach number to increase as the climb progresses but the IASICAS will decrease. 3} Climbing at a constant Mach number will result in both the lAS/CAS and the TAS decreasing as the climb progresses.

(Refer to figure 061-E54) Calibrated airspeed (CAS): The airspeed value when the lAS is corrected for pressure (system) error is calibrated airspeed, which is equal to equivalent airspeed and true airspeed in a standard atmosphere at mean sea level.

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2190 (A)

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2241 (C)

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05 In-Flight Navigation

4180. Airplane ATPL CPL Heli ATPL CPL An aircraft is climbing at a constant CAS in ISA conditions. What will be the effect on TAS and Mach number?

A) lAS increases and Mach No decreases. B) Both increase. C) Both decrease. D) lAS decreases and Mach No increases.

Now the precise solution method: • Height = sin angle x DME Range • Height =sin 3° x 25 NM • Height = 1,30 NM • Height = approx 7,800 ft Note that this height does not include the 50 ft over the runway => we have to add these 50 ft to get height above the ground => 7,850 ft.

4314. Given:

For explanation refer to question #2241 on page 132.

4293. Airplane ATPL CPL Heli ATPL CPL An aircraft at FL370 is required to commence descent at 120 NM from a VOR and to cross the facility at FL130. If the mean GS for the descent is 288 kts, the minimum rate of descent required is:

Aircraft height: ILSGP:

B) 860 ft/min.

Distance = Rate x Time 120 NM = 288 kts x 7hrs 7hrs = 120 NM + 288 kts 7= 0,4166hrs (25 min)

Airplane

AII-'L

Heli

ll-'L

AIPL

CPL

1.000 ft/min. 700 ftlmin. 900 ft/min. 800 ft/min.

Convert the angle 3,5° into an approximate gradient in % (using the 1:60 rule): • Gradient (%) = Angle (0) x 100 + 60 • Gradient (%) = 3,5°x 100 + 60 • Gradient (%) = 5,83% Now use the following formula to obtain the vertical speed: • Vertical speed = Gradient (%) x Groundspeed (kts) • Vertical speed = 5,83 (%) x 150 (kts) • Vertical speed = 874,5 ft/min

4313. Airplane ATPL CPL Heli ATPL CPL You are on ILS 3° gJideslope which passes over the runway threshold at 50 feet. Your DME range is 25 NM from the threshold. What is your height above the runway threshold elevation? (Use the 1 in 60 rule and 6.000 ft 1 NM)

=

8.010 feet. 7.450 feet. 6.450 feet. 7.550 feet.

(Refer to figures 061-E98 and 061-E99) We can solve this question using the 1:60 rule to obtain an approximate result. Note that the question specifically requests that we use this method: .Angle (0) = (Height x 60) + Distance • Height Distance x Angle (0) + 60 • Height =25 NMx3°+ 60 • Height = 1,25 NivI • Height = 7,500 ft Note that this height does not include the 50 ft over the runway => we have to add these 50 ft to get height above the ground => 7,550 ft.

=

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4180 (8)

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4293 (A)

CPL

2.500ft angle 3°

(Refer to figures 061-E98 and 061-E99) We can solve this question using the 1:60 rule to obtain an approximate result: .Angle (0) = (Height x 60) + Distance • Distance = (Height x 60) + Angle (0) • Distance (2,500 x 60) + 3° • Distance = 50,000 ft • Distance 8,22 NM

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4310 (C)

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4313 (D)

Now the precise solution method: • Height = tan angle x Ground Range • Ground Range = Height + tan angle • Ground Range =2,500 ft + tan 3° • Ground Range =47,703 ft • Ground Range'" 7,85 NM

4325. Airplane ATPL CPL Heli ATPL CPL An aircraft at FL350 is required to descend to cross a DME facility at FL80. Maximum rate of descent is 1800 ft/min and mean GS for descent is 276 kts. The minimum range from the DME at which descent should start is:

3.5° 150 kts

What is the approximate rate of descent?

A) B) C) D)

ATPL

=

ILS GP angle: GS: A) B) C) D)

Heli

=

It means that we will cover the distance of 120 NM during our descent in 25 minutes. It means that each minute during our descent we have to loose 960 ft of altitude (24.000 ft + 25 minutes). The minimum rate of descent to fulfill the requirement stated by the question is 960 ft/min.

4310. Given:

CPL

A) 14,5 NM B) 7,0 NM C) 13,1 NM D) 8,3 NM

C) 890 ftlmin.

D) 920 ft/min.

ATPL

At what approximate distance from THR can you expect to capture the GP?

A) 960 ft/min.

Cruising at 37,000 ft and need to descend to 13.000 ft = the aircraft has to loose 24.000 ftin 120 NM.

Airplane

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4314 (D)

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A) B) C) D)

79 NM 69NM 49NM 59NM

Cruising at 35.000 ft and need to descend to 8.000 ft = the aircraft has to loose 27,000 ft. If the maximum rate of descent = 1.800 ft/min, then it would take us 15 minutes (27,000 + 1.800) to loose 27,000 ft. With a GS of 276 kts we would cover a distance of 69 NM in this time: • 15 minutes = 0,25 hrs (15 + 60) • Distance 7= Rate x Time • 7=276 kts x 0,25 hrs ·7=69NM We have to start the descent 69 NM before the DME facility.

4340.

Airplane

ATPL

CPL

Heli

ATPL

CPL

An aircraft is descending down a 12% slope whilst maintaining a GS of 540 kts. The rate of descent of the aircraft is approximately: A) B) C) D)

650 ftlmin. 6.500 ftlmin. 4.500 ftlmin. 3.900 ftlmin.

Important formulas: • Gradient (in %) = (Altitude difference (ft) x 100) + Ground difference (in ft) • Approximate Gradient (%) = Climb/Descent Angle (0) x 100 + 60 • Climb/Descent angle (in 0) = Arctg (Altitude difference (ft) + Ground distance covered (in ft)) • Vertical speed (ft/min) = Gradient (%) x Groundspeed (kts) • Vertical speed (ft/min) = (Groundspeed (kts) x Gradient (ft/NM)) + 60 In our case: • Vertical speed = 12 % gradient x 540 kts • Vertical speed = 6,480 ft/min

4325 (8)

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4340 (8)

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Aviationexam Test Prep Edition 2012 4350. Airplane ATPL CPL Heli ATPL CPL An aircraft at FL390 is required to descend to cross a DME facility at FL70. Maximum rate of descent is 2.500 ft/min, mean GS during descent is 248 kts. What is the minimum range from the DME at which descent should commence?

? kts =32 NM + 0,1166 hrs ? (GS) = 274,4 kts We have determined that we need a GS of 274,4 kts to fulfill the requirement to cover a distance of 32 NM in 7 minutes. With a wind component of -25 kts (headwind) we will need a TAS of 299,4 kts (274,4 kts + 25 kts). Note: Remember that when a wind component has a negative (-) prefix, it means a headwind, whereas a positive (+) prefix means a tailwind. A good mnemonic is to realize that when you want to get to your destination a headwind is bad for you = negative; whereas a tailwind is good for you =positive.

A) 53 NM B) 58NM C) 63 NM D) 68NM Cruising at 39.000 ft and need to descend to 7.000 ft = the aircraft has to loose 32.000 ft.lf the maximum rate of descent = 2.500 ftlmin, then it would take us 12,8 minutes (32.000 + 2.500) to loose 32.000 ft. With a GS of248 kts we would cover a distance of53 NM in this time: • 12,8 minutes = 0,2133 hrs (12,8 + 60) • Distance? =Rate x Time • ?=248kts x 0,2133 ·?=52,9NM

4480. Airplane ATPL CPL Heli ATPL CPL By what amount must you change your rate of descent given a 10 knot increase in headwind on a 3° glideslope:

We have to start the descent 53 NM before the DME facility.

For a 3° glideslope the rule of thumb formula to obtain the required rate of descent (ROD) is "GS x 5" - therefore if GS is 100 kts the ROD will be 500 ftlmin. If the GS is reduced to 90 kts (e.g. due to 10 kts headwind) the ROD will be 450 ftlmin = 50 ftlmin decrease. Note that this rule of thumb is valid only fora 3°glide!

4386. Airplane ATPL CPL Heli ATPL CPL You are required to descend from FL230 to FL50 over a distance of 32 NM in 7 minutes. What will the glideslope be when you expect a Wind Component (WC) of -25 kts during the descend? A) 4,07° B) 5,29°

B) 283 kts C) 270 kts

(Refer to figure 061-E98) Cruising at 23.000 ft and need to descend to 5.000 ft = the aircraft has to loose 18.000 ft in 32 NM and 7 minutes. To obtain an approximate angle of descent we will use the following formula (1:60 rule): ·1 NM=6.080ft ·32 NM = 194.560 ft

• Angle of descent = (Height x 60) + Distance (in ft) • Angle of descent = (18.000 x 60) + 194.560 • Angle of descent =5,55° Another approximation can use the following formula:

=Angle of descent x 100 x Distance (in NM)

• Angle of descent =Height + (100 x Distance) • Angle of descent = 18.000ft+ (100 x 32 NM) • Angle of descent 5,62°

=

However, if we wanted to obtain the precise result, we would use the following formula: • Angle of descent (0) Arctg (Altitude difference (ft) + Ground dis-

=

tance covered (in ft)) ·1 NM=6.080ft ·32 NM = 194.560 ft • Angle of Descent =Arctan (18.000 ft + 194.560 ft) • Angle of Descent =Arctan 0,0925

=

• Angle of Descent 5,28°

4454. Airplane ATPL CPL Heli ATPL CPL You are required to descend from FL230 to FL50 over a distance of 32 NM in 7 minutes. What is the required TAS when you expect a Wind Component (WC) of -25 kts during the descent? 317 kts 329 kts 308 kts 300 kts

This question is very simple - all you need to do is calculate the Ground speed and apply the Wind component to obtain the TAS: 7 minutes = 0,1166 hrs (7 + 60) Distance =Rate x Time 32 NM =? kts x 0,1166 hrs

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4350 (A)

D) 309 kts (Refer to figures 061-£58 and 061-E59) Remember that to obtain the average TAS during a descent it is calculated from CAS at the altitude J.:2 of the cruising altitude. During a climb, the average TAS is calculated at % of the cruising altitude. First of all let's perform the QNH correction. Since we are climbing to a FL, we will be using a QNH of 1013. For easy calculation, let's assume that we reset the altimeter before we commence the climb - before we leave 1.500 ft of altitude. When changing from a QNH of 1032 to a QNH of 1013 we decrease the setting in the altimeter window by 79 hPa. Each hPa = 27 ft, therefore the total change in altimeter reading will be approx. 513 ft. When changing from a higher setting to a lower setting the altimeter indication decreases => the altimeter will indicate approximately 1.000 ft. In our case we are climbing from 1.000 ft Pressure Altitude to 18.000 ft Pressure Altitude => we have to climb a total of 17.000 ft => % of 17.000 ft = 11.333 ft. We add this to our initial pressure altitude of 1.000 ft to obtain the PA for TAS calculation of 12.333 ft. Standard temp at 12.333 ft is -9,66°C. Actual temp is ISA +15 => 5,34°C. Use your navigation computer to convert the CAS of 230 to an approximate TAS of283 kts.

4526. Airplane ATPL CPL Heli ATPL CPL An aircraft is descending from FL270 to FL100 following MT 054° and maintaining CAS 250 kts. Given are variation 13°E, temperatures ISA-10 °C, W/V 020/60. What is your GS?

Note: Arctan = inverse tangent function.

A) B) C) D)

4508. Airplane ATPL CPL Heli ATPL CPL What is the average TAS climbing from 1.500 ft up to FL180, given a temperature ISA +15 °C, a CAS 230 kts and QNH 1032? A) 263 kts

C) 6,25° D) 4,68°

• Height

A) 50 feet per minute increase. B) 30 feet per minute increase. C) 50 feet per minute decrease. D) 30 feet per minute decrease.

A) B) C) D)

277 kts 265 kts 325 kts 298 kts

(Refer to figures 061-E58, 061-E59, 061-E63 and 061-E64) First of all we have to find the TAS. Remember that to obtain the average TAS during a descent it is calculated from CAS at the altitude J.:2 of the cruising altitude. During a climb, the average TAS is calculated at%ofthe cruising altitude. In our case we are descending from 27.000 ft to 10.000 ft => we have to loose a total of 17.000 ft => J.:2 of 17.000 ft = 8.500 ft. We add this to our target level-off altitude of 10.000 ftto obtain the altitude for TAS calculation of 18.500 ft. Standard temp at 18.500 ft is -22°C. Actual temp is ISA-1O° => actual temp is -32°C. Use your navigation computer to convert the CAS of 250 to an approximate TASof321 kts. Second step will be finding the true track, because the wind direction is in °True. Magnetic track 054° + 013° Easterly variation = True Track of 067".

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4508 (8)

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05 In-Flight Navigation Third step will be the determination of GS using the Track (06r), TAS (321 kts) and WN (020"160 kts). Use your navigation computer to obtain a GS of approx. 277kts.

4577. Airplane ATPL CPL Heli ATPL CPL An aircraft at FL370 is required to commence descent when 100 NM from a DME facility and to cross the station at FL120. If the mean GS during the descent is 396 kts, the minimum rate of descent required is approximately: A) B) C) 0)

1.650 ft/min. 2.400 ft/min. 1.000 ft/min. 1.550 ft/min.

Cruising at 37.000 ft and need to descend to 12.000 ft = the aircraft has to loose 25.000 ft in 100 NM. Distance = Rate x Time 100 NM = 396 kts x 7 hrs 7 hrs = 100 NM -+- 396 kts 7= 0,2525 hrs (15,15 minutes)

C) 1.900 ftlmin. 0) 1.800 ftlmin. Cruising at 29.000 ft and need to descend to 8.000 ft = the aircraft has to loose 21.000 ft in 50 NM.

=

Distance Rate x Time 50NM =271 kts x 7hrs 7hrs=50NM-+-271 kts ?= 0,1845hrs (11 minutes) It means that we will cover the distance of 50 NM during our descent in 11 minutes. It means that each minute during our descent we have to loose 1.909 ft of altitude (21.000 ft -+- 11 minutes). The minimum rate of descent to fulfill the requirement stated by the question is 1.909 ft/min.

4585. Airplane ATPL CPL Heli ATPL CPL Due to pressurization problems you are request do descent with 1.000 ft/min only from FL120 down to FL50 maintaining a CAS 200 kts. What descent profile will you follow at no wind conditions and standard temperature?

A) 2°

It means that we will cover the distance of 100 NM during our descent in 15,15 minutes. It means that each minute during our descent we have to loose 1.650 ft of altitude (25.000 ft -+- 15,15 minutes). The minimum rate of descent to fulfill the requirement stated by the question is 1.650 ft/min.

4579. Airplane ATPL CPL Heli ATPL CPL You are required to descend from FL230 to FL50 over a distance of 32 NM in 7 minutes. What is the required rate of descent when you expect WC-25 during the descent? A) 2.157 ft/min. B) 2.230 ft/min. C) 2.458 ft/min.

0) 2.570 ft/min. Believe it or not, this is actually a simple question that is disguised to mislead you. Cruising at 23.000 ft and need to descend to 5.000 ft = the aircraft has to loose 18.000 ft in.7 minutes. It means that each minute during our descent we have to loose 2.571 ft of altitude (18.000 ft -+- 7 minutes). The minimum rate of descent to fulfill the requirement stated by the question is 2.571 ft/min.

4580. Airplane ATPL CPL Heli ATPL CPL An aircraft is maintaining a 5,2% gradient at 7 NM from the runway, on a flat terrain its height is approximately:

B) 2,so

C) 3° 0) 1,so First of all we need to obtain the TAS. Remember that to obtain the average TAS during a descent it is calculated from CAS at the altitude 0. of the cruising altitude (for the climb it would be at % of the cruising altitude). Therefore, we calculate the TAS from a CAS of200 kts, FL85 (mid-point ofour descent from FL120 to FL50) and temp -2°C (lSA at FL85) => use your navigation computer to obtain TAS of approx. 227 kts. We are descending from FL120 to FL50 => we have to loose 7.000 ft at a Rate of Descent of 1.000 ft/min => this descent will take us 7 minutes. With a TAS of 227 kts and no wind ourGS will also be 227 kts.ln these 7 minutes (0, 1766 hrs) we will cover a distance of26,48 NM (0,1166 hrs x 227 kts) = 161.018 ft (1NM = 6.080 ft). Now we can use trigonometry to calculate the angle: • tan angle (0) '" Height + Distance • tan angle (0) = 7.000 ft -+- 161.018 ft • tan angle (0) = 0,0434 • angle (0) = 2,49°

4613. Airplane ATPL CPL Heli ATPL CPL What is the average TAS climbing from 2.000 ft up to FL120 at standard temperatures, given a CAS 185 kts and QNH 1013? A) 188 kts

A) 680ft

B) 197 kts C) 210 kts 0) 221 kts

B) 2.210ft C) 1.890 ft 0) 3.640ft Important formulas: • Gradient (in %) =(Altitude difference (ft) x 100) -+- Ground difference (in ft) • Approximate Gradient (%) = Climb/Descent Angle (0) x 100 -+- 60 • Climb/Descent angle (in 0) = Arctg (Altitude difference (ft) -+- Ground distance covered (in ft)) • Vertical speed (ft/min) =Gradient (%) x Groundspeed (kts) • Vertical speed (ft/min) = (Groundspeed (kts) x Gradient (ft/NM)) -+- 60 In our case: ·1 NM=6.080ft ·7 NM = 42.560 ft • Gradient (in %) = (Altitude difference (ft) • 5,2 % = (7 ft x 100) -+- 42.560 ft ·7 ft = (5,2 % x 42.560 ft) -+- 100 • 7ft=2.213 ft

x 100) -+- Ground difference (in ft)

(Refer to figures 061-E58 and 061-E59) Remember that to obtain the average TAS during a descent it is calculated from CAS at the altitude 0. of the cruising altitude. During a climb, the average TAS is calculated at % of the cruising altitude. In our case we are climbing from 2.000 ft to 12.000 ft => we have to climb a total of 10.000 ft => % of 10.000 ft = 6.666 ft. We add this to our initial starting altitude of 2.000 ft to obtain the altitude for TAS calculation of 8.666 ft. Standard temp at 8.666 ft is -2,3°C. Use your navigation computer to convert the CAS of 185 to an approximate TAS of 210 kts.

4621. Airplane ATPL CPL Heli ATPL CPL Assuming zero wind, what distance will be covered by an aircraft descending 15.000 ft with a TAS of 320 kts and maintaining a rate of descent of 3.000 ft/min?

Height is approximately 2.210 ft.

A) 26,7 NM B) 19,2 NM

4583. Airplane ATPL CPL Heli ATPL CPL An aircraft at FL290 is required to commence descent when 50 NM from a VOR and to cross that VOR at FL80. Mean GS during descent is 271 kts. What is the minimum rate of descent required?

C) 38,4NM 0) 16,ONM

A) 1.700 ftlmin. B) 2.000 ft/min.

I

4577 (A)

I

4579 (D)

I

4580 (8)

I

4583 (C)

I

4585 (8)

I

If the aircraft has to loose 15.000 ft at a rate of descent of3.000 ft/min, then this descent would take 5 minutes (15.000+ 3.000). We may use the TAS in this problem because the wind is zero. With a TAS of 320 kts we would cover a distance of 53 NM in this time: ·5 minutes = 0,0833 hrs (5 + 60)

4613 (C)

I

4621 (A)

I

Aviationexam Test Prep Edition 2012 • Distance ?=Rate x Time

.? = 320 kts x 0,0833 .?=26,65NM

4733. Airplane ATPL CPL Heli ATPL CPL An aircraft is descending down a 6% slope whilst maintaining a GS of 300 kts. The rate of descent of the aircraft is approximately: A) 1.800 ft/min. B) 10.800 ft/min. C) 3.600 ft/min. 0) 900 ft/min.

In our case: • Vertical speed = 6 % gradient x 300 kts • Vertical speed = 1.800 ft/min

4746. Airplane ATPL CPL Heli ATPL CPL An aircraft at FL350 is required to commence descent when 85 NM from a VOR and to cross the VOR at FL80. The mean GS forthe descent is 340 kts. What is the minimum rate of descent required? 1.900 ftlmin. 1.800 ftlmin. 1.600 ft/min. 1.700 ft/min.

Cruising at 35.000 ft and need to descend to 8.000 ft = the aircraft has to loose 27.000 ft in 85 NM. Distance =Rate x Time 85 NM=340kts x ?hrs ? hrs =85 NM + 340 kts ?=0,25hrs It means that we will cover the distance of85 NM during our descent in 0,25 hrs (15 minutes). It means that each minute during our descent we have to loose 1.800 ft of altitude (27.000 ft + 15 minutes). The minimum rate of descent to fulfill the requirement stated by the question is 1.800 ft/min.

4749. Airplane ATPL CPL Heli ATPL CPL An aircraft at FL330 is required to commence descent when 65 NM from a VOR and to cross the VOR at FL100. The mean GS during the descent is 330 kts. What is the minimum rate of descent required? A) 1.950 ft/min. B) 1.650 ft/min. C) 1.750 ftlmin. 0) 1.850 ft/min.

Distance = Rate x Time 65 NM = 330kts x? hrs ?hrs= 65 NM .,.330kts ? = 0,1969 hrs (11,8 minutes) It means that we will cover the distance of 65 NM during our descent in 11,8 minutes. It means that each minute during our descent we have to loose 1.950 ft of altitude (23.000 ft .,. 17,8 minutes). The minimum rate of descent to fulfill the requirement stated by the question is 1.950 ft/min.

4733 (A)

Distance =Rate x Time 65 NM=240kts x ?hrs ? hrs = 65 NM +240kts ? = 0,2708 hrs (16,25 minutes) It means that we will cover the distance of 65 NM during our descent in 16,25 minutes. It means that each minute during our descent we have to loose 1.415 ft of altitude (23.000 ft + 16,25 minutes). The minimum rate of descent to fulfill the requirement stated by the question is 1.415 ft/min.

4757. Airplane ATPL CPL Heli ATPL CPL The outer marker of an ILS with a 30 glide slope is located 4,6 NM from the threshold. Assuming a glide slope height of 50 ft above the threshold, the approximate height of an aircraft passing the outer marker is (use the 1:60 rule): A) 1.400 ft B) 1.450 ft C) 1.350 ft 0) 1.300 ft (Refer to figures 061-£98 and 061-£99) We can solve this question using the 1:60 rule to obtain an approximate result. Note that the question specifically requests that we use this method: .Angle (0) = (Height x 60) + Distance • Height Distance x Angle (0) + 60 • Height = 4,6 NM x 3° + 60 • Height = 0,23 NM • Height = 1.398 ft Note that this-height does not include the 50 ft over the runway => we have to add these 50 ft to get height above the ground => 1.448 ft.

=

Now the precise solution method: Height = tan angle x Ground Distance Height = tan 3°x 4,6 NM Height = 0,24 NM Height = 1.459 ft Note that this height does not include the 50 ft over the runway => we have to add these 50 ft to get height above the ground => 1.509 ft.

4758. Given:

Airplane

I

4746 (8)

I

4749 (A)

I

4751 (A)

I

ATPL

CPL

Heli

ATPL

CPL

197 kts 240 0 180/30kts

TAS: True course: W/V:

Cruising at 33.000 ft and need to descend to 10.000 ft = the aircraft has to loose 23.000 ft in 65 NM.

I

A) 1.420 ftlmin. B) 1.630 ft/min. C) 1.270 ft/min. 0) 1.830 ft/min. Cruising at 33.000 ft and need to descend to 70.000 ft = the aircraft has to loose 23.000 ft in 65 NM.

Important formulas: • Gradient (in %) = (Altitude difference (ft) x 100) + Ground difference (in ft) • Approximate Gradient (%) = Climb/Descent Angle (0) x 700 + 60 • Climb/Descent angle (in 0) = Arctg (Altitude difference (ft) + Ground distance covered (in ft)) • Vertical speed (ft/min) = Gradient (%) x Groundspeed (kts) • Vertical speed (ft/min) = (Groundspeed (kts) x Gradient (ft/NM)) + 60

A) B) C) 0)

4751. Airplane ATPL CPL Heli ATPL CPL At 65 NM from a VOR you commence a descent from FL330 in order to arrive over the VOR at FL100. Your mean ground speed in the descent is 240 knots. What rate of descent is required?

Descent is initiated at FL220 and completed at FL40. Distance to be covered during descent is 39 NM. What is the approximate rate of descent? A) B) C) 0)

800 ftlmin. 1.400 ft/min. 950 ftlmin. 1.500 ft/min.

(Refer to figures 061-£63 and 061-£64) Use your navigation computer to obtain the GS given the values of TAS, Course and W/V. You should come up with a GS of approx. 180 kts. The descent will be made over a distance of 39 NM. At this GS the distance will be covered in 0,2166 hrs (39 NM .,. 180 kts) = 13 minutes. We have to loose a total of 18.000 ft (22.000 ft - 4.000 ft) in 13 minutes => rate of descent will be 1.385 ft/min (18.000 ft.,. 13 min) = approx. 1.400 ft/min.

4757 (8)

I

4758 (8)

I

------------

05 In· Flight Navigation 4761. Airplane ATPL CPL Heli ATPL CPL An aircraft at FL350 is required to cross a VOR/DME facility at FLll0 and to commence descent when 100 NM from the facility. If the mean GS for the descent is 335 kts, the minimum rate of descent required is: A) 1.390 ft/mi n. B) 1.340 ftlmin. C) 1.240 ft/min. D) 1.290 ft/min.

A) 124NM B) 88NM

Cruising at 35.000 ft and need to descend to 11.000 ft= the aircraft has to loose 24.000 ft in 100 NM. Distance = Rate x Time 100NM =335 kts x ?hrs ? hrs = 100 NM + 335 kts ? = 0,2985 hrs (17,9 minutes) It means that we will cover the distance of 100 NM during our descent in 17,9 minutes. It means that each minute during our descent we have to loose 1.340 ft of altitude (24.000 ft + 17,9 minutes). The minimum rate of descent to fulfill the requirement stated by the question is 1.340 ftlmin.

4873. Airplane ATPL CPL Heli ATPL CPL At 04:22 an aircraft at FL370, GS 320 kts, is on the direct track to a VOR 185 NM distant. The aircraft is required to cross the VOR at FL80. For a mean rate of descent of 1.800 ft/min at a mean GS of 232 kts, the latest time at which to commence descent is: A) 0448 B) 0445 C) 0451 D) 0454 Cruising at 37.000 ft and need to descend to 8.000 ft = the aircraft has to loose 29.000 ft. If the maximum rate of descent = 1.800 ftlmin, then it would take us 16,11 minutes (29.000 + 1.800) to loose 29.000 ft. With a descent GS of 232 kts we would cover a distance of 62,3 NM during the descent: ·16,11 minutes = 0,2685 hrs (16,11 + 60) • Distance? =Rate x Time • ? = 232 kts x 0,2685 ·?=62,3NM We have to start the descent 62,3 NM before the VOR. We are currently located at a distance of 185 NM from the VOR => we will maintain the current flight level for additional 122,7 NM at a cruise GS of 320 kts. At this speed we will cover this distance in 0,3834 hrs => approx. 23 minutes (122,7 NM + 320 kts). The time now is 04:22 UTC + 23 minutes =04:45 UTC. Descent has to be started at 04:45 UTe.

231032. Airplane ATPL CPL Heli ATPL CPL The QNH, given by a station at 2.500 ft, is 980 hPa.The elevation of the highest obstacle along a route is 8.000 ft and the OAT ISA -10°C. When an aircraft, on route has to descend the minimum indicated altitude (QNH on the subscale of the altimeter) to maintain a clearance of 2.000 ft, will be: A) 9.700ft B) 10.000ft C) 10.400ft D) 11.200ft

12,0 NM, altitude 3.000 ft 9,8 NM, altituq,.e 2.400 ft 160 kts 125 kts

The rate of descent is: A) B) C) D)

600 ft/min 570 ft/min 700 ftlmin 730 ft/min

1 4761 (8)

C) 166 NM D) 236 NM

231071. Airplane ATPL CPL Heli ATPL CPL An aircraft is departing from an airport which has an eleva- . tion of 2.000 ft and the QNH is 1003 hPa. The TAS is 100 kts, the head wind component is 20 kts and the rate of climb is 1.000 ft/min. Top of climb is FL050. At what distance from the airport will this be achived? A) B) C) D)

4,ONM 3,6 NM 4,4NM 5,4NM

Airplane ATPL CPL Heli ATPL CPL 231072. An aircraft is departing from an airport which has an elevation of 2.000 ft and the QNH is 1003 hPa. The TAS is 100 kts, the head wind component is 20 kts and the rate of climb is 500 ft/min. Top of climb is FL050. At what distance from the airport will this be achieved? A) 6,6NM B) 7,2 NM C) 8,8NM D) 10,8 NM

231073. Airplane ATPL CPL Heli ATPL CPL An aircraft is departing from an airport which has an elevation of 2.000 ft and the QNH is 1003 hPa. The TAS is 100 kts, the head wind component is 20 kts and the rate of climb is 1.000 ft/min. Top of climb is FL100. At what distance from the airport will this be achieved? A) 10,3 NM B) 11,1 NM C) 15,4NM D) 13,3 NM

231074. Airplane ATPL CPL Heli ATPL CPL An aircraft is departing from an airport which has an elevation of 2.000 ft and the QNH is 1023 hPa. The TAS is 100 kts, the head wind component is 20 kts and the rate of climb is 1.000 ft/min. Top of climb is FL100. At what distance from the airport will this be achieved? A) 11,1 NM B) 16,6 NM

231062. Airplane ATPL CPL Heli ATPL During approach the following data are obtained: DME: DME: TAS: GS:

231070. Airplane ATPL CPL Heli ATPL CPL An aircraft at FL360 is required to descent to FL120. The aircraft should reach FL120 at 40 NM from the next waypoint. The rate of descent is 2.000 ft/min. The average GS is 420 kts. The minimum distance from the next waypoint at which descent should start is:

CPL

C) 13,3 NM D) 10,3 NM

231075. Airplane ATPL CPL Heli ATPL CPL You are departing from an airport which has an elevation of 2.000 ft. The QNH is 1013 hPa. 10 NM away there is a waypoint you are required to pass at an altitude of 7.500 ft. Given a groundspeed of 100 kts, what is the minimum rate of climb? A) B) C) D)

1.080 ft/mi n 750 ft/min 590 ft/mins 920 ft/min

1 4873 (8) 1231032 (e) 1231062 (8) 1231070 (A) 1231071 (8) 1231072 (8) 1231073 (A) 1231074 (A) 1231075 (D) 1

Aviationexam Test Prep Edition 2012 231077. Airplane ATPL CPL Heli ATPL CPL You are departing from an airport which has an elevation of 1.500 ft. The QNH is 1003 hPa. 15 NM away there is a waypoint you are required to pass at an altitude of 7.500 ft. Given a groundspeed of 120 kts, what is the minimum rate of climb?

A) B) C) D)

231078.

ATPL

Heli

CPL

ATPL

CPL

A descending aircraft flies in a straight line to a DME. DME 55,0 NM, altitude 33.000 ft DME 43,9 NM, altitude 30.500 ft M 0.72, GS 525 kts, OAT ISA

=

800 ft/min 730 ft/min 870 ft/min 530 ft/min

Airplane

Given:

=

=

The descent gradient is: A) B) C) D)

4,1% 3,7% 3,5% 3,9%

05-03 Navigation in cruising flight, use of fixes to revise navigation data Airplane ATPL CPL Heli ATPL CPL Which of the following formula is correct for the calculation of maximum range? 2231.

A) B) C) D)

Maximum Range =Safe Fuel available x Specific Range Maximum Range = Block Fuel x Specific Range Maximum Range = Block Fuel/Specific Range Maximum Range'" Safe Fuel available / Specific flange

Range is the distance an aircraft can fly on the amount of fuel available. The term "5pecific Range (5R)" is used to define the number of miles per 1 unit of fuel (eg. ? NM per 1 kg of fuel) => for example 10 NM / 1 kg of fuel. If we have a total fuel on board (excluding reserves) of 100 kg, then our maximum range can be determined by dividing the total fuel by the 5R => 100 kg x 10 NM (per 1 kg) => 1.000 NM range. Range is therefore equal to Fuel available x 5pecificRange.

2242. Airplane ATPL CPL Heli ATPL CPL An aircraft at FL310, M 0,83, temperature -30°C, is required to reduce speed in order to cross a reporting point five minutes later than planned. Assuming that a zero wind component remains unchanged, when 360 NM from the reporting point Mach number should be reduced to:

A) MO,76

2) L55 remains the same = 607 kts 3) Find Mach nr. (= TA5 + L55) = 450 + 607 = 0,74

2243. Airplane ATPL CPL Heli ATPL CPL An aircraft at FL370, M 0,86, OAT -44°C, headwind component 110 kts, is required to reduce speed in order to cross a reporting point 5 min later than planned.lfthe speed reduction were to be made 420 NM from the reporting point, what Mach number is required? A) M 0,79

B) MO,73 C) MO,75 D) M 0,81 Remember that Mach number is the ratio of the aircraft's speed (TA5) to the local speed of sound (L55). Mach No. = TA5 + L55. The 5peed of 50und varies only with the temperature. As the temperature increases so does the local speed of sound. Because the temperature reduces with altitude, the speed of sound reduces as altitude increases. A formula for calculating the L55 for a given temperature is: L55 38,95 x ,jAbso/ute temperature (in °Ke/vin) ... Absolute temp in OK = °e + 273. Temperature in °e = OK - 273. We can also use an alternative formula, using 0(, but this does not yield a precise result - only an estimation: L55 = 644 + (1,2 x Temp °e).

=

To obtain the TA5 from Mach number and OAT: 1) Find OAT in OK (OAT in °e + 273) =229°K 2) Find L55 (38,95 x ,jOAT in OK) = 589,4 kts 3) Find TA5 (= Mach Nr. x L55) = 0,86 x 589,4 = 507 kts

B) M 0,74 C) M 0,78 D) MO,80 Remember that Mach number is the ratio of the aircraft's speed (TA5) to the local speed of sound (L55). Mach No. = TA5 + L55. The 5peed of 50und varies only with the temperature. As the temperature increases so does the local speed of sound. Because the temperature reduces with altitude, the speed of sound reduces as altitude increases. A formula for calculating the L55 for a given temperature is: L55 = 38,95 x ,fAbso/ute temperature (in °Ke/vin) ... Absolute temp in OK = °e + 273. Temperature in °e = OK - 273. We can also use an alternative formula, using 0(, but this does not yield a precise result - only an estimation: L55 = 644 + (1,2 x Temp 0c).

