Question

Two boreholes A and B, both inclined towards 270°, penetrate a dipping coal bed at the same point and pass through it entirely in the sub-surface as shown in the figure below. The bed dips towards 270°. The thickness of the coal bed, measured along the borehole A is 10 m and along borehole B is 8 m. The angle between the two boreholes is 20°. The orthogonal thickness \( x \) of the coal bed is ........ m. (Round off to one decimal place) 

Hint:

The orthogonal thickness of a dipping layer can be found by using the measured thicknesses along the boreholes and the angle between the boreholes, applying trigonometric relations.

Answer 1

Step 1: Understanding the diagram.
- The coal bed dips at an angle of 270°, and two boreholes, A and B, both inclined towards 270°, intersect the bed at the same point.
- The distance measured along borehole A is 10 m, and along borehole B, it is 8 m.
- The angle between the two boreholes is 20°.
- The goal is to find the orthogonal thickness \( x \), which is the true thickness of the coal bed.
Step 2: Formula for orthogonal thickness.
The orthogonal thickness \( x \) can be calculated using the following formula, based on the geometry of the intersecting boreholes: \[ x = \frac{AB}{\sin(\theta)} \] where:
- \( AB \) is the difference in the measured thicknesses along the two boreholes (the difference between the thicknesses measured along boreholes A and B).
- \( \theta \) is the angle between the two boreholes.
Step 3: Calculation of the difference in thickness.
We can use the following relation for the difference in thickness: \[ AB = \sqrt{(10^2) + (8^2) - 2 \times 10 \times 8 \times \cos(20^\circ)} \] \[ AB = \sqrt{100 + 64 - 160 \times \cos(20^\circ)} \] Using \( \cos(20^\circ) \approx 0.9397 \): \[ AB = \sqrt{100 + 64 - 150.352} = \sqrt{13.648} \] \[ AB \approx 3.7 \, \text{m} \] Step 4: Final Calculation of Orthogonal Thickness.
Now, calculate the orthogonal thickness \( x \) using the formula: \[ x = \frac{3.7}{\sin(20^\circ)} \] Since \( \sin(20^\circ) \approx 0.3420 \), we get: \[ x = \frac{3.7}{0.3420} \approx 10.8 \, \text{m} \] Final Answer: \[ \boxed{10.8} \, \text{m} \]