Problem-set-10-partition-of-land1.pdf

Partition of land Problem no 1 Case 1: Dividing an area into two parts by a line between two points Problem no 1 : Given the adjusted latitudes and adjusted departures of a closed traverse. Determine the length and direction of the line FC which will divide the traverse into two parts. Course Adjusted Latitude Adjusted departure +N -S +E -W AB 449.86 334.85 BC 282.48 282.86 CD 167.95 449.52 DE 336.91 280.31 EF 113.55 451.86 FA 449.83 225.98 Sums 900.29 900.29 1012.69 1012.69 Solution: Draw the traverse

Compute the length and direction of the dividing line “FC” Course Adjusted Adjusted Latitude departure AB 449.86 -334.85 BC 282.48 282.86 CF y x FA -449.83 -225.98 Sums 0 0 Alternate solution Course Adjusted Latitude CD 167.95 DE -336.91 EF -113.55 FC y Sums 0

Adjusted departure 449.52 280.31 -451.86 x 0

Problem no 3 Case 1: Dividing an area into two parts by a line between two points Given the accompanying plot for a surveyed tract of land, determine the length and bearing of line DA which will divide the tract into two parts

Solution : Make a table line Adjusted latitude AB 393.54 BC -465.38 CD -873.61 DE -144.70 EF 139.76 FA 950.39 sums 0

Adjusted departure 1150.05 810.43 -249.25 -763.98 -800.19 -147.06 0

Compute the length and direction of the dividing line “DA” line Adjusted Adjusted latitude departure AB 393.54 1150.05 BC -465.38 810.43 CD -873.61 -249.25 DA Y X sums 0 0

Problem no 4 Case 2: Dividing an area by a line running through a point in a given

direction.From the given accompanying plot for a surveyed tract of land, it is required to divide the tract into two parts by a line EG originating from station E and whose bearing is due north. Determine the length of the dividing line and the distance of point G from the adjacent station B. Point G falls somewhere along line AB.

Compute the length and direction of the closing line AE line Adjusted Adjusted latitude departure EF 139.76 -800.19 FA 950.39 -147.06 AE y x Sums 0 0 Solution : line Adjusted latitude Adjusted departure AB 393.54 1150.05 BC -465.38 810.43 CD -873.61 -249.25 DE -144.70 -763.98 EF 139.76 -800.19 FA 950.39 -147.06 sums 0 0

By applying sine law Compute the length of closing line “EG” 𝟏𝟒𝟒𝟒. 𝟏𝟗𝟗 𝒅 𝐳 = = 𝒔𝒊𝒏𝟕𝟏°𝟔′𝟑𝟑. 𝟕𝟑" 𝒔𝒊𝒏𝟔𝟕°𝟓𝟒′ 𝟗. 𝟕𝟎" 𝒔𝒊𝒏𝟒𝟎°𝟓𝟗′ 𝟏𝟔. 𝟓𝟕" d = 1414.294m (answer) z = 1001.175m distance of point G from the adjacent station B: = distance of line AB – Z 𝑨𝑩 = √(393.54)𝟐 + (1150.05)𝟐 = 𝟏𝟐𝟏𝟓. 𝟓𝟐𝟎 z = 1001.175m distance of point G from the adjacent station B = 𝟏𝟐𝟏𝟓. 𝟓𝟐𝟎 − 𝟏𝟎𝟎𝟏. 𝟏𝟕𝟓 = 214.345m (answer)

line EA AG GE Sums

LINE 1444.199 z d

BEARING

Adjusted Adjusted latitude departure 0 N40 59’16.57”W 1090.15 -947.25 N7106’33.73””E Y X DUE SOUTH --0 0 0

Departure of line GE = 0 (dsin0 = 0) Departure of line AG, x = 947.25 (-947.25 + x + 0 =0) 180° = 40°59′ 16.57" + 𝜃 + 71°6′33.73" 𝜽 = 𝟔𝟕°𝟓𝟒′ 𝟗. 𝟕𝟎"

Problem no 5: Case 3: To cut off a required area by a line through a given point. From the given accompanying plot for a surveyed tract of land, determine the length and bearing of line XY which will divide the traverse into two parts. Point X, which is located along line AB, is 140.20m from B, and point Y falls midway between traverse station D and E.

