2b_rivera_a3_2015-2016

Technological University of the Philippines Ayala Blvd., Ermita, Manila College of Engineering Department of Civil Engineering

CE 25-2A Elementary and Higher Surveying

Assignment No. 3 Leveling

Rivera, Ericka Shane L. 14-205-007 Date of Submission: Mar. 12, 2016

Engr. Jesus Ray M. Mansayon Instructor

PROBLEM SET NO. 3 1. TWO PEG TEST. In the two-peg test of a dumpy level the following observations were taken: with level set up near A, observed readings are a=1.855m and b=1.434m; with level set up near B, c= 1.790m and d=2.211m. Determine the correct reading on the rod held at A with the instrument still in the same position at B for a horizontal line of sight. Given: Conventional Method ROD READING On point A On point B

INSTRUMENT NEAR A a = 1.855 m b = 1.434 m

INSTRUMENT NEAR B d=2.211 m c = 1.790 m

Required: d’ Illustration:

d

c

d’ a

Solution: d’ = c + TDE TDE = DEa + DEb 2 DEa = a-b DEa = 1.855 – 1.434 DEa = 0.421 TDE = 0.421 d’ = 1.790 + 0.421 d’ = 2.211 m

DEb = d-c DEb = 2.211 – 1.790 DEb = 0.421

2. TWO PEG TEST. The two-peg test is used to determine if the line of sight of the telescope is parallel to the axis of the bubble tube. A tilting level is set up halfway between points A and B and rod readings on both points are 1.969m and 2.161 m, respectively. The level is then set up 1.50m away from A along the extension of the line AB and 53.50 from B. Then rod readings on A and B were recorded as 1.554m and 1.728m, respectively. Determine if the line of sight is inclined downward or upward and the amount by which the horizontal cross hair should be moved to adjust the line of sight.

Given: ROD READING On point A On point B

INSTRUMENT at M a = 1.969m b = 2.161 m

Required: inclination of the line of sight Illustration: d

c

d d’

c’

a

b B

A

M

P Solution: |

|

Assuming LOS upward ,

Equation 1,

Equation 2,

INSTRUMENT at P c = 1.554m d = 1.728m

,

LOS IS INCLINED DOWNWARD 3. DIFFERENCE IN ELEVATION. A dumpy level is set up and leveled approximately halfway between two points, C and D. The elevation of point C is 135.00m above the main sea level and the reading on a rod held at this point is 1.52m. If the reading on the rod held at point D is 2.74 m, determine the difference in elevation between the two points. Also, determine the elevation at point D. Given: el. of C = 135.00m rod reading at C = 1.52m rod reading at D = 2.74 m

Required: TDE and el. D Illustration:

1.52 C

2.74 Elev. at C=135 m D

Solution: |

|

4. DIFFERENCE IN ELEVATION. The backsight reading on a point is 2.25 m and the foresight reading on a second point is 0.86 m. If the elevation of the first point is 460.96 m, determine the elevation of the second point. Given: BS1 = 2.25 and FS1 = 0.86 el. 1 = 460.96 m Required: el. 2 Illustration:

TDE

0.86 2.25

TDE 460.96

Solution |

|

5. DIFFERENCE IN ELEVATION. Reading on a rod held at two point A and B, 75 m apart, are 2.965 m and 1.215m, respectively. Determine the rod reading at points on line AB which are 25 m and 45 m from A. Assume that there is a uniform gradient along line AB.

Given: AB=75 m RR at A= 2.965 m RR at B= 1.215 m A to B= 75 m A to C= 25 m A to D= 45 m Required:

RR at C, RR at D

Illustration:

B

A

TDE D C

C

Solution:

m

6. CURVATURE AND REFRACTION. Determine the combined effect of curvature and refraction on level sights of 30, 50, 100, 250, 300, 500, 1000 and 2000 m. Given: level sights 30, 50, 100, 250, 300, 500, 1000, and 2000 m Required: h Solution:

7. CURVATURE AND REFRACTION. A backsight of 3.055m is taken on a point 75m from a level station and foresight of 1.258m from the same station. Determine the difference in elevation between the two points, taking into account the combined effect of curvature and refraction.

Given: hcr1 = 9.45 m and hcr2 = 1.32m Required: K Illustration:

Solution:

8. CURVATURE AND REFRACTION. Determine the backsight or foresight distance for an instrument set up which will cause an error due to the earth’s curvature and atmospheric refraction equal to 0.0015m, 0.0575 m, 0.0986m, 0.2935 m, and 0.8750m.

