Using refraction method to find depth to bedrock

From SEG Wiki
Jump to navigation Jump to search
ADVERTISEMENT

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%.

Continue reading

Previous section Next section
Poission’s ratio from P- and S-wave traveltimes Interpreting engineering refraction profiles
Previous chapter Next chapter
Specialized techniques Introduction to Problems in Exploration Seismology and their Solutions

Table of Contents (book)

Also in this chapter

External links

find literature about
Using refraction method to find depth to bedrock
Retrieved from "http://wiki.seg.org/w/index.php?title=Using_refraction_method_to_find_depth_to_bedrock&oldid=133653"