# Relation between average and rms velocities

Series Geophysical References Series Problems in Exploration Seismology and their Solutions Lloyd P. Geldart and Robert E. Sheriff 4 79 - 140 http://dx.doi.org/10.1190/1.9781560801733 ISBN 9781560801153 SEG Online Store

## Problem 4.13a

Both the average velocity ${\displaystyle {\bar {V}}}$ and the root-mean-square velocity ${\displaystyle V_{\rm {rms}}}$ can be regarded as weighted averages along a particular raypath. Discuss the difference between the two velocities from this point of view.

### Background

For a bed of constant velocity ${\displaystyle V}$ extending from the surface down to a horizontal reflector at a depth ${\displaystyle h}$, the two-way traveltime is ${\displaystyle t=2h/V}$. If the velocity is not constant and we assume vertical raypaths, we obtain an average velocity ${\displaystyle {\bar {V}}}$ by dividing ${\displaystyle 2h}$ by the two-way traveltime, in effect replacing the actual section by a single constant-velocity layer. When the section consists of a series of horizontal constant-velocity layers, we find ${\displaystyle {\bar {V}}}$ by summing the traveltimes through each layer, and dividing the total thickness by this sum.

Dix (1955) showed that better results could be obtained by using the root-mean-square velocity ${\displaystyle V_{\rm {rms}}}$ calculated according to the formula,

 {\displaystyle {\begin{aligned}V_{\rm {rms}}^{2}=\sum _{i=1}^{n}V_{i}^{2}\Delta t_{i}/\sum _{i=1}^{n}\Delta t.\end{aligned}}} (4.13a)

This formula applies to horizontal velocity layering as the offset goes to zero.

Fermat’s principle of stationary time states that a wave travels from one point to another along a path for which the traveltime is stationary, usually a minimum, compared to that along adjacent paths. This principle can be used to derive laws involving the geometrical aspects of wave travel (see problem 6.3).

### Solution

For horizontal layering, ${\displaystyle {\bar {V}}}$ for a given depth is found by dividing the depth by the total one-way traveltime for a ray traveling vertically down to the required depth. This can be expressed as

 {\displaystyle {\begin{aligned}{\bar {V}}=h/t=(\Sigma \Delta h_{i})/\Sigma (\Delta h_{i}/V_{i}),\quad \mathrm {or} \quad h/{\bar {V}}=\Sigma (\Delta h_{i}/V_{i}),\end{aligned}}} (4.13b)

where ${\displaystyle \Delta h_{i}}$ and ${\displaystyle V_{i}}$ are the thickness and velocity of the the ${\displaystyle i^{\rm {th}}}$ th bed and ${\displaystyle h}$ is the total thickness. Thus, ${\displaystyle 1/{\bar {V}}}$ is the weighted average of the reciprocal of velocity when the weights are the layer vertical thicknesses.

Since the denominator of equation (4.13a) is the total one-way traveltime, we can write the equation in the form

{\displaystyle {\begin{aligned}V_{\rm {rms}}^{2}t=\sum \limits _{i=1}^{n}V_{t}^{2}\Delta t_{i},\end{aligned}}}

from which we see that ${\displaystyle V_{\rm {rms}}^{2}}$ is the weighted average of ${\displaystyle V_{i}^{2}}$, the weights being the one-way vertical traveltimes through each bed.

## Problem 4.13b

Calculate ${\displaystyle {\bar {V}}}$ and ${\displaystyle V_{\rm {rms}}}$ down to each of the interfaces in Figure 4.13a. Why do they differ (give a geometrical explanation)?

Figure 4.13a.  Model for the calculation of ${\displaystyle {\bar {V}}}$ and ${\displaystyle V_{\rm {rms}}}$.

### Solution

${\displaystyle {\bar {V}}}$ is found by calculating the one-way time to pass through each layer, summing the time increments to a given depth and dividing the sum into the depth; ${\displaystyle V_{\rm {rms}}}$ is found from equation (4.13a). The values are shown in Figure 4.13a.

The average velocity ${\displaystyle {\bar {V}}}$ is calculated for a straight-line vertical path and is obtained from the weighted average of ${\displaystyle 1/V_{i}}$, the weights being the layer thicknesses, whereas ${\displaystyle V_{\rm {rms}}}$ gives more weight to high-velocity beds. This is consistent with the fact that the least-time path between two points involves changes in path direction that increase travel in the high-velocity layers. Thus, using ${\displaystyle V_{\rm {rms}}}$ somewhat compensates for the failure of straight-line paths to account for ray bending at interfaces.

## Problem 4.13c

Plot ${\displaystyle {\bar {V}}}$ and ${\displaystyle V_{\rm {rms}}}$ versus depth and versus traveltime and determine the best-fit straight lines for the four cases. What are the main problems in approximating data with functional fits?

Figure 4.13b.  Linear fits of velocity-depth and velocity-time data.

### Solution

Figure 4.13b is the plot of ${\displaystyle {\bar {V}}}$ and ${\displaystyle V_{\rm {rms}}}$ versus depth and two-way traveltime. Least-squares fits (see Sheriff and Geldart, 1995, problem 15.13) are also tabulated on Figure 4.13b.

Use of any functional form involves approximations. The main problems in approximating data with functional fits are the following:

1. We must select a functional form to fit the data and the form selected may be either too simple or too complex. A simple form is easy to determine and use but it may not represent the data with sufficient precision. Use of a more complex form increases the accuracy but at the expense of increased computation time.
2. Single points far from the line of the function have a disproportionate effect on the determination of the line.
3. When the data are not evenly spaced, data bunched together have less influence on the best-fit line than more widely spaced data.
4. The accuracy of functional fits changes with offset because the relative contributions of high-velocity members increases with offset.
Table 4.14a. Depth calculations using various velocity functions.
Function ${\displaystyle t_{0}\to }$ 1.00 s 2.00 s 2.10 s 3.10 s
i) ${\displaystyle {\bar {V}}}$ 2.00 km/s 2.50 km/s 2.67 km/s 3.10 km/s
${\displaystyle z}$ 1.00 km 2.50 km 2.80 km 4.80 km
ii) ${\displaystyle V_{\rm {rms}}}$ 2.00 km/s 2.55 km/s 2.81 km/s 3.24 km/s
${\displaystyle z}$ 1.00 km 2.55 km 2.95 km 5.02 km
iii) ${\displaystyle {\bar {V}}}$ 2.02 km/s 2.54 km/s 2.59 km/s 3.12 km/s
${\displaystyle z}$ 1.01 km 2.54 km 2.72 km 4.84 km
iv) ${\displaystyle V_{\rm {rms}}}$ 2.03 km/s 2.62 km/s 2.68 km/s 3.27 km/s
${\displaystyle z}$ 1.02 km 2.62 km 2.81 km 5.07 km