Distance to the reporting point is 420 NM. With a TA5 of 507 kts and a headwind component of 110 kts our Ground 5peed (G5) will be 397 kts. At this G5 you would reach the reporting point in 63 minutes (420 NM + 397 kts = 1,06 hr = 63 minutes). The ATC want you to reach the point 5 minutes later = in 68 minutes from now (68 minutes = 1,1333 hrs). To meet this requirement your speed would have to be reduced to about 370,6 kts (420 NM.;. 1,1333 hrs). It is important to realize that this would be your reduced G5. To obtain the required TA5, aply the headwind of 110 kts => 370,6 kts G5 + 110 kts headwind = TA5 of480,6 kts. Now you just need to convert 480,6 kts to Mach number. To obtain the Mach Number from TA5 and OAT: 1) OAT in OK remains the same =229°K 2) L55 remains the same =589,4 kts 3) Find Mach nr. (= TA5 + LS5) = 480,6.;. 589,4 = 0,815

To obtain the TA5 from Mach number and OAT: 1) Find OATin OK (OATin 0(: + 273) = 243°K 2) Find L55 (38,95 x ,jOAT in OK) = 607 kts 3) Find TA5 (= Mach Nr. x L55) = 0,83 x 607 = 504 kts Distance to the reporting point is 360 NM. At the current speed of 504 kts you would reach the reporting point in 43 minutes (360 NM + 504 kts = 0,71 hr = 43 minutes). The ATe want you to reach the point 5 minutes later = in 48 minutes from now (48 minutes = 0,8 hrs). To meet this requirement your speed would have to be reduced to about 450 kts (360 NM + 0,8 hrs). Now you just need to convert 450 kts to Mach number. To obtain the Mach Number from TA5 and OAT: 1) OATin OK remainsthesame=243°K

1231077 (A) 1231078 (8) 1 2231 (A)

1 2242 (8)

1 2243 (0)

1

05 In-Flight Navigation

2244. Airplane ATPL CPL Heli ATPL CPL You are flying at a true Mach No of 0,82 in a SAT of -45 °e. At 1000 hours you are 100 NM from the POL DME and your ETA at POL is 1012. ATe ask you to slow down to be at POL at 1016. What should your new TMN be if you reduce speed at 100 NM distance to:

A) M 0,76 B) MO,72 C) MO,68 0) M 0,61 Remember that Mach number is the ratio of the aircraft's speed (TA5) to the local speed of sound (L55). Mach No. TA5 + L55. The Speed ofSound varies only with the temperature. As the temperature increases so does the local speed of sound. Because the temperature reduces with altitude, the speed of sound reduces as altitude increases. A formula for calculating the L55 for a given temperature is: L55 = 38,95 x ,fAbsolute temperature (in °Kelvin) ... Absolute temp in OK = °C + 273. Temperature in °C = OK - 273. We can also use an altemative formula, using 0[, but this does not yield a precise result - only an estimation: L55 = 644 + (1,2 x Temp °C).

=

Current distance to the POL DME is 100 NM and you commence your slowingdown at this distance. At the current speed of you would reach the POL DME in 12 minutes (0,2 hrs), therefore your current G5 is 500 kts (100 NM + 0,2 hrs). With the TA5 of 482 kts and G5 of 500 kts you can determine that you have a tailwind componentof 18 kts. TheATC wants you to reach the point 4 minutes later =in 16 minutes from now (16 minutes =0,2666 hrs). To meet this requirement your G5 would have to be reduced to about 375 kts (100 NM + 0,2666 hrs). With a G5 of 375 kts and an 18 kt tailwind component your TA5 would have to be 357 kts. Now you just need to convert a TA5 of 357 kts to Mach number. To obtain the Mach Number from TA5 and OAT: 1) Find OATin OK (OAT in °C + 273) =228°K 2) Find L55 (38,95 x ,fOAT in OK) =588 kts 3) Find Mach nr. (= TA5 + L55) = 357 + 588 = 0,61

Airplane

ATPL

ATPL

CPL

An aircraft is flying a track of 255°(M). At 2254 UTe, it crosses radial 360° from a VOR station. At 2300 UTe, it crosses radial 330° from the same station. At 2300 UTe, the distance between the aircraft and the station is: A) B) C) 0)

the same as it was at 2254 UTe. greater than it was at 2254 UTC. randomly different than it was at 2254 UTC. less than it was at 2254 UTC.

(Refer to figure 061-E48) To solve this question I recommend that you draw a sketch (refer to the attached illustration). When you finish drawing the sketch, calculate the internal angles of the reSUlting triangle. The bottom angle is clear - it is a difference between 360° and 330° = 30~ The top-right angle (22:54 position) is quite easy as well- just deduct 180° from the heading (255°) to obtain the angle of 75~ Since you now have two angles, you can easily obtain the third one - remember that the sum of the internal angles of a triangle = 180°. Therefore the top-left angle will also be 75°. We now have an isosceles triangle = two of the internal angles are the same. It means that also the lengths of the triangle sides representing the radials 330° ani:l360° will be idMtical. Therefore, thedistlirice Between the aircraft and the VOR will be identical at 22:54 and at 23:00.

4281. Given:

Airplane

ATPL

Distance A to B:

Heli

CPL

ATPL

CPL

100 NM

Fix obtained 40 NM along and 6 NM to the left of course. What heading alteration must be made to reach B1 A) 6° right. B) 9° right. C) 15° right.

I

2244 (D)

I

4044 (A)

If we alter our heading by 9° to the right (towards the original track) we will parallel our original track. Now if we wanted to apply an even larger correction and alter the heading in such a way that we re-join our original track at destination B, we would follow these steps: • TKE = (Dist off-track x 60) + Dist. to go • Dist. togo= 100NM-40NM=60NM • TKE = (6 x 60) + 60 • TKE= 6° Summary: .9° heading alteration = continuing parallel to the original track; • additional 6° heading alteration = rejoining original track at destination (= total heading alteration of 15° right).

Airplane

ATPL

ETA to cross a meridian: GS: TAS:

CPL

Heli

ATPL

CPL

21:00 UTe 441 kts 491 kts

At 20:10 UTe, ATe requests a speed reduction to cross the meridian at 21:05 UTe. The reduction to TAS will be approximately: A) B) C) 0)

60 kts 90 kts 75 kts 40 kts

First of all we need to find the distance to the meridian. We know that according to the original estimate we would cross it in 50 minutes from now (21:00 - 20:10) with a TA5 of491 kts and a G5 of441 kts.lt means that we are experiencing a headwind component of 50 kts.

Heli

CPL

(Refer to figure 061-E98) Let's start by calculating the TKE (Track Angle Error) that will give us the heading correction to parallel our intended track: • TkE = (Dist off-track x 60) + Dist. along track • TKE = (6 x 60) +40 ·TKE=9°

4285. Given:

To obtain the TA5 from Mach number and OAT: 1) Find OAT in OK (OATin °C + 273) = 22BoK 2) Find L55 (38,95 x ,fOAT in OK) = 588 kts 3) Find TA5 (= Mach Nr. x L55) = 0,82 x 588 = 482 kts

4044. Given:

0) 18° right.

Distance = Rate x Time Distance = 441 kts G5 x 0,8333 hrs Distance = 367,5 NM The ATC requests that we reduce our TA5 so that we arrive over the meridian at 21:05 = we need to cover this distance in 55 minutes (0,9166 hrs) instead of50 minutes (0,8333 hrs). 367,5 NM = ? kts x 0,9166 hrs ? kts =367,5 NM + 0,9166 hrs ? =401 kts (G5) Note that the speed 401 kts is the Ground Speed (G5), but we need to know the TA5. We have determined that we have a headwind component of50 kts => therefore to obtain the G5 of 401 kts we need to maintain a TA5 of 451 kts. Our original TA5 was 491 kts, therefore we have to reduce it by 40 kts.

4289. Airplane ATPL CPL Heli ATPL CPL The distance between positions A and B is 180 NM. An aircraft departs position A and after having traveled 60 NM, its position is pinpointed 4 NM left of the intended track. Assuming no change in wind velocity, what alteration of heading must be made in order to arrive at position B1 A) B) C) 0)

6°right. 8° right. 2° left. 4° right.

(Refer to figure 061-E98) Let's start by calculating the TKE (Track Angle Error) that will give us the heading correction to parallel our intended track: • TKE = (Dist off-track x 60) + Dist. along track • TKE = (4 x 60) + 60 • TKE=4° If we alter our heading by 4° to the right (towards the original track) we will parallel our original track. Now if we wanted to apply an even larger correction and alter the heading in such a way that we re-join our original track

I

4281 (C)

I

4285 (D)

I

4289 (A)

I

Aviationexam Test Prep Edition 2012 at destination B, we would follow these steps: • TKE = (Dist off-track x 60) + Dist. to go • Dist. togo= 180NM-60NM= 120NM • TKE = (4 x 60) + 120 ·TKE=2"

B) 6° right. C) 4° right . 0) 8° right.

Summary: .4° heading alteration =continuing parallel to the original track; • additional 2° heading alteration rejoining original track at destination (= total heading alteration of 6° right).

=

4290. Airplane ATPL CPL Heli ATPL CPL True track from A to B is 167°, and the distance is 140 NM. Variation is 12°W at A and 14°W at B. Your flight-plan WCA 8°L. When the remaining distance to B is 35 NM you find that your position is 5 NM right of the flight plan track. Since over A you have steered as flight planned. What change of heading is required at this time to bring you directly to B1

(Refer to figure 061-E98) Let's start by calculating the TKE (Track Angle Error) that will give us the heading correction to parallel our intended track: • TKE = (Dist off-track x 60) + Dist. along track • TKE = (3 x 60) + 30 ·TKE=6° If we alter our heading by 6° to the right (towards the original track) we will parallel our original track. Now if we wanted to apply an even larger correction and alter the heading in such a way that we re-join our original track at destination B, we would follow these steps: • TKE = (Dist off-track x 60) + Dist. to go • Dist. to go = 120 NM - 30 NM = 90 NM • TKE = (3 x 60) + 90 • TKE=2° Summary: .6° heading alteration =continuing parallel to the original track; • additional 2° heading alteration = rejoining original track at destination (= total heading alteration ofB· right).

A) 3° left. B) 8° left. C) 11° left. 0) 14° left. (Refer to figure 061-E98) Let's start by calculating the TKE (Track Angle Error) that will give us the heading correction to parallel our intended track: • TKE = (Dist off-track x 60) + Dist. along track ·TKE=(5x 60) + 105 • TKE = 2,9°

If we alter our heading by approx. 3° to the left (towards the original track) we will parallel our original track. Now if we wanted to apply an even larger correction and alter the heading in such a way that we re-join our original track at dp.stination 8, we would follow these steps: • TKE = (Dist off-track x 60) + Dist. to go • Dist. to go = 35 NM • TKE = (5 x 60) + 35 • TKE = 8,6° Summary: • 2,9° heading alteration = continuing parallel to the original track; • additional 8,6° heading alteration = rejoining original track at destination (= total heading alteration of 11,5° left).

4315. Airplane ATPL CPL Heli ATPL CPL The distance between two waypoints is 200 NM. To calculate compass heading, the pilot used 2°E magnetic variation instead of 2°W. Assuming that the forecast WIV applied, what will the off track distance be at the second waypoint1 A) ONM B) 7 NM C) 14NM 0) 21 NM

4411. Airplane ATPL CPL Heli ATPL CPL A ground feature was observed on a relative bearing of 325° and five minutes later on a relative bearing of 280°. The aircraft heading was 165°(M), variation 25°W, drift 10° Right and GS 360 kts. When the relative bearing was 280°, the distance and true bearing of the aircraft from the feature was: A) 30 NM; 240°. B) 40 NM; 110°. C) 40 NM; 290°. 0) 30 NM; 060°. (Refer to figures 061-E40, 061-E77 and 061-E78) First of all calculate the True Heading of the aircraft => apply the 25°W variation to the Magnetic heading of 165° to obtain a True Heading of 140°. At the second position the Relative Bearing of the feature from the aircraft is 280°. Remember that: • True Bearing (TB) RB + TH • TB = 280°+ 140°= 420°-360°= 060 0T ·TB=0600T

=

Since the True Bearing of the feature from the aircraft is 060° then the True bearing of the aircraft from the feature will be the reciprocal = 240°. SInce there is only one answer that offers 240° as a correct answer on the "bearing" part, do not worry about the calculation of the distance and use just the bearing to select the correct answer.

4412. Airplane ATPL CPL Heli ATPL CPL An aircraft obtains a relative bearing of 315° from an NOB at 08:30. At 08:40 the relative bearing from the same position is 270°. Assuming no drift and a GS of 240 kts, what is the approximate range from the NOB at 08:401

(Refer to figure 061-E98) The overall deviation from desired track = 4° (from 2°E to 2°W). This question can be easily solved using the 1:60 rule: • TKE = (Dist off-track x 60) + Dist along track ·4°= (? x 60) + 200 • ?=4+60x200

A) B) C) 0)

.?= 13,33NM Another (and more precise) method is to use the trigonometric ratios in rightangle triangles: tan a =Length of opposite side + Length ofadjacent side In our case it would be: ·tan4°= ?NM+200NM • ? NM =tan 4° x 200 NM

.?= 13,9BNM

4323. Airplane ATPL CPL Heli ATPL CPL Distance from A to B is 120 NM. After 30 NM aircraft is 3 NM to the left of course. What heading alteration should be made in order to arrive at point B1

50 NM 40 NM 60NM 30 NM

(Refer to figure 061-E47) First of all try to visualize the situation - initially the aircraft is situated SouthEast of the radio beacon (on a 315° bearing to the station). 10 minutes later we are at a position further to the North from our original position - now on a bearing 090° from the station = due East of the station. In essence, we have constructed a triangle. It can be seen from the attached illustration that the angIe a = 315° - 270·= 45~ What we have is an Isosceles triangle with a rightangle and two 45° angles = two of the internal angles are identical and two sides are of identical length. The distance «Dl» was flown in 10 minutes (= 0,1666 hours) at a speed of240 kts. Distance «Dl» = 240 kts x 0,1666 hrs = 40 NM. Distance «D2» is identical to distance «Dl» => Distance «D2» equals 40 NM. The distance of the aircraft to the station at 08:40 is 40 NM.

A) 8° left.

I

4290 (C)

I

4315 (C)

I

4323 (D)

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4411 (A)

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4412 (8)

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05 In-Flight Navigation

4442. Airplane ATPL CPL Heli ATPL CPL Halfway between two reporting points the navigation log gives the following information: TAS: 360 kts W/V: 330°/80 kts Compass heading: Deviation on this heading: Variation: What is the average groundspeed for this leg? A) B) C) 0)

373 kts 403 kts

Remember that if the values of variation or deviation are negative, they are Westerly, if they are positive, they are Easterly. We will use the deviation to convert Compass Heading (CH) into Magnetic Heading (MH): • MH = CH - Westerly deviation (or + Easterly deviation) • MH = 23r CH - SoW deviation ·MH=23r Now we need to convert the MH into True Heading (TH) using the value ofvariation: • TH = MH - Westerly variation (or + Easterly variation) • TH = 232° MH - 79° Wvariation ·TH=2W Now we have all the information to find the GS => TAS of 360 kts, W/V 330°/80 kts and TH of 273°. Use your navigation computer (for this type of problem the computer E6B or CR-3 will probably not be a suitable one => try using for example CRP-5 and make sure that you work with TH and not TC on your nav. computer) to obtain the result of 403 kts GS and a True Track of203°.

4453. Airplane ATPL CPL Heli ATPL CPL You leave A to fly to B (475 NM away) at 10:00 hours. Your ETA at B is 11:30. At 10:40 you are 190 NM from A. What ground speed is required from now on to arrive on time at B? 317 knots. 330 knots. 342 knots. 360 knots.

Let's summarize the information we have. Distance to cover from A to B is 475 NM. The actual departure time is 70:00 and the estimated time of arrival is 77:30 => estimated flight time of 7,5 hrs (90 minutes). The flight progress check at 70:40 indicated a distance covered of 790 NM = distance remaining to go of 285 NM (475 NM - 790 NM).lf our ETA is 77:30 UTC it means we have the time of 50 minutes (0,8333 hrs) to cover the distance of 285 NM. Our speed for the remainder of the flight would have to be 342 kts (285 NM + 0,8333 hrs).

4475. Airplane ATPL CPL Heli ATPL CPL An island is observed by weather radar to be 15° to the left. The aircraft heading is 1200(M) and the magnetic variation 17°W. What is the true bearing of the aircraft from the island? A) 1220 (T)

B) 302° (T) C) 088 0 (T) 0) 2680 (T) (Refer to figures 067-E24, 067-E77 and 067-E78) MB=RB+MH MB=345°+ 720° MB = 465° - 360° MB = 705° (to the island) MB = 285° (from the island)

A) 330° (T) B) 270 0 (T) C) 250 0 (T) 0) 310 0 (T)

I

4453 (C)

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4475 (0)

I

4520 (A)

Variation = 700E 0 TB MB + 10 E variation

=

TB to =330° TB from = 750°

4523. Airplane ATPL CPL Heli ATPL CPL Given: 415NM Distance A to B: 295 kts PlannedGS: ATD: 10:00 UTC At 10:20 UTC fix is obtained 110 NM along track. What GS must be maintained from the fix in order to achieve planned ETA atB? A) B) C) 0)

320 kts 268 kts 295 kts 286 kts

First of all we need to find the estimated time of arrival. With a GS of 295 kts and a distance of 475 NM it will take 7,4 hrs (84 minutes) to cover the distance from A to B.lfthe actual time of departure was at 70:00 then the ETA is at 77:24. The flight progress check at 70:20 indicated a distance covered of 770 NM = distance remaining to go is 305 NM (475 NM - 770 NM). If our ETA is 77:24 UTC it means we have the time of 64 minutes (7,0666 hrs) to cover the distance of 305 NM. Our speed for the remainder of the flight would have to be 286 kts (305 NM + 7,0666 hrs).

4552. Airplane ATPL CPL Heli ATPL CPL Given true heading 256°, VAR 13°E, relative bearing to a station is 333°. The true bearing to the station is: A) B) C) 0)

2W 229 0 252 0 320 0

(Refer to figures 067-E31, 067-E77 and 067-E78) TB=RB+ TH TB =333°+ 256° TB = 589° - 360° TB = 229° (to the station)

4554. Airplane ATPL CPL Heli ATPL CPL A ground feature appears 30° to the left of the centre line of the CRT of an airborne weather radar. If the heading of the aircraft is 355° (M) and the magnetic variation is 15°E, the true bearing of the aircraft from the feature is: 1600 220 0 310 0

130°

(Refer to figures 067-E36, 067-E77 and 067-E78) MB=RB+MH

=

4442 (0)

(Refer to figures 067-E30, 067-E77 and 067-E78) MB=RB+MH MB = 030° + 290° MB = 320° (to the island) MB = 740° (from the island)

A) B) C) 0)

Variation 7rw TB=MB-7rWvariation

I

4520. Airplane ATPL CPL Heli ATPL CPL An island is observed to be 30° to the right of the nose of the aircraft. The aircraft heading is 290° (M), variation 100E. The bearing from the aircraft to the island is:

360 kts

354 kts

(Refer to figures 067-E63, 067-E64 and 067-E97) In order to determine the Ground Speed (GS) using the navigation computer we have to obtain the True Heading and then use it together with TAS and W/V for the GS determination:

A) B) C) 0)

TBto=088° TB from = 268°

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4523 (0)

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4552 (8)

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4554 (A)

Aviationexam Test Prep Edition 2012 MB =330° + 355° MB = 685° - 360° MB =325° (to the island) MB = 145° (from the island)

4616. Airplane ATPI. (PI. Heli ATPI. (PI. An island appears 30° to the left of the centre line on an airborne weather radar display. What is the true bearing of the aircraft from the island if at the time of observation the aircraft was on a magnetic heading (MH) of 020° with the magnetic variation (VAR) 25°W?

Variation = 15°E TB=MB+ 15°Evariation TBto=340° TB from = 160°

4568. Given:

Airplane

ATPI.

Magnetic heading: Drift angle: Relative bearing of NOB:

(PI.

Heli

ATPI.

(PI.

311° 10° left 270°

What is the magnetic bearing of the NOB measured from the aircraft?

A) 211° B) 208

0

A) B) e) D)

1450 1950 205 0 325 0

(Refer to figures 061-E34, 061-E77 and 061-E78) MB=RB+MH 0 MB = 330 + 020° MB = 350° (to the island) MB = 170° (from the island)

=

Variation 25°W TB =MB - 25°W variation TBto=325° TB from = 145°

0

e) 221 D) 1800 (Refer to figure 061-E32) The drift is useless information here because the Relative Bearing is relative to the direction in which the aircraft is pointing (heading) and the direction of the line joining the aircraft and the station. MB=RB+MH MB=2700+311° MB=581°-360° MB =221° (to the station)

4576. Airplane ATPL (PI. Heli ATPI. (PI. An aircraft at FL140, lAS 210 kts, OAT -5°C and wind component minus 35 kts, is required to reduce speed in order to cross a reporting point 5 min later than planned. Assuming that flight conditions do not change, when 150 NM from the reporting point the lAS should be reduced by:

A) 25 kts B) 20 kts C) 30 kts D) 15 kts (Refer to figures 061-E58 and 061-E59) First of all we need to determine the TAS based on lAS, FL and OAT values. Use your navigation computer to obtain a TAS of approximately 262 kts. With a wind component of -35 kts (headwind) we will have the Ground Speed (GS) of227kts. Note: Remember that when a wind component has a negative (-) prefix, it means a headwind, whereas a positive (+) prefix means a tailwind. A good mnemonic is to realize that when you want to get to your destination a headwind is bad for you =negative; whereas a tailwind is good for you =positive. We need to reduce the speed so that we cross a reporting point at a distance of 100 NM 5 minutes later than planned. Our original estimate for crossing the point is: Distance =Rate x Time 150 NM = 227 kts GS x ? hrs ? hrs = 150 NM 7 227 kts ? =0,66 hrs =40 minutes Now calculate the new GS that we need to maintain to cross the point in 45 minutes (40 minutes + 5 minutes). 45 minutes = 0,75 hrs. 150 NM =? kts x 0,75 hrs ? kts = 150 NM 7 0,75 hrs ? = 200 kts (GS) Note that the speed 200 kts is the Ground Speed (GS), but we need to know the TAS and then convert it into lAS. Remember that we have a headwind component of 35 kts => therefore to obtain the GS of 200 kts we need to maintain a TAS of 235 kts. Now use your navigation computer again to convert 235 kts TAS into lAS => approximately 188 kts lAS. Our original lAS was 210 kts and we need to slow down to about 188 kts =22 kts reduction =approximately 20 kts.

4631. Airplane ATPI. (PI. Heli ATPI. (PI. The relative bearing to a beacon is 2700R. Three minutes later, at a ground speed of 180 knots, it has changed to 225°R. What was the distance of the closest point of approach of the aircraft to the beacon?

A) 45 NM B) 18 NM e) 9NM D) 3NM (Refer to figure 061-E49) First of all try to visualize the situation - initially the aircraft is situated due East of the radio beacon (on a 090° bearing from the station). 3 minutes later we are at a position further to the North from our original position - now on a bearing 045° from the station. In essence, we have constructed a triangle. It can be seen from the attached illustration that the angle a =270° - 225°= 45°. What we have is an Isosceles triangle with a right angle and two 45° angles = two of the internal angles are identical and two sides are of identical length. with a right angle. The distance «01» was flown in 3 minutes (=0.05 hours) at a speed of 180 kts. Distance «01» = 180 kts x 0,05 hrs = 9 NM. Distance «02» is identical to distance «01» => Distance «02» equals 9 NM. The distance of the closest point of approach of the aircraft to the radio beacon is therefore 9NM.

4642. Given:

Airplane

Distance A to B:

ATPI.

(PI.

Heli

ATPI.

(PI.

90NM

Fix obtained 60 NM along and 4 NM to the right of course. What heading alteration must be made to reach 8?

A) 40 left. B) 16° left. e) 120 left. D) 80 left. (Refer to figure 067-£98) Let's start by calculating the TKE (Track Angle Error) that will give us the heading correction to parallel our intended track: • TKE = (Dist off-track x 60) 7 Dist. along track • TKE= (4x 60) 760 • TKE=4°

If we alter our heading by 4° to the left (towards the original track) we will parallel our original track. Now if we wanted to apply an even larger correction and alter the heading in such a way that we re-join our original track at destination B, we would follow these steps: • TKE = (Dist off-track x 60) 7 Dist. to go • Dist. to go = 90 NM - 60 NM = 30 NM • TKE = (4 x 60) 730 • TKE=8° Summary: • 4° heading alteration = continuing parallel to the original track;

I 4568 (C) 14576(8) I 4616(A) 14631 (C) I 4642 (C) I

05 In-Flight Navigation • additional 8° heading alteration =rejoining original track at destination (= total heading alteration of 72° left).

4643. Airplane ATPL CPL Heli ATPL CPL An aircraft at FL120, lAS 200 kts, OAT _5° and wind component +30 kts, is required to reduce speed in order to cross a reporting point 5 min later than planned. Assuming flight conditions do not change, when 100 NM from the reporting point lAS should be reduced to: A) B) C) D)

169 kts 165 kts 159 kts 174 kts

(Refer to figures 061-E58 and 061-E59) First of all we need to determine the TAS based on lAS, FL and OAT values. Use your navigation computer to obtain a TAS of approximately 240 kts. With a wind component of +30 kts (tailwind) we will have the Ground Speed (GS)of270kts. Note: Remember that when a wind component has a negative (-) prefix, it means a headwind, whereas a positive (+) prefix means a tailwind. A good mnemonic is to realize that when you want to get to your destination a headwind is bad for you =negative; whereas a tailwind is good for you =positive. We need to reduce the speed so that we cross a reporting point at a distance of 100 NM 5 minutes later than planned. Our original estimate for crossing the pointis: Distance = Rate x Time 100 NM = 270 kts GS x ? hrs ? hrs = 100 NM + 270 kts ? = 0,37 hrs = 22 minutes

Note that the speed 222 kts is the Ground Speed (GS), but we need to know the TAS and then convert it into lAS. Remember that we have a tailwind component of 30 kts => therefore to obtain the GS of 222 kts we need to maintain a TAS of 192 kts. Now use your navigation computer again to convert 192 kts TAS into lAS => approximately 159 kts lAS.

4657. Airplane ATPL CPL Heli ATPL CPL You are heading 345° (M), the variation is 20O,E, and you take a radar bearing of 30 0 left of the nose from an island. What bearing do you plot? A) 1600 (T) B) 1550 (T) C) 1400 (T) D) 1800 (T) (Refer to figures 061-E35, 061-E77 and 061-E78) MB=RB+MH 0 MB = 330 + 345° 0 MB = 675°-360 MB =3W(to the island) MB = 135° (from the island)

=

TBto=335° TB from = 155°

A) 160 0

B) 130° C) 220 0 D) 1900 (Refer to figures 061-E38, 061-E77 and 061-E78) MB=RB+MH MB = 030° + 355° MB=385°-360° MB = 025° (to the island) MB = 205° (from the island)

TB to =040° TB from = 220°

4696. Airplane ATPL CPL Heli ATPL CPL An aircraft is planned to fly from position A to position B, distance 480 NM at an average GS of 240 kts. It departs A at 10:00 UTe. After flying 150 NM along track from A, the aircraft is 2 min behind planned time. Using the actual GS experienced, what is the revised ETA at B? A) B) C) D)

A) 1401 UTC B) 1333 UTC C) 1347 UTC 4657 (8)

I

4679 (8)

I

4680 (C)

1203 1206 1153 1157

First of all we need to find original estimated time at which the aircraft was scheduled to reach the point 180 NM along the track. With a GS of 240 kts and a distance of 150 NM it was planned to take 37,5 minutes (150 NM + 240 kts = 0,625 hrs). If the 150 NM point was reached 2 minutes later than planned it means it was reached in 39,5 minutes (0,6583 hrs) => at 10:40 UTe. The actual GS that was experienced was therefore approximately 230 kts (150 NM + 0,6583). With the actual GS of 230 kts the total distance of 480 NM will take 2,0869hrs (480NM +230kts) = 125 minutes. If the actual time ofdeparture was at 10:00, the revised ETA at destination is therefore at 12:05.

A)

4679. Airplane ATPL CPL Heli ATPL CPL An aircraft is planned to fly from position A to position B, distance 320 NM, at an average GS of 180 kts. It departs A at 12:00 UTe. After flying 70 NM along track from A, the aircraft is 3 min ahead of planned time. Using the actual GS experienced, what is the revised ETA at B?

I

4680. Airplane ATPL CPL Heli ATPL CPL An island appears 30° to the right of the centre line on an airborne weather radar display. What is the true bearing of the aircraft from the island if at the time of observation the aircraft was on a magnetic heading (MH) of 355 0 with the magnetic variation (VAR) 15°E?

4700. Airplane ATPL CPL Heli ATPL CPL A ground feature was observed on a relative bearing of 315° and 3 min later on a relative bearing of 270°. The W/V is calm; aircraft GS 180 kts. What is the minimum distance between the aircraft and the ground feature?

Variation 200E TB = MB + 200E variation

4643 (C)

First of all we need to find original estimated time at which the aircraft was scheduled to reach the point 70 NM along the track. With a GS of 180 kts and a distance of 70 NM it was planned to take 23,3 minutes (70 NM + 180 kts = 0,3888 hrs). If the 70 NM point was reached 3 minutes earlier than planned it means it was reached in 20,3 minutes (0,3383 hrs). The actual GS that was experienced was therefore approximately 206,9 kts (70 NM + 0,3383). With the actual GS is 206,9 kts the total distance of320 NM will take 1,5466 hrs (320 NM + 206,9 kts) = 93 minutes (1 hr 33 min). If the actual time of departure was at 12:00 UTC, the revised ETA at destination is therefore at 13:33.

Variation = 15°E TB=MB+ 15°Evariation

Now calculate the new GS that we need to maintain to cross the point in 27 minutes (22 minutes + 5 minutes). 27 minutes = 0,45 hrs. 100 NM = ? kts x 0,45 hrs ? kts = 100 NM + 0,45 hrs ? = 222 kts (GS)

I

D) 1340 UTC

I

4696 (8)

I

3NM

B) 12 NM C) 9NM D)

6NM

(Refer to figure 061-E47) First of all try to visualize the situation - you can substitute the ground feature for example with an NOB station. Initially the aircraft is situated South-East of the station (on a 315° bearing to the station). 3 minutes later we are at a position further to the North from our original position - now on a bearing 090° from the station (relative bearing 270° to the station) = due East of the station. In essence, we have constructed a triangle. It can be seen from the attached illustration that the angle a = 315° - 270°= 45°. What we have is an Isosceles

4700 (e)

I

Aviationexam Test Prep Edition 2012 triangle with a right angle and two 45° angles = two of the internal angles are identical and two sides are ofidenticallength. The distance «07» was flown in 3 minutes (= 0,05 hours) at a speed of 780 kts. Distance «07» = 780 kts x 0,05 hrs = 9 NM. Distance «02» is identical to distance «07» => Distance «02» equals 9 NM. The minimum distance of the aircraft and the ground feature is9NM.

4712. Airplane ATPL CPL Heli ATPL CPL An aircraft is planned to fly from position A to position B, distance 250 NM at an average GS of 115 kts. It departs A at 09:00 UTe. After flying 75 NM along track from A, the aircraft is 1,5 min behind planned time. Using the actual GS experienced, what is the revised ETA at B? A) B) C) D)

1110 UTC 1115 UTC 1044UTC 1050 UTC

First of all we need to find original estimated time at which the aircraft was scheduled to reach the point 75 NM along the track. With a GS of 115 kts and a distance of 75 NM it was planned to take 39,7 minutes (75 NM.,. 775 kts = 0,652 hrs). If the 75 NM point was reached 7,5 minutes later than planned it means it was reached in 40,6 minutes (0,6766 hrs). The actual GS that was experienced was therefore approximately 110,8 kts (75 NM .,. 0,6766). With the actual GS is 110,8 kts the total distance of 250 NM willtake 2,2563 hrs (250 NM.,. 110,8 kts) = 735 minutes (2 hrs 75 min). If the actual time ofdeparture was at 09:00 UTe, the revised ETA at destination is therefore at 11:75.

4835. Airplane ATPL CPL Heli ATPL CPL At 00:20 UTe an aircraft is crossing the 310° radial at 40 NM of a VOR/DME station. At 00:35 UTe the radial is 040° and DME distance is 40 NM. Magnetic variation is zero. The true track and ground speed are: A) B) C) D)

080 0 090 0 085 0 088 0 -

C) 268 0 (T)

D) 302 0 (T) (Refer to figures 067-E26, 067-E77 and 067-E78) Your Magnetic Heading is 720°. The island is observed 75° to the left. The relative bearing of the island is 345° (360°- 75°). Magnetic Bearing (to) = RB + MH MB = 345°+ 720°=465° MB = 465°-360°= 705° Now we have obtained the Magnetic bearing to the island from the aircraft, but the answers only offer True values. This is when we use the value ofvariation to convert 705° Magnetic into True. With a Westerly variation the True Bearing = 705° MB - 71" Westerly Variation = 88° True Bearing.

4846. Airplane ATPL CPL Heli ATPL CPL An island appears 60° to the left of the centre line on an airborne weather radar display. What is the true bearing of the aircraft from the island if at the time of observation the aircraft was on a magnetic heading (MH) of 276° with the magnetic variation (VAR) 100E? A) 0460 B) 0860 0 C) 226 0 D) 026 (Refer to figures 067-£33, 067-E77 and 067-E78) MB=RB+MH MB =300° + 276° MB =576° - 360° MB =276° (to the island) MB = 036° (from the island)

=

0

Variation 10 E TB = MB + 700E variation TBto=226° TB from = 046°

226 kts. 232 kts. 226 kts. 232 kts.

(Refer to figure 067-E43) First of all try to visualize the situation - it helps to draw a sketch. You will discover that the aircraft track and the radials from the VOR will form a rightangle triangle (right angle at the VOR). We know the distance of the two sides of this right angle triangle = 40 NM. We can use the Pythagorean theorem (N = B1 + (1) to determine the distance of the third side: (Distance 1)1 =401 + 401 Distance 1 =,j (1.600 + 7.600) Distance 1 =56,57 NM We have flown this distance of 57,57 NM in 75 minutes (0,25 hrs) => with this information we can easily determine that the ground speed is approx. 226 kts (56,57 NM.,. 0,25 hrs). To obtain the aircraft track you can either draw a precise sketch and then measure the angle, or you can calculate it (refer to the attached illustration). We know the distance if all 3 triangle sides + we know the value of one of the internal angles => the angle at the VOR = 90". To calculate an internal angle of a right-angle triangle we can use the formula: sin a = (Length of the opposite side.,. Length of the Hypotenuse) sin a = 40 .,. 56,57 sin a = 0,7070 a=45°

4847. Airplane ATPL CPL Heli ATPL CPL At 10:00 hours an aircraft is on the 310° radial from a VOR/DME, at 10 nautical miles range. At 10:10 the radial and range are 040°/10 NM. What is the aircraft's track and ground speed? A) B) C) D)

080/85 knots 085 / 85 knots 080 / 80 knots 085/ 90 knots

(Refer to figure 067-E44) First of all try to visualize the situation - it helps to draw a sketch. You will discover that the aircraft track and the radials from the VOR will form a rightangle triangle (right angle at the VOR). We know the distance of the two sides of this right angle triangle = 10 NM. We can use the Pythagorean theorem (N = B1 + (1) to determine the distance of the third side: (Distance 1)1 = 701 + 701 Distance 1 =,j (100 + 100) Distance 1= 74,74NM We have flown this distance of 74,74 NM in 10 minutes (0,7666hrs) => with this information we can easily determine that the ground speed is approx. 85 kts (14,74NM.,. 0,766hrs).

