# Linear increase in velocity above a refractor

Series Geophysical References Series Problems in Exploration Seismology and their Solutions Lloyd P. Geldart and Robert E. Sheriff 4 79 - 140 http://dx.doi.org/10.1190/1.9781560801733 ISBN 9781560801153 SEG Online Store

## Problem 4.21

If the velocity function in problem 4.17c applies above a horizontal refractor at a depth of 2.40 km where the refractor velocity is 4.25 km/s, (see Figure 4.21a), plot the traveltime-distance curve.

### Background

In Figure 4.18a, the refraction event does not exist to the left of point $Q$ , where $OQ$ is the critical distance (see problem 4.18).

### Solution

For a headwave to be generated, the angle of incidence $i$ must reach the critical angle $\theta _{c}$ . At the depth $z=2.40$ km, the velocity just above the refractor is $1.60+0.600\times 2.40=3.04$ km/s, so the critical angle is

{\begin{aligned}\theta _{c}=i_{c}=\sin ^{-1}(3.04/4.25)=45.7^{\circ }.\end{aligned}} To get $i_{0}$ , we have

{\begin{aligned}p=\sin i_{0}/1.60=\sin i_{c}/3.04=1/4.25=0.235.\end{aligned}} Thus

{\begin{aligned}i_{0}=\sin ^{-1}(1.60\times 0.235)=22.1^{\circ }.\end{aligned}} To get the critical distance $x^{\prime }$ , we note that $x^{\prime }$ ($OQ$ in Figure 4.18a) is equivalent to $MN+PQ=2MN=2x$ when $i=i_{c}$ . Using equation (4.17b), we have

{\begin{aligned}MN&=(1/pa)(\cos i_{0}-\cos i_{c})=(1/0.235\times 0.600)(0.926-0.698)\\&=1.62\,\mathrm {km} .\end{aligned}} The time required to travel the path $SN$ is, using equation (4.17c),

{\begin{aligned}t=(1/a)\ln \left({\frac {\tan i_{c}/2}{\tan i_{0}/2}}\right)=(1/0.600)\ln \left({\frac {\tan(45.7^{\circ }/2)}{\tan(22.1^{\circ }/2)}}\right)=1.28\,\mathrm {s} .\end{aligned}} Thus, the coordinates of the refraction traveltime curve at the critical distance are 2.56 s, 3.24 km). The refraction curve beginning at this point has the slope $(1/4.25)=0.235$ s/km. The curve (plotted in Figure 4.21b) has the equation

{\begin{aligned}t=x/4.25+t_{i}.\end{aligned}} The intercept $t_{i}$ can be found graphically by extending the refraction curve or by calculating $t$ for $x=0$ :

{\begin{aligned}t_{i}=2.56-3.24/4.25=1.80\,\mathrm {s} .\end{aligned}} The direct wave’s traveltime curve is $t=x/1.60$ . However, a diving wave with ($t,x$ ) related by equation (4.20a) arrives before this wave. Inverting equation (4.20a) gives

{\begin{aligned}t&=(2/a)\sinh ^{-1}(ax/2V_{0})=(2/0.600)\sinh ^{-1}(0.600x/2\times 1.60)\\&=3.33\sinh ^{-1}(0.188x).\end{aligned}} This equation has been used to plot the direct wave in Figure 4.21b. The crossover point is at approximately 5.9 km. However, the refraction and the direct wave do not differ appreciably in slope and hence picking the crossover point is not very accurate.