Effect of a hidden layer

**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	11
Pages	415 - 468
DOI	http://dx.doi.org/10.1190/1.9781560801733
ISBN	ISBN 9781560801153
Store	SEG Online Store

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Problem

Assume that you wish to map the 5.75 and 6.40 km/s formations in the Illinois basin. Given the velocity information shown in Figure 11.3a, what difficulties would you expect to encounter? The shale at 420–620 m and the lower velocity at 790–960 m form “hidden layers”; how much error will neglect of the hidden layers involve?

Solution

The velocity-depth data are summarized in Table 11.3a. Each of the three high-velocity layers will produce a head wave whose apparent velocity is that of the layers if the layering is all horizontal (which we assume, knowing that dips are generally gentle). We calculate the intercept times in order to plot the time-distance curves.

Because the layers are assumed to be horizontal, equation (3.1a) gives the angles of incidence for the ray that produces the head waves. For the 5150 m/s head wave,

${\begin{aligned}\sin \theta _{c1}/2650=1/5150,\qquad \theta _{c1}=31.0^{\circ }.\end{aligned}}$

Figure 11.3a. Illinois Basin interval velocity.

**Table 11.3a. Velocity-depth data.**
Depth range	Velocity
0–300 m	2650 km/s
300–420	5150
420-620	3650
620-790	5750
790-960	5000
960-1200	5750
1200-1550	6400

We use equation (4.18d) to calculate the intercept time $t_{i1}$ :

${\begin{aligned}t_{i1}=2\times 300\cos 31.0^{\circ }/2650=0.194\ {\rm {s}}.\end{aligned}}$

For the 5750 m/s head wave we have

${\begin{aligned}\left(\sin \theta _{1}\right)/2650=\left(\sin \theta _{2}\right)/5150=\left(\sin \theta _{c2}\right)/3650=1/5750;\\\theta _{1}=27.4^{\circ },\quad \theta _{2}=63.6^{\circ },\quad \theta _{c2}=39.4^{\circ }.\end{aligned}}$

Its intercept time will be

${\begin{aligned}t_{i2}&=2\times \left(300\cos 27.4^{\circ }/2650+120\cos 63.6^{\circ }/5150+200\cos 39.4^{\circ }/3650\right)\\&=0.306\ {\rm {s}}.\end{aligned}}$

To complete the time-distance curve, we have for the 6400 m/s head wave, allowing for 170 m of 5000 m/s layer that interrupts the 5750 medium (note ray direction is the same in both parts at 5750 m/s),

${\begin{aligned}\left(\sin \theta _{1}\right)/2650&=\left(\sin \theta _{2}\right)/5150=\left(\sin \theta _{3}\right)/3650\\&=\left(\sin \theta _{4}\right)/5000=\left(\sin \theta _{c3}\right)/5750=1/6400;\\\theta _{1}=24.5^{\circ },\quad \theta _{2}&=53.6^{\circ },\quad \theta _{3}=34.8^{\circ },\quad \theta _{4}=51.4^{\circ },\quad \theta _{c3}=64.0^{\circ }.\end{aligned}}$

Its intercept time will be

${\begin{aligned}t_{i3}&=2\times [300\cos 24.5^{\circ }/2650+120\cos 53.6^{\circ }/5150+200\cos 34.8^{\circ }/3650\\&\qquad \qquad +170\cos 51.4^{\circ }/5000+\left(170+240\right)\cos 64.0^{\circ }/5750]\\&=2\times \left(0.103+0.014+0.044+0.021+0.031\right)=0.426\ {\rm {s}}.\end{aligned}}$

The crossover between the 5150 and 5750 m/s head waves is given by

${\begin{aligned}0.194+x/5150=0.306+x/5750,\quad {\rm {or}}\quad x=0.112/0.0203=5.52\ {\rm {km}};\end{aligned}}$

Figure 11.3b. Time-distance plot.

and the crossover between the 5750 and 6400 m/s head waves is given by

${\begin{aligned}0.306+x/5.75=0.426+x/6.40,\quad {\rm {or}}\quad x=0.120/0.0176=6.78\ {\rm {km}}.\end{aligned}}$

The 5750 m/s curve will be responsible for first breaks for only 1.30 km.

The data are plotted in Figure 11.3b. Interpretation of this time-distance plot will be difficult because the slopes of the three head-wave curves are nearly the same. The ratios of the successive head-wave velocities in this situation are only 1.12 and 1.11; generally ratios should be 1.25 or larger to be interpreted unambiguously.

Failure to recognize a hidden layer means that the time spent in that layer will be interpreted as spent in a layer with higher velocity, which will make the depth appear too large. The shallow refraction event should be interpreted correctly because there are no hidden layers, but the depth calculated for the deeper interfaces will be too great because of the hidden layers.

If we recognize only the 5150 m/s and 6400 m/s head waves (the most probable situation unless additional information is available), that is, the 5750 m/s layer is a hidden layer, then we would calculate the thickness of the 5150 m/s layer $h_{2}$ as

${\begin{aligned}0.426=2\left(300\cos 24.5^{\circ }/2650+h_{2}\cos 53.6^{\circ }/5150\right)=2[0.103+h_{2}(0.000115)].\end{aligned}}$

This gives $h_{2}=960$ m, which, when added to the 300 m thickness of the top layer, gives the depth of the 6400 layer as 1257 m. Comparing with the correct value of 1200 m, the error is 60 m or 5%.

If we should recognize the 5750 m/s head wave, but are not aware of the 3650 m/s layer, we would calculate the thickness of the 5150 m/s layer $h_{2}$ as

${\begin{aligned}0.306=2\left(300\cos 27.4^{\circ }/2650+h_{2}\cos 63.6^{\circ }/5150\right).\end{aligned}}$

This gives $h_{2}=600$ m, which, when added to the 300-m thickness of the top layer, gives a depth of 900 m. Comparing with the correct value of 620 m, the error is 280 m or 45%.

The travel through the 170 m thick 5000 m/s layer, if it is not recognized, would probably be assumed to be at the velocity of 5750 m/s, producing a time error of only 4 ms:

${\begin{aligned}170\left(1/5000-1/5750\right)=170\left(0.000200-0.000174\right)=4\ {\rm {ms}}.\end{aligned}}$

The error is small because the difference in assumed velocities is small.

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