Add the angle a (45°) to the value of the radial at the position 2 (40°) to obtain the Magnetic track at position 2 of 85°. Note: in fact you do not have to use the trigonometry at all-just realize that we have an Isocele triangle, where the two small angles are identical- 45". Just add the 45° to the radial at position 2 to obtain the heading of 085°.

4842. Airplane ATPL CPL Heli ATPL CPL An island is observed to be 15° to the left. The aircraft heading is 120° (M), variation 17°W. The bearing from the aircraft to the island is:

To obtain the aircraft track you can either draw a precise sketch and then measure the angle, or you can calculate it (refer to the attached illustration). We know the distance if all 3 triangle sides + we know the value of one of the internal angles => the angle at the VOR = 90°. To calculate an internal angle of a right-angle triangle we can use the formula: sin a = (Length of the opposite side.,. Length of the Hypotenuse) sin a = 70.,. 74,74 sin a = 0,7072 a=45° Add the angle a (45°) to the value of the radial at the position 2 (40°) to obtain the Magnetic track at position 2 of 85°. Note: In fact you do not have to use the trigonometry at all-just realize that we have an Isocele triangle, where the two small angles are identical- 45°. Just add the 45° to the radial at position 2 to obtain the heading of 085°.

A) 122 0 (T) B) 088 0 (T)

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4712 (8)

I

4835 (e)

I

4842 (8)

I

4846 (A)

I

4847 (8)

I

05 In-Flight Navigation

4865. Airplane ATPL CPL Heli ATPL CPL An island appears 45° to the right of the centre line on an airborne weather radar display. What is the true bearing of the aircraft from the island if at the time of observation the aircraft was on a magnetic heading (MH) of 215° with the magnetic variation (VAR) 21°W? A) B) C) D)

101° 059° 239° 329°

(Refer to figures 067-E46 and 067-E98) This question can be easily solved using the 7:60 rule: • TKE = (Dist off-track x 60) + Dist along track .7°= (?x 60) +200 .?= 7 +60x200

·?=3,33NM

Variation =27°W TB = MB - 27 oW variation TBto=239° TB from = 059°

4867. Airplane ATPL CPL Heli ATPL CPL An island appears 30° to the left of the centre line on an airborne weather radar display. What is the true bearing of the aircraft from the island if at the time of observation the aircraft was on a magnetic heading of 276° with the magnetic variation 12°W? 318° 054° 234° 038"

(Refer to figures 067-E28, 067-E77 and 067-E78) MB=RB+MH MB = 330° + 276° MB = 606° - 360° MB = 246° (to the island) MB = 066° (from the island) Variation = 72°W TB = MB - 72°Wvariation TB to =234° TB from = 054°

4913. Airplane ATPL CPL Heli ATPL CPL An aircraft is at 55°30'N 036°13'W, where the variation is 15W. It is tuned to a VOR located at 53°30'N 036°13'W, where the variation is 12°W. What VOR radial is the aircraft on? A) 348

B) 012 C) 165 D) 360 Notice that the co-ordinates of the VOR and the aircraft share the same value of Longitude (036°73'W). That means the aircraft is due North of the VOR => on a True Bearing of 0° from the station. However, remember that the VOR radials are Magnetic bearings => therefore to obtain the Magnetic Bearing from a True Bearing we have to apply the value of Variation. When converting from True to Magnetic the mnemonic "West is Best, East is Least" is applicable => with Westerly variation we would add the variation => in our case the Magnetic Bearing (radial) would be 0° True Bearing + 72° Westerly Variation = 072~ The aircraft is on 071' radial from the VOR.

4964. Airplane ATPL CPL Heli ATPL CPL A pilot receives the following signals from a VOR DME station: Radial: Distance: What is the approximate error?

Another method is to use the trigonometric ratios in right-angle triangles: tan a =Length of opposite side + Length of adjacent side In our case it would be: • tan 7°=?NM+200NM ·?NM=tan 7°x200NM

·?=3,49NM

231035. Airplane ATPL CPL Heli ATPL CPL An aircraft is flying at FL100. The OAT ISA -15°C. The QNH given by a station at an elevation 3.000 ft is 1035 hPa. Calculate the approximate True Altitude. A) B) C) D)

10.000ft 10.200ft 7.200ft 9.600ft

231045. Airplane ATPL CPL Heli ATPL CPL A ground feature was observed on a relative bearing of 45° left of the nose and 3 MIN later on a relative bearing of 90° left of the nose. The WIV is calm; aircraft GS 180 kt. What is the minimum distance between the aircraft and the ground feature? A) 6NM

B) 3NM C) 9NM

D) 12 NM 231095. Airplane ATPL CPL Heli ATPL CPL The distance between A and B is 90 NM. At a distance of 15 NM from A the aircraft is 4 NM right of course. To reach destination B, the correction angle on the heading should be: A) B) C) D)

16° 3° 21° 19°

231096. Airplane ATPL CPL Heli ATPL CPL The distance betwl!en A and B is 90 NM. At a distance of 75 NM from A the aircraft is 4 NM right of course. The track angle error (TKE) is: A) B) C) D)

3°R 19°R 6°R 22°R

231097. Airplane ATPL CPL Heli ATPL CPL The True course in the flight log is 270°, the forecast wind is 045°(T)115kts and the TAS is 120 kts. After 15 minutes of flying with the planned TAS and TH the aircraft is 3 NM South of the intended track and 2,5 NM ahead of the dead reckoning position. The track angle error (TKE) is: A) SOL

B) 2°L C) 3°R D) 6°R

A) ±3,5 NM B) ± 1 NM 1 4865 (8) 1 4867 (8)

D) ±7 NM

The possible error in the position of the aircraft is therefore approximately ±3,3NM.

(Refer to figures 067-£37, 067-E77 and 067-E78) MB=RB+MH MB=045°+2W MB = 260° (to the station) MB = 080° (from the station)

A) B) C) DJ

C) ±2 NM

1 4913 (8)

1 4964 (A) 1231035 (8) 1231045 (C) 1231095 (0) 1231096 (A) 1231097 (A) 1

Aviationexam Test Prep Edition 2012

231098. Airplane ATPL CPL Heli ATPL CPL An aircraft is flying from A to B a distance of 50 NM. The True Course in the flight log is 270°, the forecast wind is 045°(T)/15kts and the TAS is 120 kts. After 15 minutes of flying with the planned TAS and TH the aircraft is 3 NM South of the intended track and 2,5 NM ahead of the.dead reckoning position. To reach destination B from this position, the correction angle on the heading should be: A) 12 0 B) 50 C)W D) 17 0

231099. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-07) After 15 minutes of flying with the planned TAS and TH the aircraft is 3 NM South of the intended track and 2,5 NM ahead of the dead reckoning position. To reach destination B from this position, the TH should be: A) B) C) D)

287 0 280 258 0 292 0

0

231100. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-06) After 15 minutes of flying with the planned TAS and TH the aircraft is 3 NM South of the intended track and 2,5 NM ahead of the dead reckoning position. To reach destination B from this position, the TH should be: A) 1000 B) 090 0 C) 078 0 D) 112 0 231101. Airplane ATPL CPL Heli ATPL CPL An aircraft is flying from A to B a distance of 50 NM. The True course in the flight log is 090°, the forecast wind is 225°(T)/15kts and the TAS is 120 kts. After 15 minutes of flying with the planned TAS and TH the aircraft is 3 NM South ofthe intended track and 2,5 NM ahead ofthe dead reckoning position. To reach destination B from this position, the correc-

tion angle on the heading should be: A) 50

B) 100 C) 12 0

D) 170

231102. Airplane ATPL CPL Heli ATPL CPL An aircraft is flying from A to B. The true course according to the flight log is 090°, the estimated wind is 225°(T)/15kts and the TAS is 120 kts. After 15 minutes of flying with the planned TAS and TH the aircraft is 3 NM South of the intended track and 2,5 NM ahead of the dead reckoning position. The Track angle error (TKE) is: A) 5°R B) 12°R

C) lrL D) 6°L 231103. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-06) An aircraft is flying according the flight log at the Annex. After 15 minutes of flying with the planned TAS and TH the aircraft is 3 NM North of the intended track and 2,5 NM ahead of the dead reckoning position. To reach destination B from this position the TH should be: A) 090 0

B) 078 0 C) 112 0 D) 1070

231104. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-07) An aircraft is flying according the flight log at the Annex. After 15 minutes of flying with the planned TAS and TH the aircraft is 3 NM North of the intended track and 2,5 NM ahead of the dead reckoning position. To reach destination B from this position the TH should be: A) 253 0

B) 258 0 C) 270 0 D) 292 0

05-04 Flight Log 2334. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-05) Complete line 4 of the FLIGHT NAVIGATION LOG, positions 'G' to 'H'. What is the HOGo (M) and ETA? A) B) C) D)

HDG 3440 HDG 3440 HDG 3540 HDG 0340 -

ETA 1303 UTe. ETA 1336 UTC. ETA 1326 UTe. ETA 1336 UTC.

2) Calculate the True Heading (TH) + Ground Speed (GS) using the navigation computer - the input variables will be the Track, W/Vand TAS. 3) Apply the magnetic variation to convert the TH into MH (Magnetic Heading). Remember the mnemonic "West is Best, East is Least" => it is applicable in this case. Westerly variation is added to TH (+), Easterly variation is subtracted from TH (-). 4) Calculate the time required to fly the leg. Use the calculator to determine time from GS and Distance. Time in hours = Distance -;- GS. 5) Determine the ETA => add the time required to fly the leg to the time of departure.

(Refer to figures 061-£51, 061-E58, 061-E59, 061-E63, 061-E64 and 061-E96) 1) Obtain the TAS => use your navigation computer to convert the CAS or Mach Nr. into TAS using the FL and OAT information. To convert the Mach number into TAs by calculation, follow the steps below: • Find OAT in oK (= OAT in °C + 273) • Find LSS (=38,95 x ,fOAT in OK) • Find TAS (= Mach nr. x LSS) 1231098 (D) 1231099 (0) 1231100 (C) 1231101 (0) 1231102 (A) 1231103 (C) 1231104 (8) 1 2334 (8)

1

05 In-Flight Navigation

2335. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-05) Complete line 5 of the FLIGHT NAVIGATION LOG, positions "J" to "K". What is the HDGo (M) and ETA? A) B) C) 0)

HOG 337° - ETA 1422 UTe. HOG 320° - ETA 1412 UTC. HOG 337° - ETA 1322 UTe. HOG 320° - ETA 1432 UTC.

A) B) C) 0)

For explanation refer to question #2334 on page 146.

2336. Airplane ATPL CPL Heli ATPL CPL (Refer to figure 061-05) Complete line 2 ofthe FLIGHT NAVIGATION LOG, positions "C" to "0". What is the HDG o (M) and ETA? A) B) C) 0)

HOG HOG HOG HOG

193°; ETA 1239 UTC. 188°; ETA .1229 UTC. 193°; ETA 1249 UTC. 183°; ETA 1159 UTC.

For explanation refer to question #2334 on page 146.

2891. Airplane ATPL CPL Heli ATPL CPL (Refer to fig u re 061-05) Complete line 3 ofthe FLIGHT NAVIGATION LOG, positions "E" to "F". What is the HDG o (M) and ETA? A) B) C) 0)

HOG 095°; ETA 1155 UTe. HOG 106°; ETA 1215 UTC. HOG 115"; ETA 1145 UTC. HOG 105°; ETA 1205 UTC.

Airplane

ATPL

(Refer to figures 061-E55 and 061-E97) When applying the drift it helps to draw a sketch or to take a look at the wind side ofyour navigation computer. For example, with a track of360° and the drift of 10° right it means that the wind is from the left, pushing you to the right => therefore, to compensate for the drift, your heading will be slightly towards the West = 360° Track - 10° drift = 350". With a drift 10 0 L it would be vice versa - your heading would be 010° (360° + 10°). When converting True Heading (TH) into Magnetic Heading (MH) using the magnetic variation, or when converting the MH into the Compass Heading (CH) using the value of Deviation, remember the mnemonic HWest is Best, East is Least H=> it is applicable in this case. Westerly variation is added to TH (+), Easterly variation is subtracted from TH (-). Also note that if the values of variation or deviation are negative, they are Westerly, if they are positive, they are Easterly. Answer A - incorrect value of MH: • Track 125° and drift 2°R => TH 123° • TH 123° and 2°Wvariation => 125° MH, but answer incorrectly states 121°

Answer C - incorrect value of MH: • Track 115° and drift 4°L => TH 121° • TH 121° and l°E variation => 120° MH, but answer incorrectly states 122°

CPL

Heli

ATPL

CPL

(Refer to figure 061-05) Complete line 6 of the FLIGHT NAVIGATION LOG, positions "L" to "M". What is the HDGo (M) and ETA? A) B) C) D)

125°; 2°R; 123°; 2°W; 121°; _4°; 117°. 115°; 5°R; 120°; 3°W; 123°; +2°; 121°. 117°;4°L; 121°; l°E; 122°; _3°; 119°. 119°; 3°L; 122°; 2°E; 120°; +4°; 116°.

Answer 8 - incorrect value of TH: • Track 115° and drift 5°R => TH 110°, but answer incorrectly states 120°

For explanation refer to question #2334 on page 146.

2894.

4709. Airplane ATPL CPL Heli ATPL CPL The flight log gives the following data: True track, Drift, True heading, Magnetic variation, Magnetic heading, Compass deviation, Compass heading. The right solution, in the same order, is:

Answer 0 - correct: • Track 119° and drift 3°L => TH 122° (119° + 3°) • TH 122° and 2°E variation => 120° MH • MH 120° and +4° deviation WE) => 116° CH

HOG 064°; ETA 1449 UTC. HOG 075°; ETA 1452 UTe. HOG 070°; ETA 1459 UTC. HOG 075°; ETA 1502 UTe.

For explanation refer to question #2334 on page 146.

4077. Airplane ATPL CPL The ICAO definition of ETA is the: A) B) C) D)

Heli

ATPL

CPL

actual time of arrival at a point or fix. estimated time of arrival at destination. estimated time of arrival at an en-route point or fix. estimated time en-route.

ICAD Doc 8400 - Definitions: ETA = Estimated time of arrival. ETD =Estimated time of departure. ETO = Estimated time over significant point. EAT = Expected Approach Time. ATD = Actual time ofdeparture. ATA =Actual time of arrival.

4509.

Airplane

ATPL

CPL

Heli

ATPL

CPL

(Refer to figure 061-05) Complete line 1 of the FLIGHT NAVIGATION LOG; positions "A" to liB". What is the HDG o (M) and ETA? A) 268°; 1114 UTe. B) 282°; 1128 UTC. C) 282°; 1114 UTe. 0) 268°; 1128 UTC. For explanation refer to question #2334 on page 146.

I 2335 (A) I 2336 (A) I 2891 (0) 12894(0) 14077(8) I 4509(A) 14709(0) I

Aviationexam Test Prep Edition 2012

Picture Supplements Questions

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Picture Supplements - Questions

FIGURE 061-01 STANDARD TIMES (Corrected to September 1993) LIST 1- PLACES FAST ON UTC (mainly those EAST OF GREENWICH) The times given below should be added to UTC to give Standard Time subtracted from Standard Time to give UTe.

Admiralty Islands Afghanistan Albania' Algeria Amirante Islands Andaman Islands Angola Armenia Australia Australian Capital Territory • New South Wales 1,' Northern Territory Queensland South Australia Tasmania Victoria' Western Australia Whitsu.nday Islands Austria Azerbaijan

.

Bahrain Balearic Islands Bangladesh Belgium' Belorussia Benin Bosnia and Herzegovina Botswana, Republic of Brunei Bulgaria Burma (Myanmar) Burundi

h m 10 04 30 01

Oli

Egypt, Arab Republic of Equatorial Guinea, Republic of Estonia' Ethiopia

04 05 30 01 04

Fiji Finland France

10 10 09 30

10 09 30·

10 10

08 10 01 04 03 01 06 01

02 01 01

02 08 02 06 30

02

Cambodia Cameroon Republic Central African Republic Chad Chagos Archipelago Diego Garcia Chatham Islands China Christmas Island, Indian Ocean Cocos Keeling Islands Comoro Islands (Comoros) Congo R.epublic Corsica Crete' Croatia Cyprus, Ercan Larnaca' Czech Republic

07

Denmark Djibouti

01 03

01 01 01 05 06 12 45

08 07 06 30 03 01 01

02 01

02 02 01

01

02 03

12

02 01

Gabon Georgia Germany Gibraltar' Greece Guam

01 03 01 01

Holland (The Netherlands) Hong Kong Hungary •

01

India Indonesia, Republic of (Bangka, Billiton, Java, West and Central Kalimantan, Madura, Sumatra Bali, Flores, South and East Kalimantan, Lombok, Sulawesi, Surnba, Sumbawa, Timor, Aru, Irian Jaya, Kai, Moluccas, Tanlmbar) Iran Iraq Israel Italy'

05 30

Japan Jordan

09

Kazakhstan Kenya Kiribati Republic 2 Korea, North Korea, Republic of (South) Kurillslands Kuwait Kyrgyzstan

06 03 12 09 09

Laccadive Islands Laos Latvia Lebanon Lesotho Libya Liechtenstein Lithuania' Lord Howe Island Luxembourg

05 30

.

Summer time may be kept in these countries. Except Broken Hill Area which keeps 09h 30m. 2 Except Kiribati Island and the Phoenix Islands which keep 10h and 11 show on UTe.

1

h m 02

02 10

08 01

07

08 09 03 30 03

02 01 01

02

.

11 03 05

07 02 02 02 02 01

02 10 30 01

Aviationexam Test Prep Edition 2012

FIGURE 061-02 STANDARD TIMES (Corrected to September 1993) LIST 111- PLACES SLOW ON UTC (WEST OF GREENWICH) continued

}

The times given below should be

Jamaica Jan Mayen Island Johnston Island Juan Fernandez Islands

h m 05 01 10 04

Leeward Islands

04

Marquesas Islands Martinique Mexico 1 Midway Islands

09 30 04 06 11

Nicaragua Niue

05 11

Panama, Republic of paragyay' Peru Puerto Rico

05 04 05 04

St. Pierre and Miquelon Salvador, EI Samoa Society Islands South Georgia Suriname

03 06 11 10 02 03

Trinidad and Tobago Trindade Island, South Atlantic Tuamotu Archipelago Tubuai Islands Turks and Caicos Islands

04 02 10 10 05

United States of America Alabama 2 Alaska 2, east of W. 169°30' Aleutian Islands 2, west of Arizona "; Arkansas 2 California 2 Colorado 2 2 Connecticut Delaware 2 District of Columbia 2 Florida 2,3

w,. 169°30'

Georgia 2 Hawaii Idaho 2,3

06 09 10 07 06

OS 07 05 05 05 05 05 10 07

subtracted from UTC to give Standard Time added to Standard Time to give UTe.

h United States of America (continued) Illinois 2 Indiana 3 Iowa 2 Kansas 2,3

.l ~.

'I

Kentucky 2, eastern part Kentucky 2, western part Louisiana 2 Maine 2 Maryland 2 Massachusetts 2 Michigan 2,3 Minnesota 2 Mississippi 2 Missouri 2 Montana 2 Nebraska 2,3 Nevada 2,3 New Hampshire 2 New Jersey 2 New Mexico 2 New York 2 North Carolina 2 North Dakota 2,3 Ohio 2 Oklahoma 2,3 Oregon 2,3

m

06 05 06 06 05 06 06 05 05 05 05 06 06 06 07 06

OS 05 05 07 05 05 06 05 06

OS

Pennsylvania 2 Rhode Island 2 ", South Carolina 2 South Dakota 2, eastern part South Dakota 2, western part Tennessee 2,3 Texas 2,3 Utah 2 Vermont 2 Virginia 2 Washington D.e. 2 Washington 2 West Virginia 2 Wisconsin 2 2 Wyoming Uruguay'

05 05 05 06 07 06 06 07 05 05 05

Venezuela Virgin Islands

04 04

Windward Islands

04

OS 05 06 07 03

Summer time may be kept in these countries. (Except the states of Sonora, Sinaloa, Nayarit and the Southern District of Lower California which keep and Northern District of Lower California which keeps OSh. 2 Summer (daylight-saving) time, one hour fast on the time given, is kept in these states from the first Sunday in April to the last h Sunday in October, changing at 02 OOm local clock time. 3 A small portion of the state is in another time zone.

oi

Picture Supplements - Questions

FIGURE 061-03 SUNRISE Lat.

November

22

19 h m

m h N 72 70 10 14 10 40 68 09 30 09 45 66 09 01 09 12 64 08 39 08 48 62 21 29 N 60 08 07 08 14 58 07 54 08 01 56 43 07 49 40 54 34 52 31 26 N 50 07 18 07 23 45 07 02 07 06 40 06 49 06 53 41 35 38 30 28 30 N 20 06 11 06 13 N 10 05 56 05 57 42 0 42 28 28 5 10 20 05 12 05 12 5 30 04 54 04 53 44 42 35 30 40 32 45 18 04 15 50 04 00 03 57 5 52 03 52 03 48 39 54 43 28 56 32 15 58 21 5 60 03 06 03 00

25 h

m

11 10 09 08

19 00 23 57 37 21 07 55 45 36 28 10 56 44 33 15 58 43 28 11 52 41 28

08 08 07

07 07 06

06 05

05 04

28 h

m

10 09 09 08 08

16 34 06 45 28 13 01 50 40 32 14 59 46 35 16 00 44 28

08 07 07 07 06

06 06 05

I h

10 09 09 08 08 08 07 07 07 06 06 06 05

05 11 05 04 52 04 40 27 04 13 04 11 04 03 54 03 51 03 03 45 03 42 03 31 35 23 19 03 10 03 05 03 02 54 02 49 02

1 m

33 45 15 52 34 19 06 55 45 36 17 02 49 38 18 01 45 29 12 51 39 26 09 49 39 29 16 02 44

4 m h

h

10 09 09 08 08

10 09 09 08

52 55 23 59 40 24

7 m

10 h m

13 h m

December 19 16 h m h m

22 h

m

25 h

m

28 h

m

31 h m

Lat.

N 72 70 68 66 64 62 N 60 58 56 54 52 N 50 45 40 35 30 N 20 N 10 0 5 10 20 5 30 35 40 45 50 5 52 54 56 58 5 60

22

19 h m

h

13 15 14 00 29 14 51 15 09 15 23 36 46 15 56 16 04 16 12 28 41 16 53 17 03 17 20 35 17 49 18 03 19 18 37 47 18 59 19 14 31 19 40 19 49 20 00 12 20 26

12 51 13 47 14 19 14 43 15 02 15 18 31 42 15 52 16 01 16 09 26 39 16 51 17 01 17 19 35 17 50 18 05 21 18 39 18 50 19 03 18 36 19 45 19 55 20 06 19 20 34

m

25 h

m

12 13 13 33 14 10 36 14 56 15 12 26 38 48 15 58 16 06 24 38 16 50 17 01 17 19 35 17 51 18 06 23 18 42 18 53 19 06 22 41 19 50 20 00 12 25 20 41

28 h

m

I

h

1 m

08 11 07 59 08 49 07 07 40 07 21 07 05 07 06 52 06 40 06 20 06 06 03 06 05 46 05 30 05 12 05 04 51 04 39 25 04 08 04 03 47 03 03 37 03 26 03 13 03 02 58 02 02 40 02

13 19

13 04

14 01 29 14 50 15 08 22 35 45 15 55 16 03 22 37 16 49 17 00 17 19 36 17 52 18 07 25 18 44 18 56 19 09 25 45 19 55 20 05 18 32 20 48

13 52

14 22 14 45 15 03 19 32 43 15 53 16 01 21 36 16 49 17 00 17 19 37 17 53 18 09 26 18 47 18 59 19 13 29 50 19 59 20 10 23 38 20 55

4 h m

h

m

11 14

05 10 14 30 09 36 05 09 10 45 08 50 29 33 15 19 03 08 07 53 07 56 43 07 47 24 27 08 07 10 54 06 57 43 45 22 06 24 04 06 06 48 05 49 31 32 13 05 13 51 04 52 39 39 25 25 07 04 07 46 03 45 36 03 35 24 23 11 03 09 55 02 53 37 02 34

10 32 09 50 22 09 00 08 43 28 15 08 10 13 07 59 08 02 08 04 07 50 07 52 07 54 29 34 32 17 07 13 15 06 59 07 01 07 03 47 06 49 06 50 06 26 06 28 06 29 06 07 06 09 06 10 05 50 05 52 05 53 35 33 34 05 14 05 15 05 17 04 52 04 53 04 55 42 40 41 27 25 26 04 07 04 08 04 08 03 45 03 45 03 46 03 34 03 34 03 35 23 22 22 03 09 03 08 03 09 02 52 02 52 02 52 02 31 02 32 02 31

10 09 09 08

22 42 15 54 37 23

10 09 09 08

28 47 19 58 40 26

10 35 10 35 09 52 09 53 24 24 09 02 09 03 08 45 08 46 30 31 17 19 08 06 011 07 07 56 07 57 35 37 18 20 07 04 07 06 06 52 06 53 06 31 06 32 06 12 06 13 05 55 05 56 37 39 05 18 05 20 04 56 04 58 43 45 28 29 04 10 04 11 03 47 03 49 03 36 03 38 24 26 03 10 03 12 02 53 02 55 02 32 02 34

10 33 09 52 24 09 03 08 46 32 19 08 08 07 58 38 21 07 07 06 54 06 33 06 15 05 58 40 05 21 04 59 46 31 04 13 03 51 03 40 28 03 14 03 57 02 37

10 30 09 51 23 09 03 08 46 32 19 08 08 07 59 38 22 07 08 06 55 06 35 06 16 05 59 42 23 05 01 04 48 34 04 16 03 54 03 43 31 17 03 01 02 41

SUNSET November

I J~n.

7 h

m

10 h m

13 h m

December 19 16 h m h m

12 47 12 27 13 44 13 37 13 30 13 26 13 22 14 17 14 12 14 08 14 05 14 04 14 41 37 34 32 31 15 00 14 57 14 55 14 54 14 53 16 15 13 15 12 15 11 15 10 29 27 26 25 25 41 38 39 38 38 15 51 50 49 48 49 16 00 15 59 15 58 15 58 15 58 20 16 19 16 19 16 19 16 19 35 35 35 35 36 16 48 16 48 16 48 16 49 16 50 17 00 17 00 17 01 17 01 17 02 17 20 17 20 17 21 17 22 17 23 37 42 38 39 41 17 54 17 55 17 56 17 58 17 59 18 10 18 12 18 14 18 15 18 17 28 30 32 34 36 18 49 18 51 18 54 18 56 18 58 19 02 19 04 19 06 19 09 19 11 16 21 23 26 18 32 36 19 39 19 41 19 44 19 54 19 57 20 01 20 03 20 06 20 04 20 07 20 11 20 14 20 17 15 19 23 26 29 28 33 37 40 20 43 20 43 20 48 20 53 20 57 21 00 21 01 21 07 21 12 21 17 21 20

22 h

m

21 13 21 14 03 14 04 32 33 14 53 14 55 15 11 15 12 26 27 39 40 49 15 51 15 59 16 01 16 20 21 37 38 16 51 16 52 17 04 17 05 17 25 17 26 17 43 17 45 18 01 18 02 18 20 37 18 39 18 59 19 01 19 12 14 27 29 19 45 19 47 20 08 20 10 20 19 20 21 31 33 20 45 20 47 21 02 21 04 21 23 21 25

13

25

28

10 25 09 48 22 09 01 08 45 31 19 08 08 07 58 38 22 07 08 06 56 "'06 36 17 06 01 05 43 25 05 03 04 51 36 04 19 03 57 03 46 34 21 03 05 02 45

I

Ja n. 3 h m

m

31 h m

13 24

13 29

13 35

13 43

14 06 35 14 57 15 14 29 42 15 52 16 02 23 40 16 54 17 07 17 28 17 46 18 04 21 18 40 19 02 15 30 19 48 20 11 20 22 34 20 48 21 05 21 25

14 10 38 14 59 15 17 31 44 15 55 16 05 25 42 16 56 17 09 17 29 17 48 18 05 23 18 41 19 03 16 31 19 49 20 12 20 22 34 20 49 21 05 21 25

14 15 14 42 15 03 20 34 47 15 58 16 07 28 44 16 58 17 10 17 31 17 50 18 07 24 18 43 19 04 17 32 19 50 20 12 20 22 34 20 48 21 05 21 24

14 21 14 47 15 07 24 38 15 50 16 01 16 10 30 16 47 17 00

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08 07

45 29 15 04 54

07

42 26 13 02 53

07

36 22 08

06

17

07 06

09 08

17

06 56 48

07 06

07 06

~l>l> 0000

:1:-

~

37 21 07

06

06 06

06

=;r

0'T1

MORNING CIVIL TWILIGHT ~

00

23 Gl 00

.... CD

'" '" '" '" '" -oJ

0 0

0 0

.... OJ 0

'" 0 0

C

~

=I 3 CI>

m

~

N72 70 68 66 64 62 N60 58 56

February

January

Lat.

29

26

1

7

4

h

m

h

m

h

m

h

m

h

m

h

m

09 08

21 54 34 18 05 54

09 08

08 44 26 11 '58 48

08

55 33 17 03 52 42

08 08 07

42 22 07 55 44 35

08 08 07

28 11 58 46 37 28

08 08 07

15 00 47 37 29 21

44 35 28

07

39 31 24

07

33 26

07

28

08 07 07

08 07

Co

08 07

m Q. ;:;: IV

7

4 h

m

h

11 10 09 09 08

27 47 20 00 42

07 06

1 m

h

;

'tJ

:;: =0' ~

",S:O

'" o ' " ~o o "'CD ~'" ~o; "'~ "'0 00 "'0 00 00 00 ~

N72 70 68 66 64 62

29

26 m

h

~ .=

February

January

Lat.

0

~Gl

~

SUNRISE

o.::~

S:C

N 0

"0

'" -oJ<>C '" 0

en;o

0 _::I: CO -10 (]I ...... G')

~

::;10

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0

N _::I: -..j -10 (]I ...... G')

N 0

,

_:1:

;0-1

en;o

N W ....

CI>

...iii'c)" = ~

~g:5-

()

()

32S

ii-

'" '"

0

0

N ;0-1

-..j

'" '"

~ ~ ii --311~ I

0

10=

~

~;o

~

....

~

!

~

g

IV

Picture Supplements - Questions

FIGURE 061-09

FIGURE 061-10 °o ~

FIGURE 061-12

FIGURE 061-11 gl

=geod~tic latitude

ICAO AERONAUTICAL CHART SYMBOL AIR TRAFFIC SERVICES

I

I

Q

Q

I I I I

1.

5. - - - -

2.

6.

f:,.

3. --------

7.

....

_._._._._-

8.

~

4. A

B

I

OBSTACLES

---------------------+---------------------

Q

,II;

I I

9.

f..,

12.

M

I I I

10.

t

13.

1.

I I I

11.

M

14.

1

16.

cb

I

I I I I

VISUAL AIDS

I

15.

c

D

"*

FIGURE 061-13

FIGURE 061-14

ICAO Aeronautical Chart Symbol Air Traffic Services

TN

1. K::1l

5.

0

2.

6.

tr

8

3.

0

4.

(ii)

7.

090°(T)

W A

B

Aviationexam Test Prep Edition 2012

FIGURE 061-15

FIGURE 061-16

Picture Supplements - Questions

FIGURE 061-17 ~

~

0

~

!!! I! 0

" l! ~

~

~

~

36

I

:I>

..

< iii' 0' ::I

/I)

><

AI

3

..;;;

~

III

"tI

"tI

m Q. ;;'

0'

::I

'" '"

g

c ·

Picture Supplements Explanations

Picture Supplements - Explanations

FIGURE 061-E02

I I--

I

I FIGURE 061-EOl

I

-

I

f.L.

\

(

I

,

-

I ~ :::!

~

0,

I

I

-

I

I

I

I

I (

-

~

I I

'\

;;, I

I

r-

I

&~

0 N

t--' :

0

60 0 25'N 03°30'E

I0

..0

'LSAXAVORD

.~ ~

(I)

u

c:

....

(l1

~ 6L

~I

~

.~

'1 -

,/

t

LSUMBURGS

I I I

I

~cP" r t ~p

I

I

~4

L

1

...

z

FIGURE 061-E03

o·

--..

I I I I I I

.

I I

~ ~~

t~

=

5J ~ tiJ, (

/

~ \

Aviationexam Test Prep Edition 2012

FIGURE 061-E04

FIGURE 061-EOS

Picture Supplements - Explanations

FIGURE 061-E06

00 :

M

r-t-+--1_ °l1 f::!.

I

: ::E n--t-t-+'---I,;~~

: z, -

-,--t-~~

FIGURE 061-E07

0 : 10

Aviationexam Test Prep Edition 2012

FIGURE 061-E08

FIGURE 061-E09

..

Picture Supplements - Explanations

FIGURE 061-Ell

FIGURE 061-El0

International Date Line 200E

------- - -------- --

\

-- ----- - ------ - - -- - ---

OO&W)

400W

0

8VOW

60 W

-----_/ NORTH

FIGURE 061-E12

I

PACIFIC

PAC I OC

OCEAN

WloE

,_

IIbCrKtateM;l,,·

I

HAWNIAII" "

E lO W · "

ISLNIDS

Hc! l 4Md

•

0

80 W

____

~

~~

o

______________2

FIGURE 061-E13

Sun

Sun

00:00

1 :00

12

11

u.s. ..

10

Aviationexam Test Prep Edition 2012

FIGURE 061-E14 21:00 1 21 :00 LMT 15 May 16 May 180 0 EIW

-- ---

"

24:00 15May .," 135°W / / OO:OOLMT / 16 May

-",

18:00 LMT 16 May 135°E

.....

,

I

FIGURE 061-E15 \

I

I

Path of the Mean Sun

I I

North Pole

,

\ \

I

,

090 0 W - -'-- - - -03:00 LMT \ 16 May \

I'

f - - - - - --.,-!:t-- 090 0 E

--1

". ~ I

15:00 LMT 16 May

Lat. 23,5'S

I

\

I

\ \

a __ ),:::J"

,

045°W 06:00LMT 16 May

, .....

.....

"

......

_-

South Pole

045°E 12:00 LMT 16 May

OOooEIW 09:00 LMT (09:00 GMT) 16 May

FIGURE 061-E16 Perihelion

Aphelion ear/yJu/y .

