# Feasibility of mapping a horizon using head waves

Series Geophysical References Series Problems in Exploration Seismology and their Solutions Lloyd P. Geldart and Robert E. Sheriff 11 415 - 468 http://dx.doi.org/10.1190/1.9781560801733 ISBN 9781560801153 SEG Online Store

## Problem 11.15

Construct the expected time-distance curve for the Java Sea velocity-depth relation shown in Figure 11.15a. Is it feasible to map the top of the relatively flat 4.25 km/s limestone at a depth of about 0.9 km by using head waves? What problems are likely to be encountered?

 Depth range (km) $V_{t}$ (km/s) $\Delta t_{T}$ (s) $t_{i}=\sum \Delta t_{i}$ $V_{t}^{2}\Delta t_{i}$ $\sum V_{t}^{2}\Delta _{t_{i}}$ $V_{\rm {rms}}$ (km/s) 0.00–0.03 1.53 0.039 0.039 0.091 0.091 0.03–0.16 1.9 0.137 0.176 0.495 0.586 0.16–0.28 1.97 0.122 0.298 0.472 1.059 0.28–0.50 2.25 0.196 0.494 0.992 2.051 0.50–0.70 2.15 0.186 0.68 0.86 2.911 0.70–0.90 2.67 0.15 0.83 1.069 3.98 2.19 ${\rm {L}}_{s}\to$ 0.90–0.97 4.25 0.033 0.863 0.596 4.576 2.30 0.97–1.10 5.27 0.049 0.912 1.361 5.937

### Solution

The time-depth data in Figure 11.15a are listed in the first two columns of Table 11.15a. We calculated the data in columns 3 ($t_{i}$ two-way traveltime through the layer) to 6 to determine reflection arrival times (column 4) and $V_{\rm {rms}}$ [using equation (4.13a)]. We take $V_{1}\approx 1.90$ km/s (see Figure 11.15a) to plot the direct wave.

We must take into account other events that might interfere, primarily the reflection and head wave from the 5.27 km/s layer. To plot the refraction curves, we need their slopes, one point on each curve, and the critical distances—where a head wave is tangent to the reflection (see Figure 4.18a). We also calculate the intercept times as a check.

The slope of the limestone refractor (assumed to be flat) is 1/4.25 s/km; taking $z=0.90$ km, $V_{1}=2.19,\theta _{c}=\sin ^{-1}(2.19/4.25)=31^{\circ }$ . From Figure 4.18a the critical distance $x'=2z\tan \theta _{c}=2\times 0.90\times \tan 31^{\circ }=1.1$ km; at this point $t'=2z/V_{1}\cos \theta _{c}=2\times 0.90/2.19\times \cos 31^{\circ }=0.96$ s. The intercept time given by equation (4.18a) is $t_{i}=2z\cos \theta _{c}/V_{1}=0.70$ s. Thus the head-wave curve passes through the point $\left(t,\;x\right)=\left(0.96,1.1\right)$ with slope 1/4.25, is tangent to its reflection at $x=1.1$ km, and the intercept time $t_{i}=0.70$ s. The reflection arrives at 0.863 s at zero offset and also passes through (0.96,1.1). These curves are shown in Figure 11.15b.

Carrying out similar calculations for the 5.27 km/s layer and using $V_{1}\approx 2.30$ km/s (estimated from Figure 11.15a), we get

{\begin{aligned}{\rm {slope}}=1/5.27,\quad \theta _{c}=26^{\circ },\quad x'=0.95\ {\rm {km}},\quad t'=0.94\ {\rm {s}},\quad t_{i}=0.86\ {\rm {s}}.\end{aligned}} The reflection arrives at zero offset at 0.912 s and also passes through (0.95, 0.94). These curves are also plotted in Figure 11.15b.

The 4.25 km/s head wave is always a second arrival. It also follows very closely the reflection from the 5.27 km/s layer. It will almost certainly not be observed as a distinctly separate arrival because later cycles of earlier events will mask it.