Sandstone attitude in oil field (feet)

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.