8 FIGURE 061-E17 New actual position

180° EIW Actual position

IRS Entered position

139°46,3E

139°46,3W (139'46 '20"W)

(139 ' 46'20"E)

New IRS position

99°32'40"W

~

<:;j

000° EIW

Picture Supplements - Explanations

FIGURE 061-E18 23,50 " Spin Axis I

Orbital Plane

, ----------------------------~------------------------- -Spring (vernal) Equinox

"

21. March Sun over the Equator

Winter Solstice 21 . December Sun oy er 23,5°S (Tropic of Capricorn)

Summer Solstice 21 . June Sun over 23,5°N (Tropic of Cancer)

Autumn Equinox 23. September Sun over the Equator

FIGURE 061-E19 NP

I

SP

Geocentric Latitude

. Aviationexam Test Prep Ed'Itlon 2012

URE 061-E

'\ I //

\

I

\

I

I

MAGNETIC

/'

,/

\

.-

\ \

/'

\ \

\ \

\

\

\

\

\ \

\ I I I

\ \

I I I I I

I I

I

\ \

/

,,

/ /' /'

URE061-

North ern Hemisp h ere

40

Initial GC Track 0

(2~

URE 061-E

tp A =228

4 RL + 4° cay

Northern H em/sphere .

Rhumb Line Track toA

a

= 224 0

(reciprocal of 044 °) 3°

4;-'"

Rhumb Line TrackA to =044 0 (040° GC + 4° cay

a

4° Initial GC Track A to = 040 0

a

WPT "A"

Rhumb Line Track A toa = 083°

Initial GC Track A to a =080° (083 ° - 3°)

WPT "8" Initial GC Track 8 toA 266° (263° + 3 °)

=

Picture Supplements· Explanations

FIGURE 061-E24

FIGURE 061-E23

MB=RB+MH MB = 345° + 120° MB = 465° • 360° MB = 105° (to the island) MB =285° (from the island)

Great Circle Track departing from WPT2

Great Circle Track arriving at WPT2

=94,33°

=85,67°

Great Circle Track change

at WPT2

= 8,66° (94,33° . 85, 67°)

Variation = 1rW TB MB . 17°W variation

Magnetic

=

North

=

TB to 088° TB from =268°

Magnetic Bearing

WPT1 60"N0300W

Rhumb Line Track = 090°

= 105

Rhumb Line Track = 090°

0

, .wPT3 50"N0100W

WPT2 60"N0200W

. . ...J..........

• Conversion angle (ca) = Y, Ch Long x sin Latitude • Conve rsion angle = Y, ( 10° x sin 60 °) • Conversion angle 4,33°

Island 15° ........... .

.........

.................

=

Relative Bearing = 345°

"ca " will be the same at all waypoints since all have the same Lat and there is the same Ch Long (10 °) between them .

................. ~

Heading 120· T

(360° · 15°)

FIGURE 061-E26

FIGURE 061-E25 65° N

Magnetic

North

Relative Bearing

51° N

=

=345

aOM 105° (Magnetic bearing of the Island from

0

.- ,

Not to scale...

~~~ ~#J# ,."

".",

.::..............

""

" .-6' ~6'

;~

,,/ Great Circle

37° N

................

Rhumb Line

FIGURE 061-E28

FIGURE 061-E27 0 0 EMJ 0

025 W

GN

Magnetic .. GN

I

North

= Grid North

.~

Island

.. MB=RB+MH MB = 330° + 276° MB = 606° • 360° MB = 246° (to the island) MB

=066° (from the island)

Variation = 12°W TB = MB · 12°Wvariation TB to = 234° TB from = 054°

Aviationexam Test Prep Edition 2012

FIGURE 061-E30

FIGURE 061-E29 70· N

80· N

•

70 · N

Island

~.'.'..

Mag. N

R e.ve l a t j'.', Bearing = 030· '."'.,'.,'.

-,.........

-"

Head ing" , 290 0 M .........................

Distance A =480 NM 01 · =60 NM 0480 NM = B· o New position = B2· N 11 OOE

,

090· E

'\

Magnetic Bearing

= 320° (to Island) ''.,

"'"

,

"

...................... ~.....

MB=RB+MH MB = 030° + 290· MB =320· (to the is/and) MB

= 140° (from the is/and)

Variation = 100E TB = MB + 100E variation

Chart converge ncy = 11 OOW o True directio n = Grid direction + 11 0° o The True direction = 154 °G + 11 0° = 264 °T

TB to =330· TB from = 150·

Distance B = 300 NM 01° =60 NM 0300 NM = 5° o Measu re 5° along a meridian and apply this distance to the line you have drawn

FIGURE 061-E31

Final position approx BooN OBooE.

MB=RB+MH MB = 045° + 215· MB = 260· (to the is/and) MB = 080° (from the is/and)

FIGURE 061-E32 Magnetic

Heading ~,311·

,,

,

..... ~....

"

Track .....

301· M

North

M

North

Variation =21·W TB = MB - 21 oW variation

...

TB to = 239° TB from = 059·

"

.

Island Relative ,/

---------------------------\ / ,' Relative Bearing = 045°

/

Bearing,,: :: r _ " _ __

)I:'

.

I

" "

Magnetic Bearing =260° (to Island)

Heading

"

,.

Magnetic Bea ring = 221 °

~ U

Magnetic

215· M

MB=RB+MH

MB =270° + 311· MB =581· - 360· MB = 221· (to the station)

Station

FIGURE 061-E33 Magnetic

FIGURE 061-E34 •

North

Island ~ Heading

~

I

\

30·

:.---./ :

020· T

/

\

I

Magnetic Bearing =350·

.."\

MB

(360· - 30·)

= 170· (from the is/and)

Variation = 25·W TB =MB - 25·W variation TB to =325· TB from = 145·

///'~-- -~~= ,/

MB=RB+MH MB = 330· + 020· MB =350· (to the is/and)

Relative Bearing = 330·

---/

;~1--T---

I

\ Mag, / \ N /

,/ :' ,/ ' • Island

MB=RB+MH MB =300· + 276° MB =576° - 360· MB = 216· (to the is/and) MB =036· (from the is/and) Variation = 100E TB = MB + 10·E variation TB to = 226° TB from = 046·

Picture Supplements - Explanations

FIGURE 061-E36

FIGURE 061-E35

• Heading : 355° T

~

Heading \ 345° M \

•

Island

~'"

,

""""',

30°

I

".. ~ II ...

\

I

...

Relative Bearing 330° (360° - 30°)

I Mag, : N

\,

=

\

'"

\

North

30° \ ~ \\

"' ",

Island

Magnetic

\ \

'"

•

..

Relative Bearing = 330° (360° - 30°)

I \\

I

\

Magnetic Bearing

= 325° Magnetic Bearing

MB MB MB MB MB

= 315°

, '\

MB=RB+MH MB =330 0 + 345 MB = 675 360 MB =315 0 (to the is/and) MB = 135 (from the is/and)

0

0

0

Variation = 15°E TB MB + 15°E variation

0

0

= RB +MH = 330 + 355° = 685°0 - 360 = 325 (to the is/and) = 145 (from the is/and)

=

0

-

TB to = 340° TB from = 160

0

Variation = 200E TB = MB + 20 0 E variation

0

TB to =335° TB from 155°

=

FIGURE 061-E38 .. Heading : 355° M

FIGURE 061-E37 True

North

•

:

Island

l

I

•

r---.... ... : .... Relative I Mag.

.... Bearing

: N I I I

:

=030

....

: :'

I I

: :'

I :' Magnetic Bearing I ..."" = 025° (to Island)

°

MB=RB -& MH MB 030 + 355 0 0 MB =385 - 360 MB 025 (to the is/and) MB 205 (from the is/and)

= = =

0

0

0 0

Variation = 15°E TB = MB + 15°E variation

=

0

TB to 040 TB from = 220

0

FIGURE 061-E40 True

North

FIGURE 061-E39

B = 35° (360 0 - 325°) a 45° (105° - 060°) V = 080° (360° - 280°)

True Bearing = 105° (140° - 35°)

,~>

=

...

~eI,Bearing =325°

" ) ...... " B(35':j····.• " .......... .. " .......... .. HOG " ••••••••• , '. 1400T

=

Distance 1 cos 30 0 x 200 NM Distance 1 = 173,2 NM Distance 2 Distance 2

=sin 30° x 200 NM = 100 NM

"

··r··:·'·······.........

=

True Bearing 060° '" 0 (140 _ 800) Rel,Bearing = 280°

' •••••••

~

: ••••~! 060° T

, HOG 1400T "

••••••••••, 105° T

Aviationexam Test Prep Edition 2012

FIGURE 061-E42

FIGURE 061-E41

__

~

t PointA ";",,n"'_' _\~:::;;L ow 130·E /

/ rQ-<" ~

/

// "l-n,; R:J~

)l. -<..\/'\.. /

,//

iF"~

=

Distance A 3° x 60 NM Distance A = 180 NM

cos 45° = Distance 1 .;. 100 NM sin 45° = Distance 2 + 100 NM

a = 180° - 120° a = 60°

/ /

,

~

Distance 1 = cos 45° x 100 NM Distance 1 = 70,71 NM

/

Point B = ?

,-.

'I

Distance 2 = sin 45° x 100 NM Distance 2 = 70,71 NM

cos a = Distance XY .;. Distance A Distance XY = Distance A.;. cos a Distance XY = 180 NM .;. cos 60° Distance XY = 360 NM

FIGURE 061-E43 x2 = 40 2 + 40 2 X = "(1.600 + 1.600) x = "3.200 x = 56,57 NM

sin a _ Length of opposite side - Length of Hypotenuse sin a = 40 NM .;. 56 ,57 NM sin a = 0,7070 a = 45° Aircraft track =40° + 45° Aircraft track = 85°

FIGURE 061-E44 x2=102+102 X = "( 100 + 100) x = "200 x = 14,14 NM

_ Length of opposite side - Length of Hypotenuse sin a = 10 NM.;. 14,14 NM sin a = 0,7072 a =45° Aircraft track = 40° + 45° Aircraft track = 85°

Picture Supplements - Explanations

FIGURE 061-E4S

FIGURE 061-E46

o :!i:

z

o o

N

II

til

C

+

"

Dist.=?

tan a = Length of opposite side.;. Length of adjacent side

True Heading = 325° Variation = 3°E

tan 1° = ? NM .;. 200 NM ? NM = tan 1° x 200 NM ? = 3,49 NM

Heading Mag. = 325° - 3° Heading Mag. = 322°

FIGURE 061-E47 FIGURE 061-E48

:~;;.1- -- ~-- ------:---------cf),m mm Radio Beacon

a

=45

I I I

_---~

23:00

~~ ~tI -- -- \.aZ'1 g~ °

HOG ... ---255°M

270° Relative

I I I I I I I • I ~ A~ I

22:54

,

, ~,

ro .-

-g

9:\

c:::

. ~,

u"

a = 255° - 180° a 75°

"b~,

=

To calculate "f?," we need to realize that the sum of the internal angles of a triangle is always 180°:

0.;---1

,, ,,

'....--\ , 30° ,, ,

,

o

f?, = 180° - 75° - 30°

B

MN

=75°

FIGURE 061-E49

;'

;' , /'

/' ,

4

270

;' ;'

45'

\

/ "

~'225·

, : :4

I

2

8 (48°30'N / 01S·W)

c

I I I I I I I I

:

Distance O

FIGURE 061-ESO

225° Relative Q) (.)

I::

.:g til

C

dJ >t:. .- _ _ _ _

a'/Q:::n/\: =

a

I

;' ~

;' a =45·

;' ,

/f)

~ :

270· Relative

i """"'~, c'"

' .... , ... L------:::-:------:----'---t.

Distance A

h

___ _

A (4S0N /

01 O°W)

Aviationexam Test Prep Edition 2012

FIGURE 061-E51 line No.

Time 1015

Coursel Track (T) 270

1

050/40

2

1050

180

320/50

3

1125

090

140/60

4

1210

360

315170

5

1245

330

240/30

6

1355

070

020/60

WN

HOG (T)

275 0

VAR - 7E

188

0

+ 5W

095

0

+ 10W

354

0

- 10E

320

0

+ 17W

065

0

+ 11W

HOG (M)

268 0

Positio n FROM TO A B

CASI MACH 210

193

0

C

D

175

105

0

E

F

M 0,82

344

0

G

H

M 0,78

337

0

J

K

150

076

0

L

M

MO,84

FL OAT 180 -20 160 -10 350 -40 310 -35 100 -10 390 -55

,

FIGURE 061-E52

'\

FIGURE 061-E53 55 0 30' N

060 0 15' W

RADIAL 360 0 T

YYRVOR

53 0 30' N {

060 0 15' W

TAS

GS

275 Ids

304 Ids

226 Ids 487 Ids 469 Ids 172 Ids 483 Ids

262 Ids 446 Ids 417 Ids 169 Ids 442 Ids

Dist 300 480 300 600 275 495

Time

ETA

59min

11:14

1hr50min

12:40

40mln

12:05

1hr26min

13:36

1hr38mln

14:23

1hr07mln

15:02

Picture Supplements - Explanations

Standard atmosphere:

TRACK

~ CAS TAS M

~ CAS TAS M

~ CAS TAS M

::J ...... :p

::J ...... :p

::J ...... :p

«

HEADING

«

«

WIND

Airspeed

Airspeed

Airspeed

Inversion layer: ~ CAS

M TAS

::J ...... :p

«

~ CAS M TAS

~ CAS

::J ...... :p

::J ...... :p

«

<4----------

M TAS

« TRACK

Airspeed

Airspeed

Airspeed

<4C Wind A orrection

FIGURE 061-E56

HEADING

------------

~-

- - - -- - -- - -- --

~-

---- - - -

~-

WIND - -- -- - - -- -----

ngle (WCA) 17° RIGHT

- - - - - --


;1- - - - - - - - -- - - --

Aviationexam Test Prep Edition 2012

FIGURE 061-E57 MAGNETIC COMPASS ·ACCELERATION ERRORS

Constant speed · heading East (090°) Compass indicates correctly => 090°

Constant speed · heading West (270°) Compass indicates correctly => 270°

Accelerating - heading East (090°)

Decelerating - heading East (090°)

Compass indicates apparent turn towards the nearer pole - towards the North => eg o060°

Compass indicates apparent turn away from the nearer pole - towards the South => ego 120°

Compass card rotates clockwise.

Compass card rotates anti-clockwise.

Accelerating - heading West (270°)

Decelerating - heading West (270°)

Compass indicates apparent turn towards the nearer pole - towards the North => eg o300°

Compass indicates apparent turn away from the nearer pole - towards the South => eg o240°

Compass card rotates anti-clockwise.

Compass card rotates clockwise.

Picture Supplements - Explanations

FIGURE 061-ES8 DETERMINE THE TAS (CR-3 computer - METHOD 1) This method is suitable only for slower speeds and lower altitudes. Cruise: FL 180 Temp: -30 · C CAS : 150 kts Calculate the True Airspeed - TAS . 1) Rotate the inner (grey) disc so that the CAS of 150 kts (on the outer - white - scale) is aligned with the Pressure Altitude of 18.000 ft (on the inner - grey - scale). 2) Use the movable green cursor to align the curved green (CT 1.0) line with the intersection of the long black curved line and the thin "wiggly" thin line representing the temperature of -30· C. 3) Read the TAS on the inner white scale under the green line => in our case TAS equals approximately 192 kts .

3

DETERMINE THE TAS and Mach number (CR-3 computer - METHOD 2) This method is more precise than the method described above - it is suitable especially for higher speeds and higher altitudes. Cruise: FL290 OAT: -46 · C CAS: 324 kts Calculate the TAS + Mach number.

1

2

1) Rotate the irner (grey) disc so that the CAS of 324 kts (on the outer - white - scale) is aligned with the Pressure Altitude of 29.000 ft (on the inner - grey - scale). 2) Read the Mach number in the "Mach Number" window inside the inner disc => in our case it is approx. Mach 0,83. 3) Refer to the small "Pressure Altitude" window on the inner disc and rotate the inner disc so that the "Mach Index" (a line inside this window with arrows pointing up and down) is aligned with the corresponding temperature - in our case -46·C. 4) Find the value of Mach number (0,83) on the perimeter scale of the inner disc and read the corresponding TAS value on the scale of the outer disc => in our case approximately 486 kts. Mach = 0,83 TAS = 486 kts

continues ...

Aviationexam Test Prep Edition 2012

FIGURE 061-E59 ... continued DETERMINE THE TAS from MACH NUMBER (CR-3 computer) Mach: 0,76 Indicated Temp: -40 °C Calculate the TAS . 1) Use the "Mach Number" window inside the inner disc and set the Mach number pointer to your Mach number => in our case it is Mach 0,76. 2) Use the movable green cursor to align the curved green line (C T 1.0) with the intersection of the long black curved line (spiral) and the thin "wiggly" thin line representing the indicated temperature of -40°C. 3) Read the TAS on the inner white scale under the green line => in our case TAS equals approximately 426 kts. Note: This method uses the indicated that includes the temperature ram-rise. If you need to convert Mach to TAS using an OAT, use the Mach Index method (see previous page): 1) Refer to the small "Pressure Altitude" window on the inner disc and rotate the inner disc so that the "Mach Index" (a line inside this window with arrows pointing up and down) is aligned with the corresponding OUTSIDE AIR temperature (True Air Temperature) - in our case -40°C. 2) Find the value of Mach number (0 ,76) on the perimeter scale of the inner disc and read the corresponding TAS value on the scale of the outer disc => in our case approximately 450 kts.

DETERMINE THE TAS (E6B computer) Cruise: FL 180 OAT: -30 °C CAS: 150 kts Calculate the True Airspeed - TAS . 1) Rotate the inner disc so that the Pressure Altitude (PA) of 18.000 ft is aligned with the value of -30°C (Air Temperature) in the PA window (underneath the Density Altitude window) . 2) On the inner perimeter scale (on the inner disc) find the value of your CAS (150 kts) => read the value of the corresponding TAS on the outer scale (outer disc) => in our case approx . 194 kts.

Picture Supplements - Explanations

FIGURE 061-E60 ALTITUDE DEFINITIONS • Indicated Altitude = the altitude reading of the altimeter, assuming it is correctly set (correct QNH in the reference window). It indicates the approximate height of the aircraft above the Mean Sea Level (MSL) . • Calibrated Altitude = indicated altitude corrected for the instrument, position and installation errors. • True Altitude = calibrated altitude corrected for non-standsrd atmospheric conditions. It is the actual height of the aircraft above the sea level. • Pressure Altitude = reading of the altimeter with 1013 mb (29 .92 inHg) set in the reference window. Pressure altitude is an important factor in determining aircraft performance . • Density Altitude = pressure altitude corrected for non-standard temperature . Aircraft performance is greatly affected by Density altitude. DETERMINE THE DENSITY ALTITUDE (CR-3 computer)" Pressure Altitude: 29.000 ft OAT: _55° C Calculate the density altitude .

<'.

1) Place the Pressure Altitude (29.000 ft) in the "Pressure Alt. " window against the reported True Air Temperature (-55°C). 2) In the small "Density Alt." window just above read the corresponding Density Altitude opposite the arrow at the top of the window (27 .500 ft) .

DETERMINE THE TRUE ALTITUDE (CR-3 computer) Pressure altitude: FL 150 overhead an airport Airport elevation: 720 ft Airport QNH: 1003 hPa OAT at FL 150: _5 °C Assuming 1 hPa = 27 ft, what is the true altitude? First of all let's obtain the Calibrated Altitude (CA). From the definition above we can see that the CA is arr indicated altitude corrected for various instrument errors. Let's disregard the instrument errors and assume that Calibrated = Indicated . Indicated Alt. reading of the altimeter with QNH set. Pressure Alt. reading of the altimeter with 1013 set as the reference. With QNH of 1003 hPa the altimeter would read a lower altitude than with 1013 => if each 1 hPa equals 27 ft then 10 hPa (1013 -1003) will represent 270 ft. Our CA = 15.000 ft - 270 ft = 14.730 ft.

=

=

1) In the "True All" window place the Pressure altitude (15.000 ft) opposite the True Air Temp·(-5°C). 2) Calculate the Calibrated Altitude (CA) above the ground (above airport elevation) => if our CA is 14.730 ft and the airport elevation is 720 ft, then our "CA above GND" will be 14.010 ft. 3) Find the value of "CA above GND" (14.010 ft) on the perimeter scale of the inner disc (inner scale) and read the corresponding True Altitude above ground (TA above GND) on -the scale of the outer disc (outer scale) => in our case approx. 14.500 ft. 4) Add the value of airport (ground) elevation to your "TA above GND" to get the True Altitude above MSL => 720 ft + 14.500 ft = TA is 15.220 ft. 5) If the ground elevation is not known , find the value of the Calibrated Altitude above MSL on the perimeter scale of the inner disc (inner scale) and read the True Altitude above MSL on the scale of the outer disc (outer scale). This method is slightly less precise than using the CA above GND and then adding the ground elevation . Practically the same methods are used on the ESB navigation computer.

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Aviationexam Test Prep Edition 2012

FIGURE 061-E61 DETERMINE THE WIND DIRECTION AND VELOCITY (CR-3 computer) True track: 095° TAS : 160 kts True heading : 08r GS : 130 kts Calculate the wind direction and velocity. 1) Set the TAS index against the TAS given in the question (160 kts). 2) Using the inner disc (green disc), set the value of True Course given in the question (095°) against the TC pointer of the middle disc (gray disc). "\ 3) If the crab angle is relatively large, determine the Effecti~e TAS by reading its value on the outer disc (white) against the crab angle value on the short black scale . In our case the difference between the TAS and the effective TAS would be ' negligible (about 1 kt) . If the crab angle is not given by the question, calculate it => find the difference between the True Heading and the True Course. In our case it would be 095° - 08r = 8°. Since the True course (track) is greater than the True Heading it means that we are experiencing a left crosswind (blowing us to the right off course) . 4) Determine the difference between the actual TAS (or the effective TAS) and the Groundspeed (GS) to obtain the value of headwind or a tailwind component. In our case it would be 160 kts TAS - 130 kts GS = 30 kts . Since the GS is slower than TAS it means we are experiencing a headwind component (of 30 kts) . 5) On the inner (green) disc draw a line representing a headwind of 30 kts (blue line). As stated above we know that we are experiencing a wind from the left, therefore we will be draing the line in the appropriate quadrant of the inner (green) disc. 6) To determine the crosswind component we have to refer to the crosswind scale of the middle disc (grey) . Remember that we have determined our crab angle (wind correction angle - WCA) = 8°. Find the value of 8° on the scale and read the corresponding corsswind value on the outer disc (white) - you will see the value of a little bit over 22 kts. 7) Now draw a line on the inner disc representing a value of 22 kts of crosswind (magenta line) . 8) Find the place where your headwind and crosswind lines intersect (yellow circle) => this is your final wind direction and velocity => in our case it is approximately 058° at 38 kts. Note: you read the wind direction on the scale of the inner disc (green) on its perimeter and you read the wind speed from the "circles" of the inner disc (green) - each circle equals to 10 kts (do not forget to count the inner-most one) .

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7

Picture Supplements· Explanations

FIGURE 061-E62 DETERMINE THE WIND DIRECTION AND VELOCITY (ES8 computer) True track: 095° TAS : 160 kts True heading: 087° GS: 130 kts Calculate the wind direction and velocity. 1) Set the True Index of the outer scale against the True Course (track) given in the question (095°). 2) Move the slide until the center of the slide (grommet) touches the line representing the Groundspeed of 130 kts . , " 3) Find the crab angle (Wind Correction Angle) if not provided by the questions. It is the difference between your True Heading and the True Track - in our case it is 8° (095° - OSr). Since our heading is less than our track then it means we are experiencing a crosswind from the left - in other words we have to crab to the left into the wind to maintain our desired track = wind from the left. 4) On the center vertical scale find the value representing your TAS (160 kts) and ' identify a horizontal line that intersects this TAS value. 5) On the horizontal line that you have identified above in step 4) draw a dot that represents your crab angle of 8° to the left => you will draw the dot on the left side . 6) Now rotate the inner disc clockwise so that the dot you have drawn aligns itself with the center vertical scale. 7) Read the resulting wind speed by counting the lines between the center of the inner disc (grommet) and the dot you have drawn - each think line represents the speed of 10 kts => in our case it is 36 kts. 8) Read the wind direction from the perimeter scale of the inner disc under the TRUE INDEX of the outer disc - in our case it is 5r. The final result is therefore wind 05r at 36 kts .

Aviationexam Test Prep Edition 2012

FIGURE 061-E63 DETERMINE THE DRIFT I WIND CORRECTION ANGLE AND GROUND SPEED (CR-3 computer) TAS: 200 kts Track: 073° (T) Wind: 210° (T) 120 kts Calculate the heading and the ground speed . 1) Using the middle disc align the "TAS" index against the TAS (200 kts). 2) Find th e tra ck on the scale on the pe rimeter of the inner (green) disc and rotate the disc so that the tra ck (073°) is aligned with the uTe" index. '\

3) Identify the wind and draw a dot on the inner (g ree n) disc that re presents the wind direction and speed . Wind direction is given by the perimeter scale of the inner (green) disc a~d the wind speed is given by the number of concentric circles (each circle represents 10 kts of wind speed). 4) Draw a vertical line (magenta) from your dot (parallel to the black lines) to intersect the scale running horizontally from the left to the right of the middle disc => read the value of the crosswind on this horizontal scale => in our case it is approx. 14 kts from the right. 5) Draw a horizontal line (bl ue) from your dot (parallel to the black lines) to intersect the scale running vertically from the bottom to the top of the middle disc => read the value of the headwind or tailwind on this vertical scale => in our case it is approx. 15 kts tailwind. 6) Since we know the crosswind component (14 kts) we can determine the crab angle by reading the value (in 0) from the middle disc against the crosswind value (14 kts) on the outer disc => in our case approx. 4°. Since the wind is from the right, our head ing (into the wind) will be greater than the track => heading will be 073° + 4° = 078° (T) . 7) For the Groundspeed (GS) calculation we need to use the value of the headwind / tailwind component that we have determined in step 5) => our ta ilwind component = approx. 15 kts , hence our GS will be TAS + tailwind component => 200 kts + 15 kts = 215 kts.

5

3 4

Picture Supplements - Explanations

FIGURE 061-E64 DETERMINE THE DRIFT I WIND CORRECTION ANGLE AND GROUND SPEED (E68 computer) TAS : 200 kts Track : 073° (T) Wind : 210° (T) 120 kts Calculate the heading and the ground speed . 1) Set the wind direction (210°) under the "TRUE INDEX". 2) Place the grommet (center of the inner disc) over the value of 100 on the vertical scale . This is just to get a convenient starting point for the wind speed. Now count the number of lines upwards from your staryiong point (100) - each line represents a wind speed of 2 kts and draw a dot that rePfesents your wind speed (20 kts => dot will be at 120). 3) Rotate the inner disc so that the value of your true co'urse (track) is aligned under the "TRUE INDEX". 4) Move the slide until the dot you have drawn in step 2) falls on a line (running from the left to right) representing your TAS (in our case 200 kts). 5) Count the number of vertical lines between the center (grommet) and your dot - each vertical line represents a crab angle of 1° - in our case the crab angle is 4° (4 vertical lines) to the right. 6) Read the value of your Ground Speed from the vertical scale at the point where the grommer (center of the inner disc) touches the vertical scale - in our case approx. 214 - 215 kts (each line represents 2 kts).

Alternatively we can use calculation: Crosswind component (XWC): XWC = sin (wind angle) x wind speed XWC = sin (210' - 073') x 20 kts XWC = 13,6 kts (from the right) Speed factor (SF): SF = TAS +60 SF =200 + 60 =3,33333 Wind Correction Angle (WCA): WCA= XWC -;- SF WCA = 13,6 + 3,33333 WCA= 4' Head/Tail wind component (HWC): HWC = cos (wind angle) x wind speed HWC = cos (210' - 073') x 20 kts HWC = 14,6 kts (in our case tailwind) Effective TAS (TAS eff): TAS eft = TAS x cos WCA TAS eft = 200 x 0,9975 TAS eft = 199,5 kts Ground speed (G5): GS = TAS eft + tailwind (or - headwind) compo GS = 199,5 kts + 15 kts tailwind GS = 214,5 kts

6 5

Aviationexam Test Prep Edition 2012

FIGURE 061-E65 DETERMINE THE DRIFT AND GROUND SPEED (EG8 computer) TAS: 190 kts True HOG: 085° WN: 110° (T) / 50kts Calculate the drift angle and Ground Speed . 1) Set the wind direction (110°) under the "TRUE INDEX". 2) Plot the wind velocity - draw a vertical line downward using the speed arcs - the length of the line corresponds to the wind velocity (50 kts). Also draw the arrow representing the wind direction (arrow pointing down). " 3) Move the slide until the grommet (center of the disc) lie q_ ~bove the TAS arc (190 kts).

4) Rotate the inner disc so that the value of your heading (085°) is aligned under the "TRUE INDEX". 5) Read the actual drift angle at the end of your wind arrow (8° Left). If necessary, apply this value to your heading to find the track. 6) Read the value of your Ground Speed from the vertical scale at the point where your wind arrow ends (146 kts) .

The result is a drift of 8° Left and GS of 146 kts.

6

Picture Supplements - Explanations

FIGURE 061-E66 DETERMINE THE WIND COMPONENTS FOR TAKEOFF AND LANDING Runway direction: 230 0 (M ) Surface WN: 280/40 kts Calculate the effective headwind and crosswind components.

1) Using the inner (green) disc set the runway heading over the "TC " index. 2) Identify the wind and draw a dot on the inner (green) disc that represents the wind direction. Wind direction is given by the perimeter scale of the inner (green) disc and the wind speed is given by the number of concentric circles (each circle represents 10 kts of wind speed) . 3) Draw a vertical line (magenta) from your dot to intersect the scale running horizontally from the left to the right of the middle disc => read the value of the crosswind on this horizontal scale => in our case it is approx. 30 kts from the right. 4) Draw a horizontal line (blue) from your dot to intersect the scale running vertically from the bottom to the top of the middle disc => read the value of the headwind or tailwind on this vertical scale => in our case it is approx. 25 kts headwind. Resulting wind components are approximately 30 kts right crosswind and approx. 25 kts headwind.

4

Note: As an alternative solution to finding the wind components you do not have to use the navigation computer and you can use the calculation method. In this case we need to determine the angle between the runway heading (230°) and the wind direction (280°) => 50° .

Headwind = cos (wind angle) x wind speed Crosswind = sin (wind angle) x wind speed

=> in our case cos 50° x 40 kts => approx. 25,7 kts headwind => in our case sin 50° x 40 kts => approx. 30,6 kts crosswind

Aviationexam Test Prep Edition 2012

FIGURE 061-E67 DETERMINE THE WIND - MULTI DRIFT METHOD (CR-3 computer) The following observations have been made while TAS was 187 kts : • With MH 015° the Drift = YOR • With MH 075° the Drift = 8°R • With MH 177° the Drift = 3°L What is the wind direction and velocity? 1) Rotate the middle disc to align the "TAS" point with the value of 187 kts on the scale of the outer disc. 2) Calculate the Track for the MH of 015° and the drift of yoR => Track will be 022° (if drifting to the dight it means we have the wind from the left, pushing us to the right) . Rotate the inner disc (green) so that the "TC" pointer is aligned with the value of 022° on the perimeter scale of the inner disc. 3) Find a value of yo (drift) on the inside scale of the middle disc and read a corresponding crosswind value of approx. 23 kts on the outside scale. 4) Draw a line representing a left crosswind of 23 kts . 5) Calculate the Track for the MH of 075° and the drift of 8°R => Track will be 083 °. Rotate the inner disc (green) so that the "TC" pointer is aligned with the value of 083° on the perimeter scale of the inner disc.

6) Find a value of 8° (drift) on the inside scale of the middle disc and read a corresponding crosswind value of approx. 26 kts on the outside scale. 7) Draw a line representing a left crosswind of 26 kts . 8) Calculate the Track for the M H of 177° and the d rift of 3° L => Track will be 174°. Rotate the inner disc (green) so that the "TC" pointer is aligned with the value of 174° on the perimeter scale of the inner disc.

9) Find a value of 3° (drift) on the inside scale of the middle disc and read a corresponding crosswind value of approx. 10 kts on the outside scale . 10) Draw a line representing a right crosswind of 10 kts . 11) Notice that your 3 lines almost intersect at one spot => this spot represents your wind => approximately 333° at 29 kts .

11

Picture Supplements - Explanations

FIGURE 061-E68 CONVERSION OF UNITS - DISTANCE , VOLUME, MASS (CR-3 computer) CR-3 Navigation computer can be used fo r converting specific values in one unit of measure into a corresponding value in other units of measure. Notice the small arrows around the perimeter of both discs along with an identification of a unit next to the arrow - ego Nautical miles between the numbers 60 and 70 and Statute miles between the numbers 70 and 80. To convert from one unit to another, simply find the arrow for the first unit of measure and rotate the inside disc to align it with the arrow of the second unit of measure on the other disc. Read the corresponding values opposite each other on the two scales .

Distance: 48 NM Convert this distance to Statute Miles (SM). 1) Find the arrow "NAUTICAL" on the scale of the outer disc and rotate the inner disc to align your NM arrow with an arrow on the inner disc that represents "STATUTE" miles. 2) On the outer disc find the value 48 (NM) and read the corresponding value in Statute miles on the inner disc => 55 SM.

CONVERSION OF UNITS - DISTANCE, VOLUME, MASS (ES8 computer) The E6B navigation computer uses the same method for solving unit conversion problems - simply align the arrows of the two units against each other and read the results from the scales on the perimeter of the circles .

Fuel amount: 85 Imperial Gallons Calculate the fuel amount in US Gallons. 1) Find the arrow "US GAL" on the scale of the outer disc and rotate the inner disc to align your "US GAL" arrow with an arrow on the inner disc that represents "IMP GAL" (Imperial Gallons). 2) On the inner disc find the value 85 (IMP GAL) and read the corresponding value in US Gallons on the outer disc => 102 US GALLONS.

Aviationexam Test Prep Edition 2012

FIGURE 061-E69 ICAO Chart symbols (ICAO Annex 4)

TOPOGRAPHY

II

I

I Con ours

I

I Gravel

1 12

!

Highest elevation on chart

2 : Approximate contours

I

I

I

9

Levee or esker

I

i 3 I Relief shown by hachures

13

I

I 4 I Bluff, cliff or escarpment

I

,

Unusual land features appropriately labelled 5

Lava flow

6

Sand dunes

7

Sand area

r:,:;-----;

! ~

I

Sm,1I VokMloes

r , I

14

Spot elevation (of doubtful accuracy)

15

Coniferous trees

16

Other trees

17

Palms

___ .... - - - ... __ J

.6397

I

,8975

6370 ±

Rock

10

' II

n

Active volcano

11

Spot elevation

Mountain pass

18 [ Areas not surveyed for contour information or relief data incomplete

7' 7"'; 7' 'C 'f 'f Caution

HYDROGRAPHY 19

Shore line (reliable)

Reservoir 30

landmark value.

Shore line (unreliable)

21

Tidal flats

22

Coral reefs and ledges

31

Large river (perennial)

24

Small river (perennial)

25

26

Falls

33

Salt lake

34

Salt pans (evaporator)

35

Swamp

36

Rice field

..... ~

f--------i

•

~ ~

Wash

41

Shoals

42

Glaciers and ice caps

43

Danger line (2 m or one fathom line)

44

Charted isolated rock

45

Rock awash

46

Unusual water features appropriately labelled

..u. -- ..u. -_ ..u.

'"

.O!:

37 Canal

40

~~~~

r~~=~=~~

.l!l

iii

lil

~

28

Dry lake bed

<

Rivers and streams (unsurveyed)

Rapids

29

~

Lakes (non-perennial)

Rivers and streams (non-perennial)

27

39 Lakes (perennial)

~

32 23

• Reservoir

Abandoned canal Note. - Dry canal having

I-..u. - ---_- -..u. -I - _ ..u. -

perennial

•

intermittent

o

Spring, well or water hole

+

I

Picture Supplements' Explanations

FIGURE 061-E70 BUILT-UP AREAS

r

Town

49

Village

MISCELLANEOUS (Cont.)

