# Second approximation for dip moveout

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 4.5

The expressions for dip in terms of dip moveout, equation (4.2b), involves the approximation of dropping higher-order terms in the expansion of equation (4.2a). What is the effect on equation (4.2b) if an additional term is carried in this expansion? What is the percentage change in dip?

### Solution

In problem 4.2b we obtained the dip equation (4.2b) by taking the first approximation of equation (4.2a), that is, using

{\displaystyle {\begin{aligned}t&=(1/V)(x^{2}+4h^{2}+4hx\sin \xi )^{1/2}\\&=t_{0}[1+r(r+2\sin \xi )]^{1/2},\end{aligned}}}

where ${\displaystyle r=x/2h}$. Expanding and taking the second approximation gives

 {\displaystyle {\begin{aligned}t=t_{0}[1+(r/2)(r+2\sin \xi )-(r^{2}/8)(r+2\sin \xi )^{2}].\end{aligned}}} (4.5a)

If we take two offsets, ${\displaystyle \Delta x}$ and ${\displaystyle -\Delta x}$, and let ${\displaystyle r=|\Delta x/2h|}$, then

{\displaystyle {\begin{aligned}\Delta t_{d}&=t_{1}-t_{2}=t_{0}(2r\sin \xi -r^{3}\sin \xi )\\&=t_{0}[\Delta x/h-(\Delta x/h)^{3}/8]\sin \xi \\&=(t_{0}\Delta x/h)\sin \xi [1-(\Delta x/h)^{2}/8]\\&=(2/V)(\Delta x\sin \xi )[1-(\Delta x/h)^{2}/8],\end{aligned}}}

{\displaystyle {\begin{aligned}{\hbox{so}}\qquad \qquad \sin \xi =(V/2)\left({\frac {\Delta t_{d}}{\Delta x}}\right)[1+(2\Delta x/h)^{2}/32].\end{aligned}}}

Comparing this result with equation (4.2b), we see that the second approximation increases the calculated dip by the fraction ${\displaystyle (2\Delta x/h)^{2}/32}$, that is, by the approximate percentage

{\displaystyle {\begin{aligned}3(2\Delta x/h)^{2}=3({\hbox{spread length/depth}})^{2}.\end{aligned}}}