---
title: "Reserve: quartzite apparent dips 1:9 and 1:12, 4.6 km2"
subject: "Engineering Geology II"
chapter: 7
key: "semester_4/engineering_geology_ii:reserve-quartzite-apparent-dips-area"
source: https://www.bctengineer.com/q/semester_4/engineering_geology_ii/reserve-quartzite-apparent-dips-area
site: BCT Engineering
---

# Reserve: quartzite apparent dips 1:9 and 1:12, 4.6 km2

Engineering Geology II · Chapter 7: Geology and Construction Materials

- Asked in: 2074 Bhadra, 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-quartzite-apparent-dips-area
- Whole chapter: https://www.bctengineer.com/md/semester_4/engineering_geology_ii/chapter-7

## Question

Apparent dip amounts of quartzite bed rock along N20°E and N65°E are 1:9 and 1:12 respectively. The vertical thickness of bedrock is 105 m. Calculate the reserve of construction material for an engineering project at an area of 4.6 km² rock quarry site.

## Short answer

True dip 6.35° (1:9.0) N23.5°E, true thickness 104.4 m. For area 4.6 km², reserve = 4.6×10⁶ m² × 104.4 m = 4.80×10⁸ m³ (0.480 km³).

## Answer

#### Step 1: true dip from the two apparent dips
For a direction of bearing $\beta$ the gradient of the bed satisfies $g_E\sin\beta + g_N\cos\beta = \tan\delta_a$ ($g_E$, $g_N$ = east and north components of the true gradient).
$$
\begin{aligned}
g_E\sin 20^\circ + g_N\cos 20^\circ &= \tfrac{1}{9} = 0.11111 \\
g_E\sin 65^\circ + g_N\cos 65^\circ &= \tfrac{1}{12} = 0.08333
\end{aligned}
$$
Solving: $g_E = 0.04434$, $g_N = 0.10211$.
$$
\tan\delta = \sqrt{0.04434^2 + 0.10211^2} = 0.1113 \Rightarrow \delta = 6.35^\circ\ (1:8.98)
$$
Dip direction $= \arctan(0.04434/0.10211) = 23.5^\circ$ (N23.5°E).

#### Step 2: true thickness
$$
t = t_v\cos\delta = 105\cos 6.35^\circ = 104.36\ \text{m}
$$

#### Step 3: volume (reserve)
Area $= 4.6\ \text{km}^2 = 4.6\times 10^6\ \text{m}^2$
$$
V = A \times t = 4.6\times 10^6 \times 104.36 = 4.80 \times 10^8\ \text{m}^3
$$

**Answer: true dip = 6.35° (1:9.0) towards N23.5°E; true thickness = 104.4 m; reserve = 4.80 × 10⁸ m³ (0.480 km³).**

## Check yourself

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

1. How is true dip found from the apparent dips 1:9 and 1:12?
   A. Add 1/9 and 1/12
   B. Take the average of the two angles
   C. Solve two gradient equations to get gE and gN, then tan δ = sqrt(gE^2 + gN^2)
   D. Use the larger apparent dip
   Answer: C. The apparent dips are components of the true gradient. (Key point: True dip method)

2. What true thickness is used?
   A. About 104.4 m, from 105 cos 6.35°
   B. About 105 m
   C. About 11.6 m
   D. About 116.9 m
   Answer: A. True dip is 6.35°, so t = 104.36 m. (Key point: True thickness)

3. What reserve results for 4.6 km²?
   A. About 4.6 x 10^6 m³
   B. About 4.80 x 10^8 m³
   C. About 1.1 x 10^8 m³
   D. About 8.7 x 10^8 m³
   Answer: B. 4.6 x 10^6 m² x 104.36 m = 4.80 x 10^8 m³. (Key point: Reserve)

## Close variants set in other exams

- [Reserve: quartzite apparent dips 1:9 and 1:12, 2.7 x 3.1 km](https://www.bctengineer.com/question.md?key=semester_4%2Fengineering_geology_ii%3Areserve-quartzite-apparent-dips-terrain) — Engineering Geology II, chapter 7, asked 2×

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## Related in other subjects

- [Strike of quartzite bed, dip 70° towards S20°E](https://www.bctengineer.com/question.md?key=semester_3%2Fengineering_geology_i%3Astrike-of-quartzite-bed-dip) — Engineering Geology I, chapter 5
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