Linear increase in velocity above a refractor

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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.


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


For a headwave to be generated, the angle of incidence must reach the critical angle . At the depth km, the velocity just above the refractor is km/s, so the critical angle is

Figure 4.21a.  Linear increase in velocity above a refractor.

To get , we have


To get the critical distance , we note that ( in Figure 4.18a) is equivalent to when . Using equation (4.17b), we have

Figure 4.21b.  Time-distance plot.

The time required to travel the path is, using equation (4.17c),

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 s/km. The curve (plotted in Figure 4.21b) has the equation

The intercept can be found graphically by extending the refraction curve or by calculating for :

The direct wave’s traveltime curve is . However, a diving wave with () related by equation (4.20a) arrives before this wave. Inverting equation (4.20a) gives

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.

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