Given:

Required: Illustration:

Solution:

√

√ √ √ √

3.600 km

9. CURVATURE AND REFRACTION. An observer standing on shore can just see the top of a lighthouse 16.5 km away. If the eye height of the observer above sea

level is 1.735m, determine the height of the lighthouse above sea level. Neglect the effect of the waves. Given: distance = 16.5 km Eye height = 1.735m Required: height of lighhouse Illustration:

1.735 m

16.5 km Solution: 2 1

10. CURVATURE AND REFRACTION. Determine how far ( in km ) out from shore an inter-island vessel will be when a red light on its deck, 9.45 m above the water, disappears from the line of sight of a child standing on shore and whose eye level is 1.32 m above the water.

Given:

hcr1 = 9.45 m and hcr2 = 1.32 m

Required: K Illustration:

1.32 9.45 Solution:

√

√

11. CURVATURE AND REFRACTION. By taking a sight across a lake 24 km wide through a pair of binoculars, determine the height of the shortest tree on the opposite shore whose tip the observer can see. Assume that the eyes of the observer are 1.675 m above the shore line on which he stands. Given: Required:

hm= 1.675m and K= 24 km height of shortest tree

Illustration:

1.675

?

24 km Solution:

√

√

12. CURVATURE AND REFRACTION. Two towers, A and B, are located on level ground and their bases have equal elevations above sea level. Determine the distance between the two towers if a person standing on top of tower A can just see the top of tower B. His eyes are 15.50 m above the ground. Given: hcr = 15.50m Required:

K

Illustration: 15.5m

B

A

Solution:

√

√

13. CURVATURE AND REFRACTION. A lifeguard is standing on a watchtower at the edge of the water such that his eye level is 3.50 m above sea level. If a life raft stars floating out to sea, determine how far out it could go before it disappears from the sight of the lifeguard. Assume that the sea is calm and a pair of binoculars is available to tract the movement of the raft.

Given: Required:

hcr=3.50m K

Illustration:

3.50m

Solution: √

√

PROBLEM SET NO. 4 1. DIFFERENTIAL LEVELLING. Complete the differential level notes shown below and perform the customary arithmetic check. Given: STA BM10 TP1 TP2 TP3 BM11 TP4 TP5 TP6 TP7 BM12 TP8 TP9 BM13

Required: Solution:

HI and Elev

BS 2.085 2.015 1.864 0.579 0.423 1.446 1.778 2.051 2.920 3.186 2.805 0.774

HI 139.535 m 140.568 m 141.004 m 140.056 m 138.056 m 137.695 m 137.748 m 137.460 m 139.375 m 140.203 m 142.013 m 141.581 m

FS 0.982 1.428 1.527 2.423 1.807 1.725 2.339 1.005 2.358 0.995 1.206 0.603

ELEVATION 137.450m 138.553 m 139.140 m 139.477 m 137.633 m 136.249 m 135.970 m 135.409 m 136.455 m 137.017 m 139.208 m 140.807 m 140.978 m

2. DIFFERENTIAL LEVELLING. In running a line of differential levels from BM1 to BM2, the following rod readings were taken in order given: 2.40, 1.30, 1.43, 0.98, 1.25, 0.87, 1.85, 1.05, 2.94, 2.08, 1.69, 2.50, 1.61, 2.71, 0.75 and 2.11 meters. If the elevation of BM1 is 445.25 m, complete the level notes and show the customary check on the computations. Given: rod readings are : 2.40, 1.30, 1.43, 0.98, 1.25, 0.87, 1.85, 1.05, 2.94, 2.08, 1.69, 2.50, 1.61, 2.71, 0.75 and 2.11 meters El. BM1 = 445.25m Required: Level notes & customary check Solution: STA BM-1 TP-1 TP-2 TP-3 TP-4 TP-5 TP-6 TP-7 BM-2

BS 2.40 1.43 1.25 1.85 2.94 1.69 1.61 0.75

HI 447.65 447.78 448.05 449.03 450.92 450.53 449.64 447.68

FS 1.30 0.98 0.87 1.05 2.08 2.50 2.71 2.11 ∑

∑ ∑

∑

ELEV 445.25 446.35 446.8 447.18 447.98 448.84 448.03 446.93 445.57

3. DIFFERENTIAL LEVELLING. Prepare and complete the differential level notes for the information shown in the companying illustration, Include the customary arithmetic check. Given: STA BM-1 TP-1 TP-2 TP-3 TP-4 TP-5 BM-2

