Question 8 of 8: Seismic Refraction — Apparent Velocities, Earth Models and Depth to Bedrock

Question 8: Seismic Refraction — Apparent Velocities, Earth Models and Depth to Bedrock (10 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. First-arrival travel times at 24 geophones, shotpoint at $x=0$; near-surface geology = glacial till over bedrock (2-layer refraction model).

Find. (a) the three apparent velocities $v_1,v_2,v_3$. (b) two earth models (bedrock flat on the left vs. flat on the right) consistent with the 3-segment graph. (c) depth to bedrock, assuming model 1. (d) the reverse-shot travel-time pattern from $x=250$ m, assuming model 1.

Approach. Fit a straight line to each of the three visually distinct segments of the time-distance graph; the reciprocal of each slope is that segment's apparent velocity. Use the direct-wave segment (slope $=1/v_1$) and the segment immediately following the first crossover (slope $=1/v_2$, extrapolated back to $x=0$ for the intercept time $T_i$) with the given critical-refraction formula to get the depth to bedrock under model 1.

1. Segment 1 — direct wave through the till ($x=10$–30 m). A least-squares line through these three points gives slope $0.625$ ms/m, i.e. $v_1=1/0.625\times1000=\boxed{1600\ \text{m/s}}$.
2. Segment 2 ($x=40$–120 m). Slope $0.2217$ ms/m, intercept $T_i=12.71$ ms, giving $v_2=1/0.2217\times1000\approx\boxed{4510\ \text{m/s}}$.
3. Segment 3 ($x=130$–240 m). Slope $0.3393$ ms/m, giving $v_3=1/0.3393\times1000\approx\boxed{2950\ \text{m/s}}$. Because $v_3\lt v_2$, the apparent velocity DROPS on the far part of the line — the signature of the bedrock refractor getting DEEPER as the survey line moves away from the shot (model 1's interpretation, part (b) below), since a deepening interface increases the intercept time faster than offset, apparently slowing the refracted arrival.
4. Earth models (b). Two extreme, equally-consistent interpretations of a two-branch refracted arrival from a single forward shot: Model 1 — bedrock is horizontal (flat, at the depth found in step 6) under the LEFT/near-shot part of the line, where segment 2's slope gives the TRUE $v_2$, then the interface steps/dips DOWN under the right-hand part, so the far branch (segment 3) records a reduced APPARENT velocity. Model 2 — the mirror image: bedrock is horizontal under the RIGHT-hand part (segment 3 gives the true $v_2$ there) and dips down toward the shot, so the near branch (segment 2) is the one recording a reduced apparent velocity relative to the true, larger $v_2$. A single forward shot cannot distinguish these two — both fit the same time-distance graph — which is exactly why part (d) asks for a REVERSE shot: a shot from the far end resolves the ambiguity because the up-dip and down-dip apparent velocities from the two directions differ predictably.
5. Depth to bedrock, assuming model 1 (c). Model 1 treats segment 2 as the TRUE horizontal-layer refraction near the shot, so its extrapolated intercept time $T_i=12.71$ ms and $v_1=1600$, $v_2=4510$ m/s go into the given formula. Critical angle: $\theta_c=\arcsin(v_1/v_2)=\arcsin(1600/4510)=20.8^\circ$. At $x=0$, $T(0)=T_i=2z\cos\theta_c/v_1$, so $$z=\frac{T_i\,v_1}{2\cos\theta_c}=\frac{(0.01271\ \text{s})(1600\ \text{m/s})}{2\cos(20.8^\circ)}\approx\boxed{10.9\ \text{m}}$$ (equivalently, the closed form $z=T_iv_1v_2/[2\sqrt{v_2^2-v_1^2}]$ gives the same 10.9 m). As a cross-check, the crossover-distance formula using $x_{cross,12}\approx30.3$ m gives $z=\tfrac{x_{cross}}{2}\sqrt{(v_2-v_1)/(v_2+v_1)}\approx10.5$ m — agreeing with the intercept-time result to within the graph-reading tolerance.
6. Reverse shot at $x=250$ m (d). Under model 1, the interface deepens moving away from the original shot, so a shot fired at the FAR end (250 m) shooting back toward the original shot is effectively shooting UP-DIP into shallower bedrock: its refracted branch should show a HIGHER apparent velocity than segment 3's forward-shot value near the far end, converging toward the same reciprocal time at the two shot points (equal total travel time along the full 0–250 m line, the standard reciprocal-time check for a valid refraction interpretation), sketched as the dashed orange line on the graph below.