7.3 Areas, Volumes and Setting Out

7.3 Areas, Volumes and Setting Out

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1. Area from coordinates

For a closed polygon with vertices $(E_1,N_1), (E_2,N_2), \dots, (E_n,N_n)$ taken in order:

$$A = \frac{1}{2}\left|\sum_{i=1}^{n}\left(E_iN_{i+1} - E_{i+1}N_i\right)\right|$$

closing the polygon with $(E_{n+1}, N_{n+1}) = (E_1, N_1)$. This is what regulation 5(E)(iv) means by computing the area "arithmetically from the coordinates of the corners".

Worked example. A four-sided parcel:

$$\begin{aligned} \sum(E_iN_{i+1} - E_{i+1}N_i) =\;& (1000.00 \times 1010.00 - 1040.00 \times 1000.00) \\ &+ (1040.00 \times 1055.00 - 1050.00 \times 1010.00) \\ &+ (1050.00 \times 1048.00 - 1005.00 \times 1055.00) \\ &+ (1005.00 \times 1000.00 - 1000.00 \times 1048.00) \end{aligned}$$ $$= (1{,}010{,}000 - 1{,}040{,}000) + (1{,}097{,}200 - 1{,}060{,}500) + (1{,}100{,}400 - 1{,}060{,}275) + (1{,}005{,}000 - 1{,}048{,}000)$$ $$= -30{,}000 + 36{,}700 + 40{,}125 - 43{,}000 = 3{,}825$$ $$A = \tfrac{1}{2} \times 3{,}825 = 1{,}912.50\text{ m}^2$$

The parcel is under two acres, so under regulation 31(j) the area is stated in square measure to the second decimal: 1,912.50 m². Note that the sign of the summation depends on whether the vertices are taken clockwise or anticlockwise; take the absolute value.

A useful check: recompute with the vertices in reverse order. The magnitude must be identical.

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2. Irregular boundaries from offsets

Regulation 9 requires that where a boundary is irregular - a stream, for instance - the traverse line follows approximately its course and offsets are taken at suitable points. The strip between the traverse line and the boundary is then computed separately and added to or subtracted from the polygon area.

Trapezoidal rule

For $n$ offsets $y_1 \dots y_n$ at equal intervals $d$:

$$A = d\left[\frac{y_1 + y_n}{2} + y_2 + y_3 + \cdots + y_{n-1}\right]$$

This is regulation 5(E)(iv)'s "series of trapezoids".

Simpson's rule

More accurate where the boundary curves smoothly, because it fits parabolic arcs rather than straight chords:

$$A = \frac{d}{3}\left[y_1 + y_n + 4(y_2 + y_4 + \cdots) + 2(y_3 + y_5 + \cdots)\right]$$

Simpson's rule requires an odd number of offsets - that is, an even number of intervals. With an even number of offsets, apply Simpson to all but the last strip and the trapezoidal rule to the remainder.

Worked example. Offsets at 10 m intervals: 3.2, 4.8, 6.1, 5.4, 3.9 m (five offsets, four intervals).

• Trapezoidal: $10\left[\frac{3.2+3.9}{2} + 4.8 + 6.1 + 5.4\right] = 10[3.55 + 16.3] = 198.5$ m²
• Simpson: $\frac{10}{3}\left[3.2 + 3.9 + 4(4.8+5.4) + 2(6.1)\right] = \frac{10}{3}[7.1 + 40.8 + 12.2] = \frac{10}{3}(60.1) = 200.3$ m²

The difference of 1.8 m² on 200 m² is about 0.9 per cent - small in absolute terms but systematic, and the trapezoidal rule always underestimates a convex boundary.

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3. Volumes from cross-sections

End-area rule

For two cross-sectional areas $A_1$ and $A_2$ a distance $L$ apart:

$$V = \frac{L}{2}(A_1 + A_2)$$

Simple, and it overestimates for a solid whose section varies non-linearly (the usual case in cut and fill).

Prismoidal formula

$$V = \frac{L}{6}\left(A_1 + 4A_m + A_2\right)$$

where $A_m$ is the area of the section at the mid-point - computed from the mid-point dimensions, not as the mean of $A_1$ and $A_2$. Computing $A_m$ as the average of the end areas reduces the prismoidal formula algebraically back to the end-area rule and defeats the purpose. This is the single most common error in earthwork computation.

Worked example. $A_1 = 24.0$ m², $A_2 = 40.0$ m², mid-section computed from mid-point dimensions as $A_m = 31.5$ m², $L = 30$ m.

• End-area: $V = \frac{30}{2}(24.0 + 40.0) = 15 \times 64.0 = 960.0$ m³
• Prismoidal: $V = \frac{30}{6}(24.0 + 4 \times 31.5 + 40.0) = 5(24.0 + 126.0 + 40.0) = 5 \times 190.0 = 950.0$ m³

The prismoidal correction of 10.0 m³ is about 1 per cent - which on a road project of many thousands of cubic metres is a commercially significant sum.

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4. Sub-division of areas

A recurring practical problem: divide a parcel so that a specified area falls on one side of a line.

Line through a fixed point. Compute the area cut off by a trial line through the point, then rotate the line and interpolate. Because area varies smoothly with the rotation angle, two trials usually bracket the answer and a third converges it.

Line of given direction. Where the dividing line must be parallel to an existing boundary - a common planning requirement - the area cut off is a simple function of the perpendicular offset, and the offset can be solved directly for a triangle or trapezium and iterated for a general polygon.

In both cases, compute the coordinates of the new corner points and set them out and demarcate them. Regulation 15 requires demarcation prior to or simultaneously with the measurements determining position, and regulation 17 requires proof by actual measurements recorded in the field book where a beacon is placed in accordance with computed data.

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5. Setting out

Setting out is the survey run backwards: instead of measuring what is there, you place what is designed.

The computation

From a control station $S$ at $(E_S, N_S)$ to a design point $P$ at $(E_P, N_P)$:

$$\Delta E = E_P - E_S, \qquad \Delta N = N_P - N_S$$ $$\text{Distance } D = \sqrt{\Delta E^2 + \Delta N^2}$$ $$\text{Bearing } \theta = \arctan\!\left(\frac{\Delta E}{\Delta N}\right) \quad\text{(resolved into the correct quadrant by the signs of } \Delta E, \Delta N)$$

The quadrant is where errors happen. $\arctan$ returns a value in a half-plane; the signs of $\Delta E$ and $\Delta N$ determine which of the four quadrants the bearing lies in, and a sign error puts the point 180 degrees out.

The scale-factor trap

Design coordinates are grid coordinates. The distance you set out on the ground is a ground distance. The reduction runs the other way:

$$D_{ground} = \frac{D_{grid}}{k_{combined}}$$

Divide, do not multiply. With $k_{combined} \approx 0.99979$ (the worked case in section 6.3), a 2,000 m grid distance corresponds to 2,000.42 m on the ground. Applying the factor in the wrong direction produces an error of twice the correction - 84 cm on 2 km in that example.

Checks

No set-out point is finished until it has been checked independently:

• set out from a second control station and compare;
• check the distance between adjacent set-out points against the design distance; or
• observe a check bearing to a third known station before and after the work.

A set-out point placed from one station by one observation, with no check, is exactly the kind of unverified value that regulation 17 refuses to accept in a survey - and it is the origin of most construction set-out disputes.