# Using refraction method to find depth to bedrock

Series Geophysical References Series Problems in Exploration Seismology and their Solutions Lloyd P. Geldart and Robert E. Sheriff 14 497 - 503 http://dx.doi.org/10.1190/1.9781560801733 ISBN 9781560801153 SEG Online Store

## Problem 14.1

To find the depth to bedrock in a damsite survey, 12 geophones were laid out at 15-m intervals from 15 to 180 m. Determine the overburden depth from the data in Table 14.1a assuming a single layer above the refractor. By how much does the depth differ if we assume two layers above the refractor?

### Background

The refraction method is discussed in problem 4.18.

Table 14.1a
$x({\rm {m}})$ $t({\rm {s}})$ 15 19
30 29
45 39
60 50
75 59
90 62
105 65
120 68
135 72
150 76
165 78
180 83

### Solution

Figure 14.1a shows the plotted data. The single layer interpretation (fine lines) gives $V_{1}=1120\ {\mbox{m/s}}$ , $V_{2}=4120\ {\mbox{m/s}}$ , $t_{i}=0.084\ {\rm {s}}$ . The critical angle $\theta _{c}=\sin ^{-1}(1120/4120)=15.8^{\circ }$ . Equation (4.18a) gives the depth to bedrock as

{\begin{aligned}h=V_{1}t_{i}/2\cos \theta _{c}=1120\times 0.084/2\cos 15.8^{\circ }=49{\mbox{ m}}.\end{aligned}} The three-layer solution (heavy lines) gives $V_{1}=790\ {\mbox{m/s}},$ $V_{2}=1490\ {\mbox{m/s}}$ , $V_{3}=4120\ {\mbox{m/s}}$ , $t_{i1}=0.020\ {\rm {s}}$ , $t_{i2}=0.084\ {\rm {s}}$ . Note that $V_{1}$ is not reliable; it could be any smaller value. If it were smaller, the first-layer values would change slightly, but it would not significantly change the values for the other layers.

For the first interface, $\theta _{c1}=\sin ^{-1}(790/1490)=32.0^{\circ }$ ,

{\begin{aligned}h_{1}=V_{1}t_{i1}/2\cos \theta _{c1}=790\times 0.020/2\times \cos 32.0^{\circ }=9.3\ {\rm {m}}.\end{aligned}} For the second interface, $\sin \theta _{1}/V_{1}=\sin \theta _{c2}/V_{2}=1/V_{3}$ , so

{\begin{aligned}\sin \theta _{1}/790=\sin \theta _{c2}/1490=1/4120;\quad \theta _{1}=11.1^{\circ },\quad \theta _{c2}=21.2^{\circ }.\end{aligned}} Thus we get from equation (4.18d)

{\begin{aligned}t_{i2}=2h_{1}\cos \theta _{1}/V_{1}+2h_{2}\cos \theta _{c2}/V_{2},\end{aligned}} so $0.084=2\times 9.3\times 0.981/790+2\times h_{2}\times 0.932/1490$ ,

{\begin{aligned}h_{2}=(0.084-0.023)\times (1490/2\times 0.932)=49\ {\rm {m}}.\end{aligned}} The total depth is $h_{1}+h_{2}=58\ {\rm {m}}$ .

The difference between the two interpretations is 9 m or about 17%.