Interval velocities

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Problem 5.15

An survey gives the stacking velocity results in Table 5.15a. Calculate the interval velocities.

Table 5.15a. survey results.
1 1.20 1.100 2.18
2 2.50 1.786 2.80
3 3.10 1.935 3.20
4 4.10 2.250 3.64

Background

In equation (4.13a), the sum is the traveltime down to the base of the layer, that is, . Therefore equation (4.13a) can be written


(5.15a)

If we write and for the rms velocity and traveltime down to the base of the layer, and for the same quantities down to the top of the layer, we can get an equation for the interval velocity in the bed by subtracting expressions for and obtained from equation (5.15a). The result is


(5.15b)

Solution

We assume horizontal velocity layering so that the values of in Table 5.15a are approximate values of (see Table 5.12b) and therefore we can use equation (5.15b) to find interval velocities. We take 2.18 km/s as the interval velocity from the surface to the first reflector. We now calculate the interval velocities to :

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