Question 13 of 21: Question 4, Part 4: Contour Map Three-Point Problem

Question 4, Part 4: Contour Map Three-Point Problem (7 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Scale bar: 1000 m = 202.1 pt on the page (0 and 1000 m ticks at x = 141.75 and 343.88 pt), i.e. 4.947 m/pt. Map positions relative to X2 (East, North): X1 (−617, +498) m at 600 m; X2 (0, 0) at 300 m; X3 (+940, +599) m at 200 m; A (+490, +105) m, ground ≈ 355 m.

Find. (a) Strike and dip of the coal layer; (b) its outcrop pattern; (c) depth of the coal layer below point A; (d) apparent dip on a N90°E section.

Approach. Treat the three X's as three points of known (Easting, Northing, elevation) on the SAME planar coal seam (the classic "three-point problem"). Fit the plane $z = z_2 + aE + bN$ through the three points; the horizontal gradient $(a,b)$ gives the dip direction (steepest descent, along $(-a,-b)$) and the true dip $\tan\delta=\sqrt{a^2+b^2}$. Strike is perpendicular to the dip direction. Extrapolate the plane to A's map location for the seam elevation there, and subtract that from A's ground elevation to get the depth.

1. (a) Fit the plane and extract strike/dip. Solving $z = 300 + aE + bN$ through X1 and X3 gives $a = -0.2737$ and $b = +0.2630$ (m per m). The seam therefore loses height toward the east and toward the south. The steepest-descent direction is $(-a,-b) = (+0.274, -0.263)$, i.e. azimuth $\approx 134^\circ$ (S46°E), and $\tan\delta = \sqrt{0.2737^2+0.2630^2} = 0.3796$, so the true dip is $\delta \approx \boxed{21^\circ}$ toward the south-east. Strike is perpendicular to that direction: $\boxed{\text{N44}^\circ\text{E}}$ (S44°W is the same line). Quick sense check: the highest point (X1, 600 m) is the north-westernmost, and the two lower points lie to its south-east (X2, 300 m) and east (X3, 200 m), so the seam must descend toward the SE.
2. (b) Outcrop pattern. The outcrop trace is the line where the seam's elevation equals the ground elevation. Draw structure contours: straight lines striking N44°E, stepping down 100 m every $100/0.3796 = 264$ m toward the SE. Mark each point where structure contour $n$ crosses topographic contour $n$, then join those points. On this map the trace runs from X1 at the top of the west hill down that hill's east and south-east flank (it crosses the 500 m and 400 m contours about 264 m and 528 m down-dip of X1), through X2 on the 300 m contour, and then wraps around the lower flanks of the east hill to X3 on the 200 m contour. The east hill is capped by rock that overlies the seam, because the seam lies below 200 m beneath it. Where the trace crosses a valley it forms a V. Because the seam dips more steeply (21°) than any stream gradient on the map, the V's point in the dip direction, i.e. toward the SE (Rule of V's).
3. (c) Depth of the coal layer at Point A. At A (E = 490 m, N = 105 m): $z_{\text{seam}} = 300 - 0.2737(490) + 0.2630(105) \approx \boxed{194\text{ m}}$. With the ground at A ≈ 355 m (interpolated between the 400 m and 300 m contours), the coal layer lies $355-194 \approx \boxed{160\text{ m}}$ below the surface at A. The answer shifts by the same amount as A's ground reading: if A is taken as right on the 400 m contour, the depth is about 205 m.
4. (d) Apparent dip on a N90°E (due-east) section. Using $\tan\delta'=\tan\delta\cos(\varphi-\alpha)$ with dip direction $\varphi\approx134^\circ$ and section azimuth $\alpha=90^\circ$: $\tan\delta' = 0.3796\cos(44^\circ) = 0.2737$ (the same as $-a$, the seam's fall per metre east), so $\delta' \approx \boxed{15^\circ}$, downward toward the east. This agrees with X3, the easternmost point, being the lowest.