Depth at fourth borehole P (coal seam)

Principle: for a plane bed and a square (or any parallelogram) layout of vertical boreholes, the depth at the fourth corner is found from the other three corners:

$$d_P = d_{\text{adjacent 1}} + d_{\text{adjacent 2}} - d_{\text{opposite}}$$

This works because the depth changes uniformly along each side, so the depth at P equals the depth at the opposite corner plus the changes in depth along the two sides that meet there.

The depths of the other three boreholes are not printed with this part of the question. This is the follow-up to the companion problem with the same square layout (side 300 m), whose data are used here: SW corner A = 15 m, SE corner B = 45 m, NW corner C = 60 m, with P at the NE corner.

Calculation

• Gradient along the south side (A to B): $\dfrac{45 - 15}{300} = 0.10$ m/m
• Gradient along the west side (A to C): $\dfrac{60 - 15}{300} = 0.15$ m/m
• Depth at P (NE corner): $d_P = 15 + 0.10 \times 300 + 0.15 \times 300 = 90$ m

Check: $d_P = d_B + d_C - d_A = 45 + 60 - 15 = 90$ m.

Attitude (for completeness)

$\tan\delta = \sqrt{0.10^2 + 0.15^2} = 0.1803 \Rightarrow \delta = 10.2^\circ$; dip direction $= \arctan(0.10/0.15) = 33.7^\circ$ (NE); strike $303.7^\circ$ (N56.3°W).

Answer: with the above data, the coal seam is met at P at a depth of 90 m. For any other data, use $d_P = d_B + d_C - d_A$ (the sum of the two adjacent corners minus the opposite corner).