Bore hole altitude of fractured bedrock

Method: The given figures are depths below M.S.L. (the top of the bed is at a smaller depth than its bottom). The top (and bottom) of the fractured bed at three non-collinear points define a plane, which is solved by the three-point (strike-line) method.

Data: B at the origin; A is 500 m due north of B; C is 400 m due east of B.

   A (top 300 m below MSL)     N
   |                           |
   | 500 m
   B (280 m) ------ C (310 m)
          400 m

Step 1: Slope of the top surface

• A is deeper than B by $300 - 280 = 20$ m over 500 m, so the bed goes down northward at $20/500 = 0.04$.
• C is deeper than B by $310 - 280 = 30$ m over 400 m, so the bed goes down eastward at $30/400 = 0.075$.

Step 2: Dip amount

$$\tan\delta = \sqrt{0.04^2 + 0.075^2} = 0.0850 \Rightarrow \delta = 4.86°$$

Step 3: Dip direction and strike

The bed gets deeper towards the north and towards the east, so it dips towards the north-east. Dip direction azimuth:

$$\theta = \arctan(0.075/0.04) = 61.93°$$

That is N61.9°E. The strike is at 90° to it: $61.93° - 90° = -28.07° = 331.93°$, i.e. N28.1°W (S28.1°E).

(Strike-line check: the depth on BC equal to that at A (300 m) lies $400 \times (300-280)/(310-280) = 266.7$ m east of B. The line joining this point (266.7, 0) to A (0, 500) is a strike line, making an angle $\arctan(266.7/500) = 28.1°$ west of north, which confirms N28.1°W.)

Step 4: Altitude of the fractured bed

At any point $(x, y)$ metres east and north of B, the depth of the top below M.S.L. is $280 + 0.075x + 0.04y$, so the altitudes are:

$$z_{top} = -(280 + 0.075x + 0.04y) \quad,\quad z_{bottom} = z_{top} - 10$$

The bed is 10 m thick vertically; its true thickness is $10\cos 4.86° = 9.96$ m (about 10 m).

Answer: The top of the jointed, fractured bed is at altitudes of −300 m at A, −280 m at B and −310 m at C (that is, 300, 280 and 310 m below M.S.L.), with the bottom 10 m deeper. Its attitude is strike N28.1°W, dip 4.9° towards N61.9°E (a gently dipping bed about 10 m thick).