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

{\displaystyle {\begin{aligned}\tan \xi =t/(t_{SR}^{2}-t_{0}^{2})^{1/2},\end{aligned}}}

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

Solution

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

{\displaystyle {\begin{aligned}(Vt_{SR})^{2}=(2x\cos \xi )^{2}+4h_{c}^{2}=(2x\cos \xi )^{2}+(Vt_{0})^{2},\end{aligned}}}

{\displaystyle {\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

{\displaystyle {\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}}}

{\displaystyle {\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}}}

Figure 4.4a.  Geometry for dipping bed.

Problem 4.4b

Using equation (4.2a), show that

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

Solution

Equation (4.2a) gives

{\displaystyle {\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

{\displaystyle {\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 ${\displaystyle V(t_{MS}+t_{MR})/4h_{c}=1}$ that is,

{\displaystyle {\begin{aligned}V(t_{MS}+t_{MR})=4h_{c}=2Vt_{0},\end{aligned}}}

{\displaystyle {\begin{aligned}{\hbox{so}}\qquad \qquad \qquad \qquad t_{0}={\frac {1}{2}}(t_{MS}+t_{MR}).\end{aligned}}}

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