BS 1.74 1.28 1.85 1.66 1.83 3.05 ∑

Required: Illustration:

HI and Elev

HI 570.49 568.32 566.85 565.72 563.63 563.97

FS 3.45 3.32 2.79 3.92 2.71 2.08 ∑

ELEV 568.75 m 567.04 565 564.06 561.8 560.92 561.89

Solution:

HI=570.49m

m 4. DIFFERENTIAL LEVELLING. The companying schematic arrangement of lines portrays a differential level route. The values indicated represent backsight and foresight readings taken at different points as labeled. Prepare and complete level notes for the information shown and perform the customary arithmetic check. Given: STA BM-1 TP-1 TP-2 TP-3 TP-4 BM-2

BS 3.25 1.75 2.08 2.63 2.22

HI 103.595 102.515 102.545 103.385 104.275

11.93 Required:

Elev of BM

Illustration: FS=1.33

FS 2.83 2.05 1.79 1.33 3.15 11.15

ELEV. 100.345 100.765 100.465 100.755 102.055 101.125

Solution:

l

5. DOUBLE-RODDED LEVELLING. Set up and complete the level notes for a double-rodded line from BM45 to BM46. In the following rod readings H refers to the stations along the high route and L refers to stations along the low route: BS on BM45=2.238, FS onTP1(H)=0.703, FS on TP1(L) = 1.252, BS on TP1(H)=2.855, BS ON TP1(L) = 3.402, FS on TP2(H) = 1.173, FS on TP2(L) = 1.558, BS on TP2(H)=2.542, BS on TP2(L)=2.932, FS on TP3(H) = 1.339, FS on TP3(L)=1.660, BS on TP3(H) = 2.390, BS on TP3(L) = 2.711, and FS on BM46=1.015 meters. Assume that the elevation of BM45 is 524.550m. Perform the customary arithmetic check. Given: STA BM45 TP1(H) TP1(L) TP2(H) TP2(L) TP3(H) TP3(L) BM46

BS (m) 2.238 2.855 3.402 2.542 2.932 2.390 2.771

HI

FS (m) 0.703 1.252 1.173 1.558 1.339 1.660 1.015

ELEV (m) 524.550

Required:

Fill up missing info

Solution:

∑ ∑ ∑ ∑ ∑

∑

= 530.345 ∑

∑

6. LOCATING STATIONS. Determine the distance from station 4 + 38.85 to the following stations: 8 + 68.42, 16 + 50.56. 2 + 73.05, 0 + 69.08 and 36 + 10.14. Given: 4 + 38.85, : 8 + 68.42, 16 + 50.56. 2 + 73.05, 0 + 69.08 and 36 + 10.14 Required: Distance Solution:

m

7. PROFILE LEVELING. Complete the following set of profile level notes and show the customary arithmetic check. Given: STA BM-a TP-1 1+00 +25 +50 +75 TP-2 2+00 +25 +50 TP-3 BM-b

BS 1.24 1.11 3.21 3.06 -

HI 235.75 233.46 234.21 234.53 -

∑

FS 3.40 2.46 2.74 2.40

IFS 4.6 2.8 3.6 2.4 1.2 3.2 2.8 -

∑

ELEV 234.51 m 232.35 228.86 230.66 229.86 231.06 231 233.01 229.81 227.01 231.47 232.13

Required: complete the table and arithmetic check Solution: ∑

∑

8. RISE AND FALL METHOD. Using the rise and fall method, complete the level notes given below and show the customary arithmetic check. Given: ROD READINGS BS 1.73

IFS 2.05 1.88

FS

DIFFRENCE IN ELEV RISE FALL 0.32 0.17

REDUCED LEVEL

STA

766.45 m 766.13 766.3

BM 13 0+00 +35

1.75 2.39 3.25 2.44 1.86 1.09 2.15 1.84

0.13 0.64 0.86 0.81 0.58 0.77 1.06 0.48

2.63 0.77 1.19 1.08 2.52

1.07 0.42 0.11 1.44 2.14

0.38 ∑

Required:

766.43 765.79 764.93 765.74 766.32 767.09 766.03 765.55 766.62 766.2 766.31 764.87 765.25