I

lL:Jo

47 1 City or large town

48

HIGHWAYS AND ROADS 1

57 Dual highway

Pipeline

~

I

rill Eil

I 70

58 Primary road

0

Oil or gas field

I

71

Tank farms

72

Nuclear power station

• • •.

• •

59 Secondary road

r

J

1 60

0

I

1

•••••

61

l 62

Road bridge

1

,-.

I

"

!--3--r-

Road tu nnel

73 Coast guard station

...

74

@

r---\

~

1

yo

Trail

I

1 50 Buildings

1

69

Lookout tower

I ~

75

Mine

~

76

Forest ranger station

:!

77

Race track or stadium

78

Ruins

79

Fort

I

RAILROADS r 51

52

Railroad (single track)

I

Railroad (two or more tracks)

MISCELLANEOUS

I

I

I

II

II

I

1- ---

Boundaries (international)

63

i

----

64 Outer boundaries 53

54 l-

55

Railroad (under construction)

I~I

Railroad bridge

Fence

66

Telegraph or telephone line (when a landmark)

-67

Dam

pD

I -

68

Ferry

j -o-i

-

Railroad station

I

--- -

I

--

--

-

-

-

-

l:l

80

Church

6

81

Mosque

6

82

Pagoda

-T- T-

~--fI-

Railroad tunnel

0 00

X- X - X

65

-

f-

t-

56

- I - --t-

©

.

83 Temple

6" .rlt

'---~

AERODROMES ,84

Civil

Land

ol

85

Civil

Water

0

.-

88 Joint civil and military

Land

()

89 Joint civil and military

Water

~

-

-

-

86

Military

Land

0

90

Emergency aerodrome or aerodrome with no facilities

0

87

Military

Water

@) I

91

Abandoned or closed aerodrome

®

-

95

Note.- Where required by the function of the chart, the runway pattern of the aerodrome may be shown in lieu of the aerodrome symbol, for exa mple:

92

93

Sheltered anchorage

Aerodrome for use on charts on which aerodrome classification is not required e.g. Enroute Charts Heliport

94 Note.-

Aerodrome for the exclusive use of helicopters

®

1

Aviationexam Test Prep Edition 2012

FIGURE 061-E71 AERODROMES (Cont.) Name of aerodrome Elevation given in the units of measurement (metres or feet) selected for use on the chart

Length of longest runway in hundreds of metres or feet (whichever unit is selected for use on the chart)

-----------

96

LIVINGSTONE

----------- 357

~

Minimum lighting - obstacles, boundary or runway lights and lighted wind indicator or landing direction indicator

-----------

L H 95 -----------

"'" ' - - - - - - - Runway hard surfaced, normally all weather '\

Note: a dash (-) is to be inserted where L or H do not apply.

,-.

AERODROME SYMBOLS FOR APPROACH CHARTS

97

Aerodromes affecting the traffic pattem on the aerodrome on which the procedure is based

:J;>--A.

98

The aerodrome on which the procedure is based

RADIO NAVIGATION AIDS* Ele<:tronic

99

Basic radio navigation aid symbol Note.-

0

This symbol may be used with or without a box to enclose the data

107

Collocated VOR and TACAN radio navigation aids

Ele<:tronic

100

Non-directional radio beacon

PLAN VIEW

B

~ ~

-" '<""'''~'''''''~y

NDB

.

Electronic

101

102

VHF omnidirectional radio range

0

0

VOR

Distance measuring equipment

®

VJ

VORTAC

108

Instrument landing system

il

ILS BACK COURSE

[J

DME

{

FRONT COURSE

-

PROFILE

103

104

Collocated VOR and DME radio navigation aids

DME distance

@

D

VORIDME

Electronic

Distance in kilometres (nautical miles) to DME - - - 15 km Identification of - - - KAV radio navigation aid Radial bearing from, and identifICation of, VOR

105

VOR radial

106

UHF tactical air navigation aid

110

Compass rose To be orientated on the chart in acoordance with the alignment of the station (normally Magnetic North)

109

\-J

0

Elliptical

~

Bone Shape

~

Rad io marker beacon

~J1~O_~AX

TACAN

"1

GLIDE PATH

Note. - Marker beacon may be shown

Compass rose to be used as appropriate in oombination with the following symbols:

8~

Note: additional points of compass may be added as required. * Note: guidance material on the presentation of radio navigation aid data is given in the Aeronautical Chart Manual (ICAO Doc 8697).

by outline, or stipple, or both .

VOR

0

VORIDME

D

TACAN

\-J

VORTAC

VJ

Picture Supplements - Explanations

FIGURE 061-E72 AIR TRAFFIC SERVICES 111

Flight information region

FIR

112

Aerodrome traffic zone

ATZ

I

I

-

.. .. .. .. ..... .... .. ..

-- --

compulsory, without radio communication requirement

.... ® .. .. . ... ® .. ..

recommended

.. ....... ..

compulsory with radio communication requirement

I

119

--- -

Visual flight path

Q)

11 3

Control area Airway Controlled route

CTA AWl

> ~ c

Reporting point

$ :;;:

--JI.,\\r--

Compulsory

.6-

On request

f::,

REP

._ ._ ._ .-

Uncontrolled route

-----wv--

~

Scale-break (on ATS route)

Ii I111I1111111

121 114

>

Q)

120

$ :;;:

Change-over point

I

COP

122 115

Advisory airspace

ADA

----

116

Control zone

CTR

--------

117

Air defence identification zone

123

ATS/MET reporting point

ADIZ

................ .................. .................

ADIZ

---------

36

To be superimposed on the appropriate route symbol at right angles to the route

124

Waypoint WPT

Compulsory

[iJ

On request

~

MRP

Flyover WPT (also used for start point and end point of a controlled turn)

©

Fly-by WPT



.~

118

Advisory route

ADR

gi

$ 111111 111111111111111 :;;:

------

125

Final approach fix

17000 10000 7000 --5000

FL 220

"Mandatory" altitude/flight level

-3000

-FL30

"Recommended" procedure altitudelflight level

5000

FL 50

Altitudelflight level "window"

"At or above" altitudelflight level "At or below" altitudelflight level 126

~

FAF

Altitudes/flight levels

"Expected" altitude Note: for use only on SID and STAR charts . Not intended for depiction of minimum obstacle clearance altitude.

--

10000 FL 70 --FL 50

--

Expect 5000 Expect FL 50

Aviationexam Test Prep Edition 2012

FIGURE 061-E73 AIRSPACE CLASSIFICATIONS

......

-

;1 Aeronautical data in abbreviated form to be used in association with airspace classification symbols:

..:. .

~

-

-

I'M

127

Airspace classifications

.

[ I

/' Name or call sign

/'

-

1m

I

TMA DONLON 11 9 .1 [I 20 0m AGL - FL 245

.-..... II II

Type

/'

-----

Radio frequency( ies)

Airspace " Vertical classification limits

g; 128 ~ 2 :;;:

uADONLONJ FL 245 200m AGL

I

I

I

119.1

AIRSPACE RESTRICTIONS 129

116~~ $#/W:lJ

Restricted airspace (prohibited. restricted or danger area)

Isr<~

Common boundary of two areas

:S::~~ ~

Note:- The angle and density of rulings may be va ri ed according to scale and the size, shape and orientation of the area ,

130

I~.-~

International boundary closed to passage of aircraft except through air corridor

OBSTACLES 131

A

Obstacle -

132

Lighted obstacle

133

Group obstacles

134

Lighted group obstacles

135

1

Exceptionally high obstacle (optional symbol)

-, I,

l\ M

, I I ...

M

136

1

Exceptionally high obstacle · lighted (optional symbol) Note:- For obstacles having a height of the order of 300 m (1 000 ft) above terrain.

52

A

137

Elevation of top (itaIiCS)- Height above specified datum ( 15; - - -(upright type in parentheses)

MISCELLANEOUS Prominent transmission line

Isogonic line or isogonal

Ocean station vessel (normal position)

- - 3 ° E- -

VISUAL AIDS Note 1-:.. Marine alternating lights are red and white unless otherwise indicated. Marine lights are white unless colours are stated.

141

Marine light Note 2:- Characteristics are to be indicated as follows:

142

Aeronautical ground light

All B F

Alternating Blue Fixed

r*T*

FI Flashing G Green Gp Group

143

1

Lightship

I O~c

SEC

Occulting Red Sector

I

Second sec Unwatched (U) W White

4

Picture Supplements - Explanations

FIGURE 061-E74

-

SYMBOLS FOR AERODROME/HELIPORT CHARTS 144

Hard surface runway

r' :':'~I

145

Unpaved runway

146

Stopway

147

Taxiways and parking areas

148

................................................. ··············.....·.... ··· .. "' .... "'··" ...·····r...·

Pierced steel plank or steel mesh runway

.:. ::/.

-';~'-!; .'., ~~':: '.:- ".~-

SWY

!,153

I;;~ 115~j @

Helicopter alighting area on an aerodrome

1I1111l11mllil1I11Il1

................................"...........................................................................····f·····

I·······..·· .. ··.....

Point light

o Obstacle light Landing

directi~~ i~~i:~~~r;lighted)

1-1-~~: Lan~in~~ir~~ti~indicator

(unlighted)

T

t.....

149

Aerodrome reference point

4-

ARP

150

VOR check-point

~

151

Runway visual range (RVR) observation site

C>

1

157

•

Stop bar

t.....

Pattern A

Runway-holding position

Pattern B

=

SYMBOLS FOR AERODROME OBSTACLE CHARTS - TYPE A, BAND C Plan 159 160 161

Tree or shrub ole, tower, spire, antenna, etc. Building or large structure

162

Railroad

163

Transmission line or overhead cable

* 0

• -T-T-

Plan

Profile Identification number

~

Terrain penetrating obstacle plane

Escarpment Stopway

SWY

Clearway

CWY

<) """""'" ]------: ------:1-------------: -------------..1

Aviationexam Test Prep Edition 2012

FIGURE 061-E75 ADDITIONAL SYMBOLS FOR USE ON PAPER AND ELECTRONIC CHARTS PLAN VIEW

Minimum sector altitude 168

Note,- This symbol may be modified to refteet

MSA

particular sector shapes.

IF

COMNG

I ~-~ 169

Terminal arrival altitude Note,- This symbol may be modified to refteet particular TAA shapes.

170

Holding pattern

171

Missed approach track

TAA

--265 0

o ~

'"

-----~ PROFILE

172

Runway

173

Radio navigation aid (type of aid and its use in the procedure to be annotated on top of the symbol)

174

Radio marker beacon (type of beacon to be annotated on top of the symbol)

175

Collocated radio navigation aid and marker beacon (type of aid to be annotated on top of the symbol)

176

DMEfix (distance from DME and the fix use in the procedure to be annotated on top of the symbol)

177

Collocated DME fix and marker beacon (distance from DME and the type of beacon to be annotated on top of the symbol)

It

V W

W

Picture Supplements - Explanations

Ii [HI);) I'@liN ICAO chart symbols used in JAA examinations

OBSTACLES

AIR TRAFFIC SERVICES 1 - Flight Information Region (FIR)

9 - Obstacle

2 - Airway (AWY) Control Area (CTA) Controlled Route

10 - Lighted obstacle

M

11 - Group obstacles

3 - Control Zone (CTR)

M

12 - Lighted group obstacles

4 - Uncontrolled Route

1

13 - Exceptionally high obstacle

5 - Advisory Airspace

1

6 - Reporting Point - non compulsory

14 - Lighted exceptionally high obstacle (height of 300 m / 1. 000 ft above terrain)

7 - Reporting Point - compulsory VISUAL AIDS 8 - Flyover Waypoint

'*

15 - Aeronautical Ground Light

c:b

16 - Lightship

NAVAIDS

I8l

1 - VOR I DME

o

2-DME

o

3-VOR

6 - TACAN

4-NDB

7 -VORTAC

o

5 - Basic, non specified navigation aid

Aviationexam Test Prep Edition 2012

FIGURE 061-E77 DIRECTIONS Direction may be measured in various ways. In all cases the direction To or From a position will be measured with reference to a particular datum , usually North. All directions are measured clockwise in degrees from 0° - 360°. There are 4 main types of direction referred to in Navigation: • Degrees Compass (OC) • Degrees Relative (OR or °Rel.)

• Degrees True (OT) • Degrees Magnetic (OM)

Magnetic North

Compass Direction

Direction 080 Rei

Radio Station

0

Direction 072° M Direction 140° T

West. variation

~

East. variation

t

Direction 225° C North Pole • = Meridians Meridians indicate the direction of the North Pole

True Direction True direction is measured with respect to True North (the direction of the Geographic North Pole from a place) . The direction ofTrue North is indicated by the meridian at the actual place where the measurement is made since meridians always run in a North/South direction. Usually navigational maps and charts are based on True directions. South Pole

Magnetic Direction The Earth may be thought of as a magnet producing an irregular magnetic field with North and South poles. These magnetic poles are in fact near to the geographic or True poles , but are not necessarily directly opposite each other. The magnetic poles move continuously, rotating in an easterly direction around the True poles , completing 360° (a full rotation) each 960 years . When a magnetised compass needle is freely suspended in the magnetic field of the Earth , it will align itself along the magnetic fie ld. The horizontal component of the magnetic field will ind icate the direction of the Magnetic Poles at a place . In doing so, the needle al igns itself along the direction of the magnetic meridian at that point. A magnetic meridian gives the direction in the horizontal plane of Magnetic North - the direction of the directive force (H). Definition : Magnetic direction is the direction of the magnetic meridian at a particular point. The magnetic meridian will point to the North and South magnetic poles . A magnetic meridian is the direction of the magnetic North Pole when measu red from a point. • At A the magnetic compass of an aircraft will point to the west of True North => variation is therefore West. • At B the magnetic compass of an aircraft will point in the same direction as True North => variation is zero. • At C the magnetic compass of an aircraft will point to the east of True North => variation is East.

Compass Direction Any magnetic or ferrous material on or near an aircraft compass will affect the local magnetic field and may move (or deviate) the compass needle away from the actual magnetic north. This may mean that the compass needle does not show magnetic north accurately. Electrical circuits and radio equipment in the aircraft also tend to cause errors in the compass readings . When a compass is affected by a local magnetic field (due to ferrous material or electrical equipment) it is pointing to a Compass North. The angle between the actual Magnetic North and the Compass North is called a "Deviation". It is a correction to be applied to a Compass heading to obtain the actual Magnetic heading. • When the compass needle points to the west of magnetic north, the deviation is "West" . • When the compass needle points to the east of the magnetic north, deviation is "East".

continues ...

Picture Supplements - Explanations

FIGURE 061-E78 ... continued Relative Direction - Relative Bearings

Tru e

North

A relative bearing is a bearing measured with respect to the nose of the aircraft, or to the aircraft heading. While the actual geographical direction of a bearing (in ' True) does not change when an aircraft changes heading , the presentation of the bearing on certain equipment, such as the ADF (Automatic Direction Finder) will change , relative to the nose of the aircraft.

Aircraft Heading 11 040' T

,,

True

Aircraft

North

Position

Relative bearings are converted into true bearings by adding true heading (and converting into the reciprocal if needed). Example: An aircraft is heading 040 ' T and obtains an ADF bearing of a radio station of 080' relative . Calculate the True bearing of the aircraft FROM the radio station. • Relative bearing =080' R (station position relative to the aircraft nose position) • True heading =040'T • True bearing =RB + TH =80' + 40' =120' T - This is the bearing of the radio station from the aircraft, which has to be converted into the bearing of the aircraft from the station . • True bearing of the station from the aircraft = 120'T Reciprocal = 120' + 180' =300' • True bearing of the aircraft FROM the station = 300'T

Reciprocals The reciprocal of a direction is the reverse (oppos ite) direction = with a difference of 180' . For example the reciprocal of 200' is 020', the reciprocal of 360' is 180', the reciprocal of 349 is 169', etc. A quick mental calculation method of finding a 0 reciprocal is to add 200' and then subtract 20' for numbers less than 180. For numbers over 180, add 20 and subtract 200' . 0

Examp les:

Calculate the reciprocal of 093': Calculate the reciprocal of 214 ' :

=293' - 20' =reciprocal =273' . =234' - 200' =reciprocal =034'.

093' + 200'

214' + 20'

Reciprocals are always in use in aviation , particularly in instrument flying . Pilots ought to be able to calculate them almost instantaneously.

Q-codes used in Navigation • QDM = Magnetic bearing TO a station • QDR = Magnetic bearing FROM a station

• QUJ = True bearing TO a station • QTE = True bearing FROM a station True

Magnetic

North

North

Aircraft Heading

Aircraft Heading

,,11

,,11

True

Magnetic

North

North

QDR: Magnetic Bearing FROM station

QTE:

True Bearing FROM station

Aviationexam Test Prep Edition 2012

FIGURE 061-E79 DIRECTION AND POSITION ON EARTH - LATITUDE AND LONGITUDE

It is generally known of course that the Earth rotates around an axis joining the North (NP) and South Poles (SP) - the Polar axis. The North and South Poles provide the datum for directional measurement on a plain sphere and on the surface of the Earth . The direction of rotation is anti-clockwise from West to East viewed from above the North Pole , although it appears to an observer on the earth as if the Sun rises in the East and moves around the Earth to set in the West. The North/South axis (polar axis) is the basis for direction on the earth and is used in describing earth pos ition. The four cardinal directional points are North , South , East and West. All directions are measured with respect to True North (also known as Due North). The units for direction are degrees true (OT) moving clockwise from True North from 0° through 360°. Direction in aviation is by convention always 0 0 expressed as a 3-fig ure group, e.g. East = 090 T, South = 180 T, 0 0 North = 360 /000 T etc ... A direction of 293°T would be measured as 293° from True North in a clockwise direction. Directions should always be labelled with reference to the type of directional system used (True, Magnetic, Compass , Relative etc) , e.g. 299°T (True) , 183°M (Magnetic) , 023°C (Compass) , 195°R (Re lative) etc. These directions (headings , bearings , tracks etc) should not be abbreviated to single or 2-fig ure groups as ambiguity may then result. An exception to this rule is made when referring to runway directions which are rounded down or up to the nearest 10° and are expressed as 2-figure groups. A runway direction of 197"M would be referred to as runway 20. Similarly, a runway direction of 023°M would be runway 02 . Runway directions are always given in degrees magnetic ("M) . The direction of Magnetic North at a particular place may change over the yea rs , because of the moveme nt of the Magnetic Poles . This means that the magnetic direction of a runway may change over a period of time, even though the actual true direction remains unaltered . Meridians Meridians are straight lines connecting the Poles . A meridian is also a semi-great circle and a meridian and its allied anti-meridian complete a Great Circle . Meridians always indicate the North/South direction as they connect the Poles . True direction is always measu red with reference to the local meridian

,, \ \

\ \

\ \

180·~-+--i--..L--( 1--+--;----'--.,.\~ 180· I I

I I

I

I

I

I

I

I

/

I

I

I

/

/

I

I I I

From the North Pole al/ places are South along a specific Meridian, al/ True directions are 180·.

360·

/////.....

;'

1/

----- -

/"",'"

I

-.... .........

'\

\ .... \ ' \

I I I

\

I

\

I

\

360· ~,I -+--i--..L--(

l--4-"';--I._"';"~360·

,

\

I

\

I

\

I \

I \

I \

\

I

~

I

I

From the South Pole al/ places are North along a specific Meridian, al/ True directions are 360·.

Meridians

Greenwich Anti-Meridian Longitude 180· EIW

180· EIW

____ Equator

Longitude

Longitude

90·W f-+-+----r--!--{ >--+--j--I--H 90·E

South Pole

O·EIW Greenwich Meridian Longitude O· EIW

(where the direction is to be measured) because this local meridian will indicate the direction of True North there . A meridian will change direction by 180° when it crosses a Pole (a north track or heading will become southerly at the North Pole and vice versa).

continues ...

Picture Supplements - Explanations

FIGURE 061-E80 ... continued Greenwich meridian The Meridian passing through Greenwich (in London) is known as the Greenwich or Prime Meridian and is now the accepted universal reference datum when measuring Longitude . An aircraft travelling from Greenwich east (or west) for 180° of longitude will arrive at the Greenwich anti-meridian , which is 180° from Greenwich (antipodean). Longitude will increase up to the anti-meridian and will decrease when passing it. North Pole (90 N) 0

Parallels of latitude A circle on the surface of the Ea rth parallel to the Equator is called a parallel of latitude. Parallels of latitude run in an EastlWest direction. They are always small circles , with the exception of the Equator, itself a great circle, which is also the reference latitude datum (zero deg rees north/south - 0° N/S).

Lat. 30·N

Equator The Equator is the Great Circle which has its plane lying at right angles (normal) to the polar axis. Movement along the Equator is east or west as the plane of the equator lies in a true east/west direction and divides the Earth into hemispheres.

Lat. 30·S

South Pole (90 0 S) North Pole

Graticule When meridians and parallels intersect to form a network (or grid), this is known as a Grati cule. Graticules are used for position reference systems on the Earth when charts are produced . The most common system in use is the "latitude/longitude" system. Other systems are the "Georef system" and various type of grid systems .

Longitude (Meridians)

Latitude (Geographic or geodetic latitude) Geographic or geodetic latitude is the angular distance between the plane of the equator and a normal to the spheroid . It is also called topographical latitude. This is in fact the chart latitude. The Geocentric latitude of a position on the earth's surface is the angle between a line to the centre of the earth and the plane of the equator. These distinctions have become necessary with the widespread use of GPS, which can give readings which differ from positions on the chart in use, if the chart has not been compiled with respect to the WGS 84 datum .

Because the earth is approximately an oblate spheroid , rather than a true sphere , geocentric and geographic differ latitudes may differ, with the maximum difference being 11.6 minutes of arc at latitude 45°. Latitude is measured in degrees North or South of the equator, which is at latitude 0°. The angular distance from the Equator to either pole is 90·. One degree of latitude is the equivalent of 60 nautical miles (by definition). Latitude is used in defining position. The latitude of a place or position is the arc of meridian from the Equator to that place . Latitude is measured in degrees and minutes North or South of the Equator. The latitude of the Equator is O· N/S; the Poles are 90· North or South of the Equator (N90/S90) respectively. Longitude Geographic or geodetic longitude of a position is the shorter arc of the Equator measured from the Greenwich Meridian to the meridian passing through that position . Longitude is expressed as degrees and minutes from the Greenwich Meridian East or West to the Greenwich anti-Meridian . The longitude of Greenwich is 000· (east or west - EfW) and the Greenwich Anti-Meridian is at longitude 180° EIW. This is chart longitude.

An aircraft proceeds on a circumnavigation of the Earth, following a westerly track (270· T) over the Equator. As the flight is along a parallel of latitude (0·) it will be a Rhumb Line track and as the Equator is by definition a Great Circle , it will also be a Great Circle track in this case . When the flight starts in the western hemisphere, westerly longitude will increase to the Greenwich anti-meridian . At this point, longitude is maximum (180°). Continuing on the same track, longitude then becomes easterly and will decrease up to the Equator. This applies regardless of start point for a flight along a parallel , which is always a Rhumb Line track as direction remains unaltered . The same holds true for a flight along any other parallel , except that the track will not be a great circle track.

continues ...

Aviationexam Test Prep Edition 2012

FIGURE 061-E81 ... continued Position on the Earth Position on the Earth can be expressed in various ways . The latitude and longitude (always given in that order) of a position is the most common method in navigational practice . This uses a reference system of lines parallel to the Equator (Latitude ) and North/South lines - Meridians (Longitude) which can define any point on the earth , or on a sphere. "Lat and Long" may be written in several ways , e.g. : 25N 128W; 25 OON 128 ~OW; N2500 W12800. Latitude is always written first. When a position is expressed in terms of latitude and longitude, it will be seen that there is only one possible location for each such position . Change of latitude (Ch Lat or D Lat) The change of latitude between two places is defined as the smaller arc of meridian between the two latitudes. It is therefore the difference in latitude between the two. It is designated North or South according to the direction of change specified e.g . the Ch Lat between 27"S and 37" S is 10° South. From 15°S 015°E to 15°N 025 °W, the change of latitude (Ch Lat) will be 30° North. Calculating change of latitude requires the subtraction of one latitude from another, or, if the latitudes are in different hemispheres, the addition of the latitudes. Change of longitude (Ch Long or D Long) Change of longitude between two places is the smaller arc of the Equator measured between the meridians of those places . It is designated East or West according to the direction of change specified. Change of Longitude (Ch Long) corresponds to change of latitude, but in terms of longitude. For example the Ch Long between 29°E and 31 °E is 2°E. Thus from 15°S 015°E to 15°N 025°W, the change of longitude (Ch Long) will be 400W (15° from 015°E to the Greenwich meridian and then 25° from Greenwich to 025°W). There is also a change of latitude, but this does not affect the Ch Long . Great circle A great circle is a circle on the surface of a sphere with the same centre and radius as that of the sphere. This applies to the Earth . Great Circles are important in navigation because the shortest distance between two places on the surface of th e Earth is the shorter arc of the great circle between those two points . Another way of expressing this is to say that the plane of a great circle connecting two places on a sphere (such as the Earth) passes through the centre of the sphere (or Earth) . Any great circle on the Earth will therefore divide the Earth into equal parts (bisect it).

There can be only one great circle connecting two places on the Earth , unless those two places are diametrically 0 0 opposite each other (antipodean) . Examples are the North and South Pole , 45°N 090 W and 45°S 090 E etc .. . In such cases , there is an unlimited number of great circles between the two points. Note: While a great circle track is the shortest distance, the direction of a great circle changes continuously because meridians are not parallel on the earth. (A flight along the Equator or along a meridian itself is an exception) . Check this statement by connecting two places on a globe with a length of string. Measure the angle at each meridian. Unless the two places are due North/South or EastlWest of each other, the direction between the two places will change from meridian to meridian. Ideally, all aircraft tracks would be great circles, as the distance would be shortest by definition.

Rhumb line A rhumb line is a line of constant direction on the surface of the Earth or a chart. This is because it crosses each meridian at a constant angle. Meridians are the datums used for measuring true direction. The Equator and Meridians are both Rhumb and great circles by definition. Parallels of Latitude must also be Rhumb lines because they are lines of constant direction since they run east/west. Places on the same parallel of latitude are therefore due east or west of each other.

Great Circle Track

Rhumb Line Track

Small circles Any circle on the surface of the Earth which is not a great circle is by definition a small circle. The only small circles which are of navigational interest are parallels of latitude. Great Circle track increases due to meridian convergency, Rhumb Line track remains constant.

Picture Supplements - Explanations

FIGURE 061-E82 CONVERGENCY (ANGLE OF CONVERGENCE) OF MERIDIANS 180· EIW

The angle between meridians is called "convergency" or "the angle of convergence". Convergency is in fact the angle of inclination between the meridians passing through two points at a given latitude. This angle is zero at the Equator, where meridians are parallel to ea ch other. As meridians leave the Equator and approach the Poles , they converge and the value 0 of convergency reaches a maximum of 1 (between successive meridians) at the Poles , where the meridians are spaced 10 apart) . It is also apparent that the distance between two successive meridians is greater at the Equator than at the poles. Therefore it can be stated that the distance be~een meridians is greatest at the Equator and least (zero,il) f.act) at the poles.

090 0 E

O·EIW

Conversion angle Conversion angle is the angle between the Great Circle and Rhumb Line tracks between two places . The Rhumb Line and Great Circle tracks between two points will always be different (except for positions on meridians or along the Equator). The Great Circle track will change direction continuously and the Rhumb Line track will always be a constant direction by definition. It can be seen by drawing that the Great Circle track will always lie closer to the Pole and the Rhumb Line closer to the Equator. On a particular chart - and this depends on the chart in use - if the Rhumb Line is a straight line, then the Great Circle will be a curved line, and vice versa. At the mid (or mean) meridian however, the Great Circle track is the same as the Rhumb Line track .

The meridians are parallel at the Equator, but converge towards the poles.

• Value of Conversion Angle (ca) = one half convergency (C).

North Pole

Equator

Calculating the Value of the Angle of Convergency The angle of convergency between two positions (i.e. two meridians) is calculated from the formula: • Convergency

=Change in Longitude x Sin of Latitude

Change in longitude (Ch Long or 0 Long) is the difference in longitude (expressed in 0) between two positions or two meridians. If the positions are at different latitudes, the average value of latitude should be used, which will give an approximate value . Example: Find the value of convergency between 30 ·S 037"E and 30·S 015·E. Solution: • Convergency • Convergency • Convergency

=Ch Long x sin Lat =2· x sin 30·

=11·

Note: when the latitudes are different, use the mean latitude.

Calculating the Value of Conversion Angle As conversion angle is one half of convergency, the formula can be written as : • Conversion angle = Y, Ch Long x sin Latitude (or sin mean Lat). • or: Conversion angle = Y, (Initial Great Circle - Final Great Circle track). If the Great Circle track between two positions can be calculated and the value of convergency is known , the Rhumb Line track between the positions can be calculated by applying conversion angle in the correct sense . Of course, the converse is also true: if the Rhumb Line track and Convergency (or conversion , angle) are known or is calculated, the Great Circle track can be determined. Note: It is usually essential to draw a diagram, bearing in mind the fact that a great circle is closer to the pole.

continues ...

Aviationexam Test Prep Edition 2012

FIGURE 061-E83 ... continued When drawing a sketch , remember that the Great Circle tracks will always lie closer to the Pole and the Rhumb Line close r to the Equator. The difference between the Rhumb Line track and the Great Circle track = conversion angle (ca) = Y, of convergency, Let's take the following example : A flight has to be made from a Waypoint 1 (30"N 040·\11 to a Waypoint 2 (30"N 020·\11, What are the Rhumb Line (RL) track and the Initial, Average and Final Great Circle (GC) tracks? '\ Solution: • Since WPT 2 is located due East of WPT 1, the Rhul7lP.. Line track would be 090', • The difference between RL track and GC track =conversion angle (ca) • Conversion Angle = Y, of Convergency • Convergency =Ch Long x sin Lat • Convergency =20 ' x sin 30' • Convergency = 10' • Conversion Angle =5'

Great Circle Track

Rhumb Line Track

We know that the difference between RL track and GC track will be 5', Now we just need to apply this correction in the right sense => this is where it is important to draw a sketch, In the Northern Hemisphere the GC track will be closer to the North Pole than the RL track and the GC track will change continuously, forming a curve: • Initial GC track: Upon leaving WPT 1 the GC track will start off as a lower value than the RL. In our case it will be 5' lower than RL => 090' - 5 ' = 085', The initial GC track will be 085', The track will then gradually increase until the mid-point between WPT 1 and WPT 2, where the GC track will be equal to the RL track, • Average GC track: mid-way between WPT 1 and WPT 2 the GC track will be parallel to the RL track => the GC track will be 090', From this point the GC track will continue to increase beyond the value of RL track, until the maximum value at WPT 2,

Northern Hemisphere Rhumb Line Track = 090· Mean GC Track = 090 ·

5' WPT2 30·N020·W

WPT1 30·N040·W

=

=

Initial GC Track 085' 090· (RL) - 5· (ca)

Final GC Track 095 · 090' (RL) + 5' (ca)

• Final GC track: Upon reaching WPT 2 the GC track will be a higher value than the RL. In our case it will be 5' higher than RL => 090' + 5' = 095', The final GC track will be 095', In the Southern Hemisphere things change a little bit. The difference between RL and GC tracks still equals to the conversion angle (ca) , but the relationship of RL and GC tracks are opposit to the relationship in the Northern Hemisphere - take a look at the sketch below (note that the GC track is also closer to the nearer pole - in this case the South Pole): • Initial GC track: Upon leaving WPT 1 the GC track will start off as a higher value than the RL. In our case it will be 5' higher than RL => 090' + 5' = 095', The initial GC track will be 095', The track will then gradually decrease until the mid-point between WPT 1 and WPT 2, where the GC track will be equal to the RL track, • Average GC track: mid-way between WPT 1 and WPT 2 the GC track will be parallel to the RL track => the GC track will be 090', From this point the GC track will continue to decrease below the value of RL track, until the minimum track value at WPT 2,

Southern Hemisphere Rhumb Line Track 090· Mean GC Track = 090·

=

WPT1 30·S 040·W

WPT2 30·S 020·W

Initial GC Track = 095 · 090· (RL) + 5· (ca)

Final GC Track = 085 · 090· (RL) - 5· (ca)

a lower value

than the RL. In our case it will be 5' lower than RL

• Final GC track: Upon reaching WPT 2 the GC track will be => 090' - 5' = 085', The final GC track will be 085',

Picture Supplements - Explanations

FIGURE 061-E84 CHART PROJECTIONS

The term "projection" describes the process of transferring the information from a globe onto a flat sheet of paper. Chart projections are an attempt to depict the surface of the earth on a flat sheet of paper - a "plane surface". Because it is impossible to show a 3-dimensional surface accurately on a flat surface , distortions of co nforma lity, shape , distance , direction, scale and area always result when making a chart. A certain type of projection may minimise some distortions but will maximise others. Some projections manage to redu ce all distortions to manageable proportions , even though inaccuracies persist. • Conformality: When the scale of a map at any point on the map is the same in any direction, the projection is conformal. Meridians (lines of longitude) and parallels (lines of latitude) intersect at right angles . Shape is preserved locally on conformal maps .

• Distance: A map is equidistant when it portrays distances from the center of the projection to any other place on the map. • Direction: A map preserves direction when azimuths (angles from a point on a line to another point) are portrayed correctly in all directions.

• Scale: Scale is the relationship between a distance portrayed on a map and the same distance on the Earth. • Area: When a map portrays areas over the entire map so that all mapped areas have the same proportional relationship to the areas on the Earth that they represent, the map is an equal-area map .

