---
title: "Volume from central depths 1.85, 2.15, 2.45"
subject: "Surveying I"
chapter: 9
key: "semester_3/surveying_i:volume-central-depths-1-85-2-15-2-45"
source: https://www.bctengineer.com/q/semester_3/surveying_i/volume-central-depths-1-85-2-15-2-45
site: BCT Engineering
---

# Volume from central depths 1.85, 2.15, 2.45

Surveying I · Chapter 9: Computation of Area and Volume

- Asked in: 2078 Bhadra, 6 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/surveying_i/volume-central-depths-1-85-2-15-2-45
- Whole chapter: https://www.bctengineer.com/md/semester_3/surveying_i/chapter-9

## Question

Find the volume by the trapezoidal and prismoidal formula with the following data.

| Chainage (m) | 0 | 30 | 60 |
|---|---|---|---|
| Central depth of cut (m) | 1.85 | 2.15 | 2.45 |

Formation width = 12 m, side slopes = 2:1, transverse slope = 6:1. Also calculate the prismoidal correction.

## Short answer

Using the given centre depths of cut, compute cross-sectional areas for cutting with side slopes 2:1, transverse slope 6:1, formation width 12 m. Trapezoidal volume = 2506.64 m³, prismoidal volume = 2504.61 m³, prismoidal correction = 2.03 m³.

## Answer

A two-level section (ground sloping across the road) is used at every chainage. Data: formation width $b = 12$ m, side slope $s:1 = 2:1$ (horizontal : vertical), transverse ground slope $n:1 = 6:1$, section spacing $L = 30$ m.

#### Step 1: Centre heights

The centre depths of cut are given directly: $h = 1.85$, $2.15$ and $2.45$ m at chainages 0, 30 and 60 m. A cutting is assumed because a depth of cut is given.

#### 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, $h = 1.85$ m:

$$
\begin{aligned} h_1 &= \frac{6 \times 1.85 + 6}{6 - 2} = 4.275\ \text{m}, & h_2 &= \frac{6 \times 1.85 - 6}{6 + 2} = 0.638\ \text{m} \\ d_1 &= 6 + 2 \times 4.275 = 14.550\ \text{m}, & d_2 &= 6 + 2 \times 0.638 = 7.275\ \text{m} \\ A &= \tfrac12(14.550 + 7.275)\left(1.85 + \tfrac{12}{4}\right) - \tfrac{144}{8} = 34.926\ \text{m}^2 \end{aligned}
$$

| Chainage (m) | h (m) | h₁ (m) | h₂ (m) | d₁ (m) | d₂ (m) | Area A (m²) |
|---|---|---|---|---|---|---|
| 0 | 1.85 | 4.275 | 0.638 | 14.550 | 7.275 | 34.926 |
| 30 | 2.15 | 4.725 | 0.862 | 15.450 | 7.725 | 41.676 |
| 60 | 2.45 | 5.175 | 1.088 | 16.350 | 8.175 | 48.831 |

#### Step 3: Volume by the trapezoidal formula

$$
V_T = \frac{L}{2}\,(A_1 + 2A_2 + A_3) = \frac{30}{2}\,(34.926 + 2 \times 41.676 + 48.831) = 2506.64\ \text{m}^3
$$

#### Step 4: Volume by the prismoidal formula

$$
V_P = \frac{L}{3}\,(A_1 + 4A_2 + A_3) = \frac{30}{3}\,(34.926 + 4 \times 41.676 + 48.831) = 2504.61\ \text{m}^3
$$

#### Step 5: Prismoidal correction

$$
C_p = V_T - V_P = 2506.64 - 2504.61 = 2.03\ \text{m}^3
$$

The trapezoidal formula overestimates the volume by 2.03 m³, so the prismoidal volume is the more accurate one.

**Answer: Volume of cutting by the trapezoidal formula = 2506.64 m³; by the prismoidal formula = 2504.61 m³; prismoidal correction = 2.03 m³.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Why is it taken as cutting?
   A. The ground is level
   B. A depth of cut is given
   C. The slope is flat
   D. Fill is not possible
   Answer: B. Central depths of cut are given. (Key point: Type of earthwork)

2. Which formula gives h1 for cut?
   A. (nh + b/2)/(n - s)
   B. (nh - b/2)/(n + s)
   C. (nh + b/2)/(n + s)
   D. nh/(n - s)
   Answer: A. High side: (nh + b/2)/(n - s). (Key point: High-ground side depth)

3. For three sections, trapezoidal volume is
   A. L/3 (A1 + 4A2 + A3)
   B. L (A1 + A2 + A3)
   C. L/2 (A1 + A3)
   D. L/2 (A1 + 2A2 + A3)
   Answer: D. End areas halved, middle fully counted. (Key point: Trapezoidal formula)

4. Cp = 2.03 m^3 is the difference between
   A. two areas
   B. cut and fill
   C. trapezoidal and prismoidal volumes
   D. two chainages
   Answer: C. 2506.64 - 2504.61 = 2.03. (Key point: Cp)

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