# User:Ageary/Chapter 14

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

## 14.1 Using refraction method to find depth to bedrock

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%.

## 14.2 Interpreting engineering refraction profiles

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}}}

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}}}

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 ${\displaystyle C}$.

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.

## 14.3 Interpretation of four-shot refraction data

Find the velocities and depths of refractors in Figure 14.3b.

Background

The four-shot method uses a refraction spread with a source at each end to provide “short-shot” data and two sources offset inline beyond the critical distance (see problem 4.18) to provide “long-shot” data that permit accurate measurement of the refraction velocities. A refraction event appearing on both profiles can be identified by the slopes of the ${\displaystyle t}$-${\displaystyle x}$ curves. Geophones at the shot points are offset when shots at ${\displaystyle A}$ or ${\displaystyle B}$ are used.

Solution

Figure 14.3b shows east-west reversed profiles with sources ${\displaystyle A}$ and ${\displaystyle B}$ providing the short-shot (SS) data and sources ${\displaystyle A^{*}}$ and ${\displaystyle B^{*}}$ (not shown) that are offset inline to provide the long-shot (LS) data. We have labeled the different segments from ${\displaystyle C}$ to ${\displaystyle J}$ and fitted each with a straight line whose slope gives the apparent velocity. The SS segments were used to get intercept times as shown in the figure. Note that fitting straight lines to the data assumes that the refractors are planar, clearly an approximation.

The data suggest that this is a three-layer situation with dip down to the east. We note that the LS data segments ${\displaystyle (I,J)}$, ${\displaystyle (E,F)}$, and ${\displaystyle (C,D)}$ show a time-offset between geophones 6 and 7, as if indicating a fault downthrown to the east. However, evidences for features at depth (such as a fault cutting the refractor) should be displaced away from the source (because the head-wave path from the refractor to a geophone is inclined, as in Figure 14.3a). However profiles ${\displaystyle I,J}$ and ${\displaystyle E,F}$ show the same anomaly at the same location even though shot from opposite directions. Because our geophones spacing is only 20 m, this suggests that they are caused by something shallower than 20 m, and the most probable cause is a near-surface anomaly—perhaps a statics error in near-surface velocity or elevation corrections. We could make an empirical surface correction, but have not done so.

Figure 14.3a.  A feature on a refractor would be observed at ${\displaystyle D}$ from source ${\displaystyle A}$ but at ${\displaystyle C}$ from source ${\displaystyle B}$.
Figure 14.3b.  Time-distance curves for four-shot data. Open squares and triangles represent short-shot data, solid squares and triangles are long-shot data. (after Milson, 1989, 173)

The SS data show a shallow refraction event (segments ${\displaystyle C}$ and ${\displaystyle G}$) that probably is a refraction from the top of an intermediate layer. The measured velocities are ${\displaystyle V_{1}=700\ {\rm {m/s}}}$, ${\displaystyle V_{A2}=1030\ {\rm {m/s}}}$, ${\displaystyle V_{B2}=1470\ {\rm {m/s}}}$ (note that neither ${\displaystyle C}$ nor ${\displaystyle G}$ is well determined, both being based on two points only), with intercept times ${\displaystyle t_{A1}=25\ {\rm {ms}}}$, ${\displaystyle t_{B1}=77\ {\rm {ms}}}$. Using equation (4.24f) we get

{\displaystyle {\begin{aligned}V_{2}&=2(1/1030+1/1470)^{-1}=1210\ {\rm {m/s}};\\\theta _{c}&=\sin ^{-1}(700/1210)=35.3^{\circ };\\h_{A}&=V_{1}t_{d1}/2\cos \theta _{c}=700\times 0.025/2\times 0.816=11\ {\rm {m}};\\h_{B}&=700\times 0.077/2\times 0,816=33\ {\rm {m}};\\\xi _{1}&=\tan ^{-1}[(33-11)/220]=5.7^{\circ }.\end{aligned}}}

Because of the small dip, we can take ${\displaystyle h_{A}}$ and ${\displaystyle h_{B}}$ as vertical depths.

Comparison of velocities shows that segments ${\displaystyle E}$ and ${\displaystyle F}$ are the LS equivalent of SS segment ${\displaystyle D}$, while ${\displaystyle I}$ and ${\displaystyle J}$ are equivalent to ${\displaystyle H}$. To interpret these events we shall use the LS velocities and the SS intercept times.

Based on our previous discussion we ignore the displacements of segments ${\displaystyle E}$ and ${\displaystyle F}$ and ${\displaystyle I}$ and ${\displaystyle J}$ and draw the best-fit straight lines as shown in Figure 14.3b, obtaining ${\displaystyle V_{3d}=1860\ {\rm {m/s}}}$, ${\displaystyle V_{3u}=5240\ {\rm {m/s}}}$. We use these values plus the intercept times ${\displaystyle t_{A}=40\ {\rm {ms}}}$ and ${\displaystyle t_{B}=165\ {\rm {ms}}}$ to get the depth and dip.

We use the method of problem 4.24b to get ${\displaystyle \theta _{c2}}$ and ${\displaystyle \xi _{2}}$:

{\displaystyle {\begin{aligned}V_{2}/V_{3d}=\sin(\theta _{c2}+\xi _{2})=1210/1860,\quad (\theta _{c2}+\xi _{2})=40.6^{\circ };\\V_{2}/V_{3u}=\sin \theta _{c2}-\xi _{2})=1210/5240,\quad (\theta _{c2}-\xi _{2})=13.4^{\circ };\end{aligned}}}

so ${\displaystyle \theta _{c2}=27.0^{\circ }}$ , ${\displaystyle \xi _{2}=13.6^{\circ }}$, ${\displaystyle V_{3}=1210/\sin 27.0^{\circ },=2670\ {\rm {m/s}}}$.

Now ${\displaystyle \sin \theta _{1}/700=\sin \theta _{c2}/1210=1/2670}$, ${\displaystyle \theta _{1}=15.2^{\circ }}$, ${\displaystyle \theta _{c2}=26.9^{\circ }}$ (the fact that this value of ${\displaystyle \theta _{c}}$ agrees so closely with the above value is because both values depend upon the velocities).

The SS intercept times of 40 and 165 ms give the following depths:

{\displaystyle {\begin{aligned}{\mbox{at }}A:0.040&=2\times 11\cos 15.2^{\circ }/700+2\ h_{A2}\cos 26.9^{\circ }/1210\\&=0.030+0.00147\ h_{A2},\quad h_{A2}=6.8\ {\rm {m}};\\{\mbox{vertical depth}}&=11+7/\cos 13.6^{\circ }=18\ {\rm {m}}\\{\mbox{at }}B:0.165&=2\times 33\cos 15.2^{\circ }/700+2h_{B2}\cos 26.9^{\circ }/1210\\&=0.091+0.00147\ h_{B2},\quad h_{B2}=50.3\ {\rm {m}};\\{\mbox{vertical depth}}&=33+50.3/\cos 13.6^{\circ }=84\ {\rm {m}};\\\tan \xi &=(84-18)/220=16.7^{\circ }.\end{aligned}}}

Note that this value of ${\displaystyle \xi }$ is based on intercept times plus velocities, whereas the value ${\displaystyle 13.6^{\circ }}$ is based on the velocities only.