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

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

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

 {\displaystyle {\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 ${\displaystyle V_{s}}$ in Table 5.15a are approximate values of ${\displaystyle V_{\rm {rms}}}$ (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 ${\displaystyle V_{2}}$ to ${\displaystyle V_{4}}$:

{\displaystyle {\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}}}