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SURCON Professional Examination — Surveyor (Nigeria)

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7.3 Areas, Volumes and Setting Out

Key Takeaways

  • The coordinate cross-product formula gives the area of any straight-sided polygon directly from the corner coordinates, and is what regulation 5(E)(iv) requires.
  • The trapezoidal rule and Simpson's rule handle irregular boundaries from offsets; Simpson's rule requires an even number of equal intervals.
  • Earthwork volumes from cross-sections are computed by the end-area rule or, more accurately, by the prismoidal formula V = (L/6)(A1 + 4Am + A2).
  • Setting out reverses the survey: computed coordinates are converted to bearings and distances from a control station, and every set-out point must be independently checked.
  • Grid distances must be converted back to ground distances before setting out, by dividing by the combined scale factor rather than multiplying.
Last updated: August 2026

7.3 Areas, Volumes and Setting Out

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=12∣∑i=1n(EiNi+1−Ei+1Ni)∣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:

PointE (m)N (m)
A1000.001000.00
B1040.001010.00
C1050.001055.00
D1005.001048.00
\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$$ The parcel is under two acres, so under regulation 31(j) the area is stated in square measure **to the second decimal**: 1,912.50 m². Note that the sign of the summation depends on whether the vertices are taken clockwise or anticlockwise; take the absolute value. A useful check: recompute with the vertices in reverse order. The magnitude must be identical. --- ## 2. Irregular boundaries from offsets Regulation 9 requires that where a boundary is irregular - a stream, for instance - the traverse line follows approximately its course and **offsets are taken at suitable points**. The strip between the traverse line and the boundary is then computed separately and added to or subtracted from the polygon area. ### Trapezoidal rule For $n$ offsets $y_1 \dots y_n$ at equal intervals $d$: $$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]$$ **Simpson's rule requires an odd number of offsets - that is, an even number of intervals.** With an even number of offsets, apply Simpson to all but the last strip and the trapezoidal rule to the remainder. **Worked example.** Offsets at 10 m intervals: 3.2, 4.8, 6.1, 5.4, 3.9 m (five offsets, four intervals). * Trapezoidal: $10\left[\frac{3.2+3.9}{2} + 4.8 + 6.1 + 5.4\right] = 10[3.55 + 16.3] = 198.5$ m² * Simpson: $\frac{10}{3}\left[3.2 + 3.9 + 4(4.8+5.4) + 2(6.1)\right] = \frac{10}{3}[7.1 + 40.8 + 12.2] = \frac{10}{3}(60.1) = 200.3$ m² The difference of 1.8 m² on 200 m² is about 0.9 per cent - small in absolute terms but systematic, and the trapezoidal rule always **under**estimates a convex boundary. --- ## 3. Volumes from cross-sections ### End-area rule For two cross-sectional areas $A_1$ and $A_2$ a distance $L$ apart: $$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. --- ## 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. --- ## 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.
Loading diagram...
Reduction and setting out run in opposite directions
Test Your Knowledge

A four-sided parcel has corners A (1000.00, 1000.00), B (1040.00, 1010.00), C (1050.00, 1055.00) and D (1005.00, 1048.00), coordinates given as (E, N) in metres. What is its area by the coordinate method?

A
B
C
D
Test Your Knowledge

In the prismoidal formula V = (L/6)(A1 + 4Am + A2), how must the mid-area Am be obtained?

A
B
C
D
Test Your Knowledge

Offsets to an irregular boundary are taken at 10 m intervals: 3.2, 4.8, 6.1, 5.4 and 3.9 m. What area does Simpson's rule give, and why is the rule applicable here?

A
B
C
D
Test Your Knowledge

A design point is 2,000.000 m from a control station in grid distance, and the combined scale factor for the line is 0.99979. What distance is set out on the ground?

A
B
C
D
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