---
title: "Sandstone attitude in oil field (feet)"
subject: "Engineering Geology II"
chapter: 6
key: "semester_4/engineering_geology_ii:bh-oil-field-sandstone-feet"
source: https://www.bctengineer.com/q/semester_4/engineering_geology_ii/bh-oil-field-sandstone-feet
site: BCT Engineering
---

# Sandstone attitude in oil field (feet)

Engineering Geology II · Chapter 6: Measurement, analysis and interpretation of structural geological data

- Asked in: 2069 Bhadra, 5 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/bh-oil-field-sandstone-feet
- Whole chapter: https://www.bctengineer.com/md/semester_4/engineering_geology_ii/chapter-6

## Question

Bore hole B in an oil field is 5000 feet due north of bore hole A and bore hole C is 10,000 feet due east of bore hole A. The tops and bottoms of a key sandstone bed are reached at the following altitudes relative to sea level in the three holes: A, -2500 and -2700 feet; B, -2800 and -3000 feet; and C, -3000 and -3200 feet. What is the attitude of the sandstone and how thick is it?

## Short answer

Using three-point method with boreholes A at origin, B 5000 ft north, C 10000 ft east and top-surface depths -2500, -2800, -3000 ft: attitude strike N50.2°W, dip 4.5° towards NE (40°), true thickness = 200 ft × cos(4.47°) = 199.4 ft.

## Answer

A bed is a plane, so its level changes uniformly in plan: $z = z_A + a x + b y$, where $a$ and $b$ are the rates of change of level per foot east and north. This is the analytical form of the three-point (strike-line) method.

#### Data (A at origin, feet)
A = (0, 0); B = (0, 5000) (north); C = (10000, 0) (east). Levels of the top: A = -2500 ft, B = -2800 ft, C = -3000 ft. The thickness in every hole is 200 ft (e.g. -2500 to -2700), so $t_v = 200$ ft.

#### Gradients of the top surface
$$
\begin{aligned}
b &= \frac{z_B - z_A}{5000} = \frac{-2800 - (-2500)}{5000} = -0.06 \\
a &= \frac{z_C - z_A}{10000} = \frac{-3000 - (-2500)}{10000} = -0.05
\end{aligned}
$$
So the level falls both northward and eastward: the bed dips towards the NE.

#### Dip, dip direction and strike
$$
\tan\delta = \sqrt{a^2+b^2} = 0.0781 \Rightarrow \delta = 4.47^\circ
$$
The bed descends towards the direction $\arctan(0.05/0.06)$ = **39.8°** (NE), so the strike is $39.8^\circ - 90^\circ = -50.2^\circ = 309.8^\circ$, i.e. **N50.2°W**.

#### True thickness
All three boreholes show the same vertical thickness, 200 ft, so
$$
t = t_v\cos\delta = 200\cos 4.47^\circ = 199.39\ \text{ft}
$$

**Answer: attitude of the sandstone = strike N50.2°W, dip 4.5° towards the NE (dip direction 40°); true thickness = 199.4 ft.**

## Check yourself

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

1. With A at origin, B 5000 ft north and C 10000 ft east, how is the northward gradient b found?
   A. b = (zC - zA)/10000 = -0.05
   B. b = (zB - zA)/5000 = -0.06
   C. b = (zB - zA)/10000 = -0.03
   D. b = 5000/(zB - zA)
   Answer: B. B lies due north of A, so the level change per foot north is (-2800 + 2500)/5000 = -0.06. (Key point: Gradients)

2. The level falls both northward and eastward. Which way does the bed dip?
   A. Towards the SW
   B. Towards the NW
   C. Towards the SE
   D. Towards the NE
   Answer: D. Both gradients are negative, so the bed descends toward north-east. (Key point: Dip direction)

3. What dip and strike result?
   A. Dip 7.8°, strike N39.8°E
   B. Dip 4.5°, strike N50.2°W
   C. Dip 4.5°, strike N39.8°E
   D. Dip 0.0781°, strike N50.2°W
   Answer: B. tan(dip) = sqrt(0.05^2 + 0.06^2) = 0.0781 so 4.47°; strike = 39.8° - 90° gives N50.2°W. (Key point: Dip and strike)

4. The sandstone is 200 ft thick vertically in each hole. What is the true thickness?
   A. About 15.6 ft, from tv tan(dip)
   B. About 200.6 ft, from tv / cos(dip)
   C. Exactly 200 ft, as dip does not matter
   D. About 199.4 ft, from t = tv cos(dip)
   Answer: D. 200 x cos 4.47° = 199.4 ft. (Key point: True thickness)

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