# Reflector dip in terms of traveltimes squared

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 4.4a

Using the dip-moveout equation (4.2b) and the result of problem 4.3, verify that

{\begin{aligned}\tan \xi =t/(t_{SR}^{2}-t_{0}^{2})^{1/2},\end{aligned}} where $\xi ={\rm {dip}}$ , $t=t_{SM}-t_{MR}$ , $t_{SR}=$ traveltime for path ${\rm {SR}}'{\rm {R}}={\rm {S}}'{\rm {R}}'{\rm {R}}$ , $t_{0}=2h_{c}/V=$ traveltime to receiver at $M$ (Figure 4.4a).

### Solution

Since $t$ in problem 4.3 equals $t_{SR}$ here, we have

{\begin{aligned}(Vt_{SR})^{2}=(2x\cos \xi )^{2}+4h_{c}^{2}=(2x\cos \xi )^{2}+(Vt_{0})^{2},\end{aligned}} {\begin{aligned}{\hbox{or}}\qquad \qquad \cos \xi =\left(V/2x\right)(t_{SR}^{2}-t_{0}^{2})^{1/2}.\end{aligned}} From equation (4.2b) we get

{\begin{aligned}\sin \xi =\left(V/2\right)\left(\Delta t_{d}/\Delta x\right)=\left(V/2x\right)\left(t_{SM}-t_{MR}\right),\end{aligned}} {\begin{aligned}{\hbox{or}}\qquad \qquad \tan \xi =\left(t_{SM}-t_{MR}\right)/(t_{SR}^{2}-t_{0}^{2})^{1/2}=t/(t_{SR}^{2}-t_{0}^{2})^{1/2}.\end{aligned}} ## Problem 4.4b

Using equation (4.2a), show that

{\begin{aligned}\sin \xi =V^{2}\left(t_{SM}^{2}-t_{MR}^{2}\right)/8h_{c}x.\end{aligned}} ### Solution

Equation (4.2a) gives

{\begin{aligned}(Vt_{SM})^{2}=x^{2}+4h_{c}^{2}+4h_{c}x\sin \xi ,\\(Vt_{MR})^{2}=x^{2}+4h_{c}^{2}+4h_{c}x\sin \xi ,\\{\hbox{thus}}\qquad \qquad \sin \xi =V^{2}\left(t_{SM}^{2}-t_{MR}^{2}\right)/8h_{c}x.\end{aligned}} ## Problem 4.4c

Under what circumstances is the result for part (b) the same as equation (4.2b) and also consistent with part (a)?

### Solution

The result in part (b) can be written

{\begin{aligned}\sin \xi =\{[V(t_{MS}-t_{MR})]/2x\}\{[V(t_{MS}+t_{MR})]/4h_{c}\}.\end{aligned}} The expression in the first curly brackets is the same as the right-hand side of equation (4.2b). Hence, for the above to be the same as equation (4.2b), we must have $V(t_{MS}+t_{MR})/4h_{c}=1$ that is,

{\begin{aligned}V(t_{MS}+t_{MR})=4h_{c}=2Vt_{0},\end{aligned}} {\begin{aligned}{\hbox{so}}\qquad \qquad \qquad \qquad t_{0}={\frac {1}{2}}(t_{MS}+t_{MR}).\end{aligned}} In part (a) we got $\tan \xi$ by finding $\sin \xi$ and $\cos \xi$ . The expression for $\cos \xi$ involves no approximation, so the only approximation is that used in the derivation of equation (4.2b) to get $\sin \xi$ . Thus, results in (a) and (c) are consistent provided we take $t_{0}={\frac {1}{2}}(t_{MS}+t_{MR})$ .