Offsets area: trapezoidal, Simpson, coordinate

Given: 8 offsets at a common interval $d = 5$ m, so the length of the chain line = 7 × 5 = 35 m.

(i) Trapezoidal rule

$$\text{Area} = d\left[\frac{o_0 + o_n}{2} + o_1 + o_2 + \dots + o_{n-1}\right]$$

Sum of the intermediate offsets = 36.69 m; (first + last)/2 = (0.00 + 0.90)/2 = 0.450 m.

$$\text{Area} = 5\times\left[0.450 + 36.69\right] = 185.700\ \text{m}^2$$

(ii) Simpson's one-third rule

There are 8 offsets, i.e. 7 (an odd number of) intervals, so Simpson's rule cannot be used for the whole length. It is applied to the first 6 intervals (offsets 0 to 6), and the last strip is found by the trapezoidal rule.

$$\text{Area}_1 = \frac{5}{3}\left[(0.00 + 4.75) + 4(20.12) + 2(11.82)\right] = 181.450\ \text{m}^2$$ $$\text{Area}_2 = 5\times\frac{4.75 + 0.90}{2} = 14.125\ \text{m}^2$$ $$\text{Total area} = 181.450 + 14.125 = 195.575\ \text{m}^2$$

(iii) Coordinate method

Take the chain line as the x-axis with the first offset at chainage 0. The corner points of the area, taken in order round the boundary, are:

$$\text{Area} = \frac{1}{2}\left|\sum \left(x_i\,y_{i+1} - x_{i+1}\,y_i\right)\right|$$

Evaluating the sum with these points gives Area = 185.700 m². (The coordinate method, with straight lines between the offsets, gives the same figure as the trapezoidal rule, since both treat the boundary as a series of straight segments.)

Answer: trapezoidal = 185.70 m², Simpson's = 195.58 m², coordinate method = 185.70 m².