Computation of Area and Volume

Given: formation width $b = 9$ m, side slope $s:1 = 2:1$ (horizontal : vertical), transverse (cross) slope of ground $n:1 = 6:1$, length 60 m with sections at 0, 30 and 60 m. The formation falls 1 in 40, so the fall in 30 m = 30/40 = 0.75 m.

Step 1: Depth of cutting at the centre line

Formation level at 30 m = 1178.25 m, at 60 m = 1177.50 m.

$$h = \text{GL} - \text{formation level}:\quad h_0 = 2.50,\ h_{30} = 3.55,\ h_{60} = 4.90\ \text{m}$$

Step 2: Area of a two-level section

For ground rising at $1$ in $n$ on one side, the horizontal distances of the two cut edges from the centre line are

$$x_1 = \frac{n\left(\frac{b}{2} + s h\right)}{n - s},\qquad x_2 = \frac{n\left(\frac{b}{2} + s h\right)}{n + s}$$

and the depths of the cut at these edges are $h_1 = h + x_1/n$ and $h_2 = h - x_2/n$. The area is

$$A = \frac{b\,h}{2} + \frac{\left(\frac{b}{2} + s h\right)(h_1 + h_2)}{2}$$

Example for the section at chainage 0: $h = 2.50$, $\frac{b}{2} + sh = 4.5 + 2(2.50) = 9.50$, $x_1 = 6(9.50)/4 = 14.250$, $x_2 = 6(9.50)/8 = 7.125$, $h_1 = 4.875$, $h_2 = 1.3125$, $A_0 = \frac{9(2.50)}{2} + \frac{9.50(6.1875)}{2} = 40.641$ m².

Step 3: Volumes

(a) Trapezoidal (average end area) formula with 30 m between sections:

$$V_t = \frac{d}{2}\left[A_0 + 2A_1 + A_2\right] = \frac{30}{2}\left[40.641 + 2(65.565) + 104.901\right] = 4150.07\ \text{m}^3$$

(b) Prismoidal formula (three sections, i.e. two equal intervals):

$$V_p = \frac{d}{3}\left[A_0 + 4A_1 + A_2\right] = \frac{30}{3}\left[40.641 + 4(65.565) + 104.901\right] = 4078.01\ \text{m}^3$$

(c) Prismoidal correction

$$C_p = V_t - V_p = 4150.07 - 4078.01 = 72.06\ \text{m}^3$$

The trapezoidal formula overestimates the volume by this amount, so the prismoidal volume is the more accurate value.

Answer: volume of cutting by trapezoidal formula = 4150.1 m³; by prismoidal formula = 4078.0 m³; prismoidal correction = 72.1 m³ (to be subtracted from the trapezoidal volume).

A two-level section (ground sloping across the road) is used at every chainage. Data: formation width $b = 8$ m, side slope $s:1 = 1.5:1$ (horizontal : vertical), transverse ground slope $n:1 = 6:1$, section spacing $L = 30$ m.

Step 1: Formation levels and depths at the centre line

The formation rises at 1 in 40, so it gains $30/40 = 0.75$ m in every 30 m: $FL = 380.00 + 0.75 \times (\text{number of intervals})$. The centre depth of cut is $h = GL - FL$.

Step 2: Cross-sectional areas

Along the cross-section the ground rises on one side and falls on the other. The depths of cut at the two side slope ends (where the side slope meets the ground) are

$$h_1 = \frac{nh + b/2}{n - s} \quad (\text{high-ground side}), \qquad h_2 = \frac{nh - b/2}{n + s} \quad (\text{low-ground side})$$

The horizontal distances from the centre line to these points are $d_1 = b/2 + s h_1$ and $d_2 = b/2 + s h_2$, and the area of the section is

$$A = \frac{1}{2}(d_1 + d_2)\left(h + \frac{b}{2s}\right) - \frac{b^2}{4s}$$

(the triangle between the two side slopes extended and the ground line, minus the triangle below the formation width).