+60 +80 1+00 +15 +40 +90 2+00 TP-1 2+45 +68 3+00 +50 BM 14

∑

Complete level notes and arithmetic check

Solution: ∑

∑

9. RISE AND FALL METHOD. Given below is the schematic arrangement of lines of a level route run from BM-1 to BM-2 I which intermediate foresights to a number of points have been observed, it is desired to determine the elevation of BM-2 and also the elevations of the intervening points. Prepare and complete the appropriate format of level notes for the portrayed information using the rise and fall method of leveling. Also show the customary arithmetic check. Given: ROD READINGS BS 2.363

IFS 1.234 2.168 3.229 1.784

FS

DIFFRENCE IN ELEV RISE FALL 1.129 0.934 1.061 1.445

REDUCED LEVEL

STA

362.45 m SS 362.645 361.584 363.029

BM-1 a b c d

2.335 1.765

0.551 0.899

1.436 1.102

0.663 1.128 1.776 2.305 1.763 2.986

0.465 0.648 0.529 0.542 3.280

1.517

1.541 2.010 2.058 3.519 1.022

1.445 0.469 0.048 1.461 2.497 2.695

1.673 ∑

∑

362.478 363.914 365.016 364.551 363.903 363.374 363.916 362.399 363.844 363.375 363.327 361.866 364.363 362.690

e CP1 f g h i j CP2 k l m n o BM-2

Required: elev of BM-2 and arithmetic check Solution: ∑

∑

10. RECIPROCAL LEVELING. Reciprocal leveling between points A and B, located on opposite banks of a wide river gives the following readings in meters. From the first setup near A: on A, 0.993; on B, 2.076, 2.077, and2.078. For the set up near B the readings are: on B, 2.549; on A, 1.463, 1.462 and 1.463. If the elevation of A is 925.28 meters, determine the true difference in elevation between the two points and elevation of B. Given: a=0.993 m a’=2.076, 2.077 and 2.078 m b=1.463, 1.462 and 1.463 m b’=2.549 m Required: Elev.B and TDE

Illustration:

TDE

Elev=925.28 A

B

Solution:

Elev B = Elev of A + TDE Elev B = 925.28 – 0.4485 Elev B = 924.8315 11. RECIPROCAL LEVELING. In reciprocal leveling across a deep ravine the following sets of od readings were taken: Instr Near A: BS on A = 1.719m, Average FS on B = 2.578m Instr Near B: Average BS on A= 1.335 m, FS on B = 2.176m Point A is one side of the ravine and has a known elevation of 197.428 m. Point B is on the other side. Determine the elevation of point B Given: el A = 197.428 Instr Near A: BS on A = 1.719m, Average FS on B = 2.578m Instr Near B: Average BS on A= 1.335 m, FS on B = 2.176m Required: Elev B Illustration:

TDE

Elev=197.428 A

B

Solution: Instrument Set-up Near A STA BS FS a 0.993 b(ave) 1.335

DE

-0.342

Instrument Set-up Near B STA BS FS b’ 2.176 a’ 2.578

DE

0.402

12. TRIGONOMETRIC LEVELING. Two points, A &B, are 3,134.50 meters apart. From the third point, C, on the line between A and B, and 1,992.25 meters from A, the measured vertical angle is +35°28’ and that to B is -15°14’. Determine the difference in elevation between A and B making due allowance for the effects of curvature and atmospheric refraction. Given and Illustration:

A DEAC DEAB

+35°28’ -15°14’

DECB

C B 1 992.25 m 3 134.50 m

Required: DEab Solution:

(

)

(

)

13. TRIGONOMETRIC LEVELLING. In problem 12, and still considering the effects of curvature and refraction, determine the difference in elevation between the same two points if the measured vertical angles of A and B are instead +28°30’ and +6°10’, respectively.

Given and Illustration:

A DEAB

DEAC

B +28°30’

DECB

+6°10’

C

1 992.25 m

3 134.50 m

Required: DEAB Solution:

(

)

(

)

14. ALTIMETER SURVEY. Given the following data gathered from an altimeter survey: Elevation of the high base, 818m; elevation of the low base 422 m, altimeter readings at the high base is 6579; and altimeter reading at the low base

3333. If the altimeter reading at a field station is 4775, determine its corresponding elevation. Given and Illustration:

RDG=6579 m

DiffHL

DiffFL

DiffHF

RDG=4775 m

HB DEHF

FS

DEHL=818 m DEFL

RDG=3333 m

LB Elev.LB=422 m

Required:

Elev of FS

Solution:

15. INVERSE LEVELING. A line of inverse levels was run into a shaft for a copper mine. All of the stations in the accompanying tabulations were located in the shaft’s ceiling and rod readings were taken by inverting the leveling rod. Complete the given level notes and perform the customary arithmetic check.

