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

### Solution

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

{\displaystyle {\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 ${\displaystyle V_{1}=790\ {\mbox{m/s}},}$ ${\displaystyle V_{2}=1490\ {\mbox{m/s}}}$, ${\displaystyle V_{3}=4120\ {\mbox{m/s}}}$, ${\displaystyle t_{i1}=0.020\ {\rm {s}}}$, ${\displaystyle t_{i2}=0.084\ {\rm {s}}}$. Note that ${\displaystyle V_{1}}$ 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, ${\displaystyle \theta _{c1}=\sin ^{-1}(790/1490)=32.0^{\circ }}$,

{\displaystyle {\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, ${\displaystyle \sin \theta _{1}/V_{1}=\sin \theta _{c2}/V_{2}=1/V_{3}}$, so

{\displaystyle {\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)

{\displaystyle {\begin{aligned}t_{i2}=2h_{1}\cos \theta _{1}/V_{1}+2h_{2}\cos \theta _{c2}/V_{2},\end{aligned}}}

so ${\displaystyle 0.084=2\times 9.3\times 0.981/790+2\times h_{2}\times 0.932/1490}$,

{\displaystyle {\begin{aligned}h_{2}=(0.084-0.023)\times (1490/2\times 0.932)=49\ {\rm {m}}.\end{aligned}}}

The total depth is ${\displaystyle h_{1}+h_{2}=58\ {\rm {m}}}$.

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