Open-Pit Volume from Cross-Sectional Areas: 18-Geom-B1 Digital Terrain Modelling, December 2017, Q6

Question 6: Open-Pit Volume from Cross-Sectional Areas (8 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Seven horizontal cross-sections at a constant depth interval $h = 15\text{ m}$, with areas $A_0\ldots A_6 = 220, 180, 160, 72, 64, 42, 11\text{ m}^2$ (6 intervals).

Find. The total excavated volume of the pit.

Cross-sectional area plotted against depth. The volume is the area under this curve — the sum of the prisms between successive sections.

Approach. Treat the pit as a stack of prisms between successive sections and integrate the area–depth profile; the average-end-area (trapezoidal) rule gives the standard estimate, with the prismoidal (Simpson) rule as a refinement.

Average-end-area (trapezoidal) rule. Each prism between two sections has volume $\tfrac{h}{2}(A_i+A_{i+1})$; summing telescopes to $$V = h\left[\frac{A_0}{2} + A_1 + A_2 + A_3 + A_4 + A_5 + \frac{A_6}{2}\right]$$
Substitute the areas. $$V = 15\left[\frac{220}{2} + 180 + 160 + 72 + 64 + 42 + \frac{11}{2}\right] = 15\,[110 + 518 + 5.5] = 15(633.5)$$ $$\boxed{V \approx 9\,502.5\ \text{m}^3}$$
Prismoidal (Simpson 1/3) cross-check. With 6 intervals (even), Simpson’s rule applies directly and better models the curved taper: $$V = \frac{h}{3}\big[A_0 + 4(A_1+A_3+A_5) + 2(A_2+A_4) + A_6\big] = \frac{15}{3}\big[220 + 4(294) + 2(224) + 11\big] = 5(1855)$$ $$V \approx 9\,275\ \text{m}^3$$ The two estimates agree to within about 2.5%, confirming the result; the prismoidal value is the more accurate where the section areas change curvilinearly.

Method	Volume
Average end-area (trapezoidal)	$9\,502.5\ \text{m}^3$
Prismoidal (Simpson 1/3) refinement	$9\,275\ \text{m}^3$

Concept & Theory

Topic. Volume from a series of parallel cross-sections (numerical integration of area vs. distance).

Key relations. End-area: $V = h\big[\tfrac{A_0+A_n}{2} + \sum_{i=1}^{n-1} A_i\big]$; prismoidal (Simpson, even $n$): $V = \tfrac{h}{3}\big[A_0 + 4\!\sum_{\text{odd}}A_i + 2\!\sum_{\text{even interior}}A_i + A_n\big]$.

Why this works. Volume is the integral of cross-sectional area along the axis; the end-area rule approximates each slab as a trapezoidal prism (area varies linearly), while Simpson’s rule fits a parabola through triples of sections, exactly integrating the second-degree variation of a true prismoid.

Common pitfall. Applying Simpson’s 1/3 rule with an odd number of intervals (it needs an even count), and mishandling the half-weighting of the two end sections in the end-area sum.

Source. Ghilani & Wolf, Elementary Surveying, ch. 26 (volumes); Li, Zhu & Gold, ch. 7 (DTM volume applications).

Question 6 of 12: Open-Pit Volume from Cross-Sectional Areas

Question 6: Open-Pit Volume from Cross-Sectional Areas (8 marks)