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Rock layer attitude: A 700 m N, C 600 m W

Short answer

Three-point method using top depths (-410, -380, -430 m) for boreholes A (0,700), B (0,0), C (-600,0) yields strike N27.2°E, dip 5.4° towards WNW (297°), true thickness = 20 m × cos(5.35°) = 19.9 m.

Bore hole A is 700 m due north of bore hole B and bore hole C is 600 m due west of bore hole B. The tops and bottoms of a rock layer are reached at the following altitudes relative to the sea level in three holes. Bore hole A: -410 m and -430 m Bore hole B: -380 m and -400 m Bore hole C: -430 m and -450 m Find the attitude and thickness of the rock layer.

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

Data and coordinates (B at origin)

A = (0, 700), C = (-600, 0). Levels of the top: A = -410 m, B = -380 m, C = -430 m. The layer is 20 m thick in each hole (-410/-430, -380/-400, -430/-450), so tv=20t_v = 20 m.

Gradients of the top surface

b=−410−(−380)700=−0.0429a=−380−(−430)600=0.0833\begin{aligned} b &= \frac{-410 - (-380)}{700} = -0.0429 \\ a &= \frac{-380 - (-430)}{600} = 0.0833 \end{aligned}

The level falls northward and rises eastward, so the layer dips towards the NW.

Dip, dip direction and strike

tan⁡δ=a2+b2=0.0937⇒δ=5.35∘\tan\delta = \sqrt{a^2+b^2} = 0.0937 \Rightarrow \delta = 5.35^\circ

The bed descends toward the direction arctan⁡(−a/−b)\arctan(-a/-b) = 297.2° (WNW), so the strike is  297.2∘−90∘=207.2∘\ 297.2^\circ - 90^\circ = 207.2^\circ, i.e. N27.2°E.

True thickness

All three boreholes show the same vertical thickness, 20 m, so

t=tvcos⁡δ=20cos⁡5.35∘=19.91 mt = t_v\cos\delta = 20\cos 5.35^\circ = 19.91\ \text{m}

Answer: attitude of the layer = strike N27.2°E, dip 5.4° towards WNW (dip direction 297°); true thickness = 19.9 m.

✓ Answer checked by Claude Opus 5.5 (high effort)

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