The different ways of making chart projections can be categorised in two types : • "Direct" projections which are constructed by straight lines from a point through the surface of the reduced earth onto the projection plane. The point of projection may be located at infinity (orthographic) , at the centre of the reduced earth (gnomic) or on the surface of the reduced earth opposite the projected point (stereographic) • "Indirect" projections which are constructed mathematically. Constructing Charts When referring to the term "Reduced Earth" (RE) it means a scale model of the earth on which the chart projection is based . To create a "1:500 000" chart, the earth is scaled down to one 500 oooth scale model. In practice most charts are constructed from mathematical models , but it may help to understand projections to assume that a transparent model of the Earth (the "Reduced Earth") is used by the mapmaker. All charts therefore are known as "projections" which are usually named after the cartographer who first produced the projection . Well-known examples are "Mercator's" and "Lambert's" Projections .

This reduced earth has a light source ("globe") often in the centre , with an overprinted graticule. When the "globe" is lit, an image of the surface of the reduced earth is projected onto a sheet of paper placed next to the reduced earth. The paper sheet may be flat or rolled into a cylinder or cone. The image of the reduced earth is projected onto the chart as . form of graticule. The presentation of the image will vary with the position of the paper and the reduced earth. The chart which results is called a "projection". Three main types of projection are used in constructing aviation charts:

Planar

• Plane or azimuthal projection - an azimuthal projection is the projection of the Reduced Earth onto a plane. A plane surface (a chart) is placed tangential to the Reduced Earth at a point, with a light source in the centre of the Earth or at the other pole (as in the polar stereographic projection) . • Cylindrical projection - a cylinder is wrapped around the Reduced Earth , tangential to a specified para llel or meridian, or intersecting at specified parallels , projecting points on the Reduced Earth onto the cylinder. • Conic projection - the projection plane is a cone arou nd the Reduced Earth , either at a tangent or cutting the earth at two specified parallels (as in the Lambert's conical orthomorphic projection).

Cylindrical or conical projections can be further classified into projection types depending on the position of the rotational axis of the cone or cylinder in relation to the Earth axis :

---

• Normal projections - tangential to the Equator (e .g. the Mercator projection) • Transverse projections - tangential to a chosen meridian (e .g. the Transverse Mercator) • Oblique projections - projections which are neither normal nor transverse (e.g. the oblique Mercator.)

continues ...

Aviationexam Test Prep Edition 2012

uel J;J:c.r3':t:fl ... continued Standard Parallel Most charts will have a "standard parallel". This is a latitude on the reduced earth where it touches or is tangential to the chart. Since the reduced earth is in direct contact with the paper this point, the values on the chart: convergency (convergence), shape, scale etc must be identical to those on the reduced earth where the paper and the Reduced Earth touch. Scale of the charts "Scale" is the relationship of the distance between two positions on a chart and the distance between the same positions on the earth. Scale = Chart length + Earth distance.

There are 3 forms of displaying the scale on the charts: • A simple statement Scale may be expressed in a simple statement, such as "one inch equals five kilometres, or "one centimetre to five kilometres". • Graduated lines Scale may also be shown as a graduated scale line (often seen on topographical charts). This is a printed scale, usually at the bottom or on the side of the chart, marked in selected distances, often giving the nautical and statute mile scales and metric equivalents. These graduated lines are quite accurate for normal use, remembering that no aviation chart has a 100% correct and constant scale. • A representative fraction Scale can also be expressed as a representative fraction (RF). This is the same as a statement such as "one in five hundred thousand" but it is written as "1 : 500 000". This means that one centimetre (or inch) on the chart equals 500 000 cm (or inches) on the earth. "One inch equals 5 nautical miles" is the same as saying 1 in = 5 NM, or 1 in = 5 NM x 6080 ft x 12 in = 364 800 in.

Scale problems are often combined with distance, departure and/or speed problems. The easiest method of solving problems is always to use the formula "Scale = Chart length + Earth distance".

Picture Supplements - Explanations

FIGURE 061-E86 LAMBERT's PROJECTION

The Lambert's Projection is a conical projection derived mathematically from the Simple Conic projection. A cone is placed tangential to a chosen latitude so that it lies on the North/South axis of the Reduced Earth, with the cone apex above the pole. The cone is considered as if it cuts the earth at two standard parallels instead of one. This is achieved mathematically. The Latitude halfway between the two standard parallels is termed the Parallel Of Origin. It can be compared to the parallel of origin (standard parallel) of the simple conic projection . The origin is the centre of the reduced earth.

Simple Conic Projection

When the projection is developed, it is cut along a meridian and then lies flat. The shape changes from a cone to a circle with a missing segment. The angular extent of the chart reduces from 360° to a lesser value . The smaller angle represents 360° on the Earth. The ratio of the smaller value and 360° is known as the constant of the cone which will always be less than unity. The constant of the cone (a lso called "n") equals the sine of the parallel of origin of the cone . If the smaller angle is 300°, then the constant of th e cone is 300° .;. 360° ::: 0,8333. Inverted Sin 0,8333 ::: 56°26'. The parallel of origin is therefore 56°26 '. Properties of the Lambert's Projection The chart is orthomorphic by construction as meridians and parallels intersect at 90° and bearings are correct on the chart. Scale expands at the same rate in all directions over small distances. Distance measurement is straightforward along a meridian (latitude scale) in the region of the track flown.

Apex

Lambert's Conical Orthomorphic Projection

• Meridians are shown as straight lines radiating from the pole . • Parallels of latitude are arcs of concentric circles convex to the Equator, spaced almost equally. • Rhumb lines are curved on a Lambert's chart on the equatorial side of the great circle. • An exact Great Circle results when two positions on the Parallel of Origin are connected by a straight line. Elsewhere, a straight line is only an approximate great circle (it is a curve concave to the parallel of origin). As the difference is so small and of no importance for practical purposes , any straight line on the chart may be taken as a great circle. Great circles are easily measured at any meridian, rhumb line tracks are measured at mid-meridian. This is useful to remember when plotting radio bearings. It is therefore much easier to plot radio bearings on a Lambert's chart than a Mercator chart. Special techniques are used when flying great circle tracks on a Lambert's chart, as the track direction changes continually. One method is to fly mean (average) tracks over short distances. The angle between the rhumb line and great circle is equal to conversion angle . • Convergency is that of the parallel of origin and is constant on the chart. The convergency between two positions is the Change in Longitude x sine of the Parallel of origin. This value of convergency is also known as the chart convergency factor. The value of chart convergency (CC) between any two meridians on the Lambert's projection may be calculated in either of the following ways , depending what data are known: CC CC CC CC

(chart (chart (chart (chart

convergency) convergency) convergency) convergency)

::: ::: ::: :::

Ch Ch Ch Ch

long long long long

x sin parallel of origin

x constant of the cone x "n" x Chart Convergency Factor (CCF)

It can be seen that each formulae is basically the same as the constant of the cone::: "n" ::: CCF ::: sin parallel of origin . Any of the values may be noted on the chart, or may be calculated from a knowledge of the standard parallels. continues ...

Aviationexam Test Prep Edition 2012

FIGURE 061-E87 ... continued Example: Calculate the value of "n". when the standard parallels of a Lambert's chart are 3S' N and 43' N. Solution: The parallel of origin, by definition, lies midway between the standard parallels (3S'N and 43'N) . The parallel of origin is therefore 39'N (half way between the standard parallels) . The value of "n" is equal to the sin of 39 ' N =0,62932. The angular extent of the chart would therefore be 227' (0,62932 x 360') . The CCF would also be equal to 0,62932.

Scale Scale on the Lambert's projection is variable. Where the cone cuts the standard pa ra llels, scale is the same as the redu ced Earth and is therefore corre ct along those parallels . The scale contracts between the standard parallels and expands outside them . By design, the rate of expansion and contraction is small and depends on the standard para llels chosen . The standard parallels are marked on the chart and often match the chart limits. So that scale is relatively constant throughout th e chart, the "1/6th rule" is used. When the chart is constructed , standard parallels are selected with the distance between them approximately 2/3 of the North/South coverage of the chart. Each standard parallel will co nsequently sit at approximately 1/6 of the chart N/S extent from the top or bottom of the page . This method gives a scale error of 1% maximum so that the chart is considered as having a constant scale. If the chart is pla nned to cover an area between , say, 33°N and 45°N, the standard parallels to comply with the 1/6 th rule would be 35°N and 43°N. If the "1/6 th rule" is followed , shapes and areas are not distorted and Lambert's charts are very suitable for map reading .

/t

Uses of the Lambert's Projection • Plotting charts • Topographical charts • Synoptic meteorological charts • Radio/nav charts. • General Navigation Version

Chart Fit & Limits Charts constructed to the same scale and the ' same standard parallels will fit east/west. Charts will only fit north/south if they are part of a larger projection . The differences will be noticed if several charts are joined as part of a larger area , such as on a wall chart.

1; 6th \\

/,'

,,

/i

/

"

,

'

\ 43°N " '" \ , 39°N , \

11"" \<~.N

The parallel of 3soN and 43°N are each 1/6th of the chart coverage from the vertical limits, thus complying with the 1/6th rule .

Plotting Position on the Lambert's Chart The easiest way to plot a position is to plot Longitude first and then Latitude - the opposite method to that used on the Mercator (Latitude / Longitude). It is easier because the parallels of latitude are curved . A "local meridian" has to be plotted first, then the latitude is measured up or down the meridian. Rhumb line tracks are measured at mid-meridian = halfway along track. Great circle tracks are measured at the initial and fina l meridians. Plotting a Range and Bearing on the Lambert's Projection Plotting a position in terms of a range and bearing on the Lambert's is straightforward . The local meridian and local latitude scale near the source of the range and bearing should be used for best accuracy. The lower the latitude, the more accurate the position will be. Ranges are accurate for approximately 300 NM, depending on the latitude. DME ranges are easy to plot from the origin of the bearing , using a convenient meridian . Distance is measured with the latitude scale of the chart using the area of the flight for the best accuracy. General Remarks on Plotting on the Lambert's Chart Plotting on the Lambert's is not difficult. Great circle tracks are straight lines. In areas where latitude is low (e .g. equatorial countries) there is little difference between rhumb line and great circle tracks because convergency is small. Always measure the mean track angle at the mid-point or at the mid-meridian, where the great circle and rhumb line directions are the same. On flights with long sectors between positions , it will be found easier to use distances of approximately 350 400 NM and measure the average great circle track between these positions . Distance is always measured in NM along a convenient meridian , near to, or central to the track in use. On the chart on the right, parallel spacing in this case is 1° = 60 NM apart. The space between each successive "tick" along a meridian is 5 NM . The best area to use for scale measurement is between the position "U" and "W' - either will be accurate enough. It will not make sense to use an area outside this range because of scale expansion / contraction .

Picture Supplements - Explanations

FIGURE 061-E88 MERCATOR PROJECTION

The Mercator projection is a cylindrica l projection with the plane tangential to the equator of the reduced earth and the light source at the centre of the reduced earth. It was first produced by Gerhardus Mercator, a Flemish mapmaker of the 16th century. The main purpose of the chart was to enable a constant heading to be steered by mariners and for directions to be plotted as straight lines.

75° North -,---,------,------,-,-,--,--.

II

60· North

+-+-+--+--t--t--k

60· South +--+--+---+--+-+--1'

II

75· South

-'...-'----'----'---'L-L--L-j.

Properties of the Mercator Projection The Mercator projection is immediately recognisable as it is the only aviation chart with a completely rectangular graticule.

• The graticule is rectangular • The chart is orthomorphic (conformal) • Meridians are straight lines equal distances apart on the chart and parallel to each other. • The true direction of rhumb lines is easily measured at any convenient meridian . • Great circles , which are complex curves , cannot be plotted with confidence . • Parallels of latitude are parallel straight lines at varying distances apart. As latitude increases towards the poles , the distance between successive latitudes increases so rapidly (proportional to the secant of the latitude) that the chart becomes useless for navigation in polar reg ions (it is impossible to show the poles on Mercator charts) • The chart is tangential to the earth at the Equator, so every condition on the reduced Earth will be the same as on the chart at the Equator. At the Equator: • convergency is correct and constant • shapes and areas are represented correctly • scale is correct and constant. (Away from the Equator, the above properties change). On the reduced earth , meridians converge towards the poles . On the Mercator chart, meridians are parallel , so a straight line on a Mercator chart will be a rhumb line since it crosses all meridians at the same angle . A great circle track on the Mercator chart is a complex curved line which cannot be plotted with confidence . The great circle track will always be closer to the pole than the rhumb line, as on the surface of the earth . Chart convergency (convergence) is the angle between meridians on the chart. As meridians are parallel , there is no angle between them . There is therefore no convergency on a Mercator chart. Alternatively, convergency is zero and constant everywhere on the chart. Since the value of convergency at the Equator on earth is also zero , then convergency on a Mercator chart is only correct at the Equator. It is clear that scale on a Mercator chart is not constant because of the increasing distance between parallels of latitude when moving North or South from the Equator. Scale is correct at the Equator and expands as the secant of the latitude. If a scale is given for a Mercator chart, it must always relate to a specific latitude - e.g. "scale at 35°S = 1:500000". As scale is constantly changing , care must be taken when measuring distances. The best way is to use the mean latitude covering the route portion. In practice , the measurement of distance is not difficult and speedily improves with time and practice. At low latitudes the shapes and areas of land masses are not distorted, but severe distortion occurs in high latitudes because of the difference between chart convergency and Earth convergency. This has the result that countries in high latitudes, Greenland for example , will have a disproportionate size and shape on a Mercator chart. Charts will fit together North/South and EastlWest if the same scale is used at the joining latitude. Chart coverage is usually limited to between 75°N - 75°S. Advantages of the Mercator Projection

• Rhumb lines are straight lines • Plotting is uncomplicated and speedy. Disadvantages of the Mercator Projection

• The varying scale means distance measurement is not straightforward • The chart is useless for high latitude • Great circles are not easily plotted (except the Equator & meridians )

continues ...

Aviationexam Test Prep Edition 2012

FIGURE 061-E89 ... continued Plotting Position on the Mercator The rectangu lar graticule means that plotting position is easy on the Mercator. Normally the latitude of a position is plotted first on two adjacent meridians which are then joined horizontally. The longitude scale is constant on the chart, so the longitude is measured at a convenient positi on and trans ferred along the constructed horizontal line to obtain the position . Rhumb line tracks are measured at any meridian. Exact great circle tracks cannot be plotted or measured on this projection. To fly a great circle track, it must first be plotted on a su itable chart, such as the Lambert's conical orthomorphic and va rious points picked off and plotted on the Mercator, giving a composite great circle track. Distance is measured with the latitude scale of the chart using the area of the flight for best accuracy. Plotting a position in terms of a range and bearing on the Mercator is straightforward . The local meridian and local latitude scale should be used for best accuracy. Ranges are accurate for approximately 300 NM , depending on the latitude. The lower the latitude, the more accurate the position will be . DME ranges are plotted from the origin without correction .

Uses of the Mercator Projection • Plotting charts • Topographical charts in equatorial regions • Synoptic meteorological charts • Radio/nav charts .

Plotting Radio Bearings Radio bearings must be converted before plotting on a Mercator. As radio bearings are great circle bearings and a great circle bearing is a complex curve on this chart (except when the aircraft is north or south of the radio station) it is almost impossible to plot it accurately on the chart as a great circle. Instead, the great circle bearing is changed into a rhumb line bearing by applying conversion angle and is then plotted as a rhumb line. (Conversion angle is the difference between a great circle and a rhumb line bearing). If conversion angle (or the convergency) between two positions is known , then a rhumb line may easily be converted into a great circle and vice versa. The rationale is that both the rhumb line and great circle bearings will pass through the aircraft position - as one of the bearings is difficult to plot, the other is plotted instead. The easiest way to decide which way to apply the correction is to draw a diagram . Note: Apply all corrections to bearings , including conversion angle , where the bearing is measured. This is perhaps best remembered by the rule:

"Apply conversion angle & variation where the "work" is done" The "work" refers to where the bearing is measured: • at the VOR station for VOR bearings • at the VDF station for VDF bearings (QTEs or QUJs) • at the aircraft position for ADF/ RMI bearings.

Examples of Mercator Chart Problems QTE =True Bearing from a station and is what is plotted on the chart. QUJ = True Direction to a station from an aircraft. It is the reciprocal of a QTE. Example 1: In the northem hemisphere, an aircraft receives a OTE of 070' from a VDF station. The conversion angle (CA) between aircraft and the radio station is 6' . Calculate the Rhumb Line (RL) bearing to plot on a Mercator chart. Solution : N Hemisphere OTE =070' GC, CA =6'. By inspection of a diagram, CA is added => 070' CG + 6' CA = 076' RL. Plot 076' Rhumb Line bearing from the VDF station.

AC position

Radio station

Remember that the Great Circle (GC) bearing will always be nearer to the closer pole and the Rhumb Line (RL) will always be nearer to the Equator. This means that the Rhumb Line is always a straight line and the Great Circle (a curved line) will be nearer and convex to the closer pole.

continues ...

Picture Supplements - Explanations

FIGURE 061-E90 ... continued Example 2: A pilot in the southern hemisphere receives a QTE of 255' from a VOF station. Convergency between aircraft and station is S'. Calculate the bearing to plot on a Mercator chart. Solution: Note that a Conversion Angle (CA) = 112 of Convergency. S Hemisphere QTE = 255' GC, CA = 8'. By inspection of a diagram , CA is added => 255' CG + 4' CA 259' RL. Plot 259' Rhumb Line bearing from the VOF station.

=

Example 3: An aircraft in the southern hemisphere measures a relative bearing of an NOB of 030'. The aircraft is on heading 048'T and conversion angle between the AC and the NOB is 6' . Calculate the bearing to plot on a Mercator chart.

Solution: The radio bearing (a relative bearing) of the NOB is measured inside the aircraft, so all corrections are applied at the aircraft DR (estimated) position. First the relative bearing is converted into a great circle bearing. Then it is converted into a rhumb line bearing. As this bearing is the bearing of the station from the aircraft, it must be converted into the bearing of the aircraft from the station (recipro cal) before it can be plotted .

=

~/

AC position

NOB Meridian

Aircraft Meridian

• 030' RB + 048' TH 078' TB (GC) 0078' TB (GC) - 6' CA 072 ' TB (Rh .Line) ,072' TB (Rh.Line) + 180' = 252 0 TB (RL)

=

Calculation of difference between Rhumb Line and Great Circle The difference between the Rhumb Line (RL) and a Great Circle (GC) is called the "Conversion Angle" (CA). To calculate the value of Conversion Angle, the formula is used: Convergency = Change in Long x sin Latitude . Conversion angle is 1/2 of Convergency. 45'E 65'E Example: Calculate the difference between the directions of the Rhumb Line and the Great Circle connecting the stations ABC (30'N 045'E) and XYZ (30'N 065'E) .

85'

=

95'

90'

Solution: o Convergency = Ch.Long x Sin Lat o Change in Longitude 20' o Sin 30' 0,5 o Convergency = 10' • Conversion Angle = 5' (112 Convergency). o The RL direction is therefore 090' (along the meridian) . • The GC direction at 045' E is 085' , at 065'E it is 095'.

=

Northern hemisph. Great Circle closer to the North pole

~~--~~R~h-u-m~b~li~ne--=~09~0~·~------I~~~­ XYZ Ch. Long

=20'

Calculation of scale at different Latitudes When the scale at one Latitude on a Mercator chart is known , the scale at any other La!. may be calculated by a formula: Scale Denominator "A" x cosine " B" = Scale denominator "B" x cosine "A" ... where "A" and "B" are the Latitudes . The formula is best remembered as "ABBA" (A x B = B x A). Example: Scale atthe Equator on

a Mercator chart is 1 : 2.000.000. calculate the scale at 40'S.

Solution : o Let ':4 " be the Equator (0') and "B" 40'S. o "A BBA " => Scale denom. "A" x cos "B" =Scale denom "B" x cos "A" o 2.000.000 x cos 40' =Scale denom "8 " x cos 0' o Scale denom "8 " (2.000.000 x cos 40') .;. cos 0' o Scale denom "8 " = 1.532.089 => the scale at 40'S is "1: 1.532.089"

=

Aviationexam Test Prep Edition 2012

FIGURE 061-E91 POLAR STEREOGRAPHIC PROJECTION

The Polar Stereographic projection is an azimuthal projection . The earth graticule is projected onto a chart which is tangential to the selected pole . The point of projection is at the opposite pole. Meridians are drawn from the pole with angu lar spacing equal to the actual change of longitude. With the Pole as centre, circles are drawn with latitude radius 2R tan Y:, co-I at. (R radius of the "Reduced Earth", co-I at 90° minus latitude). A complete hemisphere may be shown on one chart, although the accuracy suffers at latitudes lower than 70°. The chart is orthomorp hic as sca le varies at the same rate in all directions and meridians 1200E 1000E 80 and parallels intersect at 90°. The projection can be re cog nised immediately as the graticule looks like a wagon wheel with meridians (spokes) radiating from the centre with parallels as concen160 0E _ tric circles. It is symmetrical about any meridian.

=

=

0

E

60 0 E

40 0E

Properties of the Polar Stereogr. Projection

Meridians are equally spaced straight lines radiating from the selected pole which is at the centre of the projection .

160 0W -

- 200W

The angles shown are the true angles between meridians. • Parallels of latitude are a series of unequal concentric circles centred 1400W 1200W 1000W on the selected pole . The spacing between parallels increases with distance from the pole . The radius of the parallels is 2R tan 1/2co-lat.

80 0 W

60 0 W

400W

• Only one pole can be shown on the chart. • Meridians and parallels intersect at 90°. • Except for meridians, Rhumb Lines are curves concave to the nearer pole which are difficult to measure accurately. • For practical purposes , all Great Circles within the area covered are considered straight lines. The chart is constructed so that great circles are approximately straight lines, but in fact they are slightly concave to the pole . The closer a great circle is to the centre of the projection (the pole) , the more it will appear as a straight line. This is completely true only for the lines crossing the central meridian at 90°. Great circles can be measured at any convenient meridian. The Equator is a circle, furthest from the pole . • Convergency is constant and correct at the pole , incorrect elsewhere , with a value of 1. Chart convergency is equal to change of longitude. Scale Scale is correct at the pole on ly, expanding away from the pole as sec 2 y:'(co-Iat). Distances can be measured with reasonable accuracy, utilising the latitude scale on a convenient meridian in the are where plotting is being done. Shapes and areas are approximately correct, with some slight distortion away from the pole .

The method of solving scale problems is similar to that used for the Mercator projection - the "ABBA" formula is used , with the substitution of cos 2 y:'(co-Iat). (co-Iat = 90° -latitude) . Example: Scale at the Pole is 1: 2.000.000. Calculate the scale at 70W Solution: • The pole is taken as latitude "A", latitude 70 0N is "B". 2 • Scale denominator "A" x cos Y:,(co-Iat "B') = scale denominator "B" x cos2 y:'(co-Iat "A') • 2.000.000 x cos 2 Y:, (90'_70°) = scale denominator "B" x cos2 Y:, (90°-90°) 2 2 • 2.000.000 x cos Y:, (20°) =scale denominator "B" x cos Y:, (0') 2 • 2.000.000 x cos 10' scale denominator "B" x coi 0' • 1.939.692,6 =scale denominator "B" • Scale at 70'N is 1 : 1.939.692,6

=

Note: To find cos 2 10' , find cos 10· and square the result (cos 10' x cos 10·)

continues ...

Picture Supplements - Explanations

FIGURE 061-E92 ... continued Advantages of the Polar Stereographic Projection o o

o

The chart is excellent for flights in polar re gions Great circles are approximately straight lines The chart is suitable for map-read ing near the pole

Disadvantages of the Polar Stereographic Projection o

o

Rhumb lines are curves Scale expands away from the poles

Uses of the Polar Stereographic Projection o o o

Polar navigational charts Topographical charts Meteorologica l charts .

"

Plotting Positions By Latitude And Longitude Plotting a position on the polar stereographic chart is the same as that for a Lambert's chart. Normally the latitude of a position is plotted first on two adjacent meridians which are then joined horizontally. The longitude scale is constant on the chart, so the longitude is measured at a convenient position and transferred along the constructed horizontal line to obtain the pos ition. To derive the co-ordinates of a position on the chart, the reverse procedure applies . Plotting a Range and Bearing on the Polar Stereographic Projection Plotting a position in terms of a range and bearing on the polar stereographic projection is straightforward. The local meridian and local latitude scale near the source of the range and bearing should be used for best accuracy. Ranges are accurate for approximately 200 NM , depending on the latitude. The higher the latitude, the more accurate the position will be . Accuracy falls away below latitude 70°. Deriving the latitude and longitude from a position is the reverse of the procedure. Rhumb line tracks are measured at the mid-meridian. Great circle tracks are measured at the initial and final meridians. Distance is measured using the latitude scale of the chart as convenient. Plotting Radio Bearings on the Polar Stereographic Projection A great circle is approximately a straight line, therefore VDF bearings measured by a ground station are plotted as true bearings without correction . VOR radials are converted to true bearings and plotted from the stafi on meridian. ADF bearings , which are taken at the aircraft meridian, need to be corrected for convergency. On the polar stereographic projection , convergency is equal to change of longitude. The method used is to transfer the aircraft meridian through the station and then plot the reciprocal bearing. In polar regions , where longitude changes quickly, the method may produce inaccuracies . These may be minimised by using the "curve of equal bearing" when plotting NOB bearings . At positions 1, 2 and 3 each aircraft will measure an identical bearing of the NOB.

Curve of Equal Bearing Aircraft in different positions which obtain an identical radio bearing of a distant NOB will not in fact all be on the same great circle bearing . They will lie along another line known as the curve of equal bearing . On the other hand , aircraft situated at different points along a single radio bearing (that is along a great circle) will measure different readings , since the direction of the radio bearing (a great circle bearing) will be continually changing because of convergency.

The curve of equal bearing (CEB) is the locus of points Angle "cr" has the same where the direction of a distant radio station is identical. value for each aircraft. An interesting feature of the CEB is that it lies on the side of the rhumb line nearest to the Equator and differs from the rhumb line bearing by the amount of conversion angle . The rhumb line therefore lies between the great circle (nearest the pole) and the CEB (nearest the Equator).

Polar Stereographic Calculations Example 0

0

Example: A straight line joins position X (71°N 090 E) to Y (71°N 050 E). Calculate the great circle bearing of Yfrom X. Solution: o If two positions are on the same parallel of latitude, lines joining them and the pole will form an isosceles triangle. o The angle at the pole is change of longitude between the positions i.e. Ch long at the pole is 40'. o A line from the pole bisecting the angle at the Pole will be perpendicular to the track between X and Y The two angles at the pole are therefore 20° (Y> Ch.Long) , the angles at X and Yare each 70' (there are 180 degrees in a triangle and the angle at Z is 90°). o The track is (360 ' - 70') == 290' GC.

EDI

Aviationexam Test Prep Edition 2012

FIGURE 061-E93 GRID NAVIGATION Great circle tracks are always preferred because of distance savings. When flying near the poles however, convergency changes so rapidly that it is almost impossible to fly great circle tracks . Instead , a technique known as "Grid Navigation" is applied , using a grid for measuring directions. The advantage of measuring direction with respect to Grid North is that direction is the same everywhere on the chart. This technique gives a practical means of flying great circle tracks. Savings can also be made when flying great circle tracks over long distances at lower latitudes with the grid technique . In very high latitudes magnetic compasses are useless and a directional gyro is used as a heading reference , utilising specialised techniques .

TN

GN

GN

GN

GN GN

GN

GN GN

GN

Grid drawn on a Lambert's Chart

True Track Changes Rapidly Near the Pole

Constructing the Grid The plotting chart used for "gridding" must be orthomorphic. It should also have a relatively constant scale in the plotting region . A great circle track should be approximately a straight line. Suitable chart projections for polar regions are :

• Lambert's Conformal Conic • Transverse Mercator • Polar Stereographic. A series of equally-spaced , parallel lines is drawn on the chart parallel to a reference (datum) meridian. These parallel lines may be considered as "false" meridians . The datum meridian is chosen for convenience as the datuQ'l meridian fo r that particular chart. It will often be the central meridian of the chart or sector covered for the flight. It may be the Greenwich Meridian. The space between meridians is arbitrary, but in order to make measuring distance straightforward , 60 NM spacing is preferred . A series of "false parallels" can also be drawn perpendicular to the false meridians which forms the GRID. If the same spacing between the parallels is used , then the grid is square . The chart now has two reference directions: grid north and true north. Grid north is a constant direction but True North varies as convergency of the meridians. Definition The difference between True and Grid directions on the chart is Grid Convergency (or convergence). The value of convergency is that between the local meridian , where true direction is measured and the datum meridian on which the chart grid is based .

If the direction of true north at the local meridian is west of grid north, convergency is west and vice versa . Therefore , in the northern hemisphere, if the local meridian itself is west of the datum meridian, then convergency is east (True North is east of grid north). If a meridian is east of the datum' meridian, convergency is west. At position A convergency is east: Grid = True + Convergency G = T+C True = Grid - Convergency T = G-C

At Position B convergency is west:

GN

GN

GN ""\ Grid Track

Grid = True - Convergency G =T - C True = Grid + Convergency T = G + C

""\ True Track ""\ Convergency B

AtA Convergency is East

Datum Meridian

AtB Convergency is West

Grid Convergency in the Northern Hemisphere

continues ...

Picture Supplements - Explanations

FIGURE 061-E94 ... continued

In the southern hemisphere the picture changes. If the local meridian is west of the grid datum , convergency is west (a sketch always resolves uncertainty). GN

GN

""\ Grid Track

Datum Meridian

""\ True Track ""\ Convergency

AtD Convergency is West

AtE Convergency is East South Pole

Grid Convergency in the Southern Hemisphere

Converting Grid to True and Vice Versa The easiest way of converting grid to true and vice versa is to follow the rhyme: • Convergency East - True Track least • Convergency West - True Track best

Plotting on the Chart Wind directions are usually in degrees true and must therefore be converted into grid. The value of convergency is found using the mid-meridian . Grid values are used for all directions when grid techniques are employed , in the same way that true directions are always used in ordinary plotting , Note: A straightforward technique for solving grid navigation problems is to draw an accurate diagram with the Douglas protractor and measure the angles. It is however, always a good idea to draw a brief sketch for all grid problems to visualise the situation.

Grivation Grid values must often be converted into magnetic values (and vice versa) when a magnetic compass is used. The difference between Grid and Magnetic directions is GRIVATION (Grid variation) . Grivation is the algebraic sum of convergency and variation. Therefore: • Grid = Magnetic ± Variation ± Convergency. • Grid = Magnetic ± Grivation. The rule for applying Grivation is the same as that for variation, substituting "Grivation" for" Variation": • "Grivation East - Magnetic least" • "Grivation West - Magnetic Best".

Variation

100E

Grid Convergency

25 °E

Grivation

35 °E

Isogrivs Lines joining places of equal grivation are known as Isogrivs. On most grid navigation charts , the isogrivs are already plotted . Isogonals are also normally printed on charts as well as lines joining positions with the same convergency. On a Lambert's chart these latter may be assumed to be meridians , which will give only a small error. At or near the pole, the same assumption applies for polar stereographic and transverse mercator charts. It takes a long time to calculate and plot isogrivs and is laborious. Pilots are not usually required to do it.

Grid Navigation Techniques Make all calculations with respect to grid north and measure all directions in degrees grid. o

• Convert all grid directions to true wind directions • Calculate heading and groundspeed in the usual way using grid directions • When wind velocities are found , the results will be in grid direction. They may have to be converted to true direction. • Remember "Convergency east, true track least". • Plot OTEs in the usual way in degrees true at the local meridian of the station • Convert ODMs into true bearings by applying station variation , or into grid bearings by applying station grivation • Convert ADF bearings to grid bearings by adding grid heading and plotting the reciprocal from the station • Plot ground positions with reference to the lat/long graticule • Use grid directions for tactical problems .

Aviationexam Test Prep Edition 2012

FIGURE 061-E95 MAGNETIC DEVIATION Deviation is solely a compass error which varies from aircraft to aircraft. It varies in many ways , such as with heading and latitude. Deviation is caused by localised magnetic fields in the aircraft which are originate in different ways , such as magnetic fields introduced into the aircraft structure during manufacture and electrical fields produced by electrical and radio equipment.

COMPASS NORTH

MAGNETIC NORTH

COMPASS NORTH DEV = Deviation HDG = Heading M = Magnetic C

= Compass

Deviation is defined as the angle between Magnetic north ' '\ and Compass north. It is a correction to be applied tQ..a · Compass heading to obtain the Magnetic heading . Any magnetic or ferrous material on or near an aircraft compass will affect the local magnetic field and may move (or deviate) the needle away from Magnetic north . This may mean that the compass needle does not show Magnetic north accurately (navigation computers near the compass will also have a similar effect). Electrical circuits and radio equipment in the aircraft also tend to cause errors in the compass readings. When the compass needle pOints to the west of magnetic north, the deviation is "West". When the needle points to the east of the magnetic north, deviation is "East". Application And Use Of Deviation A useful mnemonic to remember this property of compasses is very similar to that for variation: Deviation East = Compass Least Deviation West = Compass Best. This mnemonic should also be memorised and used whenever deviation is applied. Deviations affecting aircraft compasses are measured during a procedure known as a "compass swing" . Any corrections noted are then processed mathematically and tabulated on a compass deviation card, which is placed in the aircraft. This card should be checked at any large change of heading for any corrections to apply in order to find the correct aircraft heading to steer. Deviation and variation must never be confused . Two aircraft in the same position will each be subject to the same value of variation , but the value of deviation in each case may very well be quite different since deviation will vary from aircraft to aircraft, as well as from heading to heading. As a summary: CH = MH + Westerly deviation (or - Easterly deviation) MH = CH - Westerly deviation (or + Easterly deviation) Deviation correction The deviations in an aircraft can be minimised to a certain extent after a procedure known as a "Compass Swing". During this procedure, compass readings are taken on different headings and then compared to an accurate datum. Errors are noted and processed mathematically and are then compensated by means of corrector coils ("micro-adjusters" or "corrector magnets") which are part of the compass itself. The corrector coils neutralise much of the permanent magnetism and associated effects such as "soft iron" magnetism associated with it. They do so by applying a magnetic field in the opposite direction to the residual fields affecting the compass which can effectively neutralise most of the deviation errors

Compass Card For

Steer

000 045 090 135 180 225 270 315

001 043 092 136 180 223 267 312

RadiOS ON A typical compass card. The headings on the left are designed magnetic leadings. On the right are compass leadings to steer to obtain the required heading. Intermediate values are interpolated. An individual compass card is prepared for each aircraft at certain levels. The deviation card for an aircraft is only valid for that aircraft.

Not all deviations can be removed from an aircraft however and those remaining after completing the swing are noted and a "Compass Correction Card" (Deviation Card) is produced. The corrections noted on the card are applied by the pilot in flight to assist in steering the desired magnetic heading.