Given: STA BM 66 TP-1 TP-2 TP-3 BM 67 TP-4 TP-5 BM 68

BS 2.84 2.43 2.76 2.19 3.05 1.45 0.96

HI 101.46 100.68 100.4 98.91 96.81 97.02 98.14

FS 1.65 2.48 0.70 0.95 1.66 2.08 3.06

∑

ELEV 104.30 m 103.11 103.16 101.1 99.86 98.47 99.1 101.2

∑

Required: Elev of BM68 Solution: ∑

∑

16. INVERSE LEVELING. It is desired to determine the elevation of the bottom of the protruding concrete beam. In so doing, the instrumentman observes a backsight reading of 1.25 m on a rod held in a normal position on a bench mark whose known elevation is 225.26m. The rod is next is held upside down against the bottom of the beam and foresight reading of 2.32 m is observed. Determine the required elevation. Given: BS=1.25 and FS=2.23 Required: elev of B Illustration: B

A Elev.=225.26

Solution:

17. INVERSE LEVELING. A backsight of 1.47 m is read on a rod held upside down against BM-1 (elev=315.15 m) on the ceiling of a mine. Then a foresight of 1.52 m is taken on TP-1 on the floor with the rod in normal position. The instrument is next transferred and setup at another station and a backsight of 1.45 m is read on TP-1. Determine the elevation of BM-2 set on the ceiling if the foresight of 2.12 m is read on it with the rod again held upside down. Given and Illustration: Elev.BM-1=315.15 m

BM-2

BM-1

TP-1

Required:Elev of BM-2 Solution: STA BM-1 TP-1 BM-2

BS 1.47 1.45

HI 313.68 313.61

FS 1.52 2.12

ELEV 315.15 m 312.16 315.73

Elev.BM-2=315.73 m 18. LEVELS OVER ONE ROUTE. Starting a BM-42 whose known elevation is 810.15 m, a closed level circuit was run to set the elevations of several benchmarks as shown in the accompanying tabulation. Adjust these observed elevations. Given: Elevation observe= 809.90 Elevation known= 810.15

Required:

Adjusted values of each

Solution: STA

BM 42 BM 43 BM 44 BM 45

DISTANCE FROM BM 42 0.000 km 2.525 6.428 8.050

OBSERVED ELEVATION

STA

810.15 m 843.29 876.40 793.85

BM 46 BM 47 BM 48 BM 42

DISTANCE FROM BM 42 10.005 km 16.112 18.234 20.500

OBSERVED ELEVATION 805.22 m 815.14 813.75 809.90

=

= ( ) (

)

(

)

(

)

(

)

(

)

(

)

(

)

so:

19. LEVELS OVER DIFFERENT ROUTES. Six lines of levels are run over different routes to establish the elevation of BM-X as shown in the accompanying tabulation. Determine the most probable elevation of the bench mark.

Given: ROUTE

LENGTH

OBS ELEV OF BM-X

ROUTE

LENGTH

a b c

4.15 5.20 4.76

246.85m 246.63 246.70

D E R

5.82km 6.03 4.98

OBS. ELEV OF BM-X 246.75m 246.56 246.69

Required: most probable elevation Solution:

(

) (

( )

) (

(

)

(

)

)

20. LEVELS OVER DIFFERENT ROUTES. By Route 1 (3.0 km long) point b is 21.200 meters higher than point A, By route 2 (4.0 km long) B is 20.972 meters above A and by route 3(6.0 km long) B is 21.261 meters above A. Determine the following: a. The most probable value of the difference in elevation.

b. The elevation of point B, if the elevation of point A is 532.462 meters above mean sea level.

