View source for Interpretation of four-shot refraction data

{{publication
 | name    = Problems in Exploration Seismology and their Solutions
 | image   = [[File:897.Def.Lz-cropped.png|frameless|180px|link=Problems in Exploration Seismology and their Solutions]]
 | width   = 120px
 | series  = Geophysical References Series
 | title   = [[Problems in Exploration Seismology and their Solutions]]
 | part    = 
 | chapter = 14
 | pages  = 497 - 503
 | author  = [[Lloyd P. Geldart]] and [[Robert E. Sheriff]]
 | doi     = 10.1190/1.9781560801733
 | pdf     = 
 | store   = https://shop.seg.org/shop/products/detail/897
 | isbn    = ISBN 9781560801153
}}
== Problem 14.3 ==
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 <math>t</math>-<math>x</math> curves. Geophones at the shot points are offset when shots at <math>A</math> or <math>B</math> are used.

=== Solution ===
Figure 14.3b shows east-west reversed profiles with sources <math>A</math> and <math>B</math> providing the short-shot (SS) data and sources <math>A^{*}</math> and <math>B^{*}</math> (not shown) that are offset inline to provide the long-shot (LS) data. We have labeled the different segments from <math>C</math> to <math>J</math> 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 <math>(I,J)</math>, <math>(E,F)</math>, and <math>(C,D)</math> 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 <math>I,J</math> and <math>E,F</math> 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.

[[file:Ch14_fig14-3a.png|thumb|{{figure number|14.3a.}} A feature on a refractor would be observed at <math>D</math> from source <math>A</math> but at <math>C</math> from source <math>B</math>.]]

[[file:Ch14_fig14-3b.png|thumb|{{figure number|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 <math>C</math> and <math>G</math>) that probably is a refraction from the top of an intermediate layer. The measured velocities are <math>V_{1} =700\ {\rm m/s}</math>, <math>V_{A2} =1030\ {\rm m/s}</math>, <math>V_{B2} =1470\ {\rm m/s}</math> (note that neither <math>C</math> nor <math>G</math> is well determined, both being based on two points only), with intercept times <math>t_{A1} =25\ {\rm ms}</math>, <math>t_{B1} =77\ {\rm ms}</math>. Using equation (4.24f) we get

<math>
\begin{align}
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{align}
</math>

Because of the small dip, we can take <math>h_{A}</math> and <math>h_{B}</math> as vertical depths.

Comparison of velocities shows that segments <math>E</math> and <math>F</math> are the LS equivalent of SS segment <math>D</math>, while <math>I</math> and <math>J</math> are equivalent to <math>H</math>. 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 <math>E</math> and <math>F</math> and <math>I</math> and <math>J</math> and draw the best-fit straight lines as shown in Figure 14.3b, obtaining <math>V_{3d} =1860\ {\rm m/s}</math>, <math>V_{3u} =5240\ {\rm m/s}</math>. We use these values plus the intercept times <math>t_{A} =40\ {\rm ms}</math> and <math>t_{B} =165 \ {\rm ms}</math> to get the depth and dip.

We use the method of problem 4.24b to get <math>\theta _{c2}</math> and <math>\xi _{2}</math>:

<math>
\begin{align}
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{align}
</math>

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

Now <math>\sin\theta _{1} /700=\sin\theta _{c2} /1210= 1/2670</math>, <math>\theta _{1} =15.2^{\circ}</math>, <math>\theta _{c2} =26.9^{\circ}</math> (the fact that this value of <math>\theta _{c}</math> 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:

<math>
\begin{align}
\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/\cos13.6^{\circ} =18\ {\rm m}\\
\mbox{at }B:0.165&=2\times 33\cos15.2^{\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/\cos13.6^{\circ} =84\ {\rm m};\\
\tan \xi &=(84-18)/220=16.7^{\circ}.
\end{align}
</math>

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

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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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