Worked section at chainage 0, $h = 0.80$ m:

$$\begin{aligned} h_1 &= \frac{6 \times 0.80 + 4}{6 - 1.5} = 1.956\ \text{m}, & h_2 &= \frac{6 \times 0.80 - 4}{6 + 1.5} = 0.107\ \text{m} \\ d_1 &= 4 + 1.5 \times 1.956 = 6.933\ \text{m}, & d_2 &= 4 + 1.5 \times 0.107 = 4.160\ \text{m} \\ A &= \tfrac12(6.933 + 4.160)\left(0.80 + \tfrac{8}{3}\right) - \tfrac{64}{6} = 8.562\ \text{m}^2 \end{aligned}$$

Step 3: Volume by the trapezoidal formula

$$V_T = L\left[\frac{A_1 + A_{5}}{2} + A_2 + \dots + A_{4}\right] = 30\left[\frac{8.562 + 33.714}{2} + 17.782 + 25.687 + 21.974\right] = 2597.43\ \text{m}^3$$

Step 4: Volume by the prismoidal formula

$$V_P = \frac{L}{3}\left[A_1 + A_{5} + 4(A_2 + A_4) + 2A_3\right] = \frac{30}{3}\left[8.562 + 33.714 + 4(17.782 + 21.974) + 2(25.687)\right] = 2526.74\ \text{m}^3$$

Step 5: Prismoidal correction

$$C_p = V_T - V_P = 2597.43 - 2526.74 = 70.69\ \text{m}^3$$

The trapezoidal formula overestimates the volume by 70.69 m³, so the prismoidal volume is the more accurate one.

Answer: Volume of cutting by the trapezoidal formula = 2597.43 m³; by the prismoidal formula = 2526.74 m³; prismoidal correction = 70.69 m³.

Given: formation width $b = 9$ m, side slope $s:1 = 2:1$ (horizontal : vertical), transverse (cross) slope of ground $n:1 = 6:1$, length 60 m with sections at 0, 30 and 60 m. The formation falls 1 in 40, so the fall in 30 m = 30/40 = 0.75 m.

Step 1: Depth of cutting at the centre line

Formation level at 30 m = 278.25 m, at 60 m = 277.50 m.

$$h = \text{GL} - \text{formation level}:\quad h_0 = 2.50,\ h_{30} = 3.55,\ h_{60} = 4.90\ \text{m}$$

Step 2: Area of a two-level section

For ground rising at $1$ in $n$ on one side, the horizontal distances of the two cut edges from the centre line are

$$x_1 = \frac{n\left(\frac{b}{2} + s h\right)}{n - s},\qquad x_2 = \frac{n\left(\frac{b}{2} + s h\right)}{n + s}$$

and the depths of the cut at these edges are $h_1 = h + x_1/n$ and $h_2 = h - x_2/n$. The area is

$$A = \frac{b\,h}{2} + \frac{\left(\frac{b}{2} + s h\right)(h_1 + h_2)}{2}$$

Example for the section at chainage 0: $h = 2.50$, $\frac{b}{2} + sh = 4.5 + 2(2.50) = 9.50$, $x_1 = 6(9.50)/4 = 14.250$, $x_2 = 6(9.50)/8 = 7.125$, $h_1 = 4.875$, $h_2 = 1.3125$, $A_0 = \frac{9(2.50)}{2} + \frac{9.50(6.1875)}{2} = 40.641$ m².

Step 3: Volumes

(a) Trapezoidal (average end area) formula with 30 m between sections:

$$V_t = \frac{d}{2}\left[A_0 + 2A_1 + A_2\right] = \frac{30}{2}\left[40.641 + 2(65.565) + 104.901\right] = 4150.07\ \text{m}^3$$

(b) Prismoidal formula (three sections, i.e. two equal intervals):

$$V_p = \frac{d}{3}\left[A_0 + 4A_1 + A_2\right] = \frac{30}{3}\left[40.641 + 4(65.565) + 104.901\right] = 4078.01\ \text{m}^3$$

(c) Prismoidal correction

$$C_p = V_t - V_p = 4150.07 - 4078.01 = 72.06\ \text{m}^3$$

The trapezoidal formula overestimates the volume by this amount, so the prismoidal volume is the more accurate value.

Answer: volume of cutting by trapezoidal formula = 4150.1 m³; by prismoidal formula = 4078.0 m³; prismoidal correction = 72.1 m³ (to be subtracted from the trapezoidal volume).

A two-level section (ground sloping across the road) is used at every chainage. Data: formation width $b = 9$ m, side slope $s:1 = 2:1$ (horizontal : vertical), transverse ground slope $n:1 = 6:1$, section spacing $L = 30$ m.

Step 1: Formation levels and depths at the centre line

The formation rises at 1 in 40, so it gains $30/40 = 0.75$ m per 30 m: $FL = 278.00 + 0.75 \times (\text{number of intervals})$. The centre depth of cut is $h = GL - FL$.

