# Interval velocities

Series Geophysical References Series Problems in Exploration Seismology and their Solutions Lloyd P. Geldart and Robert E. Sheriff 5 141 - 180 http://dx.doi.org/10.1190/1.9781560801733 ISBN 9781560801153 SEG Online Store

## Problem 5.15

An $X^{2}-T^{2}$ survey gives the stacking velocity results in Table 5.15a. Calculate the interval velocities.

Table 5.15a. $X^{2}-T^{2}$ survey results.
$i$ $z\ ({\hbox{km}})$ $t_{i}\ ({\hbox{s}})$ $V_{s}\ ({\hbox{km/s}})$ 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 $\mathop {\sum } \nolimits _{i=1}^{n}\Delta t_{i}$ is the traveltime down to the base of the $n^{th}$ layer, that is, $t_{n}$ . Therefore equation (4.13a) can be written

 {\begin{aligned}V_{\rm {rms}}^{2}t_{n}=\mathop {\sum } \limits _{i=1}^{n}V_{i}^{2}\Delta t_{i}.\end{aligned}} (5.15a)

If we write $V_{L}$ and $t_{L}$ for the rms velocity and traveltime down to the base of the $n^{th}$ layer, $V_{U}$ and $t_{U}$ for the same quantities down to the top of the layer, we can get an equation for the interval velocity $V_{n}$ in the $n^{th}$ bed by subtracting expressions for $V_{L}$ and $V_{U}$ obtained from equation (5.15a). The result is

 {\begin{aligned}V_{n}^{2}=\left(V_{L}^{2}t_{L}-V_{U}^{2}t_{U}\right)/\Delta t_{n}.\end{aligned}} (5.15b)

### Solution

We assume horizontal velocity layering so that the values of $V_{s}$ in Table 5.15a are approximate values of $V_{\rm {rms}}$ (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 $V_{2}$ to $V_{4}$ :

{\begin{aligned}V_{2}^{2}=\left(V_{L}^{2}t_{L}-V_{U}^{2}t_{U}\right)/\Delta t_{2}=\left(2.80^{2}\times 1.786-2.18^{2}\times 1.100\right)/0.686,\\V_{2}^{2}=3.57\ {\rm {km/s}},\\V_{3}^{2}=\left(3.20^{2}\times 1.935-2.80^{2}\times 1.786\right)/0.149,\;V_{3}=6.25\ {\rm {km/s}},\\V_{4}^{2}=\left(3.64^{2}\times 2.250-3.20^{2}\times 1.935\right)/0.315\;,\;V_{4}=5.63\ {\rm {km/s}}.\end{aligned}} 