Using DMD and DPD Methods in Lot Area Computation

Example Problems in Elementary Surveying

Example: Using DMD and DPD Methods in Lot Area Computation

Example:

Determine the area of the lot whose description is given below. Use (a) the DMD method and (b) the DPD method

Line / Length (m) / Bearing
1-2 / 42.48 / S 32°47’ W
2-3 / 118.93 / N 56°5’ W
3-4 / 13.72 / N 2°55’ E
4-5 / 67.82 / N 73°38’ E
5-6 / 57.31 / S 77°49’ E
6-1 / 51.37 / S 0°7’ W

Solution:

a)  by DMD Method

Line / Latitude / Departure / DMD / 2A = DMDx Lat
1-2 / - 35.71 / - 23.00 / - 23.00 / + 821.33
2-3 / + 66.36 / - 98.69 / - 144.69 / - 9,601.63
3-4 / + 13.70 / + 0.70 / - 242.68 / - 3,324.72
4-5 / + 19.11 / + 65.07 / - 176.91 / - 3,380.75
5-6 / - 12.09 / + 56.02 / - 55.82 / + 674.86
6-1 / - 51.37 / - 0.10 / + 0.10 / - 5.14
Total / 0.00 / 0.00 / - 14,816.05

Therefore 2A = |-14,816.05|, or A = 14,816.05/2 = 7,408.03 sq.m.

b)  by DPD Method

Line / Latitude / Departure / DPD / 2A = DPD x Lat
1-2 / - 35.71 / - 23.00 / - 35.71 / + 821.33
2-3 / + 66.36 / - 98.69 / - 5.06 / + 499.37
3-4 / + 13.70 / + 0.70 / + 75.00 / + 52.50
4-5 / + 19.11 / + 65.07 / + 107.81 / + 7,015.20
5-6 / - 12.09 / + 56.02 / + 114.83 / + 6,432,78
6-1 / - 51.37 / - 0.10 / + 51.37 / - 5.14
Total / 0.00 / 0.00 / 14,816.04

Therefore 2A = 14,816.04 or A = 14,816.04/2 = 7,408.02 sq.m.

Elementary Surveying Notes of A.M. Fillone, DLSU-Manila