Step 2: Cross-sectional areas

Along the cross-section the ground rises on one side and falls on the other. The depths of cut at the two side slope ends (where the side slope meets the ground) are

$$h_1 = \frac{nh + b/2}{n - s} \quad (\text{high-ground side}), \qquad h_2 = \frac{nh - b/2}{n + s} \quad (\text{low-ground side})$$

The horizontal distances from the centre line to these points are $d_1 = b/2 + s h_1$ and $d_2 = b/2 + s h_2$, and the area of the section is

$$A = \frac{1}{2}(d_1 + d_2)\left(h + \frac{b}{2s}\right) - \frac{b^2}{4s}$$

(the triangle between the two side slopes extended and the ground line, minus the triangle below the formation width).

Worked section at chainage 0, $h = 3.50$ m:

$$\begin{aligned} h_1 &= \frac{6 \times 3.50 + 4.5}{6 - 2} = 6.375\ \text{m}, & h_2 &= \frac{6 \times 3.50 - 4.5}{6 + 2} = 2.062\ \text{m} \\ d_1 &= 4.5 + 2 \times 6.375 = 17.250\ \text{m}, & d_2 &= 4.5 + 2 \times 2.062 = 8.625\ \text{m} \\ A &= \tfrac12(17.250 + 8.625)\left(3.50 + \tfrac{9}{4}\right) - \tfrac{81}{8} = 64.266\ \text{m}^2 \end{aligned}$$

Step 3: Volume by the trapezoidal formula

$$V_T = \frac{L}{2}\,(A_1 + 2A_2 + A_3) = \frac{30}{2}\,(64.266 + 2 \times 53.077 + 49.551) = 3299.57\ \text{m}^3$$

Step 4: Volume by the prismoidal formula

$$V_P = \frac{L}{3}\,(A_1 + 4A_2 + A_3) = \frac{30}{3}\,(64.266 + 4 \times 53.077 + 49.551) = 3261.25\ \text{m}^3$$

Step 5: Prismoidal correction

$$C_p = V_T - V_P = 3299.57 - 3261.25 = 38.32\ \text{m}^3$$

The trapezoidal formula overestimates the volume by 38.32 m³, so the prismoidal volume is the more accurate one.

Answer: Volume of cutting by the trapezoidal formula = 3299.57 m³; by the prismoidal formula = 3261.25 m³; prismoidal correction = 38.32 m³.

A two-level section (ground sloping across the road) is used at every chainage. Data: formation width $b = 10$ m, side slope $s:1 = 2:1$ (horizontal : vertical), transverse ground slope $n:1 = 6:1$, section spacing $L = 30$ m.

Step 1: Formation levels and depths at the centre line

The formation falls at 1 in 40, so it loses $30/40 = 0.75$ m per 30 m: $FL = 538.20 - 0.75 \times (\text{number of intervals})$. The centre depth of cut is $h = GL - FL$.

Step 2: Cross-sectional areas

Along the cross-section the ground rises on one side and falls on the other. The depths of cut at the two side slope ends (where the side slope meets the ground) are

$$h_1 = \frac{nh + b/2}{n - s} \quad (\text{high-ground side}), \qquad h_2 = \frac{nh - b/2}{n + s} \quad (\text{low-ground side})$$

The horizontal distances from the centre line to these points are $d_1 = b/2 + s h_1$ and $d_2 = b/2 + s h_2$, and the area of the section is

$$A = \frac{1}{2}(d_1 + d_2)\left(h + \frac{b}{2s}\right) - \frac{b^2}{4s}$$

(the triangle between the two side slopes extended and the ground line, minus the triangle below the formation width).

Worked section at chainage 0, $h = 2.50$ m:

$$\begin{aligned} h_1 &= \frac{6 \times 2.50 + 5}{6 - 2} = 5.000\ \text{m}, & h_2 &= \frac{6 \times 2.50 - 5}{6 + 2} = 1.250\ \text{m} \\ d_1 &= 5 + 2 \times 5.000 = 15.000\ \text{m}, & d_2 &= 5 + 2 \times 1.250 = 7.500\ \text{m} \\ A &= \tfrac12(15.000 + 7.500)\left(2.50 + \tfrac{10}{4}\right) - \tfrac{100}{8} = 43.750\ \text{m}^2 \end{aligned}$$

Step 3: Volume by the trapezoidal formula

$$V_T = \frac{L}{2}\,(A_1 + 2A_2 + A_3) = \frac{30}{2}\,(43.750 + 2 \times 69.856 + 110.710) = 4412.58\ \text{m}^3$$

Step 4: Volume by the prismoidal formula

$$V_P = \frac{L}{3}\,(A_1 + 4A_2 + A_3) = \frac{30}{3}\,(43.750 + 4 \times 69.856 + 110.710) = 4338.84\ \text{m}^3$$

Step 5: Prismoidal correction

$$C_p = V_T - V_P = 4412.58 - 4338.84 = 73.74\ \text{m}^3$$

The trapezoidal formula overestimates the volume by 73.74 m³, so the prismoidal volume is the more accurate one.

