# Relationship for a dipping bed

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 4.3

Show that, for a dipping reflector and constant velocity, equation (4.2a) becomes (see Gardner, 1947)

 {\begin{aligned}(Vt)^{2}=(2x\cos \xi )^{2}+4h_{c}^{2},\end{aligned}} (4.3a)

where $h$ in equation (4.2a) is replaced by $h_{c}$ , the slant depth at the midpoint $M$ between source $S$ and receiver $R$ , and $t=t_{SR}$ in Figure 4.3a.

### Solution

Equation (4.2a) is based on Figure 4.2a where the receiver is down dip from the source, the offset being $x$ ; in Figure 4.3a the up-dip receiver $R$ is offset $2x$ from source $S$ , so that the dip $\xi$ is negative; thus equation (4.2a) becomes

{\begin{aligned}(Vt)^{2}=(2x)^{2}+(2h)^{2}-4h\left(2x\right)\sin \xi .\end{aligned}} Replacing $h$ with $h_{c}$ where $h=h_{c}+x\sin \xi$ , we obtain

{\begin{aligned}(Vt)^{2}&=4x^{2}+4(h_{c}+x\sin \xi )^{2}-8x\left(h_{c}+x\sin \xi \right)\sin \xi \\&=4x^{2}+4\left(h_{c}^{2}+2xh_{c}\sin \xi +x^{2}\sin ^{2}\xi \right)-8xh_{c}\sin \xi -8x^{2}\sin ^{2}\xi \\&=4x^{2}\left(1+\sin ^{2}\xi -2\sin ^{2}\xi \right)+4h_{c}^{2}\\&=(2x\cos \xi )^{2}+4h_{c}^{2}.\end{aligned}} 