# Interpreting engineering refraction profiles

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 14.2a

What can you conclude from the time-distance data in Figure 14.2a?

### Background

Refraction interpretation is discussed in problems 4.18 and 4.24.

### Solution

While the alignments for ${\displaystyle V_{1}}$ are short, we have four of them, giving the velocities 780, 700, 820, 800 m/s. Because three values agreed so closely, we discarded the 700 value and took ${\displaystyle V_{1}=800\ {\mbox{m/s}}}$. The other velocities were obtained using the intercept times.

From the reversed profiles with sources ${\displaystyle A}$ and ${\displaystyle B}$, we measure

{\displaystyle {\begin{aligned}V_{d2}=3460\ {\mbox{m/s}},\quad V_{u2}=3600\ {\mbox{m/s}},\quad t_{d2}=0.007\ {\rm {s}},\quad t_{u2}=0.008\ {\rm {s}}.\end{aligned}}}

Figure 14.2a.  Engineering refraction profile.

We ignore the short alignment at the left end of ${\displaystyle V_{u1}}$ with intercept at ${\displaystyle B}$ of 16 ms, because it depends entirely on one point on the vertical axis at ${\displaystyle A}$, and there is no matching event elsewhere.

Using equation (4.24f), ${\displaystyle V_{2}\approx 2(1/V_{d1}+1/V_{u1})^{-1}=3530\ {\rm {m/s}}}$, so

{\displaystyle {\begin{aligned}\theta _{c}=\sin ^{-1}(800/3530)=13.1^{\circ }.\end{aligned}}}

From equation (4.24b), we get

{\displaystyle {\begin{aligned}h_{A}&=V_{1}t_{id}/2\cos \theta _{c}=800\times 0.007/2\times 0.974=2.9\ {\rm {m}},\\h_{B}&=800\times 0.008/2\times 0.974=3.3\ {\rm {m}},\\\xi &=\tan ^{-1}[(3.3-2.9)/90]\approx 0.3^{\circ }.\end{aligned}}}

The reversed profiles with sources ${\displaystyle B}$ and ${\displaystyle C}$ yield the following values:

{\displaystyle {\begin{aligned}V_{d2}=3330\ {\rm {m/s}},\quad V_{u2}=3460\ {\rm {m/s}},\quad t_{d2}=0.008\ {\rm {s}},\quad t_{u2}=0.009\ {\rm {s}}.\end{aligned}}}

Equation (4.24f) now gives

{\displaystyle {\begin{aligned}V_{2}&=2(1/3330+1/3460)^{-1}=3390\ {\rm {m/s}};\\\theta _{c}&=\sin ^{-1}(800/3390)=13.6^{\circ };\\h_{B}&=800\times 0.008/2\times 0.972=3.3\ {\rm {m}};\\h_{C}&=800\times 0.009/2\times 0.972=3.7\ {\rm {m}};\\\xi &=\tan ^{-1}(3.7-3.3)/90=0.3^{\circ }.\end{aligned}}}

## Problem 14.2b

Apply equation (4.18c) to get approximate thicknesses of the second layer.

### Solution

From the profiles with sources ${\displaystyle A}$ and ${\displaystyle C}$ extending to offsets of 180 m we observe a high-velocity event giving the following measurements:

{\displaystyle {\begin{aligned}V_{3d}&=4740\ {\mbox{m/s}},\quad t_{d3}=0.012\ {\rm {s}};\\V_{3u}&=6210\ {\mbox{m/s}},\quad t_{u3}=0.021\ {\rm {s}};\\V_{3}&=2(1/4740+1/6210)^{-1}=5380\ {\mbox{m/s}}.\end{aligned}}}

We need also the value of ${\displaystyle V_{2}}$ for the reversed profiles ${\displaystyle AC}$ and ${\displaystyle CA}$:

{\displaystyle {\begin{aligned}{\mbox{for }}AC,V_{d2}=3460\ {\mbox{m/s}}=V_{u2}{\mbox{ for }}CA,{\mbox{ so }}V_{2}=3460\ {\mbox{m/s}}.\end{aligned}}}

Assuming 800 m/s for ${\displaystyle V_{1}}$ (since it is not determined), we have

{\displaystyle {\begin{aligned}\sin \theta _{1}/800=\sin \theta _{2}/3460=1/5380,\quad \theta _{1}=8.6^{\circ },\quad \theta _{2}=40.0^{\circ }.\end{aligned}}}

We get the total thickness at ${\displaystyle A}$ using data from part (a) and equation (4.18c):

{\displaystyle {\begin{aligned}0.012&=2h_{1A}\cos \theta _{1}/V_{1}+2h_{2A}\cos \theta _{2}/V_{2},\\0.012&=2\times 2.9\times 0.989/800+h_{2A}(2\times 0.766/3460),\\h_{2A}&=(0.012-0.007)(3460/1.53)=11.3{\mbox{ m}}={\mbox{thickness at }}A.\end{aligned}}}

## Problem 14.2c

What is the dip of the deeper interface?

To get the dip of the deeper interface, we get the total depths at ${\displaystyle A}$ and ${\displaystyle C}$. Using data from (a) and (b), the depth at ${\displaystyle A}$ is

{\displaystyle {\begin{aligned}(h_{1A}+h_{2A})=(2.9+11.3)=14.2{\mbox{ m}}.\end{aligned}}}

The thickness of the lower bed at ${\displaystyle C}$ is obtained as in part (b):

{\displaystyle {\begin{aligned}0.021&=2\times 3.7\times 0.989/800+h_{2C}(2\times 0.766/3460),\\h_{2C}&=(0.021-0.009)\times 3460/2\times 0.766=27.1\ {\rm {m}}.\end{aligned}}}

The total thickness at ${\displaystyle C}$,

{\displaystyle {\begin{aligned}h_{1C}+h_{2C}=3.7+27.1&=31\ {\rm {m}},\\\xi =\tan ^{-1}[(31-14)/180]&=5.4^{\circ },\end{aligned}}}

down from ${\displaystyle A}$ to ${\displaystyle C}$.

## Problem 14.2d

Why are the answers in parts (b) and (c) approximate?

### Solution

Some of the intercept times have only one significant figure so that the calculated depths in part (a) are not accurate. Also we used equation (4.18c) in both (b) and (c) to get depths. Since equation (4.18c) is valid only for zero dip, these values are approximate for this reason also.