Given: ROUTE 1 2 3

LENGTH

ELEV

3.0 km 4.0 km 6.0 km

21.200 m 20.972 m 21.261 m

Required: MPV Elev of B Illustration:

Route 1

A

B

Route 2 Route 3 Solution:

ΣW = W1 + W2 + W3 ΣW =0.747 Σweighted elevation = EL1W1 + EL2W2 + EL3W3 = 15.790 MPVDE = 21.14 m ELEVB = ELEVA + DE = 532.462 + 21.14 = 553.602 m

Unit Exam No. 3 A 1. Leveling is ths process of directly or indirectly measuring vertical distances to determine the elevation of points or their a) differences in elevation b) clearances above mean sea level c) horizontal distances from each other d) relationship to natural and man-made features e) respective positions with respect to the true meridian C

2. Mean sea level is an imaginary surface of the sea which a) is below the level of natural lakes and reservoirs b) comes very close to high tide c) is usually found midway between high and low tides d) deviates by about 0.5 m to 1.0 m from the lowest tide level e) is determined by averaging the height of the sea's surface for all its tide stages over a period of one month

E

3. The elevation of a point is the vertical distance a) below mean sea level b) above or below the level line c) measured from the lowest point on the ground d) above mean sea level or any other selected datum e) above or below mean sea level or any other selected datum

A 4. Direct leveling is the commonly employed method of determining the elevation of points some distance apart by a series of setups of a leveling instrument along a selected route. It is said to be the most precise method of leveling and is used when a) a higher degree of accuracy is required b) quick measurements are needed c) it is necessary to obtain a profile of the existing ground surface d) vertical angles and three intercepts on a rod can be obtained e) leveling across a wide river or a deep ravine E 5. If in a given locality the elevation of points A and B are 247.50 m and 200.00 m, respectively, the value 47.50 m represents their a) datum difference b) elevation factor c) datum correction d) vertical clearance e) difference in elevation E 6. Reciprocal leveling is commonly employed when leveling across a wide river, a deep ravine, or across canyons and gullies where it would be difficult or impossible to a) locate a reference bench mark b) determine the north-south line

c) set up an advantageous instrument station d) read vertical angles and differences in elevation e) maintain a foresight and a backsight distance of nearly equal lengths D 7. Any surveying instrument may be employed in stadia leveling as long as it can be used to measure vertical angles and is equipped with the standard cross hairs and a) an inverting eyepiece b) a compensator c) a Roelof's prism d) stadia hairs e) an erecting eyepiece B 8. The dumpy level is very identical to the wye level. The only distinct difference between these two instruments is in the manner by which a) theh are used in the field b) their telescopes are attached to the supporting level bar c) the image sighted appears on the eyepiece d) the instruments are set up and leveled e) their level vials are positioned B 9. This small device is attached to the leveling rod when extremely long sights make direct reading of the rod difficult or impossible. It may be moved up or down the rod under the direction of the instrumentman. The device referred to is called a a) rod ribbon b) target c) rod level d) hand level e) laser system C 10. The cross hairs consists of a pair of lines which are perpendicular to each other. They are used to a) determine stadia intercept b) project the center of the instrument c) define the instrument's line of sight d) check for effects of parallax and refraction e) determine the effect of the earth's curvature A 11. A backsight reading of 2.73 m is taken on point A and subsequently a foresight reading of 1.35 m is taken on point B from a level station established midway between the two points. The difference in elevation between points A and B is a) 1.38 m b) 4.08 m c) 2.73 m d) 4.11 m e) 1.35 m

Given: BS = 2.73m and FS = 1.35m Required:

Solution:

E 12. In Question 11, if the elevation of A is 386.70 m above mean sea level, the elevation of B should be a) 389.43 m b) 388.05 m c) 390.78 m d) 390.81 m e) 388.08 m Given: el A = 386.70m Required:

Solution:

C 13. The two-peg test is used to determine if the line of sight of an engineer's level is in adjustment, and the following rod readings are taken: with instrument setup near A, backsight on A is 1.623 m, foresight on B is 2.875 m; with instrument setup near B, backsight on B is 1.622 m, and foresight on A is 0.362 m. The correct rod reading on A to give a level line of sight with the instrument still setup near B should be a) 1.256 m b) 2.878 m c) 0.366 m d) 1.614 m e) 1.630 m Given: a=1.623, b=2.875, c=1.622 and d=0.362 Required:

Solution:

B 14. In Question 13, in order to adjust the line of sight, thr horizontal cross hair should be moved a) downward by 0.894 m b) upward by 0.004 m c) upward by 2.516 m d) downward by 1.252 m e) downward by 0.0004 m Given:

Required:

Solution:

A 15. In the two-peg test of a dumpy level, the following obersvations are taken: with level setup at M (midway between A and B), the rod readings on A and B are 1.155 m and 1.770 m, respectively. With level at P, the rod reading on A is 2 423 m and that on B is 2 999 m. If point P is located 1.67 m from A along the extension of line AB and 76.910 m from B, the correct rod reading (d') on the far rod for a horizontal line of sight (with the level still setup at P) should be a) 3.039 m b) 2.424 m c) 2.959 m d) 2.422 m e) 2.999 m Given: a=1.155, b= 1.770, c= 2.423, d=2.999, x= 1.67m and x+d=76.910m Required:

Solution:

C 16. In Question 15, the correct rod reading (c') on the near rod for a horizontal line of sight (with level still setup at P) should be a) 2.959 m b) 2.422 m

c) 2.424 m d) 3.039 m e) 2.999 m Given: a=1.155, b= 1.770, c= 2.423, d=2.999, x= 1.67m and x+d=76.910m Required:

Solution:

C

17. In Question 15, the line of sight is a) horizontal b) inclined upward c) inclined downward d) inclined to the left e) inclined to the right

E 18. The combined effect of the earth's curvature and refraction on a level sight of 1200 m is a) 0.972 m b) 0.081 m c) 0.054 m d) 0.005 m e) 0.097 m Given: K = 1200m Required:

Solution:

A 19. A fishing vessel with a green light on its deck, 7.5 m above the water, disappears from the sight of a person on shore whose eye level measures 1.30 m above the water. Neglecting the effect of waves, the distance of the vessel from the man on shore is a) 14.93 km b) 130.37 km c) 4.39 km d) 10.54 km e) 6.15 km

Given: hcr1 = 1.30m and hcr2=7.50m Required:

Solution: √

√

√

√

A 20. A woman on shore standing close to the sea water's edge looks toward the direction of the sea. If her eyes measure 1.525 m above sea level, her visible horizon, neglecting the effect of waves, should be about a) 4.75 km away b) 0.21 km away c) 0.04 km away d) 22.73 km away e) 1.63 km away

Unit Exam No. 4 E 1. Differential leveling is the process of determining the difference in elevation between two or more points some distance apart. It requires a series of set ups of the instrument along a general route and, for each set up, a rod reading is taken back to the point of known elevation and forward to a a) permanent bench mark b) temporary bench mark c) change point d) terminal point e) point of unknown elevation D

2. A bench mark is a fixed point of reference whose a) elevation is determined by trigonometric leveling b) elevation is always below mean sea level c) location is marked by range poles d) elevation is either known or assumed e) position is used as the first instrument station in any leveling operation

D 3. A backsight is a reading taken on a rod held on a point of known or assumed elevation. It is a measure of the vertical distance from the established line of sight to the point sighted. Backsights are frequently referred to as plus sights since they are added to the elevation of points being sighted to a) locate a turning point b) obtain the difference in elevation c) select the next suitable level station d) determine the height of the instrument e) detemine the elevation ofbthe terminal point A 4. A reading taken on a rod held on a point whose elevation is to be determined is called a a) foresight b) backsight c) height of instrument d) release point e) sight reading D 5. A turning is an intervening point between two bench marks upon which point foresight and backsight rod readings taken. It is sometimes referred to as a a) level point b) reference point c) BS-FS point d) change point e) elevation point

D 6. The height of instrument, which is sometimes called the height of collimation, is the elevation of the line of sight of an istrument above or below a selected reference datum. It is determined by a) subtracting the rod reading on the backsight from the elevation of the point on which the sight is taken b) adding the rod reading on the foresight to the elevation of the point on which the sight is taken c) subtracting the rod reading on the turning point from the elevation of the bench mark d) adding the rod reading on the backsight to the elevation of the point on which the sight is taken e) measuring the height of the instrument above the ground where it is set up A 7. A profile is a curved line which graphically portrays the intersection of a vertical plane with the surface of the earth. It depicts a) ground elevations of selected critical points along a surveyed line and the horizontal distances between these points b) stationings and the location of points backsighted as well as those foresighted c) horizontal exaggeration to accentuate the location of points d) all natural and man-made features along the surveyed route e) the correct position of the reference datum used C 8. The process of drawing the vertical scale for a profile much larger than the horizontal scale in order to accentuate the differences in elevation is referred to as a) scale uplifting b) vertical enlargement c) vertical exaggeration d) elevation highlighting e) horizontal exaggeration C 9. The aneroid barometer functions by using an airtight box which responds to change in air pressure. It is well suited for ordinary leveling purposes where only approximate elevations are required. They are, however, not desirable when a) used near metallic objects b) temperatures are below freezing point c) pressures in the area are changing rapidly d) used in rough or mountainous terrain e) pressures in the area change very slowly E