Answer: Volume of cutting by the trapezoidal formula = 4412.58 m³; by the prismoidal formula = 4338.84 m³; prismoidal correction = 73.74 m³.

Latitudes sum to 0.090 m and departures to 0.000 m. The latitude misclosure of 0.090 m is very small compared with the perimeter, so the figures are used as given.

DMD method

DMD rule (taking the meridian through the starting station):

• DMD of the first line = departure of the first line.

• DMD of any line = DMD of the previous line + departure of the previous line + departure of the line itself.

• DMD of the last line = departure of the last line with opposite sign (a check).

• Double area = Σ (latitude × DMD), and area = ½ |Σ L × DMD|.

• DMD(AB) = D(AB) = -33.990

• DMD(BC) = -33.990 + (-33.990) + (-13.911) = -81.891

• DMD(CD) = -81.891 + (-13.911) + (18.867) = -76.935

• DMD(DE) = -76.935 + (18.867) + (28.608) = -29.460

• DMD(EA) = -29.460 + (28.608) + (0.426) = -0.426

• Check: DMD(EA) = -0.426 = −D(EA) = -0.426, so the DMDs are correct.

$$\text{Area} = \frac{|4056.821|}{2} = 2028.411\ \text{m}^2$$

Answer: Area = 2028.41 m² (0.2028 ha).

The traverse closes: ΣL = −370 + 240 − 260 + 390 = 0 and ΣD = 220 − 400 − 300 + 480 = 0.

DMD method

DMD rule (taking the meridian through the starting station):

• DMD of the first line = departure of the first line.

• DMD of any line = DMD of the previous line + departure of the previous line + departure of the line itself.

• DMD of the last line = departure of the last line with opposite sign (a check).

• Double area = Σ (latitude × DMD), and area = ½ |Σ L × DMD|.

• DMD(AB) = D(AB) = 220

• DMD(BC) = 220 + (220) + (-400) = 40

• DMD(CD) = 40 + (-400) + (-300) = -660

• DMD(DA) = -660 + (-300) + (480) = -480

• Check: DMD(DA) = -480 = −D(DA) = -480, so the DMDs are correct.

$$\text{Area} = \frac{|-87400|}{2} = 43700\ \text{m}^2$$

Coordinate method

Taking the first station as the origin (N = 0, E = 0) and adding latitudes and departures successively gives the total coordinates. The area is

$$2A = \sum N_i\,(E_{i+1} - E_{i-1})$$ $$\text{Area} = \frac{|87400|}{2} = 43700\ \text{m}^2$$

Both methods agree.

Answer: Area = 43700.00 m² (4.3700 ha).

The traverse closes: ΣL = 0 and ΣD = 0.

DMD method

DMD rule (taking the meridian through the starting station):

• DMD of the first line = departure of the first line.

• DMD of any line = DMD of the previous line + departure of the previous line + departure of the line itself.

• DMD of the last line = departure of the last line with opposite sign (a check).

• Double area = Σ (latitude × DMD), and area = ½ |Σ L × DMD|.

• DMD(AB) = D(AB) = 202

• DMD(BC) = 202 + (202) + (80) = 484

• DMD(CD) = 484 + (80) + (-332) = 232

• DMD(DE) = 232 + (-332) + (-301) = -401

• DMD(EA) = -401 + (-301) + (351) = -351

• Check: DMD(EA) = -351 = −D(EA) = -351, so the DMDs are correct.

$$\text{Area} = \frac{|-163938|}{2} = 81969\ \text{m}^2$$

Answer: Area = 81969.00 m² (8.1969 ha).

The traverse closes: ΣL = 0 and ΣD = 0.

DMD method

DMD rule (taking the meridian through the starting station):

• DMD of the first line = departure of the first line.

• DMD of any line = DMD of the previous line + departure of the previous line + departure of the line itself.

• DMD of the last line = departure of the last line with opposite sign (a check).

