Tunnel alignment from three boreholes

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

Coordinates (BH#1 at origin)

$$\begin{aligned} \text{BH\#2}&: x = 1000\sin 32^\circ = 529.92\ \text{m},\ y = 1000\cos 32^\circ = 848.05\ \text{m} \\ \text{BH\#3}&: x = 800\sin 107^\circ = 765.04\ \text{m},\ y = 800\cos 107^\circ = -233.90\ \text{m} \end{aligned}$$

Levels of top of bedrock: BH#1 = -200 m, BH#2 = -300 m, BH#3 = -500 m.

Plane of the top of bedrock

$z = -200 + a x + b y$:

$$\begin{aligned} -300 &= -200 + 529.92a + 848.05b \\ -500 &= -200 + 765.04a - 233.90b \end{aligned}$$

Solving: $a = -0.3595$, $b = 0.1067$.

Dip, dip direction and strike

$$\tan\delta = \sqrt{a^2+b^2} = 0.3750 \Rightarrow \delta = 20.56^\circ$$

The bed descends toward the direction $\arctan(-a/-b)$ = 106.5° (ESE), so the strike is $\ 106.5^\circ - 90^\circ = 16.5^\circ$, i.e. N16.5°E.

So the top of the bedrock dips 20.6° towards 107° (ESE); strike N16.5°E.

Selecting the alignment

• Along the strike (about N16.5°E - S16.5°W) the surface of the bedrock is horizontal, so the depth of overburden and the rock cover above the tunnel stay constant, the tunnel stays in the same stratum, and the beds meet the tunnel roof uniformly. This gives the most uniform conditions and easiest support design.
• Along the dip (towards 107°) the bedrock drops $0.375$ m for each metre (about 20.6° slope), so the tunnel would pass from deep rock to shallow cover or soil, and would cross different beds.

Answer: the bedrock surface strikes N16.5°E and dips 20.6° towards ESE (107°). A tunnel driven along the strike direction (N16.5°E) at a constant level below the bedrock surface is most suitable.