Using refraction method to find depth to bedrock

Problems in Exploration Seismology and their Solutions

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

Figure 14.1a.  Plot of the data.

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%.

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Also in this chapter

• Using refraction method to find depth to bedrock
• Interpreting engineering refraction profiles
• Interpretation of four-shot refraction data

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