• Double area = Σ (latitude × DMD), and area = ½ |Σ L × DMD|.

• DMD(PQ) = D(PQ) = 450

• DMD(QR) = 450 + (450) + (110) = 1010

• DMD(RS) = 1010 + (110) + (-380) = 740

• DMD(SP) = 740 + (-380) + (-180) = 180

• Check: DMD(SP) = 180 = −D(SP) = 180, so the DMDs are correct.

$$\text{Area} = \frac{|506200|}{2} = 253100\ \text{m}^2$$

Answer: Area = 253100.00 m² (25.3100 ha).

A two-level section (ground sloping across the road) is used at every chainage. Data: formation width $b = 10$ m, side slope $s:1 = 2:1$ (horizontal : vertical), transverse ground slope $n:1 = 8:1$, section spacing $L = 25$ m.

Step 1: Formation levels and depths at the centre line

The formation falls at 1 in 50, so it loses $25/50 = 0.50$ m per 25 m: $FL = 1081.50 - 0.50 \times (\text{number of intervals})$. The centre height of fill is $h = FL - GL$.

Step 2: Cross-sectional areas

For an embankment on cross-sloping ground, the heights of fill at the two side slope ends are

$$h_1 = \frac{nh - b/2}{n + s} \quad (\text{high-ground side}), \qquad h_2 = \frac{nh + b/2}{n - s} \quad (\text{low-ground side})$$

The horizontal distances from the centre line to these points are $d_1 = b/2 + s h_1$ and $d_2 = b/2 + s h_2$, and the area of the section is

$$A = \frac{1}{2}(d_1 + d_2)\left(h + \frac{b}{2s}\right) - \frac{b^2}{4s}$$

(the triangle between the two side slopes extended and the ground line, minus the triangle above the formation width).

Worked section at chainage 0, $h = 1.00$ m:

$$\begin{aligned} h_1 &= \frac{8 \times 1.00 - 5}{8 + 2} = 0.300\ \text{m}, & h_2 &= \frac{8 \times 1.00 + 5}{8 - 2} = 2.167\ \text{m} \\ d_1 &= 5 + 2 \times 0.300 = 5.600\ \text{m}, & d_2 &= 5 + 2 \times 2.167 = 9.333\ \text{m} \\ A &= \tfrac12(5.600 + 9.333)\left(1.00 + \tfrac{10}{4}\right) - \tfrac{100}{8} = 13.633\ \text{m}^2 \end{aligned}$$

Step 3: Volume by the trapezoidal formula

$$V_T = \frac{L}{2}\,(A_1 + 2A_2 + A_3) = \frac{25}{2}\,(13.633 + 2 \times 16.705 + 32.641) = 996.05\ \text{m}^3$$

Step 4: Volume by the prismoidal formula

$$V_P = \frac{L}{3}\,(A_1 + 4A_2 + A_3) = \frac{25}{3}\,(13.633 + 4 \times 16.705 + 32.641) = 942.45\ \text{m}^3$$

Step 5: Prismoidal correction

$$C_p = V_T - V_P = 996.05 - 942.45 = 53.60\ \text{m}^3$$

The trapezoidal formula overestimates the volume by 53.60 m³, so the prismoidal volume is the more accurate one.

Answer: Volume of filling by the trapezoidal formula = 996.05 m³; by the prismoidal formula = 942.45 m³; prismoidal correction = 53.60 m³.

Meridian distance (M.D.)

The meridian distance of a line is the perpendicular distance of the middle point of the line from the reference meridian (a N-S line). Its value is positive on the east of the meridian. For convenience, the reference meridian is taken through the most westerly station of the traverse so that all meridian distances are positive.

Double meridian distance (D.M.D.)

The D.M.D. of a line is twice its meridian distance. It is used because it avoids fractions and halves. For the line AB, M.D. = 1/2 (departure of AB) from the starting station, so D.M.D. = departure of AB.

Calculation of D.M.D.

The latitudes and departures are first adjusted (so that $\Sigma L = 0$ and $\Sigma D = 0$). Then:

1. D.M.D. of the first line = its departure.
2. D.M.D. of any line = D.M.D. of the previous line + departure of the previous line + departure of the line itself (with signs; east +, west -).
3. D.M.D. of the last line = its departure with the opposite sign (check on the work).

   reference
   meridian        B
      |          /
      |     m  /        M.D. = distance of the
      |<-----*/          mid-point m of AB from
      |    /             the meridian
      |  A

Area of a traverse by the D.M.D. method

The area enclosed between each line and the meridian is a trapezium; its area equals the meridian distance of the line (the mean width) multiplied by its latitude, that is, half of D.M.D. times latitude. The sum of these areas, with signs, gives the area of the traverse:

$$\text{Area} = \frac{1}{2}\sum \left(\text{D.M.D.} \times \text{Latitude}\right)$$

(The sum is taken with the signs of the products; the numerical value is the area. North latitude +, south -.)

