---
title: "Bore hole altitude of fractured bedrock"
subject: "Engineering Geology I"
chapter: 5
key: "semester_3/engineering_geology_i:bore-hole-altitude-of-fractured"
source: https://www.bctengineer.com/q/semester_3/engineering_geology_i/bore-hole-altitude-of-fractured
site: BCT Engineering
---

# Bore hole altitude of fractured bedrock

Engineering Geology I · Chapter 5: Structural Geology

- Asked in: 2066 Bhadra (old course), 8 marks
- Video: [What is Strike and Dip and How is it Indicated on a Map?](https://www.youtube.com/watch?v=-yPcPlIHDl0) — Ruppert Rocks Geology
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/engineering_geology_i/bore-hole-altitude-of-fractured
- Whole chapter: https://www.bctengineer.com/md/semester_3/engineering_geology_i/chapter-5

## Question

Three boreholes A, B and C are drilled in a dam site of a hydropower project. Bore hole A lies 500 m due north of Bore hole B and Bore hole C lies 400 m due east of Bore hole B. A highly jointed and fractured bedrock is encountered in the following depths from M.S.L. as below. Find out the altitude of the jointed and fractured bed rock.

| | Top | Bottom |
|---|---|---|
| Bore hole A | 300 | 310 m |
| Bore hole B | 280 | 290 m |
| Bore hole C | 310 | 320 m |

## Short answer

Using the three-point method, the top of the fractured bed is at −300 m (A), −280 m (B), −310 m (C) relative to MSL; bottom 10 m deeper. The bed strikes N28.1°W and dips 4.9° towards N61.9°E.

## Answer

**Method**: The given figures are **depths below M.S.L.** (the top of the bed is at a smaller depth than its bottom). The top (and bottom) of the fractured bed at three non-collinear points define a plane, which is solved by the three-point (strike-line) method.

**Data**: B at the origin; A is 500 m due north of B; C is 400 m due east of B.

| Borehole | Top depth (m) | Bottom depth (m) | Altitude of top (m) | Altitude of bottom (m) | Thickness (m) |
|---|---|---|---|---|---|
| A | 300 | 310 | −300 | −310 | 10 |
| B | 280 | 290 | −280 | −290 | 10 |
| C | 310 | 320 | −310 | −320 | 10 |

```text
   A (top 300 m below MSL)     N
   |                           |
   | 500 m
   B (280 m) ------ C (310 m)
          400 m
```

#### Step 1: Slope of the top surface
- A is deeper than B by $300 - 280 = 20$ m over 500 m, so the bed goes down northward at $20/500 = 0.04$.
- C is deeper than B by $310 - 280 = 30$ m over 400 m, so the bed goes down eastward at $30/400 = 0.075$.

#### Step 2: Dip amount
$$
\tan\delta = \sqrt{0.04^2 + 0.075^2} = 0.0850 \Rightarrow \delta = 4.86°
$$

#### Step 3: Dip direction and strike
The bed gets deeper towards the north and towards the east, so it dips towards the north-east. Dip direction azimuth:

$$
\theta = \arctan(0.075/0.04) = 61.93°
$$

That is **N61.9°E**. The strike is at 90° to it: $61.93° - 90° = -28.07° = 331.93°$, i.e. **N28.1°W (S28.1°E)**.

(Strike-line check: the depth on BC equal to that at A (300 m) lies $400 \times (300-280)/(310-280) = 266.7$ m east of B. The line joining this point (266.7, 0) to A (0, 500) is a strike line, making an angle $\arctan(266.7/500) = 28.1°$ west of north, which confirms N28.1°W.)

#### Step 4: Altitude of the fractured bed
At any point $(x, y)$ metres east and north of B, the depth of the top below M.S.L. is $280 + 0.075x + 0.04y$, so the altitudes are:

$$
z_{top} = -(280 + 0.075x + 0.04y) \quad,\quad z_{bottom} = z_{top} - 10
$$

The bed is 10 m thick vertically; its true thickness is $10\cos 4.86° = 9.96$ m (about 10 m).