10. Given the following differential level notes: STA BS HI BM-A 1.73 TP 1 2.50 TP 2 2.76 TP 3 3.15 BM-B 1.93

FS 1.85 1.05 1.20 0.74

ELEVATION 346.50 m

TP 4 TP 5 BM-C

1.67 2.12 -

0.98 1.84 1.77

The station or point along the level circuit which has the highest elevation is a) BM-A b) TP 5 c) BM-B d) TP 3 e) BM-C B

11. In Question 10, the difference in elevation between BM-B and BM-C is a) 5.30 m b) 1.13 m c) 6.43 m d) 0.78 m e) 0.97 m

C 12. The following set of level notes is to be compeleted by employing the rise and fall method ROD READINGS REDUCED ROD RISE FALL LEVEL STA BS IFS FS BM-X 1.73 a 1.05 b 2.45 2.60 223.75 m 2.38 c 3.02 d BM-Y Based on the data given in the accomanying tabulation, the sum of all values to be tabulated under the "RISER" and "FALL" columns should, respectively, be equal to a) 2.45 and 2.60 m b) 8.18 m and 5.05 m c) 1.82 and 1.97 m d) 1.97 m and 1.82 m e) 8.18 m and 3.79 m E 13. In Question 12, the difference between the reduced level of point a and the reduced level of point d is a) 0.57 m b) 0.72 m c) 0.30 m d) 0.42 m e) 1.29 m

B 14. If a turning point along a profile level route measures 205.73 m beyond station 10 + 33.5, its stationing be indicated as a) 2 + 39.23 b) 12 + 39.23 c) 1+ 39.23 d) 8 + 27.77 e) 1 + 827.77 E 15. In the accompanying sketch, a vertical angle of -14*30' is read to a target 1.75 m above point M. The measured inclined distance (S) is 644.15 m and the elevation of point K is 320.70 m above mean sea level. If the height of instrument at K is 1.27 m, the elevation of M, considering the effects of the earth's curvature and atmospheric refraction, is a) 482.43 m b) 162.47 m c) 478.93 m d) 158.91 m e) 158.97 m A 16. In leveling across a river, reciprocal level readungs were taken between two points A and B. From setup near A: on A, 0.994 and 0.992 m; on B, 2.076, 2.077, 2.078, and 2.077 m. At the setup near B: on B, 2.550 and 2.548 m; on A, 1.462, 1.463, 1.464, and 1.463 m. From the observed data it can be determined that a) B is 1.085 m below A b) B is 1.085 m above A c) A is above B by 1.088 m d) A is below B by 1.084 m e) A and B are of the same elevation E 17. In Question 16, if the elevation of point A is 246.850 m, the elevation of point B should be a) 247.935 m b) 245.764 m c) 247.938 m d) 245.763 m e) 245.765 m B 18. An inverse method of leveling is executed between two existing concrete structures and across a concrete wall located midway between the two structures. All rod readings, as indicated in the accompanying sketch, were taken with the held upside down. From the data portrayed, if the elevation of A (bottom of the concrete beam of the first structure) is 240.05 m, the elevation of the bottom of the concrete beam of the second structure (indicated as C) should be

a) 234.68 m b) 240.42 m c) 239.68 m d) 238.78 m e) 242.40 m A 19. A closed loop of differential levels was run tl establish the elevations of several bench marks and the following results were obtained DISTANCE FROM OBSERVED POINT BM-1 ELEVATION BM-1 0.0 km 186.60 m BM-2 3.2 196.03 BM-3 4.8 199.10 BM-4 9.6 183.57 BM-5 12.8 177.43 BM-6 16.0 188.95 BM-1 22.4 186.75 The differene in elevation between the adjusted elevations of BM-6 ans BM-4 is a) 6.38 m b) 6.33 m c) 0.05 m d) 21.73 m e) 9.26 m B 20. Several lines of levels are run over different routes from BM-A to establish the elevation of BM-B. The lengths of these routes are shown in the accompanying sketch and the observed elevations of BM-B through each route were recorded as follows: Route 1, 379.855; Route 2, 379.904 m; Route 3, 379.927 m; Route 4, 379.775 m; and Route 5, 379.910 m. a) 379.874 m b) 379.898 m c) 379.895 m d) 379.795 m e) 379.798 m