Example

Consider the closed traverse with stations A (E 0, N 0), B (E 80, N 30), C (E 100, N 100), D (E 20, N 90). The westmost station is A, so the reference meridian passes through A.

The D.M.D. of the last line DA is 20, equal to the departure of DA (-20) with the sign reversed, which checks the work.

$$\text{Area} = \frac{1}{2}\left|12000\right| = 6000.0\ \text{m}^2$$

This agrees with the area from the coordinates, 6000.0 m².

Given: 8 offsets at a common interval $d = 20$ m, so the length of the chain line = 7 × 20 = 140 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 = 26.27 m; (first + last)/2 = (0.00 + 6.55)/2 = 3.275 m.

$$\text{Area} = 20\times\left[3.275 + 26.27\right] = 590.900\ \text{m}^2$$

(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.

$$\text{Area}_1 = \frac{20}{3}\left[(0.00 + 4.30) + 4(12.28) + 2(9.69)\right] = 485.333\ \text{m}^2$$ $$\text{Area}_2 = 20\times\frac{4.30 + 6.55}{2} = 108.500\ \text{m}^2$$ $$\text{Total area} = 485.333 + 108.500 = 593.833\ \text{m}^2$$

Answer: trapezoidal rule = 590.90 m²; Simpson's rule = 593.83 m².

There are 12 offsets, so there are 11 equal strips of width $d = 20$ m, and the first offset is $O_1 = 0$ and the last is $O_{12} = 4.19$ m.

(i) Trapezoidal rule

$$A = d\left[\frac{O_1 + O_n}{2} + O_2 + O_3 + \dots + O_{n-1}\right]$$

Sum of the intermediate offsets ($O_2$ to $O_{11}$):

$$\sum O_{mid} = 50.62\ \text{m}$$ $$A = 20\left[\frac{0.00 + 4.19}{2} + 50.62\right] = 20 \times 52.715 = 1054.30\ \text{m}^2$$

(ii) Simpson's rule

Simpson's one-third rule needs an odd number of offsets (an even number of strips). Here there are 12 offsets, so apply Simpson's rule to the first 11 offsets (10 strips) and the trapezoidal rule to the last strip.

$$A = \frac{d}{3}\left[O_1 + O_{11} + 4(O_2 + O_4 + O_6 + O_8 + O_{10}) + 2(O_3 + O_5 + O_7 + O_9)\right] + d\,\frac{O_{11} + O_{12}}{2}$$

• Sum of even-numbered offsets: $O_2 + O_4 + O_6 + O_8 + O_{10} = 26.13$ m
• Sum of odd-numbered intermediate offsets: $O_3 + O_5 + O_7 + O_9 = 18.24$ m
• First and last offset of the Simpson part: $O_1 + O_{11} = 6.25$ m

$$\begin{aligned} A_{1-11} &= \frac{20}{3}\left[6.25 + 4(26.13) + 2(18.24)\right] = \frac{20}{3} \times 147.25 = 981.67\ \text{m}^2 \\ A_{last} &= 20 \times \frac{6.25 + 4.19}{2} = 104.40\ \text{m}^2 \\ A &= 981.67 + 104.40 = 1086.07\ \text{m}^2 \end{aligned}$$

Answer: Trapezoidal area = 1054.30 m²; Simpson's area = 1086.07 m².

The traverse closes: ΣL = 0.00 and ΣD = 0.00.

DMD method

DMD rule (taking the meridian through the starting station):

• DMD of the first line = departure of the first line.

• DMD of any line = DMD of the previous line + departure of the previous line + departure of the line itself.

• DMD of the last line = departure of the last line with opposite sign (a check).

• Double area = Σ (latitude × DMD), and area = ½ |Σ L × DMD|.

