Question 4 of 7: Relative Dating and Map

Question 4: Relative Dating and Map (20 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.

1) Relative dating of the cross-section

The cross-section is read using the standard relative-dating principles — original horizontality, superposition, cross-cutting relationships, inclusions, and unconformity — applied in the order that makes the observed geometry consistent. Working from oldest to youngest:

1. Event 1 – deposition of the lower sequence. The right-hand panel of dipping, parallel-bedded strata (hachured/patterned units) was deposited originally as horizontal layers (Principle of Original Horizontality), with the lowest bed the oldest (Principle of Superposition).
2. Event 2 – tilting. Tectonic deformation rotated this sequence to its present inclined attitude; tilting must post-date deposition of every bed in the tilted panel, but pre-date anything that cuts the tilted panel at a discordant angle or that lies flat across its truncated edges.
3. Event 3 – intrusion of the tapered light-coloured body. The coarse, unbedded, wedge-shaped body (granite/pluton texture) cuts discordantly across the tilted beds, so by the Principle of Cross-Cutting Relationships it is younger than the tilting. It does not continue up into the flat-lying cap, so it was emplaced, and then truncated by erosion, before the cap sequence was deposited.
4. Event 4 – erosion to an angular unconformity. Uplift and erosion bevel both the tilted sequence and the intrusion to a single surface, removing the top of the tilted panel; this erosion surface becomes the angular unconformity separating the tilted lower sequence from the flat-lying upper sequence.
5. Event 5 – deposition of the upper, flat-lying sequence. The horizontally-bedded units above the unconformity (including the brick-patterned resistant cap at the very top) were deposited after erosion levelled the surface; by Superposition, the lowest of these flat beds is the oldest of the group.
6. Event 6 – the small enclosed fragment near the base of the flat sequence. The isolated angular fragment sits wholly within a flat-lying bed, near the unconformity. By the Principle of Inclusions, a fragment enclosed in a host rock must be older than the host: it is debris reworked from the underlying tilted/intruded sequence into the base of the new sedimentary cover, confirming (independently of the cross-cutting argument) that the tilted sequence and its intrusion predate the flat sequence.
7. Event 7 – the discordant dike. The narrow, steeply-oriented, texturally distinct band cuts across the tilted sequence, the wedge intrusion, the unconformity, AND the flat-lying cap sequence, reaching the present erosion surface. By Cross-Cutting Relationships it is younger than every unit and every contact it crosses — the youngest rock-forming event recorded in the section.
8. Event 8 – present-day erosion. Erosion of the dike-cut sequence produces today's topographic (ground) surface, the youngest surface shown, truncating all of the above.

2) Three-point problem – strike, dip, outcrop trace, depth and apparent dip

Given.

Map (E, N) coordinates for the three outcrop points were read from the map using the printed 0-1000 m scale bar (isotropic scale assumed): X1 ≈ (870, -449), X2 ≈ (1496, -948), X3 ≈ (2426, -357), in metres east/north of the map's local origin.

Find. Strike and dip of the coal layer; its outcrop trace; the depth to the layer below point A; and the apparent dip in the N90°E (due-east) direction.

Approach. Three points on the same planar bed, each with a known map position and elevation, fully determine the bed's attitude (the classic three-point problem): fit the plane through the three points, read its strike and true dip from the plane's normal vector, then use the plane equation to find the layer's elevation anywhere else on the map (including under point A) and its apparent dip in any specified direction.

1. a) Fit the plane through X1, X2, X3. Forming the two edge vectors P1P2 and P1P3 (in E, N, Z) and taking their cross product gives the plane's normal vector n = (n_E, n_N, n_Z); the true dip angle follows from $$ \alpha = \arctan\!\left(\frac{\sqrt{n_E^2+n_N^2}}{n_Z}\right) $$ and the dip-direction azimuth from the direction of the plane's steepest descent (the down-gradient horizontal direction). Carrying out the cross product and arctangent gives $$ \boxed{\alpha \approx 21^{\circ}} $$ dipping toward an azimuth of about 134°, i.e. to the SE, with the strike perpendicular to the dip direction: $$ \boxed{\text{strike} \approx N44^{\circ}E \ (\text{or equivalently } S44^{\circ}W)} $$
2. b) Outcrop trace. A structure contour (a line of equal elevation on the coal layer itself) through X1, X2 and X3 is a straight line on this planar bed, trending along the N44°E strike direction, and every other structure contour on the layer is parallel to it, offset by 100 m of elevation for each 100/tan(21°) ≈ 260 m of horizontal distance measured in the down-dip (S46°E) direction. Where the coal layer's surface elevation equals the ground elevation, the layer outcrops; because the bed dips gently (21°) compared with the steep, closed hillside contours, the outcrop trace on each hill crosses the topographic contours in the classic "V" pattern, with the V pointing in the down-dip (SE) direction wherever the trace crosses a valley/re-entrant, and pointing up-dip (opposite, i.e. NW) wherever it crosses a ridge or spur — the same rule used to trace any dipping contact on a topographic map. The trace links X1 (600 m outcrop point) through progressively lower elevations on the coal layer to X2 (300 m) and on to X3 (200 m), always following its own 100 m structure contours rather than the ground's topographic contours.
3. c) Depth at point A. Using the fitted plane equation z_layer = z_X1 + a·(E_A-E_X1) + b·(N_A-N_X1) (with a, b the plane's E- and N-gradients from step a), the coal layer's elevation projected beneath point A works out to about 194 m. With the ground surface at A read from the map as ≈ 385 m: $$ \text{depth} = z_{ground} - z_{layer} \approx 385 - 194 = \boxed{191\ \text{m}} $$
4. d) Apparent dip toward N90°E. The angle in the horizontal plane between the true dip direction (azimuth ≈134°) and the desired N90°E (due-east, azimuth 90°) direction is θ = 134-90 ≈ 44°. Applying the given formula: $$ \alpha^\prime = \arctan(\cos\theta \times \tan\alpha) = \arctan(\cos 44^{\circ} \times \tan 21^{\circ}) \approx \boxed{15^{\circ}} $$ — smaller than the true dip, as any apparent dip measured off the true dip direction must be.