- Establish the DMD of each course by successive
applications of Eq. 3
Thus, DMDDE - 143.6 ft (43.77 m); DMD^ = 143.6 + 143.6 + 246.7 = 533.9 ft
(162.73 m); DMD^ = 533.9 + 246.7 + 21.3 - 801.9 ft (244.42 m); DMD 45 = 801.9 +
21.3 - 135.6 = 687.6 ft (209.58 m); DMDBC = 687.6 - 135.6 - 77.5 = 474.5 ft (144.62 m);
DMDCD - 474.5 - 77.5 - 198.5 = 198.5 ft (60.50 m). This is acceptable.
- Calculate the area of the tract
Use the following theorem: The area of a tract is numerically equal to the aggregate of the
projection areas of its courses. The results of this calculation are
Course Latitude x DMD = 2 x Projection area
AB -132.3 687.6 -90,970
BC +9.6 474.5 +4,555
CD +97.9 198.5 +19,433
DE +161.9 143.6 +23,249
EF -35.3 533.9 -18,847
FA -101.8 801.9 -81,634
Total -144,214
Area = ^(144,2U) = 72,107 ft^2 (6698.74 m^2 )
PARTITION OFA TRACT
The tract in the previous calculation procedure is to be divided into two parts by a line
through E, the part to the west of this line having an area of 30,700 ft
2
(2852.03 m
2
). Lo-
cate the dividing line.
Calculation Procedure:
- Ascertain the location of the dividing line EG
This procedure requires the solution of an oblique triangle. Refer to Fig. 5. It will be nec-
essary to apply the following equations, which may be readily developed by drawing the
altitude BD:
Area = Vibe sin A (4)
„ c sin A ...
tan C=- (5)
b-c cos A
In Fig. 6, let EG represent the dividing line of this tract. By scaling the dimensions and
making preliminary calculations or by using a planimeter, ascertain that G lies between B
and C.
- Establish the properties of the hypothetical course EC
By balancing the latitudes and departures of DEC. latitude of EC = -(+161.9 + 97.9) =
-259.8 ft (-79.18 m); departure of EC = -(+143.6 - 198.5) = +54.9 ft (+16.73 m); length
of EC = (259.S^2 + 54.9^2 )^05 = 265.5 ft (80.92 m). Then DMDDE = 143.6 ft (43.77 m);