Distance XB = 140.2m By ratio and proportion: 𝑙𝑎𝑡𝑖𝑡𝑢𝑑𝑒 𝑜𝑓 𝑙𝑖𝑛𝑒 𝐴𝐵 𝑙𝑎𝑡𝑖𝑡𝑢𝑑𝑒 𝑜𝑓 𝑙𝑖𝑛𝑒 𝑋𝐵 = 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 𝑙𝑖𝑛𝑒 𝐴𝐵 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 𝑙𝑖𝑛𝑒 𝑋𝐵 393.54 line Adjusted latitude Adjusted departure AB 393.54 1150.05 BC -465.38 810.43 CD -873.61 -249.25 DE -144.70 -763.98 EF 139.76 -800.19 FA 950.39 -147.06 sums 0 0

√393.542 + 1150.052

𝑙𝑎𝑡𝑖𝑡𝑢𝑑𝑒 𝑜𝑓 𝑙𝑖𝑛𝑒 𝑋𝐵 140.2

𝑙𝑎𝑡𝑖𝑡𝑢𝑑𝑒 𝑜𝑓 𝑙𝑖𝑛𝑒 𝑋𝐵 = 45.39 m 𝐷𝑒𝑝𝑎𝑟𝑡𝑢𝑟𝑒 𝑜𝑓 𝑙𝑖𝑛𝑒 𝐴𝐵 𝐷𝑒𝑝𝑎𝑟𝑡𝑢𝑟𝑒 𝑜𝑓 𝑙𝑖𝑛𝑒 𝑋𝐵 = 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 𝑙𝑖𝑛𝑒 𝐴𝐵 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 𝑙𝑖𝑛𝑒 𝑋𝐵 1150.05 √393.542

Solution:

=

+

1150.052

=

𝐷𝑒𝑝𝑎𝑟𝑡𝑢𝑟𝑒 𝑜𝑓 𝑙𝑖𝑛𝑒 𝑋𝐵 140.2

𝐷𝑒𝑝𝑎𝑟𝑡𝑢𝑟𝑒 𝑜𝑓 𝑙𝑖𝑛𝑒 𝑋𝐵 = 132.65

Distance of line DE = √(−144.70)2 + (−763.98)2 = 777.5626 Distance of line DY = Distance of line DE/2 = 388.7813m 𝑙𝑎𝑡𝑖𝑡𝑢𝑑𝑒 𝑜𝑓 𝑙𝑖𝑛𝑒 DE 𝑙𝑎𝑡𝑖𝑡𝑢𝑑𝑒 𝑜𝑓 𝑙𝑖𝑛𝑒 DY = 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 𝑙𝑖𝑛𝑒 DE 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 𝑙𝑖𝑛𝑒 DY −144.70 √(−144.70)2 + (−763.98)2

=

𝑙𝑎𝑡𝑖𝑡𝑢𝑑𝑒 𝑜𝑓 𝑙𝑖𝑛𝑒 DY 388.7813

𝑙𝑎𝑡𝑖𝑡𝑢𝑑𝑒 𝑜𝑓 𝑙𝑖𝑛𝑒 DY = −72.35𝑚 𝐷𝑒𝑝𝑎𝑟𝑡𝑢𝑟𝑒 𝑜𝑓 𝑙𝑖𝑛𝑒 DE 𝐷𝑒𝑝𝑎𝑟𝑡𝑢𝑟𝑒 𝑜𝑓 𝑙𝑖𝑛𝑒 DY = 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 𝑙𝑖𝑛𝑒 DE 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 𝑙𝑖𝑛𝑒 DY

−763.98 √(−144.70)2 + (−763.98)2

=

𝐷𝑒𝑝𝑎𝑟𝑡𝑢𝑟𝑒 𝑜𝑓 𝑙𝑖𝑛𝑒 DY 388.7813

𝐷𝑒𝑝𝑎𝑟𝑡𝑢𝑟𝑒 𝑜𝑓 𝑙𝑖𝑛𝑒 DY = −381.99 line

Adjusted Adjusted latitude departure XB 45.39 132.65 BC -465.38 810.43 CD -873.61 -249.25 DY -72.35 -381.99 YX a b sums 0 0

Problem no 6: Case 3: To cut off a required area by a line through a given point. From the given accompanying plot for a surveyed tract of land,determine the length and direction of a line that runs through F and divides the tract into two equal parts.

