Borehole attitude: square 360 m, M N O P

Method: the bed is a plane; its depth varies linearly with position. Take N as origin, x east, y north.

Data

• M is west of N: M = (-360, 0), depth 160 m
• N = (0, 0), depth 60 m
• P is north of N: P = (0, 360), depth 240 m
• O is north of M: O = (-360, 360)

Depth gradients

$$\begin{aligned} \text{east-west:}\quad & a = \frac{60 - 160}{360} = -0.2778\ \text{m/m (depth decreases eastward, M to N)} \\ \text{north-south:}\quad & b = \frac{240 - 60}{360} = 0.5\ \text{m/m (depth increases northward)} \end{aligned}$$

Depth $d = 60 - 0.2778\,x + 0.5\,y$ (m) - here $x,y$ in metres. Check M: $60 + 0.2778 \times 360 = 160$ m. Check P: $60 + 0.5 \times 360 = 240$ m.

Attitude

$$\tan\delta = \sqrt{0.2778^2 + 0.5^2} = 0.572 \Rightarrow \delta = 29.8^\circ$$

The bed deepens toward the direction $(-0.2778\ \text{east},\ +0.5\ \text{north})$, i.e. $\arctan(0.2778/0.5) = 29.05^\circ$ west of north: dip direction $= 330.95^\circ$ (N29°W). The strike is perpendicular: $240.95^\circ$, i.e. N61°E (to S61°W).

Attitude of the quartzite: strike N61°E, dip 29.8° towards N29°W (NNW).

Depth at O

$$d_O = 60 - 0.2778(-360) + 0.5(360) = 60 + 100 + 180 = 340\ \text{m}$$

Check using the parallelogram rule: $d_O = d_M + d_P - d_N = 160 + 240 - 60 = 340$ m.

Answer: attitude N61°E/29.8° NNW (dip direction 331°); borehole O meets the quartzite at 340 m depth.