Question 8 of 8: Reversed Refraction Profile – Dipping Interface

Question 8: Reversed Refraction Profile – Dipping Interface (equal value)

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.

(i) Qualitative interpretation. Both Ta and Tb show exactly ONE change of slope: a steep DIRECT-wave segment close to each source breaks to a gentler REFRACTED (head-wave) segment beyond the crossover. That is the signature of a SINGLE refracting interface (a 2-layer earth); a third layer would add a second break to a still gentler slope, which neither curve shows. The two profiles are NOT mirror images, and that is the evidence for DIP. Ta's refracted branch ($x\approx70$–300 m, about 0.59 ms/m) is clearly STEEPER than Tb's ($x\approx10$–190 m, about 0.35 ms/m), so the two shooting directions record different apparent velocities on the same refractor, which is only possible if the refractor is inclined. Shooting from A gives the lower apparent velocity, so A fires DOWN-dip: the interface deepens from A toward B. The crossover distance is also shorter from A (about 60 m) than from B (about 110 m), consistent with a shallower interface under A. The two direct-wave segments have matching slopes, so layer 1 is uniform. One inconsistency is worth noting: for a single refractor the reciprocal times should agree, but Ta at 300 m is about 204 ms while Tb's refracted branch extrapolated to 0 m gives about 184 ms, a 20 ms gap. The analysis below therefore uses each branch's own slope and intercept.

(ii) Quantitative analysis. The velocities, dip and depths are derived below from the same digitized data.

Given. Reversed refraction line, length 300 m, source A at $x=0$ (times $T_a$) and source B at $x=300$ m (times $T_b$); first-arrival times digitized at 10 m stations (selected values below).

Find. Layer velocities $V_1,V_2$, the dip $\beta$ of the interface, and the vertical depth to the interface beneath each shot point.

Approach. Least-squares fit the direct-wave segment and the refracted-branch segment of EACH profile separately; average the two direct-wave slopes for a single robust $V_1$; combine the two refracted-branch apparent velocities through the standard dipping-layer pair of equations $\theta_c+\beta=\arcsin(V_1/V_A)$ (down-dip, lower apparent velocity) and $\theta_c-\beta=\arcsin(V_1/V_B)$ (up-dip, higher apparent velocity) to solve simultaneously for the critical angle $\theta_c$ (hence true $V_2$) and the dip $\beta$; then use each profile's own extrapolated intercept time to get the perpendicular, then vertical, depth beneath its own shot point.

1. Direct-wave velocity, $V_1$. A least-squares fit of $Ta(x\le60)$ gives slope $1.045$ ms/m, so $V_1=957$ m/s from source A; fitting $Tb(x\ge200)$ gives slope $-1.048$ ms/m, so $V_1=955$ m/s from source B. The two agree to within 0.2%, confirming a uniform layer 1; their average is $V_1=\boxed{956\ \text{m/s}}$.
2. Refracted-branch apparent velocities. Fitting $Ta(x\ge70)$ gives slope $0.587$ ms/m, so the apparent velocity from source A is $V_A=1704$ m/s (the LOWER one: A fires down-dip). Fitting $Tb(x\le190)$ gives slope $-0.354$ ms/m, so $V_B=2823$ m/s (the HIGHER one: B fires up-dip). The two differ by about 66%, far beyond reading error, which confirms the dip.
3. Critical angle and dip. $\theta_c+\beta=\arcsin(V_1/V_A)=\arcsin(956/1704)=34.1^\circ$ and $\theta_c-\beta=\arcsin(V_1/V_B)=\arcsin(956/2823)=19.8^\circ$. Solving together: $\theta_c=\tfrac12(34.1+19.8)=27.0^\circ$ and $\beta=\tfrac12(34.1-19.8)=\boxed{7.2^\circ}$, dipping down from A toward B.
4. True refractor velocity. $V_2=V_1/\sin\theta_c=956/\sin27.0^\circ=956/0.453=\boxed{2108\ \text{m/s}}$.
5. Depth beneath each shot point. Each refracted branch, extrapolated back to its own shot point, gives an intercept time: $t_{iA}=27.8$ ms at $x=0$ and $t_{iB}=78.2$ ms at $x=300$ m. The perpendicular depth below a shot is $h_\perp=V_1t_i/(2\cos\theta_c)$ and the vertical depth is $z=h_\perp/\cos\beta$. Beneath source A: $z_A=\dfrac{956\times0.0278}{2\cos27.0^\circ\cos7.2^\circ}=\boxed{15\ \text{m}}$. Beneath source B: $z_B=\dfrac{956\times0.0782}{2\cos27.0^\circ\cos7.2^\circ}=\boxed{42\ \text{m}}$. The interface is shallow beneath A and deepens toward B, consistent with A being the down-dip shooting direction. As a check, the depth difference over the 300 m line, $(42-15)/300$, corresponds to a slope of about $5^\circ$, the same order as the $7.2^\circ$ dip; the difference reflects the 20 ms reciprocal-time inconsistency in the plotted data.