Computation of area volumes.

Problem 05

(A) The Following Offsets Were Taken from A Chain Line to An Irregular Boundary Line at an Interval Of 10 M.: 3.10, 4.20, 5.35, 6.45, 7.15, 8.25, 7.95, and 5.20m.

Compute the area between the chain line, the irregular boundary line, and the end offsets by

• Trapezoidal rule
• Simpson’s rule

(B) The table below gives the lengths and the bearing of the lines of a traverse ABCDE, the length and bearing of EA having been omitted.

Calculated the length and bearing of the line EA.

Solutions.

(A) Given offsets, 3.10, 4.20, 5.35, 6.45, 7.25, 8.25, 7.95, 5.20. the values can be written in form of yo, y1, y2, y3 … yn.

Hence the distance, h = 10m

By trapezoidal rule, Computation of area volumes.

Area = (h / 2) [ (y0+ yn ) + 2(y1+y2+y3+…..+yn−1)] = (10/2) ( (3.1+5.2) + 2(4.2+5.35+6.45+7.15+8.25+7.95)) = 435 m².

By Simpson’s rule Computation of area volumes.

Area = (h / 3) x [ (y1+ yn ) + 4(y2+y4+ ….. yn-2) + 2 x (y3 + y5 + y7 + … yn-1)] = (10/3) x ((3.1+5.2) + 4 x (4.2 + 6.45+ 8.25) + 2 x (5.35+ 7.15+ 7.95)) = 416 m2.

(B) For a correct closed traverse, there are some checks.

∑ Latitude = 0.

∑ Departure = 0.

Latitude = L x Cosθ . Departure = L x Sin θ

θ is the magnetic bearing of line.

∑ Latitude = (204 x Sin 87,30) +( 226 x Sin 20,20) +( 187 x Sin 280) + (192 x Sin 210,30) + (L x Sin  θ) = 0

=> L x Sinθ = -98.371

∑ Departure = (204 x Cos 87,30) +( 226 * Cos 20,20) +( 187 * Cos 280) + (192 x Cos 210,30) + (L x Cos θ) = 0

=> L Cos θ = – 87.855

(L Sin θ / L cos θ) = 1.119 => Tan θ = 1.119 => θ = Tan^-1 (1.119) = 48.214 degrees. or (180+48.214) =228.214 degrees.

Because Lsinθ is negative

L = – 98.371 / Sin 228.214 = 131.928m

=> L = 131.928m.

Computation of area volumes.