• DMD(AB) = D(AB) = 405.85

• DMD(BC) = 405.85 + (405.85) + (0.00) = 811.70

• DMD(CD) = 811.70 + (0.00) + (-151.55) = 660.15

• DMD(DE) = 660.15 + (-151.55) + (-148.10) = 360.50

• DMD(EA) = 360.50 + (-148.10) + (-106.20) = 106.20

• Check: DMD(EA) = 106.20 = −D(EA) = 106.20, so the DMDs are correct.

$$\text{Area} = \frac{|155128.97|}{2} = 77564.49\ \text{m}^2$$

Answer: Area = 77564.49 m² (7.7564 ha).

The traverse closes: ΣL = 0.00 and ΣD = 0.00.

DMD method

DMD rule (taking the meridian through the starting station):

• DMD of the first line = departure of the first line.

• DMD of any line = DMD of the previous line + departure of the previous line + departure of the line itself.

• DMD of the last line = departure of the last line with opposite sign (a check).

• Double area = Σ (latitude × DMD), and area = ½ |Σ L × DMD|.

• DMD(AB) = D(AB) = 227.26

• DMD(BC) = 227.26 + (227.26) + (417.26) = 871.78

• DMD(CD) = 871.78 + (417.26) + (-196.47) = 1092.57

• DMD(DA) = 1092.57 + (-196.47) + (-448.05) = 448.05

• Check: DMD(DA) = 448.05 = −D(DA) = 448.05, so the DMDs are correct.

$$\text{Area} = \frac{|-326035.30|}{2} = 163017.65\ \text{m}^2$$

Answer: Area = 163017.65 m² (16.3018 ha).

When the coordinates (northing $N$ and easting $E$, or $x$ and $y$) of the stations of a closed traverse or boundary are known, the area is found directly from them.

Formula

Number the stations $1, 2, \dots, n$ in order round the figure (clockwise or anticlockwise), with the coordinates $(x_i, y_i)$ and $(x_{n+1}, y_{n+1}) = (x_1, y_1)$.

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

or, in the form with the coordinate differences,

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

Procedure

1. Compute the adjusted latitudes and departures, and the coordinates of every station from a convenient origin (preferably so that all coordinates are positive).
2. Write the coordinates in order in two columns and repeat the first station at the end.
3. Multiply each $x$ by the next $y$ (downward products) and each $y$ by the next $x$ (upward products).
4. Take the difference of the two sums and halve it. The sign depends on the direction of travel; the numerical value is the area.

Example

Stations: A (0, 0), B (80, 30), C (100, 100), D (20, 90) in (E, N).

$$\text{Area} = \frac{1}{2}(17000 - 5000) = 6000\ \text{m}^2$$

The method is exact for a polygon and is suitable for a computer or calculator, with no need for dividing the figure into parts.

A prismoid is a solid having two parallel end faces (the two end sections, areas $A_1$ and $A_2$) joined by plane or ruled surfaces. The earthwork between two cross-sections is generally treated as a prismoid. The prismoidal formula gives its volume accurately.

Derivation

Let $L$ be the distance between the end sections, and let $x$ be the distance measured from the first end. The cross-sectional area at distance $x$ varies as a second-degree (quadratic) function of $x$ (since each linear dimension varies linearly with $x$):

$$A(x) = a + b x + c x^2$$

The areas at the ends and at the middle are

$$A_1 = A(0) = a,\quad A_m = A\left(\tfrac{L}{2}\right) = a + \frac{bL}{2} + \frac{cL^2}{4},\quad A_2 = A(L) = a + bL + cL^2$$

The volume is

$$V = \int_0^L A(x)\,dx = aL + \frac{bL^2}{2} + \frac{cL^3}{3}$$

Now $A_1 + 4A_m + A_2 = 6a + 3bL + 2cL^2$, so

$$\frac{L}{6}\left(A_1 + 4A_m + A_2\right) = aL + \frac{bL^2}{2} + \frac{cL^3}{3} = V$$ $$\boxed{V = \frac{L}{6}\left(A_1 + 4A_m + A_2\right)}$$

where $A_m$ is the area of the middle section (not the mean of $A_1$ and $A_2$; its dimensions are the mean of the end dimensions).

Prismoidal rule for a number of sections

If there are an odd number of sections at equal spacing $d$ along the line (that is, an even number of intervals), apply the formula to each pair of intervals (length $2d$, middle section is the one between):

$$V = \frac{d}{3}\left[A_1 + A_n + 4(A_2 + A_4 + \dots) + 2(A_3 + A_5 + \dots)\right]$$

This is the same as Simpson's one-third rule. If the number of sections is even, the last interval is treated by the trapezoidal formula.