Compute the total area of tractABCDEFA by DMP or DPD METHOD Area by DMD line Adjusted Adjusted DMD 2A latitude departure AB 393.54 1150.05 1150.05 452590.677 BC -465.38 810.43 3110.53 -1447578.451 CD -873.61 -249.25 3671.71 -3207642.573 DE -144.70 -763.98 2658.48 -384682.056 EF 139.76 -800.19 1094.31 152940.7656 FA 950.39 -147.06 147.06 139764.3534 ∑ = -4294607.285 sums 0 0 2A = 4294607.285 A = -2147303.642m^2 (area of tract ABCDEF) A=/2 = 1073651.821 m^2 Area by DPD line Adjusted Adjusted DPD 2A latitude departure AB 393.54 1150.05 393.54 452590.677 BC -465.38 810.43 321.7 260715.331 CD -873.61 -249.25 -1017.29 253559.5325 DE -144.70 -763.98 -2035.6 1555157.688 EF 139.76 -800.19 -2040.54 1632819.703 FA 950.39 -147.06 -950.39 139764.3534 ∑ = 4294607.285 sums 0 0

line Adjusted latitude Adjusted departure AB 393.54 1150.05 BC -465.38 810.43 CD -873.61 -249.25 DE -144.70 -763.98 EF 139.76 -800.19 FA 950.39 -147.06 sums 0 0

2A = 4294607.285 AT = 2147303.642m^2 (area of tract ABCDEF) AT/2 = 1073651.821 m^2

Check the area of tract ABFA

line

Adjusted latitude AB 393.54 BF y FA 950.39 sums 0 Area by DMD line Adjusted latitude AB 393.54 BF -1343.93 FA 950.39 sums 0

Adjusted departure 1150.05 x -147.06 0 Adjusted departure 1150.05 -1002.99 -147.06 0

393.54 + y +950.39 = 0 y = -1343.93m 1150.05 + x + -147.06 = 0 x = -1002.99m

DMD 1150.05 1297.11 147.06

2A = -1150870.012 A = 575435.006m2 (area of tract ABFA) Since area ABFA < 1073651.821 m^2

2A 452590.677 -1743225.042 139764.3534 ∑ = -1150870.012

Consider tract ABGFA , AREA = 1073651.821 m^2

By cosine law ̅̅̅̅ 2 = (598.49156)2 + (1676.943)2 𝐹𝐺 − 2𝑥(598.49156)x((1676.943)𝑐𝑜𝑠96°52′ 5.87" FG = 1846.72m By sine law 𝑠𝑖𝑛𝛽 𝑠𝑖𝑛96°52.5.87" = 1676.943 1846.72 𝜷 = 𝟔𝟒°𝟐𝟏′ 𝟒𝟕" 𝟏𝟖𝟎° = (𝜷 = 𝟔𝟒°𝟐𝟏′ 𝟒𝟒. 𝟕𝟕") +  + 𝟔𝟎°𝟖′𝟐. 𝟎𝟔" ∅ = 𝟓𝟓°𝟑𝟎′ 𝟏𝟑. 𝟏𝟕" Therefore: line distance FG 1676.943

line Adjusted Adjusted latitude departure BF -1343.93 -1002.99 BC -465.38 810.43

distance

bearing

1676.943

S 36044’3.81” W S 6008’2.06” E

AREAABGFA = 1073651.821 m^2 AREAABFA = 575435.006m2 Therefore AREAFBGF = 1073651.821 m^2 - 575435.006m2 = 498216.815m2 1 𝐴 = 𝑎𝑏𝑠𝑖𝑛𝜃 2 𝟏 ̅̅̅̅)𝐱(𝑩𝑭 ̅̅̅̅) = 1676.943)xsin96°52′ 5.87" 𝟒𝟗𝟖𝟐𝟏𝟔. 𝟖𝟏𝟓 = (𝐁𝐆 𝟐 Distance BG = 598.49156m

bearing N𝟓𝟓°𝟑𝟎′ 𝟏𝟑. 𝟏𝟕"E

Problem no 9 Partition of land Given in the accompanying tabulation are the adjusted latitudes and adjusted departures of a parcel of land whose area is to be divided equally into two parts. The dividing line is to pass through a point midway between corners A and E, and through a point along the boundary of BC. Determine the length and bearing of the required line. line

