---
title: "Aggregate reserve: sandstone N42W/34, 80 m"
subject: "Engineering Geology II"
chapter: 7
key: "semester_4/engineering_geology_ii:reserve-sandstone-n42w-34"
source: https://www.bctengineer.com/q/semester_4/engineering_geology_ii/reserve-sandstone-n42w-34
site: BCT Engineering
---

# Aggregate reserve: sandstone N42W/34, 80 m

Engineering Geology II · Chapter 7: Geology and Construction Materials

- Asked in: 2078 Chaitra, 2 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/engineering_geology_ii/reserve-sandstone-n42w-34
- Whole chapter: https://www.bctengineer.com/md/semester_4/engineering_geology_ii/chapter-7

## Question

The attitude of sandstone is N42°W/34°. The difference of top and bottom of bed rock is 80 m. Calculate the reserve of aggregate in 2.5 km strike length and 0.5 km dip length of rock quarry site.

## Short answer

True thickness t = 80 cos34° ≈ 66.3 m. Area on bed = 2.5 km × 0.5 km = 1.25×10⁶ m². Volume = 1.25×10⁶ m² × 66.3 m = 8.29×10⁷ m³.

## Answer

**Principle:** the area of the quarry on the bed surface = strike length $\times$ dip length; reserve = area $\times$ true thickness, with $t = t_v\cos\delta$.

#### Given
Attitude N42°W/34° (dip $34^\circ$), vertical thickness (top to bottom) $t_v = 80$ m, strike length 2.5 km, dip length 0.5 km (taken along the dip of the bed).

#### Calculation
$$
\begin{aligned}
t &= 80\cos 34^\circ = 66.32\ \text{m} \\
A &= 2500 \times 500 = 1.25 \times 10^6\ \text{m}^2 \\
V &= 1.25 \times 10^6 \times 66.32 = 8.29 \times 10^7\ \text{m}^3
\end{aligned}
$$

**Answer: reserve of aggregate = 8.29 × 10⁷ m³ (about 82.9 million m³ = 0.083 km³).**

## Check yourself

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

1. How is quarry area found here?
   A. Strike length x vertical thickness
   B. Strike length + dip length
   C. Dip length squared
   D. Strike length x dip length = 2500 x 500 = 1.25 x 10^6 m²
   Answer: D. The area lies on the bed surface, along strike and dip. (Key point: Area on bed)

2. What is the true thickness for tv = 80 m and dip 34°?
   A. About 44.7 m
   B. About 66.3 m
   C. About 96.5 m
   D. About 80 m
   Answer: B. 80 cos 34° = 66.32 m. (Key point: True thickness)

3. What reserve results?
   A. About 1.0 x 10^8 m³
   B. About 5.6 x 10^7 m³
   C. About 1.25 x 10^6 m³
   D. About 8.29 x 10^7 m³
   Answer: D. 1.25 x 10^6 x 66.32 = 8.29 x 10^7 m³. (Key point: Reserve)

## Close variants set in other exams

- [Aggregate reserve: sandstone N44W/36, 82 m](https://www.bctengineer.com/question.md?key=semester_4%2Fengineering_geology_ii%3Areserve-sandstone-n44w-36) — Engineering Geology II, chapter 7, asked 2×

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