# Maximum array length for given apparent velocity

Series Geophysical References Series Problems in Exploration Seismology and their Solutions Lloyd P. Geldart and Robert E. Sheriff 8 253 - 294 http://dx.doi.org/10.1190/1.9781560801733 ISBN 9781560801153 SEG Online Store

## Problem

Reflections in the zone of interest have apparent velocities around 6300 m/s, whereas the velocity just below a uniform LVL is 2100 m/s. If we wish to avoid severe attenuation below 80 Hz when using an array, what is the maximum inline array length?

### Background

The LVL is discussed in problem 4.16. Apparent velocity is defined in problem 4.2d. See problem 8.6 for a discussion of array response. Note that the array length, $n\Delta x$ , is larger than the distance between the first and last geophones, $(n-1)\Delta x$ .

### Solution

To avoid deterioration, the response curve for the array length $n\Delta x$ , where n is the number of geophones and $\Delta x$ the geophone interval, must not extend beyond the first null (see Figure 8.6b). Equation (8.6b) shows that the first null occurs when $n\gamma /2=\pi$ or $n=2\pi /\gamma$ , where $\gamma$ is the phase difference between adjacent geophones. Problem 8.6a gives $\gamma =2\pi \left(\Delta x/\lambda \right)\sin \alpha$ , so

 {\begin{aligned}n&=2\pi /2\pi \left(\Delta x/\lambda \right)\sin \alpha ,\\{\mbox{and }}\qquad \qquad n\Delta x&=\lambda /\sin \alpha =\lambda _{a}=V_{a}/f,\end{aligned}} (8.5a)

where $V_{a}=$ apparent velocity and $\lambda _{a}=$ apparent wavelength. Thus.

{\begin{aligned}n\Delta x=2100/80{\mbox{ m}}=26{\mbox{ m}}.\end{aligned}} The geophone interval must not exceed $\Delta x=26/n$ .