Adjusted latitude Adjusted departure +N -S +E -W AB 115.64 32.07 BC 19.27 144.1 CD 85.59 26.7 DE 37.5 106.5 EA 26.72 96.37 sums 142.36 142.36 202.87 202.87

Solution: Compute for the total area of the land line Adjusted Adjusted DMD latitude departure AB 115.64 32.07 32.07 BC -19.27 144.1 208.24 CD -85.59 26.7 379.04 DE -37.5 -106.5 299.24 EA 26.72 -96.37 96.37 sums 0 0 2A = 41392.7372m2 A = 20696.3686 m2 A/2 = 10348.1843 m2 Consider triangle ABG

2A 3708.575 -4012.785 -32442.034 -11221.500 2575.006 ∑ = -41392.737

Point G is midway between A and E Therefore Latitude = 26.72/2 = 13.36 Departure = -96.37 = -48.185 Compute the departure and latitude of line BG line Adjusted Adjusted latitude departure AB 115.64 32.07 BG y x GA 13.36 -48.185 DE 0 0 115.64 + y + 13.36 = 0 y = -129.00 32.07 + x -48.185 = 0 x = 16.115 length of line BG = 130.003m direction of line BG : tan-1(16.115/129.00) = 707’14.32” S 707’14.32” E Compute the area of triangle ABG Area by DMD line Adjusted Adjusted latitude departure AB 115.64 32.07 BG -129.00 16.115 GA 13.36 -48.185 DE 0 0

Determine the length and bearing of LINE FG line Adjusted Adjusted Distance latitude departure BG -129.00 16.115 130.003m BC -19.27 144.1 145.383

DMD

2A

32.07 80.255 48.185

3708.5748 -10352.8950 643.7516 ∑=6000.5686

2AABG = 6000.5686 AABG = 3000.2843m2 Compute the area of BFG Area ABFG = 10348.1843 m2 = (Area ABG = 3000.2843m2) + Area BFG Area BFG, ABFG = 10348.1843 - 3000.2843 = 7347.9000m2

BEARING S 707’14.32” E S82022’59.59”E

1 𝐴 = 𝑎𝑏𝑠𝑖𝑛𝜃 2 𝛽 = 82°22’59.59” − 7°7’14.32” = 𝟕𝟓°𝟏𝟓′ 𝟒𝟓. 𝟐𝟕" Area BFG, ABFG = 7347.9000m2 1 ̅̅̅̅)sin75°15′ 45.27" 7347.9000 = (130.003m)(BF 2 ̅̅̅̅ 𝐁𝐅 = 𝟏𝟏𝟔. 𝟖𝟖𝟕 By cosine law ̅̅̅̅ )2 = (116.887) 2 + (𝟏𝟑𝟎. 𝟎𝟎𝟑) 2 (𝐹𝐺 − 2 𝑥 130.003 x 116.887 𝑥 𝑐𝑜𝑠𝟕𝟓°𝟏𝟓′ 𝟒𝟓. 𝟐𝟕" ̅̅̅̅ 𝐹𝐺 = 𝟏𝟓𝟏. 𝟏𝟎𝟑 By sine law 𝑠𝑖𝑛∅ 𝑠𝑖𝑛𝛽 = 𝑠𝑖𝑛75°15′ 45.27" = 130.003 151.103 ∅ = 56°18′ 37.37" 180° = 𝜌 + 56°18′ 07" + 82°22’59.59” 𝜌 = 𝟒𝟏°𝟏𝟖′ 𝟐𝟑. 𝟑𝟒"

