Volume of area computations.
Problem 4.
A, B, C, and D are the four clockwise considered boundary points of a plot. Sides AB and BC are straight, while sides CD and DA are irregular, with < CBA = 900, A straight line CD does not cross the plot, while a straight line DA passes inside the plot. Other measurements along the straights are as follows: AB = 114m, BC = 99m, with line AC not crossing any boundary.
The offsets measured from lines CD and DA to irregular boundaries are given in Tables I & II respectively.
| Chainages (m) | 0 (C) | 15 | 30 | 45 | 60 | 75D |
| Offsets (m) | 0 | 1.8 | 2.8 | 3.88 | 1.9 | 0 |
| Chainages (m) | 0 (D) | 20 | 40 | 60 | 80 (D) |
| Offsets (m) | 0 | 1.02 | 3.05 | 1.22 | 0 |
Calculate the area of the plot in hectares, applying Simpsons and Trapezoidal rules in tables I & II respectively.
Solutions:
Using Simpson’s rule (Volume of area computations.)
Area of plot = area of ∆1 + area of ∆2 – Offset area1 + Offset area2
Area of ∆1 = ½ bh = ½ x 114 x99 = 5643m2
Area of ∆2 = √S (S-a) (S-c) (S-d)
AC = √ (AB2 + BC2) = √ (1142 + 992) = 150.99m
Where: S = (a + c + d)/2
S = (75 + 80 + 150 .99)/2 = 152.99m
Area of ∆2 = √152.99(152.99- 75) (152.99-80) (152.99- 150.99) =1319.77m2
Offset Area1 i.e., Tables 1 by Simpsons rule but offsets are even.
So, we shall use Simpsons rule for A1-5 & then Trapezoidal for A5-6
A = L/3 [01 + 0N + 4∑even offset + 2∑odd offsets]
A1-5 = L/3 [(01 + 05 + 4 (02 + 04) + 2 (03)]
A1-5 = 15/3 [0 + 1.95 + 4 (1.8 +3.88) + 2(2.8)] = 151.35m2
Using trapezoidal rule. (Volume of area computations.)
A5-6 = L/2 (06 + 05) = ½ (0 + 1.95) x15 = 14.63m2
Offset area 1 = 151.35 +14.6 3 = 165.98m2
Offset area 2 i.e., table II by Trapezoidal rule.
A = L/2 [(01 + 0N + 2 (02 + 03 + ……..+ 0N-1)]
= 20/2 [0 + 0 + 2 (1.02 + 3.05 + 1.22)] = 105.8m2
Total Area of the plot = 5643 + 1319.77 – 165.98 + 105.80
= 6902.59m2
= 0.6902hect.
