# Relationship for a dipping bed

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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)

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

where ${\displaystyle h}$ in equation (4.2a) is replaced by ${\displaystyle h_{c}}$, the slant depth at the midpoint ${\displaystyle M}$ between source ${\displaystyle S}$ and receiver ${\displaystyle R}$, and ${\displaystyle t=t_{SR}}$ 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 ${\displaystyle x}$; in Figure 4.3a the up-dip receiver ${\displaystyle R}$ is offset ${\displaystyle 2x}$ from source ${\displaystyle S}$, so that the dip ${\displaystyle \xi }$ is negative; thus equation (4.2a) becomes

{\displaystyle {\begin{aligned}(Vt)^{2}=(2x)^{2}+(2h)^{2}-4h\left(2x\right)\sin \xi .\end{aligned}}}

Replacing ${\displaystyle h}$ with ${\displaystyle h_{c}}$ where ${\displaystyle h=h_{c}+x\sin \xi }$, we obtain

Figure 4.3a.  Geometry for dipping bed.

{\displaystyle {\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}}}