Picture Supplements - Explanations

FIGURE 061-E96 MAGNETIC VARIATION The Magnetic Poles are in continuous movement around the geographic poles. This movement is mon itored and it can be predicted . Since Magnetic Poles are not situated at the True (Geographical) Poles , the direction of a magnetic compass needle will vary according to the aircraft position. The difference between True North and Magnetic North (the true and magnetic meridians) at a place is called magnetic variation :

MAGNETIC NORTH

TRUE NORTH

MAGNETIC NORTH VAR = Variation HOG = Heading T = True M =Magnetic

,

" When the magnetic North Pole lies to the East of-!he True Meridian , variation is easterly. .

When the magnetic North Pole lies to the West of the True Meridian, variation is westerly.

Isogonals Lines on a map or chart joining positions with equal magnetic variation are termed isogonals. A line joining places with zero variation is termed an agonic line. When variation is zero , magnetic and true directions are the same . Isogonals are shown as broken lines on the chart, with the value of variation marked alongside . As stated , variation will change continuously, because of Magnetic Pole movement "precession of the poles" as well as in other ways . The rate of change should always be printed on a chart, either with arrows showing the amount and direction of the annual movement, or in a table . The chart date is usually shown , so the precise value of variation can be calculated for any year. Note: When plotting, it is useful to highlight isogonals on a chart to distinguish them from meridians, especially near the Greenwich meridian. This can avoid reading errors by confusing variation with longitude .

Application And Use Of Variation Navigational calculations are usually made using the True directions (OT) as a datum. The aircraft compass (a magnetic compass) will only indicate magnetic direction (OM). In order to calculate a Magnetic heading which can be steered in the aircraft, any True heading calculated must be converted accordingly. This is achieved by applying variation to the True heading, which will then give the Magnetic heading for a pilot to steer. If True head ing is requ ired and Magnetic heading is known , then the Magnetic heading can be converted into True head ing, also by applying variation (in the correct sense). When Magnetic north lies to the west of True north, (variation "West"), then a Magnetic heading will be greater than a True heading. Similarly, if variation is "East", then Magnetic heading will be less than True heading. This rule is easily remembered as follows : Variation West = Magnetic Best Variation East Magnetic Least

=

This mnemonic should be remembered and used every time when Magnetic/True heading conversions/calculations are made. As a summary: MH TH

=TH + Westerly variation (or. Easterly variation) =MH • Westerly variation (or + Easterly variation)

Aviationexam Test Prep Edition 2012

FIGURE 061-E97 CONVERSION OF MAGNETIC AND TRUE HEAD INGS

There is frequently a need to convert Magnetic head ings to True head ings and vice versa . When converting from one to the other, it is important to apply the values of Variation and Deviation in the correct sense. Deviation and va riation may be given in the form of negative ("+") or pos itive ("-") values , but the rule is still the same : 1) apply the val ue of Deviation to Compass to obtain Magnetic , 2) apply the va lue of Va ri ati on to Mag netic to obtai n True. Noti ce th at Easterl y Deviation as well as Easterly Va ri ation is a positive value (+ °E). Westerly Deviati on as well as Westerly Va ri ation is a negative value (- oW). This will be clear if you re member the definiti on of deviati on: "Deviation is a correcti on which is applied to Compass heading to obtain Magnetic heading". '\

COMPASS NORTH

MAGNETIC NORTH

TRUE NORTH

MAGNETIC NORTH

COMPASS NORTH

MAGNETIC NORTH

DEV : Deviation HOG: Heading M: Magnetic

VAR : Variation HOG: Heading

C: Compass

M: Magnetic

T: True

COMPASS DEVIATION

MAGNETIC VARIATION

A useful way of remembering how to apply variation and deviation is to use th e mnemonic:

"C D M V T" = Can Dead Men Vote Twice? Compass

Deviation

Magnetic

Variation

True

When applying the values of Deviation and Variation we usually start off with the True heading => we work from the right to the left using the above "C DMVT" mnemonic. When applying the Deviation and Variation values we have to apply it in the correct sence - either with a "plus" or a "minus" sign . Remember the mnemonic "West is Best, East is Least" = in other words West plus sign , East minus sign . This mnem onic works when starting with True heading and working your way towards the left though the Magnetic heading to the Compass heading. When working the other way around = starting with the Compass heading and wishing to obtain the Magnetic heading or the True heading , you have to reverse this logic => West = minus sign and East = plus sign.

=

=

Example 1: True heading = 98°, Deviation = 3°E, Variation = 1Q

0

w. Calculate the Compass heading.

Remember that in this case you have to start from the right and work your way to the left.

M

?

108°M <= 105° Compass

V 100W

T + 98°T

Example 2: Compass heading = 209°, Deviation = _4°, Variation = 20°W. Calculate the True heading. Remember that in this case you have to start from the left and work your way to the right. Remember that a negative value of Deviation (_4°) Westerly deviation 4°W

=

D M V 4°W => 205°M - 200W

T

7"T

?

= 185° True

=

Picture Supplements - Explanations

FIGURE 061-E98 1 :60 RULE - TRACK ANGLE ERROR An aircraft will often drift off-track because of unexpected wind effect. The actual wind en-route may turn out to be stronger or lighter than expected , or may blow from a different direction , or a combination of the two may occur, causing an error which is known as "Track Angle Error (TKE)". This error (TKE) is expressed in degrees left or right of the desired track. TKE can be calculated quickly by a very simple method - the "One in 60 Rule" (1 :60 rule) . The rule has several applications in navigation. Once found , the TKE can be used to alter heading to parallel required track or to alter heading direct to the next turning point en route , at the end of the leg. Using the 1:60 rule , The TKE is calculated first, then the correction required to regain track is calculated . The rule is approximate only and usually the figures are rounded off to whole degrees. It is nevertheless a very useful method of calculating a systematic error when the aircraft is found ~o be off-track , providing quick (or "snap''') changes of heading" instead of long and more complicated methods for correcting track errors. When a fix shows that the aircraft is not on the required track, the distance "off' track is measured a';{d compared with the distance run "along" track. The 1 : 60 rule is:

1 NM off-track for each 60 NM flown along track equals 1° track angle error.

In practice the following formula is used: TKE

Example: After flying for 83 NM a fix places the aircraft 11 NM off the intended track. Find the Track Angle Error (TKE).

Distance off Track Distance Flown Along Track

60 TKE =

"'{acY- _---\)a\ \ -p.ct _---

~------\~;E = ?O

Distance off Track x 60 Distance Flown Along Track

-=-.,------:,.---::-:----:=----,--

~

.t+ i , i

~1

Distance off-Track 11 NM

Planned Track 83 NM

TKE = (dist off-track x 60) .;. dist along track TKE = (11 x 60) .;. 83 TKE= 8°

1 :60 RULE - CORRECTIVE ACTION The first action is to stop the off-track error accumulating and an alteration of heading is immediately made onto the parallel track by turning towards track by the number of degrees off-track. In the case above, turn right 8° , making the new heading 8° greater than the previous heading. Altering Heading To Destination If an immediate turn to destination is required , then a further calculation is necessary and the distance to destination must be known. The distance to go should be available from the chart distance markings. The method is approximate as it assumes distances along track made good are the same as distances along planned track. Nevertheless, it is accurate enough for practical navigation and, again , the method works very well in practice. First, calculate the heading required to parallel track as above. Calculate the additional alteration of heading, using the same rule, but distance to go is substituted for distance along track in the formula. In effect, a second "track angle error" is calculated. Example: After flying 75 NM the aircraft is 5 NM off-track. Calculate the total alteration of heading required to reach destination if there are 150 NM to run.

TKE = (dist off-track x 60) .;. dist along track TKE = (5 x 60) .;. 75 TKE = 4° (a 4' alteration of heading will parallel the planned track)

...{acY\ -p.ct \la'_--

-""""!---: ----

_ ---~_---\ TKE = ?O

~

+

The further alteration to reach the destination is: TKE = (dist off-track x 60) .;. dist along track TKE = (5 x 60) .;. 150 TKE 2° (an additional 2' alteration of heading must be made to reach the destination =total of 4'+2' =6°)

Planned Track 75 NM

i Distance

i off-Track . ) 5 NM

=

- - - ___ _ ---_ TKE =

?: ]---- ______ •

Distance to destination 150 NM

Aviationexam Test Prep Edition 2012

FIGURE 061-E99 ESTIMATING DISTANCE FROM A FEATURE It is sometimes poss ible to obtai n an accurate range positi on line from a feature which is recognisable , but is too distant for accurate distance estimation by sight alone. The method would work well , for example , for a sing le easily identified mo untai n top wh ich is well below the aircraft. Example. While cruising at 6.000 ft MSL, an obstacle is seen on the right hand side of the aircraft. A sighting is taken giving a reading of 18' depression of the obstacle. Calculate the range of the obstacle from the aircraft.

t ......... :" ::--___-.----______

t----_

='"

i

i

Height

180)

---_

G.OOOft !

----_

:

---_

i~

! 90°

>

Tan 18° = Height.;. Range Tan 18' =6.000 ft.;. Range Range =6.000 ft.;. 0,3249 Range =18.467 ft =3 NM

--

----_

1,

1

,I" (180---_ - t-- -A/.\

Range to obstacle = ?

HEIGHTS AND RANGES ON THE GLiDEPATH Heights and ranges on the ILS glidepath are calculated using a standard 3° glideslope. Descent on the glideslope assumes a standard figure of 300 ft per nautical mile . Thus , at an outer marker at 5 NM from touchdown , the height AGL will be 1.500 ft. In order to stay on the glideslope , corresponding reductions of altitude must match the descent rate of 300 ftlNM , giving 4 NM at 1.200 ftAGL, 3 NM at 900 ftAGLetc. Ageneral rule of thumb to calculate the rate of descent necessary to achieve this figure is to multiply the Groundspeed by 5. Thus a Groundspeed of 130 kts requires a rate of descent of 650 feet per minute (130 kts x 5). For the glides lope with an angle other than 3° use the following method: Example: You are on ILS 4' glideslope which passes over the runway threshold at 50 feet. Your DME range is 25 NM from the threshold. What is your height above the runway threshold elevation? Height = sin angle x DME Range Height = sin 4' x 25 NM Height = 1,74 NM Height = approx 10.500 ft

t ......... -. . ~

Note that this height does not include the 50 ft over the runway => we have to add these 50 ft to get height above the ground.

Height = ? ft

--

Note: Remember that the DME range is the slant-range and not the ground range!

-----_ -_ 0""1: ran - __

~

ge:::<~

- - _

"IV""

- - - - -[ ;.- - - - - _ ....50 ft above .................................................................................................... .... Runway

If we were given the ground distance instead of the DME distance, we would use the following formula: Height = tan angle x Ground Range Height = tan 4' x 24,9 NM Height = 1,74 NM Height = approx 10.500 ft

t ......... :" ~

Height =?ft

................... .... ......... .... ....

~

..........

.... .........

. (~> . . . . . . . . . . .

----------------------------... Ground Range to obstacle = 24,9 NM

Runway

Picture Supplements - Explanations

FIGURE 061-El00 RHUMB LINE DISTANCE BASED ON LAT/LONG CO-ORDINATES Two points on the same line of Longitude

0 0 Let's say we have point A (60 N 20°W) and point B (50 N 200W) and need to find the distance between them - see the figure below. Since both points are located on a common line of Longitude (same Meridian - 200W) the solution is very easy. We can assume that 1 minute of Latitude 1 NM (1 ° of Latitude 60 NM) . Just calculate the difference in Latitude and apply this rule.

=

=

Distance Y: Y = Difference in Latitude (in minutes) • We can assume that 1 minute = 1 NM (60 minutes

CI)

= =60 NM )

1° 0 0 • In our case 60 N - 50 N = 10° difference = 600 minutes

u

c

~

'6

';

.-.

Y= 600 NM

II

>-

Two points on the same line of Latitude

0 0 Let's say we have point A (50 N 20°W) and point B (50 N 20° E) and need to find the distance between them - see the figure 0 below. Since both points are located on a common line of Latitude (same Parallel - 50 N) the solution is quite easy. We can assume that 1 minute of Latitude at the Equator = 1 NM (1 ° of Latitude = 60 NM). To apply this rule at "our" Latitude we also have to use a Cosine adjustment to this rule - see the procedure below:

Distance Z: Z

=distance

0 - - - - - - -0 B (50 0N 20° E)

A (50 0 N 200W)

Z = Change in Longitude (in minutes) x Cosine Latitude • In our case the total change in Longitude is 40° (200W - 0° - 200E) = 2.400 minutes (60 minutes = 1°) 0 • Latitude = 50 N Z = 2.400 x Cosine 50° Z = 2.400 x 0,6427

Z

=1.543 NM

Two points· no common line of Latitude or Longitude

0 0 Let's say we have point A (60 N 20°W) and point B (50 N 20° E) and need to find the distance between them - see the figure below. We can relatively easily calculate the distance between A and B using the Pythagorean theorem which would (in this case) be represented by a formula: X2 = y2 + Z2. In order to use this formula , we must first find the distance Y (between point A and the imaginary point C) and the distance Z (between point B and the imaginary point C):

Distance Y: Y = Difference in Latitude (in minutes)

A (SOON 20 0 W)

CI)

g

• We can assume that 1 minute = 1 NM (60 minutes = 1° =60 NM) 0 0 • In our case 60 N - 50 N = 10° difference = 600 minutes Y = 600 NM

.t~ (fist.

~ ---

ilI1Ce

v.

'6

Mean Latitude = 55W

Distance Z:

.q.~

II

>C

Z = distance CoB

=Imaginary point 0 (50 N 200W)

B (50 0 N 200E)

Z = Change in Longitude (in minutes) x Cosine Mean Latitude • In our case the total change in Longitude is 40° (200W - 0° - 200E) = 2.400 minutes (60 minutes = 1°) 0 0 • Mean Latitude = average between 50 N and 60 N = 55°N Z = 2.400 x Cosine 55° Z =2.400 x 0,573 Z = 1.375 NM

Distance X: y2 + Z2 X2 = 600 2 + 1.375 2

X2

=

X = " (360.000 + 1.890.625) X = " (2.250 .625) X = 1.500 NM

Aviationexam Test Prep Edition 2012

FIGURE 061-El0l TIME There is a solid and convenient association between movement of the Sun and time on Earth as the average rotational speed of the sun is 360° of longitude each day (one complete revolution of the Earth). For time-keeping purposes , this period is assumed to be standard and unchanging, ignoring the irregular speed of the sun in orbit. The sun therefore covers 15° of longitude per hou r, or one degree of longitude every four minutes. This useful rate of change of time versus longitude (arc/time) is used frequently in time calculations , although a table is given in Air Almanac which serves the same purpose . Time References Time is expressed in four main systems: • Local Mean Time (LMT) • Universal Co-ordinated Time (UCT/UTC)

'\

• Universal Time (UT) • Loca l Standard Time (LST) Local Mean Time (LMT) Time at a place needs to be related directly to the position (i.e. the longitude) of the sun in order to follow the regular routines of daily life. These routines require that a start to the day is made at or near sunrise and that the middle of the day will be coincident with the Sun passing overhead the local meridian. Similarly, the time for sleep is when the sun is over the horizon (at a different long itude west of the observer). Local Mean Time (LMT) is the time at a position (i.e . longitude) relative to sun pos ition . When the mean Sun transits the anti-meri dian of a position , the local day starts at 00:00 hours LMT. The sun transits the actual meridian of the position at local midday - 12:00 LMT. Crossing the anti-meridian again occurs at 24 :00 LMT, the end of the day. Simultaneously, 00:00 signifies the start of another day. It will be clear that all places on any given meridian have the identical LMT as time is regulated by the position of the Sun in relation to the local meridian. This is not the case for positions on Earth at different longitudes. LMT and Longitude The Local Mean Time at a position is required when calculating precise sunrise/sunset or twilight times at a particular longitude. Even though Local Standard Time (LST) may be the same at two pl
r-.. .s co!:: 0> co

t: C\j o ~ -J..c:

1---+-+--....... .

.c:"I' () II

Greenwich Mean Time (GMT) Greenwich Mean Time (GMT) is local time on the Greenwich meridian based on the hypothetical mean sun . As the Earth's orbit is elliptical and OOEfW its axis is tilted , the actual position of the sun against the background of stars appears slightly ahead or behind the expected position. The accumulated timing error varies through the year in a smoothly periodic manner by up to 14 minutes slow in February to 16 minutes fast in November. The use of a hypothetical mean sun removes this effect. GMT was used generally until 1925. At that time the reference point was changed from noon to midnight and it was recommended that, to avoid confusion , the term "Universal Time" should be used for the "new GMT". Universal Time (UT) Universal Time (UT) now has three separate definitions (UTO, UT1 , UT2) depending on which corrections have been applied to the Earth's motion. Authorities do not agree whether GMT equates with UTO or UT1 , however the differences between the two are of the order of thousandths of a second . GMT is no longer used for scientific pu rposes .

continues ...

Picture Supplements - Explanations

FIGURE 061-El02 ... continued Universal Co-ordinated Time (UTC) Universal co-ordinated time (UTCITUC ) is a time standard more accurately calculated than the local mean time at the Greenwich meridian (Greenwich Mea n Time) , but in practice it am ounts to the same thing . UTC is regulated against International Atomi c Time . Flight arrival and departure times are usually given in as local times (Local Standard Time) for passenger convenience . In civil aviation , however, all operational times are given using a common time datum - UTC - for air safety and to prevent ambiguity. Position reporting and estimating is always made using UTC. While this can result in apparent anomalies if UTC is compared with local time at the operating base , it is essential and is followed throughout civil aviation by all operators and agencies . The international time datum used is the Local Mean Time at Greenwich, in London . The international datum for longitude also passes through Greenwich. The time stahdard is known as Universal Co-ordinated Time (UTC) . UTC is the Local Mean Time at Greenwich and is identical fBrall aircraft everywhere. That is , no matter what local time at a place is , local mean time at Greenwich is fixed and is the same for all aircraft, no matter where they may be operating . All LMTs can be referred from the Greenwich meridian. LMT at Greenwich was for many years known as "Greenwich Mean Time" or GMT. The official term to be used now is "Universal Co-ordinated Time" - UTC. "GMT" is still in use at some places , but it means the same time. "Z" (Zulu) time refers to the time zone in which Greenwich lies and is also interchangeable with "GMT and "UT" . In actual practice, while it may not be strictly accurate from a scientific viewpoint, the terms "Greenwich Mean Time", Universal Time", "Universal Co-ordinated Time" and "Zulu" time , or their abbreviations , are used interchangeably as they all mean the same time effectively.

Converting UTC to and From LMT Converting UTC into LMT and vice versa is straightforward , using the fixed relationship between long itude and the time of 15° difference per hour. Any difference in LMT between places is caused by their difference in longitude. When Local Mean Time at Greenwich is 12:00 (12:00 UTC), Local Mean Time at longitude 015°W will be one hour earlier (11 :00 LMT - the sun has not yet reached the meridian of 015°W) , but UTC at the position is still 12:00. A good, reliable and straight-forward rule to use when converting to and from UTC and or LMT is: • Longitude east - UTC least (smaller) • Longitude west - UTC best (greater) When converting from one time to another, places east of Greenwich have an LMT greater (best) than Greenwich as the sun arrives there earlier. The time difference will therefore be added to UTC to obtain LMT (and vice versa). Similarly, places west of Greenwich have an LMT earlier (least) than Greenwich because the sun has yet to reach them. To find LMT of a place west of Greenwich , subtract the arc/time difference from UTC. Example: LMTat Hong Kong (114°E) is 17:15 on 25 December. Calculate the Universal Time . Solution: • Universal time is the LMT at Greenwich. Convert LMT at Hong Kong to LMT at Greenwich. • Hong Kong: 25 day 17:15 LMT • Longitude 114°E""" 15° - 07:36 oflongitude to time difference • Greenwich: 25 day 09:39 LMT at Greenwich (UTC) . • Hong Kong (114°E) is east of Greenwich, so in this case, the time difference is subtracted from LMT to obtain UTC (LMT at Greenwich) . (Longitude east =UTC least).

Conversion of LMT at One Position to LMT at Another Position In order to convert LMT at one position to LMT at another Vancouver position, change the first LMT into UT and then change 01 Oct 11:21 LMT this UT into LMT at the second position, using the "arc to 01 Oct 19:32 UTC time" rule at 15° hour. This method avoids uncertainty in deciding whether to add or subtract the corrections (arc converted to time). When additional calculations are necessary, such as sunrise or sunset etc, always perform the calculations in UTC and then convert to LMT or LST as appropriate at the end of the calculation . This techn ique has many advantages, one of which is that it avoids pondering the question of the International Date Line, which will be referred to later. It will also reduce the chance of making errors and simplify calculations. Example: LMT at Nairobi (037"E) is 22:00 on 01 October. Calculate LMT at Vancouver BC (122°55'V\? at the same time.

Nairobi 01 Oct 22:00 LMT 01 Oct 19:32 UTC

continues ...

Aviationexam Test Prep Edition 2012

FIGURE 061-El03 ... continued Solution: 01 day 22:00 LMT at Nairobi • Oateltime group at Nairobi: - 02:28 Arc of longitude to time • Longitude 37'E.;. 15 • (longitude east, UTC least) • Greenwich: 01 day 19:32 LMT at Greenwich (UT) . • Longitude 122'55'W.;. 15 ·08:11 Arc oflongitude to time • (longitude west, UT best) 01 day 11:21 LMT at Vancouver. The LMT at Vancouver is therefore 11:21 on 01 October.

Local Standard Time (LST) Local Standard Time (LST) is a convenient system where everywhere in a country, or in a time zone in a country keeps the same time as everywhere else . If Local Mean Time were '\used throughout a country, particularly one with a large east/west spread , life would be very difficult if not impos'S'ible (except for places north/south of each other) as all LMTs would be different. The system to counteract this problem is called Local Standard Time. Standard time is the time kept by a country. It is decided by the government of that country alone . It can also be the time in a particular zone of a country with a large change of longitude. Usually, it is the LMT of the centre of the country adjusted to a convenient figure ahead or behind UTC . The value chosen will usually reflect the passing of the Sun at midday. Time Zones The earth is theoretically divided into 24 time zones, each 15° of longitude wide , straddling a meridian divisible by 15 (15, 30 , 45 , 60 etc). Each zone is numbered with a letter. Thus time zone "Z" (Zulu) straddles the Greenwich meridian , from 00r30'E to 00r30'w. The next zone, "A", lies eastwards from 00r30'E to 022°30'E. East of this zone is zone "8" , also 15° of longitude wide . Nairobi in Kenya lies in time zone 37,5°W "C", which is 3 hours ahead of Greenwich (the time at Greenwich is three hours earlier). Standard time in Kenya is therefore three hours later than UTC . This is the same as saying that LST in Kenya is three hours fast on UTC.

7,5°W

OOOOEIW

7,5°E

Most countries adopt the standard time of the time zone in which the country is situated , but exceptions may exist. Countries with a wide change east/west spread within the national boundaries will have several time zones in the country. Examples are the USA and Russia . In practice , LST will often be very close to LMT, as the standard time in use corresponds to the sun's passage over the country. Standard Time Factor (STF) Another way of converting to LST is by using a Standard Time Factor (STF).The standard time factor is applied to LST to obtain UTC or vice versa : • LST = UTC +/- STF. The Standard Time Factor may be provided or can be taken from the tables in the Air Almanac. These tables are self-explanatory and easy to use, and advise whether to add or subtract the STF to LMT. Examination of the tables shows that STF need not be an exact number of hours. The STF for Nepal fo r example is 5 hours 45 minutes. If a country does not appear in the most obvious list, check the other pages , as political or geographic requirements may not find it on the obvious page. Example: An aircraft is scheduled to depart Johannesburg at 17:55 UTC. STF Solution: 'ETD ·STF ·ETD

=2 hours. Calculate the LST of departure.

17:55 UTC 02:00 (longitude east, UTC least) 19:55 LST

Standard time corrections should be checked from current documents such as the AlP, NOTAMs, Air Almanac or AERAD / Jeppesen publications. Daylight Saving Time In the summer months in certain countries , daylight saving time may be in force. Daylight saving time is usually one hour faster than standard time. Details for individual countries will be found in the appropriate AlP or NOTAMs.

continues ...

Picture Supplements - Explanations

FIGURE 061-El04 ... continued Time Terminology Civil aviation is notorious for the use of abbreviations, many of which may be unfamiliar. Certain abbreviations are in common use in matters dealing with time in the aviation world and they should be known and recognised. • Estimated Time of Departure = ETD • Actual Time of Departure = ATD • Estimated Time of Arrival ETA • Actual Time of Arrival ATA

=

=

International Atomic Time Atomic time is based on counting cycles of a high-frequency electrical signal which is kept in resonance with an atomic transition. The fundamental unit of atomic time is the SI second. The time scale based on atomic clocks is International Atomic Time (TAl). This time scale is maintained by the International Bureau of Weights and Measures (BIPM) and is based on the intercomparison of about 200 frequency standards located around the world which are combined to form a standard time scale.

Co-ordinated Universal Time (UTC) Although TAl provides a continuous, uniform and precise time scale for scientific reference purposes, it is not convenient for everyday use because it is not in step with the Earth's rate of rotation. A time scale that corresponds to the alternation of day and night is much more useful, and since 1972, all broadcast time services distribute time scales based on Co-ordinated Universal Time (UTC). UTC is an atomic time scale which is kept in agreement with Universal Time. Leap seconds are occasionally added or subtracted from UTC to keep it within 0.9 seconds of UT1. As mentioned previously, GMT, UT, UCT and Z times may be regarded by practical aviators as the same for aviation purposes.

The International Date Line (IDL)

0 The local date of a place always changes at 24:00 LMTILST. It will also change at longitude 180 EIW (the International Date Line) when the sun crosses the Greenwich (Prime) meridian. The IDL follows the general direction of the Greenwich anti-meridian. When an aircraft crosses the line, the local date for passengers and crew changes. If the flight is from the western hemisphere to the eastern hemisphere (say from 1700W to 170° E) on the 4 October, the date will change to 5 October when crossing the Greenwich anti-meridian. If the flight is from 1700E to 1700W and the flight departs on the 5 October, the date will change to 4 October when crossing the Greenwich anti-meridian. It is perhaps fortunate for some countries that the Greenwich ante-meridian runs mainly through the Pacific Ocean with a few zigzags to keep all islands in a group under the parent country or the main island. This avoids a country having two different days in different parts of the state. For example, many Pacific south sea islands are part of the New Zealand time system for mutual convenience. A traveller eastbound crossing the notional IDL from one island to another does have to reset his watch as he stays inside the same time zone. A properly labelled diagram will always help clarify the situation. Fortunately, however, this rather tricky situation becomes immaterial if all calculations are made in UTC, as the problem of which particular day it is will be automatically covered. Example: The scheduled time for a flight from an airfield at longitude 160 0 30'E to an airfield at 167°15'W is 6 hrs 15 mins. ATD is 09:00 LMT on 10 October. Calculate the ETA (LMT) for the destination airfield. Solution: 'ATD

09:00 LMT 24:00 ... it is clear that UT is on the preceding day 09 day 33:00 LMT • 160 30'E 10:42 - Longitude east UT least 'ATD 09day 22:18UTC • Flight time 06: 15 + Sector time • ETA 09 day 28:33 UTC • 167°15W 11:09 - (Longitude west UT best) • ETA 09 day 17:24 LMT • The ETA appears to be the day before departure. This is in fact the case, because the LMT is that of the destination airfield. This situation exists because the International Date Line (IDL) is crossed during the flight, which is eastbound (into the sun). 0

10 day

Aviationexam Test Prep Edition 2012

FIGURE 061-El05 SUNRISE, SUNSET, TWILIGHT Sunrise and Sunset Sunrise or sunset occurs when the upper edge of the sun is on the observer's horizon. Certain atmospheric conditions can affect the times of sunrise and sunset and the length of twilight. If the air is clear, the sun will rise or set quickly. When the air is thick with dust or water particles, it is occasionally possible to see the sun after it has "officially" set or before the official time of sunrise because of refraction of the sun's rays. The amount of refraction will determine how early or late a sunrise or sunset will be and how long a sunrise or sunset will last. It will be noted from a study of the SUnrise 1 Sunset tables that the amount of daylight, as well as twilight, will vary from season to season at different latitudes. Thus, for example, in January at high northern latitudes, there is no daylight at all in extreme latitudes. This is because the sun's rays do not reach beyond the arctic circle then because of the angle of inclination of the earth's polar axis to the sun. Another example - in very high northern latitudes the sun does not set for certain periods in the northern summer. This is because declination of the sun is high and the Arctic circle receives the sun's rays for 24 hours every day during this time.

c:::=J )

and when the sun stays below the The tables indicate when the sun is overhead for 24 hours by a clear box ( horizon by a solid black box ( _ ). A series of oblique lines ( / / / I) indicates that twilight lasts all night. Sunrise and sunset vary little at the equator and occur every day at 06:00 118:00 LMT plus or minus a few minutes. Tables for Sunrise (SR), Sunset (SS) and Twilight Times Times of sunset and sunrise and the beginning of civil twilight are tabulated in the Air Almanac. The times are given in UTC for the Greenwich meridian at three-day intervals. The times may be taken, without significant error, as the LMT of the phenomena for any other meridian, since UTC is LMT at Greenwich. To obtain the UTC of a phenomenon at a particular place, the longitude must be converted to time (by the rule of 15° per hour) and applied in the correct sense to the UTC time. The times are given for a particular latitude and date. This is because the sun rises (and sets) at the same LMT for all places on the same latitude. The times of SR and SS at any place will not change greatly over successive days, except at high latitudes. The times tabulated in the tables for specific latitudes may therefore be taken as the same for all longitudes. The LMT for a particular date and place (latitude) is extracted and then converted to the time standard (LST/LMT) required. Sunrise, sunset and twilight calculations are time calculations based on the LMT of sunrise, sunset or twilight. Using the Tables The major difficulty when using these tables is usually the monotony of interpolating. Dates are listed every three days and at latitude increments varying from 2° at high latitudes to 10° at low latitudes. Usually it will be sufficiently accurate to use the times given for the nearest tabular date, or, to find the following or preceding phenomenon, it is sufficiently accurate to add or subtract 24 hours, changing the Greenwich date by one day. It will be seen that values in high latitude change much faster than in lower latitudes and linear interpolation will then be necessary. Depending on the accuracy required, it is usually sufficient to interpolate to the nearest minute. Method: • • • •

Select the correct page for extracting data Select the nearest date for the phenomenon Extract data against latitude and interpolate as necessary. Correct for longitude and UT/LST/LMT as necessary.

Example: Calculate the UTC of the beginning of sunrise, sunset and morning and evening civil twilight on 1 January 2000 at position 33°20'N 135°39'E. Solution: • Use the Air Almanac - the column for January 2nd as the nearest date. Latitude 35°N 0 30 N

Twilight 020640 020630

Sunrise 020708 020656

Twilight 021728 021738

Sunset 021700 021712

03°20'

0007

0008

0007

0008

33°20'N ArcJtime

0206:37 - 09:03 0121:34

0207:04 - 09:03 0122:01

0217:31 - 09:03 0208:28

0217:04LMT -09:03 UTC 0208:01

• The first two times are in fact on the previous day (I January) at E135 39. To find the times on 2 January, add 24 hours and change the date to 2 January. The four times therefore will be: 21:34, 22:01, 08:28, 08:01 aI/ falling on 2 January. A short sketch with the position of the sun will resolve any difficulty. If local standard time is required, then apply the standard time factor for the country or zone concerned to UTC.

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Picture Supplements - Explanations

FIGURE 061-El06 ... continued Twilight There are three types of twilight: • Civil twilight, when the sun is 6° below the horizon • Nautical twilight, when the sun is 12° below the horizon • Astronomical twilight, when the sun is 18° below the horizon. Twilight is an arbitrary condition, it is a period when there is sufficient light to distinguish general outlines of ground objects, although the horizon may not be visible. Evening twilight is a time to exercise great caution when airborne, because the visibility is continuously reducing with time and can be deceptive.

Duration of Civil Twilight Civil twilight is the value used in civil aviation. Morning civil twilight starts when the sun is 6° below the horizon and lasts until sunrise, when the upper limb of the sun rises to the horizon of the observer. Evening civil twilight is of course the reverse condition. These are the tabulated times in the Air Almanac.

Calculation of the Duration of Civil Twilight The duration of civil twilight is from sunset to the beginning of civil twilight in the evening and from the start of civil twilight to sunrise in the morning. It can readily be calculated from a knowledge of these times. Civil twilight starts

Sunrise

Duration of twilight

Civil twilight starts

Sunset

Duration of twilight

06:45

07:04

19 minutes

18:55

19:17

22 minutes

In June, at the summer solstice, when northern declination is maximum and the sun is overhead the Tropic of Cancer, twilight lasts all night at certain latitudes in the Arctic. The reverse happens in the southern hemisphere (Antarctic). At the December solstice, the converse happens and the situation is reversed in the hemispheres. The interval between sunrise and sunset will vary with the seasons and at different latitudes.

Aviationexam Test Prep Edition 2012

FIGURE 061-El07 POINT OF EQUAL TIME (PET) I CRITICAL POINT (CP) The Point of Equal Time (PET) or sometimes referred to as the Critical Point (CP) is a point along the route of flight from which it will take the same time to either continue to the destination airport or to return to the departure airport. It will allow the pilot to make a quick decision for example in case of a medical emergency on board or any technical problem such as an engine or a system failure. From PET it will take the same time to either return back to departure or to proceed to the destination. This is the easiest way to visualize this point - however, it does not always have to involve the departure and destination airports - it could be a point along the route of flight from which it will take the same time to fly to any two suitable airports - for example en-route alternates. When considering a frlight from point A to point B, the PET will not be right in the middle between A and B, because we are most likely going to experience some sort of wind during the flight, headwind or a tailwind. The PET will be precisely calculated with respect to this wind. If an emergency arises before reaching the PET it will be faster to return to point A. If an emergency arises after passing PET it will be faster to continue the flight to B.

Calculation of "All engines operating" PET between 2 points For the calculation of PET between points A and B we will use the following abbreviations: Dist GS Out GS Home

= Distance between A and B = Groundspeed out (from A to B) = Groundspeed back home (from PET back to A)

To calculate the PET we will use the following formula: Distance from A to PET = (Dist x GS Home) + (GS Out + GS Home) Time from A to PET = Distance to PET + GS Out

Calculation of "One engine inoperative" PET between 2 points For the calculation of PET, at which an engine failure occurs, between points A and B we will use the same variables as above, but we have to realize that at the point of the engine failure our speeds will change => with one engine inop both the speeds to continue to destination and to return to departure airports will be lower. Dist = Distance between A and B GS OutSE = Single Engine Groundspeed from PET continuing to B GS HomeSE = Single Engine Groundspeed from PET back to A To calculate the Single Engine PET we will use the following formula:

=

Distance from A to SE PET (Dist x GS HomeSE) + (GS OutSE + GS HomeSE) Time from A to PET = Distance to PET + GS Out ( ... in this case we use the all-engine GS Out)

Effect of wind on PET In still air conditions the PET would be located in the middle between point A and point B. However, in reality we almost never have absolutely still air. Therefore, if we experience a headwind on a flight from A to B, the PET will move closer to the the destination (closer to B) - into the wind. If we experience a tailwind on a flight from A to B, the PET will move closer to the departure (closer to A). In both cases, the stronger the wind, the greater the PET move away from the mid-point towards the particular direction.