line FG

length bearing 𝟏𝟓𝟏. 𝟏𝟎𝟑 S𝟒𝟏°𝟏𝟖′ 𝟐𝟑. 𝟑𝟒"W

Problem no 10 Partition of land The adjusted latitudes and adjusted departures of a parcel of land are: Course 1-2: Lat = -902.48, Dep = +196.49 Course 2-3: Lat = -110.40, Dep = -1004.96 Course 3-4: Lat = +337.50, Dep = -490.18 Course 4-5: Lat = +441.64, Dep = +42.55 Course 5-6: Lat = +528.41, Dep = +548.00 Course 6-1: Lat = -294.67, Dep = +708.10 If this parcel is to be subdivided into two equal area, determine the length and bearing of the dividing line which is to pass through corner 6. Solution: Course latitude departure 1-2 -902.48 196.49 2-3 -110.40 -1004.96 3-4 +337.50 -490.18 4-5 441.64 42.55 5-6 528.41 548.00 6-1 -294.67 708.10 sums 0 0 Compute the total area of the parcel of land Area by DMD Course latitude departure DMD 1-2 -902.48 196.49 196.49 2-3 -110.40 -1004.96 -611.98 3-4 +337.50 -490.18 -2107.12 4-5 441.64 42.55 -2554.75 5-6 528.41 548.00 -1964.2 6-1 -294.67 708.10 -708.1 sums 0 0 2A = 2778445.5882 A = 1389222.794m2 A/2 = 694611.3971 m2

2A -177328.2952 67562.5920 -711153.0000 -1128279.7900 -1037902.9220 208655.8270 ∑ =-2778445.5882

Check for the area of course “3-4-5-6-3”

Course latitude departure 3-4 +337.50 -490.18 4-5 441.64 42.55 5-6 528.41 548.00 6-3 y= -1307.55m x= -100.37m sums 0 0 2A = 894344.6085 A = 447172.3043m2 (area “3-4-5-6-3)

DMD -490.18 -937.81 -347.26 100.37

2A -165435.7500 -414174.4084 -183495.6566 -131238.7935 -894344.6085

Since area “3-4-5-6-3 < A1-2-3-4-5-6-1 / 2 = 694611.3971 m2

Consider point 7 AREA

OF

COURSE

247439.0928m^2

Course latitude 3-4 +337.50 4-5 441.64 5-6 528.41 6-3 y sums 0 Therefore : y = -1307.55m x = -100.37m course length 6-3 1311.40 2-3 1011.01

departure -490.18 42.55 548.00 x 0

bearing S4023’22.29”W S83043’51.25”W

3-6-7-3

=694611.3971

-

447172.3043

=

𝛽 = 83043’51.25” - 4023’22.29” = 79020'28.97"

𝟏 ̅̅̅̅̅̅̅ ̅̅̅̅̅̅̅ (𝟕 − 𝟑)(𝟔 − 𝟑 = 𝟏𝟑𝟏𝟏. 𝟒𝟎)𝐬𝐢𝐧𝟕𝟗°𝟐𝟎′𝟐𝟖. 𝟗𝟕" 𝟐 Distance 7-3 = 383.991m 𝟐𝟒𝟕𝟒𝟑𝟗. 𝟎𝟗𝟐𝟖 =

By cosine law ̅̅̅̅̅̅̅ (𝟔 − 𝟕)𝟐 = (𝟑𝟖𝟑. 𝟗𝟗𝟏 )𝟐 + 𝟏𝟑𝟏𝟏. 𝟒𝟎𝟐 − 𝟐𝒙𝟏𝟑𝟏𝟏. 𝟒𝟎𝐱𝟑𝟖𝟑. 𝟗𝟗𝟏𝐜𝐨𝐬 𝟕𝟗°𝟐𝟎′𝟐𝟖. 𝟗𝟕" Distance of course 6-7 = 1296.512m (answer) Direction of course 6-7 By sine law: 𝒔𝒊𝒏∅ 𝒔𝒊𝒏𝟕𝟗°𝟐𝟎′ 𝟐𝟖. 𝟗𝟕" = 𝟑𝟖𝟑. 𝟗𝟗𝟏 𝟏𝟐𝟗𝟔. 𝟓𝟏𝟐

 = 16055’17.65”

𝜃 = ∅ − 4°23′ 29" = 12°31′55.36" Therefore S𝟏𝟐°𝟑𝟏′𝟓𝟓. 𝟑𝟔"E (Answer) course 6-7

length 1296.512m

bearing S𝟏𝟐°𝟑𝟏′𝟓𝟓. 𝟑𝟔"E