# Quick-look velocity analysis and effects of errors

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 5.13a

Velocity analysis usually results in a plot of stacking velocity against traveltime. Bauer (private communication) devised a “quick look” method of determining the interval velocity, assuming horizontal layering and that the stacking velocity equals the average velocity. The method is shown in Figure 5.13a. A box is formed by the two picks between which the interval velocity is to be picked; the diagonal that does not contain the two picks when extended to the velocity axis gives the interval velocity. Prove that the method is valid and discuss its limitations.

### Solution

We extend the diagonal of the box as shown in Figure 5.13a, thus giving $V_{m}$ . The interval velocity is given by

 {\begin{aligned}V_{i}=\Delta z/\Delta t=\left(V_{2}t_{2}-V_{1}t_{1}\right)/\left(t_{2}-t_{1}\right)\\=\left[\left(V_{1}+\Delta V\right)t_{2}-V_{1}t_{1}\left]/\Delta t=\right[V_{1}\left(t_{2}-t_{1}\right)+t_{2}\Delta V\right]/\Delta t\\=\left(V_{1}\Delta t+t_{2}\Delta V\right)/\Delta t=V_{1}+t_{2}\left(\Delta V/\Delta t\right).\end{aligned}} (5.13a)

In Figure 5.13a the triangle with apices at the points $V_{m}$ , $V_{1}$ , and $A$ is similar to the triangle with sides $\Delta V$ and $\Delta t$ , so we have

{\begin{aligned}\Delta V/\Delta t=\left(V_{m}-V_{1}\right)/t_{2}.\end{aligned}} Substituting in equation (5.13a), we get

{\begin{aligned}V_{i}=V_{1}+\left(V_{m}-V_{1}\right)=V_{m}.\end{aligned}} Thus the method gives the interval velocity provided the stacking velocity equals the average velocity. For horizontal velocity layering the stacking velocity is often about 2% higher than the average velocity, but the two may differ considerably if the reflectors are dipping.

## Problem 5.13b

This method can be used to see the influence of measurement error. Discuss the sensitivity of interval-velocity calculations to

1. errors in picking velocity values from this graph,
2. errors in picking times,
3. picking events very close together, and
4. picking events late.

### Solution

1. Errors in $V_{1}$ or $V_{2}$ change the slope of the diagonal and hence change $V_{m}$ and $V_{i}$ ; the error is proportional to $t/\left(t_{2}-t_{1}\right)$ where $t$ is either $t_{1}$ or $t_{2}$ .
2. Changes in $t_{1}$ or $t_{2}$ has an effect similar to that in (i).
3. When both $\Delta V$ and $\Delta t$ are small, the slope of the diagonal and $V_{i}$ are very sensitive to errors.
4. Picking each event late by the same amount $\Delta \tau$ will increase $V_{i}$ by an amount proportional to $\Delta \tau .$ We now derive mathematical expressions for the changes in (i) to (iv).

i) Equation (5.13a) is

{\begin{aligned}V_{1}=(V_{2}t_{2}-V_{1}t_{1})/(t_{2}-t_{1}).\\{\rm {Thus}},\qquad \qquad \qquad \qquad {\rm {d}}V_{i}/{\rm {d}}V_{1}=-t_{1}/(t_{2}-t_{1})=-t_{1}/\Delta t,\\{\rm {and}}\qquad \qquad \qquad \qquad {\rm {d}}V_{i}/{\rm {d}}V_{2}=t_{2}/(t_{2}-t_{1})=t_{2}/\Delta t.\end{aligned}} Hence the error is directly proportional to either $t_{1}$ or $t_{2}$ and inversely proportional to $(t_{2}-t_{1})$ ; it increases rapidly as $t_{1}$ approaches $t_{2}$ .

ii) {\begin{aligned}{\frac {{\rm {d}}V_{i}}{{\rm {d}}t_{1}}}=-{\frac {V_{1}}{\left(t_{2}-t_{1}\right)}}+{\frac {\left(V_{2}t_{2}-V_{1}t_{1}\right)}{(t_{2}-t_{1})^{2}}}={\frac {-V_{1}\left(t_{2}-t_{1}\right)+\left(V_{2}t_{2}-V_{1}t_{1}\right)}{(t_{2}-t_{1})^{2}}}\\={\frac {(V_{2}-V_{1})t_{2}}{(t_{2}-t_{1})^{2}}}.\end{aligned}} Likewise,

{\begin{aligned}{\frac {{\rm {d}}V_{i}}{{\rm {d}}t_{2}}}={\frac {-\left(V_{2}-V_{1}\right)t_{1}}{(t_{2}-t_{1})^{2}}}.\end{aligned}} Since $t_{2}>t_{1}$ , errors in $t_{1}$ are more serious than errors in $t_{2}$ ; also, the errors increase rapidly as $t_{1}$ approaches $t_{2}$ .

iii) Let $t_{2}=t_{1}+\Delta t$ , $\Delta t\approx 0$ . Then using equation (5.13a),

{\begin{aligned}V_{t}=\left[V_{2}\left(t_{1}+\Delta t\right)-V_{1}t_{1}\right]/\Delta t=V_{2}+\left(V_{2}-V_{1}\right)\left(t_{1}/\Delta t\right).\end{aligned}} In general $t_{1}\gg \Delta t$ , so the error in $V_{i}$ depends mainly on the factor $t_{1}/\Delta t$ .

iv) We assume that both events are late by the same amount $\Delta \tau$ . Then equation (5.13a) becomes

{\begin{aligned}V_{i}+\Delta V_{i}=\left[V_{2}\left(t_{2}+\Delta \tau \right)-V_{1}\left(t_{1}+\Delta \tau \right)\right]/\left(t_{2}-t_{1}\right)\\=V_{i}+\left(V_{2}-V_{1}\right)\Delta \tau /\left(t_{2}-t_{1}\right),\\{\rm {so}}\qquad \qquad \qquad \qquad \Delta V_{i}=\Delta \tau \left[\left(V_{2}-V_{1}\right)/\left(t_{2}-t_{1}\right)\right]=\Delta \tau \tan \zeta ,\end{aligned}} where $\tan \zeta$ is usually fairly constant over a moderate range of depths; in this case the error in $V_{i}$ is proportional to the error $\Delta \tau .$ 