Prismoidal correction

The trapezoidal (average end area) formula gives $V_t = \frac{L}{2}(A_1 + A_2)$. The difference between the trapezoidal and prismoidal volumes is the prismoidal correction:

$$C_p = V_t - V_p = \frac{L}{3}\left(A_1 + A_2 - 2A_m\right)$$

For a level section with side slope $s:1$ and end depths $h_1$ and $h_2$, this becomes

$$C_p = \frac{s\,L}{6}\,(h_1 - h_2)^2$$

It is always positive, so the prismoidal volume = trapezoidal volume - $C_p$. The correction is large only where the depths $h_1$ and $h_2$ differ greatly.

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 = 195.57 m², coordinate method = 185.70 m².

The offsets are at unequal intervals (5.5, 7.2, 12.8 and 15.0 m), so Simpson's rule cannot be used directly. Two methods that work for unequal intervals are the trapezoidal (strip) method and the coordinate method.

Method 1: Trapezoidal rule (strip by strip)

Each strip between two successive offsets is a trapezium with area $w(o_1 + o_2)/2$.

Area by trapezoidal rule = 206.493 m².

Method 2: Coordinate method

Chain line is the x-axis; the points in order round the figure are:

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

Area = 206.493 m², which agrees with the first method.

Answer: area enclosed between the chain line, the hedge and the first and last offsets = 206.49 m² (by both methods).

Given: 4 offsets at a common interval $d = 10$ m, so the length of the chain line = 3 × 10 = 30 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 = 16.80 m; (first + last)/2 = (4.60 + 6.40)/2 = 5.500 m.

$$\text{Area} = 10\times\left[5.500 + 16.80\right] = 223.000\ \text{m}^2$$

(ii) Simpson's one-third rule

There are 4 offsets, i.e. 3 (an odd number of) intervals, so Simpson's rule cannot be used for the whole length. It is applied to the first 2 intervals (offsets 0 to 2), and the last strip is found by the trapezoidal rule.

$$\text{Area}_1 = \frac{10}{3}\left[(4.60 + 9.60) + 4(7.20) + 2(0.00)\right] = 143.333\ \text{m}^2$$ $$\text{Area}_2 = 10\times\frac{9.60 + 6.40}{2} = 80.000\ \text{m}^2$$ $$\text{Total area} = 143.333 + 80.000 = 223.333\ \text{m}^2$$

Answer: trapezoidal rule = 223.00 m²; Simpson's rule = 223.33 m².

Check by Simpson's three-eighth rule (valid for 3 intervals): Area $= \frac{3d}{8}[o_0 + 3o_1 + 3o_2 + o_3] = \frac{3 \times 10}{8}[4.6 + 3(7.2) + 3(9.6) + 6.4] = 230.250\ \text{m}^2$.

The chainages are at unequal intervals, so the area is found strip by strip with the trapezoidal rule (the strip between two offsets is a trapezium of width equal to the chainage difference).

Check by coordinates: with the points (0,0), (52,0), (52,6.8), (42,9.9), (34,10.2), (25,9.8), (20,7.6), (12,6.9), (0,0) the formula

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

gives 396.800 m², the same value.

Answer: area = 396.80 m².

Given: 9 offsets at a common interval $d = 5$ m, so the length of the chain line = 8 × 5 = 40 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 = 41.30 m; (first + last)/2 = (0.00 + 0.00)/2 = 0.000 m.

$$\text{Area} = 5\times\left[0.000 + 41.30\right] = 206.500\ \text{m}^2$$

(ii) Simpson's one-third rule

The number of intervals (8) is even, so the rule applies to the whole length:

$$\text{Area} = \frac{d}{3}\left[(o_0 + o_n) + 4(\text{odd offsets}) + 2(\text{even offsets})\right]$$

Odd-numbered offsets sum = 21.00, even-numbered (intermediate) offsets sum = 20.30.

$$\text{Area} = \frac{5}{3}\left[(0.00 + 0.00) + 4(21.00) + 2(20.30)\right] = 207.667\ \text{m}^2$$

Answer: trapezoidal rule = 206.50 m²; Simpson's rule = 207.67 m².

Method. The chain line is taken as the x-axis and the first offset position as the origin. The ends of the offsets and the two chain-line corners give a closed polygon whose area is found from its coordinates:

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

Coordinates (taken in order round the area)

Calculation

$$\text{Area} = \frac{1}{2}\times |76.9575| = 38.4787\ \text{m}^2$$

Check (trapezoidal strips): the four strips give 38.4787 m².

Answer: area = 38.479 m².