Important thing to realize when calculating PET First of all we need to realize that we have to use the groundspeeds for the PET calculations. Many JAA questions will give you the lAS or CAS together with altitude, temperature and wind information. You need to calculate your TAS first and then apply the wind correction to obtain your GS. If the question does not state the specific headwind or tailwind component, but only the course and wind info, we can easily determine the wind components using the fomula: cos (wind angle) 0 x wind speed". For example, if we have a true course of 350 0 and the wind is reported from 320 at 40 kts, it means that 0 0 our wind angle is 30 (headwind from the left) => cos 30 x 40 kts => approx. 35 kts headwind.

Picture Supplements - Explanations

liW1J;JI,@O§N:J POINT OF SAFE RETURN TIME (PSR) I POINT OF NO RETURN (PNR) Imagine that you are flying a long transcontinental flight without any en-route or destination alternate airport available. Some time into the flight you find out that the weather conditions at your destination airport have suddenly deteriorated. You have to make a decision whether to continue the flight to the destination airport or to return back to your departure airport. Do you have enough fuel to return back? This type of problem can be solved using the determination of the Point of Safe Return (PSR). At any time before reaching this point you will be able to return to the departure airport (or departure alternate airport) and land with the sufficient fuel reserve still in your tanks - that means within the Safe Endurance limits of your aircraft. The term "Safe Endurance" means the time for which you can fly without using the minimum required fuel reserves. Safe Endurance is typically listed in terms of time, althought in the JAA questions you may encounter it as a fuel quantity and a fuel flow value from which you can calculate the time. Do not confuse "Safe Endurance" with "Total Endurance". Total Endurance represents the time for which the aircraft can remain airborne - at the end of this time the tanks are completely empty, without any kind of fuel reserve. Note: This point has been previously referred to as the Point of No Return (PNR) , but this was inferred to mean a return with no fuel reserve. Nevertheless one was accustomed to hearing "The PNR for a return with a fuel reserve of "x". The expression "Point of Safe Return" is now preferred.

Calculation of PSR between 2 points using Endurance given as time For the calculation of PSR between points A and B we will use the following abbreviations: Endur GS Out GS Home

= Endurance in Hrs (flight time to having only reserve fuel available) = Groundspeed out (from A to B) = Groundspeed back home (from PSR back to A)

To calculate the PSR we will use the following formula: Time from A to PSR = (Endur x GS Home) + (GS Out + GS Home) Dist from A to PSR = Time to PSR x GS Out

Calculation of PSR between 2 points using variable fuel flow values For the calculation of PSR between points A and B we might be given the total quantity of fuel on board (in kg) along with different fuel flow values for the flight to PSR and for the return from PSR. In this case we can use the following procedure to solve these problems: Fuel Qty FC Out FC Home

= Fuel quantity on board in Kilograms, excluding the required reseves. = Fuel Consumption from A to PSR - in terms of kg of fuel per NM = Fuel Consumption from PSR back to A - in terms of kg of fuel per NM

To calculate the PSR we will use the following formula: Dist from A to PSR = Fuel Qty + (FC Out + FC Home) Time from A to PSR = Dist to PSR + GS Out

Effect of wind on PSR In still air conditions the PSR would be located furthest from the departure airport. However, in reality we almost never have absolutely still air. Therefore, if we experience any kind of wind component - irrespective whether it is head or tail wind, the distance from departure airport to PSR will be smaller (PSR will move closer to the departure airport => you will have to turn back sooner if it becomes necessary). The greater the wind, the greater the distance by which the PSR moves closer to the departure airport.

Important thing to realize when calculating PSR First of all we need to realize that we have to use the groundspeeds for the PSR calculations. Many JAA questions will give you the lAS or CAS together with altitude, temperature and wind information. You need to calculate your TAS first and then apply the wind correction to obtain your GS. If the question does not state the specific headwind or tailwind component, but only the course and wind info, we can easily determine the wind components using the fomula: cos (wind angle) 0 x wind speed. For example, if we have a true course of 350° and the wind is reported from 320 at 40 kts, it means that our wind angle is 30° (headwind from the left) => cos 30° x 40 kts => approx. 35 kts headwind.

Aviationexam Test Prep Edition 2012

FIGURE 061-E109 AIRCRAFT MAGNETISM AND COMPASS SWING The aircraft magnetism which produces the deviation error in th e compass is a combination of hard iron magnet ism (permanent magnetism) and soft iron magnetism (tem pora ry magnetism). During a compass sw ing proce dure th ese errors are resolved into two coeffi cients: "8 " and "C". A furthe r error known as Coefficient "A" ca n exist w hich is dealt with during th e compass swingi ng procedu re.

Hard Iron (Permanent) Magnetism Aircraft hard iron magnetism is permanent magnetism mostly induced by the earth's magnetic field , especially at the aircraft assembly plant during construction . During the assembly process , the aircraft might be subject to a certain amount of beating and riveting as part of the process , although this feature has been greatly redu ced with modern construction te chniques. Any vibration or banging on the aircraft structure while the aircraft is stationary will align the aircraft magnetic molecules with the earth field in the same way that happens to an ordinary magnet. The result of the banging and vibrating is that any hard metal in the aircraft will be magnetised permanently to a certain extent. The polarity and strength of this permanent magnetism depends on the aircraft's heading during manufacture. That part nearest the local Magnetic North will have a red pole and that part nearest the local Magnetic South will have a blue pole . If the aircraft is coincident with the local magnetic meridian , the nose and front half will be the north-seeking (red pole) and the rear will be the blue pole .

Component P (longitudinal axis)

Hard iron aircraft magnetism can be separated into three components . Each component can be compensated fo r separately: • Longitudinal (nose <=> tail) axis: component "P" • Lateral axis (wing tip <=> wing tip): component "Q" • Vertical axis: component "R" Note: It is normal for the aircraft and the compass to be level in horizontal flight, so component "R" (the vertical component) has no effect on an aircraft in level flight and is therefore ignored.

The amount and effect of aircraft hard iron magnetism varies with several factors : • magnetic latitude (angle of dip) at the assembly plant • aircraft heading during assembly • the amount of magnetic material in the aircraft. • the vibration and banging during assembly.

Component Q (lateral axis)

Soft Iron (Temporary Magnetism) Soft iron magnetism mainly occurs when soft iron type metal in the aircraft becomes magnetised by the earth 's magnetic field . A secondary cause is from the aircraft radio and electrical circuits producing electro-magnetic fields . This magnetism can be further analysed into three components : • Longitudinal (nose <=> tail) axis: component "cl" • Lateral axis (wing tip <=> wing tip): component "fl" • Vertical axis : component "kl"

Combined Soft + Hard Iron Magnetism Combining hard and soft Iron components produces the following total aircraft magnetism co-efficients : • Longitudinal (nose <=> tail) axis : P + cZ = Coefficient B • Lateral axis (wing tip <=> wing tip): Q + fZ = Coefficient C • Vertical axis: R + kZ = Coefficient D Note: Coefficient "0 " is not compensated for as it is in the vertical plane.

Coefficients Band C During a compass swing , aircraft magnetism is separated into its horizontal components : • co-efficient 8 (longitudinal = fore-and-aft) • co-efficient C (lateral = athwartships) . After calculation and analysis of deviation in the compass system various corrections are made to the compass system to

continues ...

Picture Supplements - Explanations

FIGURE 061-El10 ... continued compensate for the errors produced by co-efficients Band C. The corrections are made by means of micro-adjusters, which are small magnets attached to the compass itself and which exert a force in he opposite plane of the deviation which annuls it. The co-efficients are the calculated quantities of deviation caused by P and Q respectively. The effects of vertical soft iron are also included. Co-efficient B is the algebraic value of the calculated deviation produced by component P and the allied vertical soft iron when heading East or West. The effect of co-efficient B on compass heading on North and South is zero as B is the fore-and-aft component and have no effect on these headings. In the same way, co-efficient C, the athwartships (lateral) effect, is maximum on North/South and minimum on EastlWest headings.

Correcting for Coefficients Band C Coefficient B is calculated from the deviations observed on East and West. The value of B is maximum on these headings while the value of coefficient C is zero.

Compass Card

Example:

For

000 090 180 270

Coefficient B = (deviation on East - deviation on West).;. 2 Coefficient B (E-W) + 2

=

Coefficient C is calculated from the deviations observed on North and South. The value of C is maximum on North/South headings while the value of Coefficient B is zero. Coefficient C = (deviation on North - deviation on South) .;. 2 Coefficient C (N-S) + 2

=

Steer Deviat. _4° 004 +4° 086 +8° 172 _2° 272

=

Coefficient B (dev. on E - dev. on Coefficient B = ((+4°) - (_2°)) + 2 Coefficient B W + 2°) + 2 Coefficient B =6° + 2 =3°

II\.? + 2

=

Coefficient C = (dev. on N - dev. on S) + 2 Coefficient C = ((_4°) - (+8°)) + 2 Coefficient C = (_4° - 8°) + 2 Coefficient C =_12° + 2 =_6°

The Micro-Adjuster The micro-adjuster incorporates two small permanent magnets which are mounted one above the other. In the neutral position, one pair of magnets lies fore and aft and the other pair lies athwartships. The adjuster is so arranged that when an adjuster key is inserted and turned, opposing bevel gears in the mechanism rotate in opposite directions opening the magnets and producing a corrective field. Coefficient A A further coefficient can exist in the compass - coefficient A. The error from this coefficient will be constant on all headings and can be resolved into two components: "real A" and "apparent A". "Real A" (Magnetic) Real A is produced by horizontal soft iron indistinguishable from apparent A, but is seldom of any appreciable value. It is maximum where "H" is maximum => that is at the magnetic Equator. Real A is zero at the poles as "H" is zero therefore, the compass is unusable. "Apparent A" (Mechanical) Apparent A can be a significant error but can easily be removed. It is a mechanical error caused by a mis-aligned lubber line (reference mark) which should be aligned exactly with the aircraft fore and aft axis. Apparent A gives constant deviations on all headings. Coefficient A is found by averaging the sum of the deviations on eight headings. It may also be calculated on 4, 6 or 8 headings but they must be reciprocal headings. In this case, the result would not be so accurate as that obtained from 8 headings.

Correcting for Coefficient A Coefficient A is caused by an incorrectly aligned lubber line and is corrected by rotating the compass back or forward against the lubber line by the required number of degrees. RULE: • For positive values of A, turn the compass clockwise. • For negative values of A, turn the compass anti-clockwise. • Any heading can be used to alter the lubber line setting.

MH

CH

Deviation

005 045 084 130

007 048 089 132

_2° _3°

177

177

225 270 320

223 268 320

S _2° 0° +2° +2° 0°

Total deviations are _8°. Coefficient A = _8° + 8 = -1°. The compass is adjusted on 320° to a new heading compensating for co-efficient A => anti-clockwise to 319°.

Aviationexam Test Prep Edition 2012

FIGURE 061-Elll COMPASS ERRORS In the search for accuracy of an indicating system, it is often found that the methods used to counter an undesirable error under one set of circumstances create other errors under different circumstances. This is precisely what happens when the compass system is made pendulous to counteract the effect of magnetic dip by displacing the CG and thus making the instrument effective over a greater latitude band. Unfortunately, having done this, any manoeuvre which introduces a component of aircraft acceleration either east or west from the aircraft's magnetic meridian will produce a torque about the magnet system's vertical axis, causing it to rotate in azimuth to a false meridian.

Acceleration Errors These errors occur during airspeed changes and are most apparent on headings of East and West. They are caused by a combination of inertia and magnetic dip. As the aircraft accelerates, the compass card, acting like a pendulum, tilts slightly during the acceleration because of the card's inertia. The force applied by an aircraft when accelerating or decelerating on a fixed heading is applied to the magnet system at the pivot, which is the magnet's only connection with the remainder of the instrument. The reaction to the force must be equal and opposite and must act through the CG which is below and offset from the pivot (except at the magnetic equator). The two forces thus constitute a couple which, dependent on heading, will cause the magnet system to change the angle of dip or to rotate in azimuth. Figure "A" on the right shows the forces affecting a compass needle when an aircraft accelerates on a Northerly heading. Since both the pivot "P" and CG are in the plane of the local magnetic meridian, the reactive force "R" will cause the northern end of the system to dip further, thus increasing the angle of dip without any needle rotation. Conversely, when the aircraft decelerates on north, figure "8", the reaction tilts the needle down at the south end. The opposite of these reactions will be observed when accelerating/decelerating - on north along the meridian in the Southern Hemisphere.

~

N

I

..

DIP Angle

___ _

P

- - --..S

C.G."

• R

A - Acceleration on East Heading

N---~ -------- -

P

- __- . . . -

..

"

- - S

................... ...

R~.C.G.

8 - Deceleration on West Heading

N When an aircraft flying in either hemisphere changes speed on headings other than north or south, the change will result in azimuth rotation of the magnet system, and hence there will be errors in heading indication. When an aircraft flying in the Northern Hemisphere, accelerates on an easterly heading / P (decelerates on a Westerly heading), as in figure "A" on the -"E R right, the accelerating force will act through the pivot "P", and, I c.G. unless the value of "Z" is zero, the reaction "R" will act through the CG. The two forces will now form a couple, turning the / needle in a clockwise direction. Action of "R" will also cause z Figure "A" Figure "8" the magnet system to tilt in the direction of acceleration, and thus the pivot and CG will no longer be in line with the magnet meridian. The magnets will come under the influence of "Z", as shown in figure "8" above, providing a further turning moment in the same direction as the force "P/R" couple. When the aircraft decelerates on east, the action and reaction of "P" and "R" respectively will have the opposite effect, causing the assembly to turn anti-clockwise with all forces again turning in the same direction.

Summary of Acceleration I Deceleration errors

• • • • •

Heading

Speed

Needle turns

Visual Effect

East West East West

Increase Increase Decrease Decrease

Clockwise Anti-clockwise Anti-clockwise Clockwise

Apparent turn Apparent turn Apparent turn Apparent turn

to to to to

north north south south

In the Southern Hemisphere, errors are in opposite sense. Similar errors can occur in turbulent flight conditions. No error on North or South headings as reaction force acts along the needle. No error on magnetic equator, as value of "Z" is zero and hence pivot and CG are co-incident. Good mnemonic is "ANDS" => Accelerate North, Decelerate South.

Picture Supplements - Explanations

FIGURE 061-E112 Turning Errors When an aircraft executes a turn, the compass pivot is carried with it along the curved path of the turn. The centre of gravity (of the magnet system) being offset from the pivot to counter the effect of "Z" is subject to the force of centrifugal acceleration acting outwards from the centre of the turn. Further, in a correctly banked turn the magnet system will tend to maintain a position parallel to the athwartship (wing tip to wing tip) plane of the aircraft and will therefore now be tilted in relation to the earth's magnetic field. As before, the pivot and CG will no longer be in the plane of the local magnetic meridian. The needle will be subject to a component of "Z", as shown in figure below, causing the system, when in the northern hemisphere, to rotate in the same direction as the turn and further increase the turning error.

4fC-=~~~ N

A

~~~-=~-~~

Direction of Turn

N

c

B

z

z

The extent and direction ofTurning Error is dependent upon the aircraft heading, the angle of bank (degree of tilt of the magnet system) and the local value of "Z" (dip). However, turning errors are maximum on north/south and are of significance within 35° of these headings. For example, an aircraft flying on a northerly heading in the Northern Hemisphere. The north-seeking end of the compass needle is coincident with the lubber line. The aircraft now turns east. As soon as the turn is commenced, centrifugal acceleration acts on the system CG. This causes it to rotate in the same direction as the turn. Since the magnet system is now tilted, the earth's vertical component "Z" exerts a pull on the northern end, causing further rotation of the system. The same effect will occur if the heading change is from north to west in the Northern Hemisphere. The speed of system rotation is a function of the aircraft's bank angle and rate of turn, and, depending on those factors, three possible indications may be registered by the compass: • Turn in a correct sense, but smaller than that carried out - magnet system turns at a slower rate than the aircraft. • No turn - magnet system turns at the same rate as the aircraft. • A turn in the opposite sense - magnet system turns at a faster rate than the aircraft. When turning from a southerly heading in the Northern Hemisphere onto east or west, the rotation of the system and indications registered by the compass will be the same as when turning from north, except that the compass will over-indicate the turn. In the Southern Hemisphere the south magnetic pole is dominant and, in counter-acting its downward pull on the compass magnet system, the CG moves to the northern side of the pivot. If an aircraft turns from a northerly heading eastward, the centrifugal acceleration acting on the CG causes the needle to rotate more rapidly in the opposite direction to the turn, thus indicating a turn in the: correct sense but of greater magnitude than that actually carried out. The turn will thus be over-indicated. Turning from a southerly heading onto east or west in the Southern Hemisphere will, because the CG is still north of the compass pivot, result in the same effect as turning through north in the Northern Hemisphere. Summary of Turning errors Turn Direction Through north Through south

Needle Movement Same as aircraft Opposite to aircraft

Visual Effect Under indication Over indication

Liquid Swirl Adds to error Reduces error

Corrective Action Turn less than needle shows Turn more than needle shows

• Good mnemonic is "UNOS" => Undershoot North, Overshoot South • In the "N" Hemisph, when turning through a N heading => pilot must undershoot. • In the "N" Hemisph, when turning through a Sheading => pilot must overshoot. • When turning through an E or W heading, the turning error is zero.

Liquid Swirl Errors So far no mention has been made regarding motion of the liquid in the compass bowl. Ideally it should remain motionless to act as an effect damping medium preventing compass oscillation (aperiodicity). Regrettably, this is not so and the liquid turns with and in the same direction as the turn; its motion thus adds to or subtracts from needle error, depending on relative movement.

Aviationexam Test Prep Edition 2012

FIGURE 061-El13 POSITION LINES (P/Ls) A position line is a line along which the aircraft was at a particular time. It can be straight, curved or even an irregular line if it is a river or a road. It is not an definite position itself but it marks a line of position where the aircraft is orwas at a particular time. For example, crossing a railway line, a canal or a highway or the coast would give a line of position which can then be plotted on the chart. What can be said is that the aircraft was somewhere along the position line at the time of the observation. This is true even though it is not exactly certain whereabouts on the position line it actually was. Although a P/L is not a fix, it is very useful for navigation. When a P/L is plotted on the chart, it is usually marked with a single arrow at each end indicating it is a position line. Radio Position Lines (Radio Bearings) The most common P/L used in navigation is the radio P/L. A radio P/L can be a bearing from or to a VOR or VDF radio station. It can also be an NDB bearing measured by ADF or RMI. Except for a QTE from a VDF station some work will have to do be done on the bearing before it can be plotted. Only QTEs are plotted in degrees true without change. DME Position Lines (Range P/Ls) Another type of radio P/L is a DME reading, or range position line. A DME P/L is plotted as the arc of a circle, i.e. the radius of the DME distance from the DME station.

1

o

[:] ............... .

Uses of Position Lines P/Ls can be used for many purposes: • Groundspeed checking When a P/L crosses track at 90° (±100) it can be used for checking the average groundspeed since the last fix or groundspeed check. The new groundspeed found is then used for checking the latest ETA and any other groundspeed purposes until a more recent groundspeed is found. Method: The distance "run" along track is measured from the most recent fix to the point on track where the P/L cuts at 90°. This distance is measured and used for a groundspeed check.

• Revising ETA When a P/L crosses track at 90° (±100) it can be used for checking the average groundspeed since the last fix or groundspeed check. The new groundspeed found is then used for checking the latest ETA and any other groundspeed purposes until a more recent groundspeed is found. Method: The distance "run" along track is measured from the most recent fix to the point on track where the P/L cuts at 90°. This distance is measured and used for a groundspeed check. Technique: - Plot the P/L to cut the estimated track. - Measure the distance "run" along the estimated track - Enter in the nav log - Divide this distance by the time since the last fix - Calculate groundspeed - The new GS must be used to calculate future times and ETAs as it is the most recent information.

• Track checking When a radio station is overflown, a bearing from the station will give the Track Made Good (TMG). This is the actual track of the aircraft as compared with the planed or desired track and is known as a back bearing. Method: At 10:00Z aircraft is overhead "ABC", heading 145°M, Var 9°W. At 10:15Z "ABC" bears 183° Relative. TMG in degrees true? Heading 145°M =136°T. Heading 136°T + 183°Rel =Bearing 319°T (to the station) =139°T from =TMG.

• Fixing aircraft position with two or more position lines. Fixing Position With Position Lines Position lines can be used to construct fixes. Two or more position lines may be used. The simplest form and most convenient form is when two simultaneous position lines are obtained and the ideal combination is when they intersect at right angles. This gives the greatest accuracy. When position lines cut at a shallow angle, there is a greater "band of error". A common example is a radio bearing and reading from a DME station both at the same position ("co-located").

continues ...

Picture Supplements - Explanations

FIGURE 061-E114 ... continued Transferring Position Lines It is unusual to be able to obtain simultaneous position lines. It is more normal to use two or three successive P/Ls. Three lines give greater accuracy. These may be from a beacon when flying near to or abeam it or from other sources.

When position lines are obtained at different times, a technique known as "transferring the position line" is used. All P/Ls are transferred along track to the time of the last P/L. If all is well, they will intersect at one position with no error. If there is error, a small triangle is formed called a "cocked hat" because of the shape. The intersection, or centre of the cocked hat is the position of the fix.

A Good Fix (No Cocked Hat)

Method: The time chosen for the fix will be given or can be the time of the latest (most recent) position line. The position line or lines obtained before the final line of the fix are moved "forward" along the aircraft DR track, at the DR groundspeed to the time of the latest position line. This distance is called the "run". In other words, if position lines are obtained at 14:30,14:40 and 14:50, the P/Ls for 14:30 and 14:40 are "run" along the estimated track at the estimated GS to 14:50. This type of fix is sometimes called a "running fix". Example: Overhead position XY at 17:05 en route to "DEF". Estimated GS 390 kts. The following radio bearings are obtained: 17:14 ODR from "ABC" of 206° Transferred 17:17 ODR from "ABC" of 239° Position 17:20 ODR from "ABC" of 276°. Lines Plot the aircraft position at 17:20. Variation 8°W

'i

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, .........

/~

"fo

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Solution: ,; .-;;P • Convert the ODRs to OTEs and plot the three position lines, ....... 17:20 writing the time alongside each • Var 8° W: OTEs are 199°, 231° and 268°. • Observe that the 17:17 position line cuts track at approximately 90°. It must therefore be used for a groundspeed check, based ••••••••••:tI' on the principle of using the most recent information available. • Distance XY (at 17:05) to 17:17 position line is 60 NM. Time elapsed is 9 minutes. GS is 400. • Use the GS found at 1717 to transfer the position lines for 17:14 and ./ 17:17 17:17to time 17:20. This Pos. line transfelTed i • The 17:14 PIL moved along the track for 6 minutes at 400 kts = 40 NM. 20 NM along the track i • The 17:17 PIL moved along the track for 3 minutes at 400 kts =20 NM. from the 17:17 point i • Position lines intersect at the fix position for 17:20 N5323 W00402. • The fix at 17:20 could be used for wind-finding, ETA revision or for 17:14 altering heading - whichever is required. This Pas. line transferred 40 NM along the track from the 17:14 point • Note: A PIL which has been transferred is marked with two alTows.

=

Cocked Hat The accuracy of the fix depends very much on how accurate the groundspeed used for transferring the position lines is. This groundspeed is based on forecast winds. If the groundspeed used to calculate the run is out, then the three position lines will not intersect at the same point. They will form a small hollow triangle which is called a "cocked hat". If your cocked hat is on the large side, say the sides are about 80 NM, the chances are that there is a serious error. Recheck your calculations and measurements. The fix position is always taken as the centre of the cocked hat.

Transferring the Station An alternative method to transferring the position line along track is to transfer the actual radio station itself. Whether or not to use this method is a matter of personal preference. The radio station is moved along a track parallel to the planned aircraft track. Each position line is then plotted normally from the transferred station or "base".

When a DME position line has to be transferred, the DME station must be transferred - there is no other way to do it.

I

Xy

The Cocked Hat

20 minutes of GS along a track parallel to the aircraft track

Aviationexam Test Prep Edition 2012

FIGURE 061-E115 THE VECTOR TRIANGLE The vector triangle is most important as it is the basis of all air navigation and affects the aircraft on every flight. For success in plotting, as well as in every other field of navigation, a thorough understanding of the vector triangle is necessary. Most plotting techniques rely intrinsically on the properties of the vector triangle. The vector triangle is also called: • the navigation triangle, • the triangle of velocities, • the wind triangle.

Speed, Velocity and Vectors Speed is the ratio of distance against time. Velocity is speed in a stated direction. A vector quantity has length (or speed) and direction. If two vectors are added in the correct sense (vectorially) - that is with one acting from the end of the other, their combined effect produces another vector. This third vector is called the "resultant". Thus the result of adding a vector of 100 kts in direction 090° and a vector of 100 kts in direction 000° "head-to-tail" is the vector of 141,4 kts in a direction 045°.

Vector A 100 kts along 090·

Knowledge of two vectors enables the third vector to be found. The solution for any vector problem can be found by drawing and measuring or by trigonometry. Note: Vectors must be added (combined) in the correct sense the sense of the vector quantities will be lost.

=the head of one vector must lie off the tail of another vector otherwise

The Vector Triangle The vector triangle is composed of three vectors. Each vector has two components. The three vectors are: Vector • Air Vector • Ground Vector • Wind Vector

Components True Heading (TH) + True Airspeed (TAS) True Track (TT) + Groundspeed (GS) True Wind Direction + Wind speed (WN)

The components of each vector are always coupled, it is not possible to have say, groundspeed and true heading in one vector. If any 4 of the 6 vector components are known, the other 2 can always be calculated.

The Air Vector The 2 components of the Air Vector are the True Heading (TH) and the True Air Speed (TAS). The air vector is the actual path of the aircraft as it moves through the air expressed in terms of direction and velocity. In still air conditions (zero wind), heading and track are the same as are TAS and groundspeed. The distance travelled along the air vector is measured in "air nautical miles" (also called the "still air distance" or "nautical air miles" or "NAM"). In fact the two components of the air vector (true heading and true airspeed) are not normally available directly except for aircraft fitted with True Air Speed Units (TASUs) and a direct output of true heading from certain advanced types of compass. Usually TAS and TH may be calculated from: • Compass or Magnetic heading • Variation / Deviation • Rectified airspeed or Mach number • Pressure altitude • Outside air temperature.

) By convention the Air Vector carries one arrow pointing in the direction of aircraft heading.

The air vector is the product ofTAS and true heading = it represents how far the aircraft flies through the air on a certain heading and at a certain airspeed. Any change in TAS or TH will alter the length and or the direction of the air vector. True Heading If true heading (Hdg T) is not known or given, as is usually the case, it can be calculated from the magnetic heading (MH) or the compass heading (CH) by applying variation and/or deviation as appropriate. Remember the C - D - M - V - T) mnemonic rule. True Airspeed If True Air Speed (TAS) is not given, it can be calculated from knowledge of the pressure altitude (flight level), the (ambient) actual outside air temperature (OAT) and rectified airspeed (RAS/CAS), or Mach number and OAT. Density altitude is seldom encountered today. The common abbreviation for Knots True Air Speed is KTAS.

continues ...

Picture Supplements - Explanations

FIGURE 061-E116 ... continued The Ground Vector The 2 components of the Ground Vector are True Track (TT) and Groundspeed (GS). The True Track (TT) is a line connecting two positions. A straight line is usually drawn between two actual positions on the chart (e.g. departure and destination airfields, or points on the route - "waypoints"). The line is the track required to be flown, which is sometimes called the "course" in the USA. It may be a Great Circle Track (on a Lambert's chart) or a Rhumb Line Track (on a Mercator chart).

By convention the Ground Vector carries two arrows pointing in the direction of aircraft track.

A line joining two positions which have been overflown is called the "Track Made Good (TMG)". This is the actual track over the ground or water which has been followed by the aircraft, it is the result of the actual wind and the heading and TAS flown. A line representing the desired track to fly, which has been calculated or estimated is called the "DR" (or planned) Track.

=

Note: The expressions "Course" and "Track" should not be confused. Some years ago "course" used to mean heading the direction in which an aircraft points, but it is sometimes used now to mean the desired path over the surface (= required track). This is especially so in the USA. To avoid ambiguity it is perhaps better to avoid the expression completely and use the terms 'Track" as defined above and "Heading": the direction in which an aircraft is pointing, or the fore and aft axis of the aircraft.

The Groundspeed (GS) is the speed of the aircraft over the ground or water. It can be the estimated or the actual groundspeed. Actual groundspeed can be found by noting the time taken to overfly two ground positions (fixes), measuring the distance between them and dividing by the time taken to obtain the average groundspeed. Groundspeed can also be calculated by drawing, or using a navigation computer to solve the Vector Triangle. When groundspeed is calculated before the flight or a sector of the flight it is known as the DR groundspeed or estimated groundspeed. Some aircraft are fitted with navigation equipment which gives an instantaneous reading of track and groundspeed. The Wind Vector - Wind Speed and Direction (WN) The 2 components of the Wind Vector are Wind Direction and Wind Speed. By convention the Wind Vector carries three arrows pointing Wind is the movement over the surface of an air mass in which an aircraft is flying. down wind. The effect of wind at altitude is not easily appreciated at first. Wind velocities of 200 kts are not at all uncommon in the higher levels. These winds can make a huge difference to the expected performance of an aircraft if they differ from predicted values. Navigation can become difficult if the existence of such winds is not soon appreciated, although it is also true to say that knowledge of the wind is important to accurate navigation anywhere.

The components of the wind vector are the wind direction in ° True (the direction from which air is moving) and the wind speed in Kts. If wind velocity is not given, it can be calculated from knowledge of HOG, TAS, GS and Track or Drift. Practical Examples of the Vector Triangle use in Navigation In the vector triangle, the wind vector acts on the air vector and the resultant is the ground vector. If there is no wind (no wind vector therefore) the ground vector and the air vector are the same because there is no "wind effect". It could be said that the wind "blows" the aircraft from its air position to its ground position. When the hand-held navigation computer, or an electronic computer is used to solve for wind or another related problem such as track made good etc, this is actually solving the vector triangle. Finding the Track I Drift and Groundspeed - knowing the Heading, TAS, WN Example: WN = 090·/25 kts, Heading = 340· T, TAS = 180 kts. Find Track and GS. Using units of a suitable scale, the true heading is drawn to represent one hour of flight. The wind vector is then applied in the correct sense to the end of the air vector. The resulting ground vector is measured and calculated to give track and groundspeed. Select an appropriate scale e.g. 1 cm = 10 kts 110 NM. Draw in a long line in direction heading 340° T (heading) from the origin (point A). Mark off point B = 1 hour of flight (180 NM) at 18 cm from A. From the end of the air vector A-B, draw the wind vector downwind (towards 270°). Mark off point C = the wind effect for one hour at 25 kts from point B (2,5 cm). Connect the end of the wind vector B-C and the beginning of the air vector (point A) with a line - this is the ground vector A-C. 7) Measure the resultant ground vector (A-C) in scale units => this is the groundspeed => 19 cm = GS of 190 kts. 8) Measure the direction of the ground vector (A-C) => this is the true track resulting from the action of the wind on the aircraft (air vector) => 333° True Track (7°L Drift angle).

True N

1) 2) 3) 4) 5) 6)

A

continues ...

Aviationexam Test Prep Edition 2012

FIGURE 061'-E117 ... continued Finding the Heading and Groundspeed - knowing the Track, TAS, WN Example: WN = 030°/20 kts, Track = 090° T, TAS = 150 kts. Find Heading and GS. A vector triangle is constructed using units for one hour. First, the track line (direction only) is drawn. From a point at the end of the track line, the wlv is plotted upwind for one hour. From this point an arc, radius one hour ofTAS is drawn to cut the track line. By measuring the result, the heading required and the groundspeed can be determined.

True !II

~_~_~~------

Hading (?J" + TAS 1

e _____________>--------

.....----A

Track 090· + GS (?) kts

__ C WN 030°/20 kts

B

Select an appropriate scale e.g. 1 cm = 10 kts 110 NM. Draw a long line in direction 090° T. Select a point at the end of this line (point B) and draw the wind vector upwind (towards 030° T). Measure B-C at 20 kts (2 cm) along this line, towards 030°, from the track direction line. This one hour of wind velocity. 5) From point C, describe an arc radius 15 cm (one hour at TAS 150 kts) to cut the track line at point A. 6) A-B is now the ground vector, A-C is the air vector. 7) Measure the direction of A-C => this is the heading => 083°T. 8) Measure the length of A-B, this is groundspeed => 139 kts. 9) The angle between the line A-B and the line A-C is the drift => - 7° (R).

1) 2) 3) 4)

Finding the Wind Direction and Speed - knowing the Heading, Track, TAS, GS. Example: Heading 235°, TAS 280 kts. After 30 miutes of flight a distance of 165 NM has been covered and the Track Made Good (TMG) is found to be 245° T. Find Wind Direction and speed.

=

=

The air vector and the ground vector are constructed. Since the ground vector is where the aircraft is after 30 minutes of wind, the direction from air vector to ground vector represents the wind direction and the distance along this line is the wind effect for 30 minutes.

True N

=

1) Select an appropriate scale e.g. 1 cm 10 kts 110 NM. 2) Draw a line on a heading of 235°T from a suitable point => this becomes the air vector. 2) Measure 140 NM (30 minutes at 280 KTAS) = 14 cm along the air vector => this is the air position. 3) Draw in the track made good of 245°T from the start point => this is the ground vector. ---~---4) Measure 165 NM (16,5 cm) along the ground vector. This is 30 WN minutes of flight along the track made good (TMG) => ground ? position after 30 minutes. 5) Connect air and ground positions. 6) Measure the direction from air vector to ground vector along this line => wind direction of 107°. 7) Measure the distance along this line from air vector to ground vector (3,65 cm). This is 30 minutes worth of wind effect - 36% kts => multiply by 2 => wind velocity of 73 kts.

Summary These techniques are the same techniques as are used on the navigation computer, either the hand-held "whizz wheel" or an electronic version. The computer simply solves the navigation (vector) triangle, maybe a little faster. These techniques can be used (with modifications) to solve any part of the vector triangle, so long as four of the quantities are known. The vector triangle may be compared to a swimmer crossing a fast-flowing stream. If he points towards the opposite bank and swims directly towards it, he will be swept downstream. This is rather like simply flying a heading and seeing where the wind will take you. If the swimmer aims upstream in the right direction however, he will eventually cross to the opposite bank. This is just like flying a heading in order to achieve a particular track.

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