# 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, ${\displaystyle n\Delta x}$, is larger than the distance between the first and last geophones, ${\displaystyle (n-1)\Delta x}$.

### Solution

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

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

{\displaystyle {\begin{aligned}n\Delta x=2100/80{\mbox{ m}}=26{\mbox{ m}}.\end{aligned}}}

The geophone interval must not exceed ${\displaystyle \Delta x=26/n}$.