**Answer: The top of the jointed, fractured bed is at altitudes of −300 m at A, −280 m at B and −310 m at C (that is, 300, 280 and 310 m below M.S.L.), with the bottom 10 m deeper. Its attitude is strike N28.1°W, dip 4.9° towards N61.9°E (a gently dipping bed about 10 m thick).**

## Check yourself

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

1. How is the fractured bed plane obtained from three boreholes?
   A. By averaging the three depths only
   B. The three non-collinear points of known altitude define a plane, solved by the three-point method
   C. By using only the deepest borehole
   D. By assuming the bed is horizontal
   Answer: B. Three points not on a line fix a plane. (Key point: Method)

2. Borehole A (300 m) is 500 m north of B (280 m). The northward rise is
   A. 20/500 = 0.04
   B. 300/500 = 0.6
   C. 20/400 = 0.05
   D. 10/500 = 0.02
   Answer: A. Elevation difference 20 m over 500 m. (Key point: Northward rise)

3. C (310 m) is 400 m east of B (280 m). The eastward rise is
   A. 30/500 = 0.06
   B. 310/400 = 0.775
   C. 10/400 = 0.025
   D. 30/400 = 0.075
   Answer: D. Difference 30 m over 400 m. (Key point: Eastward rise)

4. The bed rises to the north and east, so it dips
   A. towards the north-east
   B. towards the south-east
   C. towards the south-west
   D. towards the north-west
   Answer: C. Dip direction is down the slope, opposite to the rise. (Key point: Dip direction)

## Related in Engineering Geology I

- [Three-point problem: A 600 m N, C 300 m E (500, 100, 400)](https://www.bctengineer.com/question.md?key=semester_3%2Fengineering_geology_i%3Athreepoint-problem-a-600-m-2) — Engineering Geology I, chapter 5, asked 4×
- [Three-point problem: A 600 m N, C 300 m E (500, 400, 400)](https://www.bctengineer.com/question.md?key=semester_3%2Fengineering_geology_i%3Athreepoint-problem-a-600-m) — Engineering Geology I, chapter 5
- [Short note: Bore hole log](https://www.bctengineer.com/question.md?key=semester_3%2Fengineering_geology_i%3Ashort-note-bore-hole-log) — Engineering Geology I, chapter 1
- [Three-point problem: A 800 m N, C 400 m W (500, 100, 300)](https://www.bctengineer.com/question.md?key=semester_3%2Fengineering_geology_i%3Athreepoint-problem-a-800-m-n) — Engineering Geology I, chapter 5, asked 4×
- [Site investigation for dam and reservoir](https://www.bctengineer.com/question.md?key=semester_3%2Fengineering_geology_i%3Asite-investigation-for-dam-and) — Engineering Geology I, chapter 1, asked 2×
- [Three-point problem: A 800 m N, C 600 m E (600, 200, 500)](https://www.bctengineer.com/question.md?key=semester_3%2Fengineering_geology_i%3Athreepoint-problem-a-800-m) — Engineering Geology I, chapter 5, asked 4×

## Related in other subjects

- [Bedrock attitude: A 700 m N23E, C S71W](https://www.bctengineer.com/question.md?key=semester_4%2Fengineering_geology_ii%3Abh-hydropower-a700-n23e) — Engineering Geology II, chapter 6
- [Tunnel alignment from three boreholes](https://www.bctengineer.com/question.md?key=semester_4%2Fengineering_geology_ii%3Atunnel-alignment-three-boreholes) — Engineering Geology II, chapter 6
- [Limestone true thickness: A 600 m N28E, C 400 m S10W](https://www.bctengineer.com/question.md?key=semester_4%2Fengineering_geology_ii%3Abh-limestone-true-thickness-a600) — Engineering Geology II, chapter 6, asked 2×
- [High dam hydropower project components](https://www.bctengineer.com/question.md?key=semester_8%2Fpower_plant_design%3Ahigh-dam-components) — Power Plant Design, chapter 3