# Resolution of cross-dip

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 4.9a

Sources $B$ and $C$ are, respectively, 600 m north and 500 m east of source $A$ . Traveltimes to zero-offset geophones at $A$ , $B$ , and $C$ for a certain reflection are 1.750, 1.825, and 1.796 s. What are the dip and strike of the horizon, the average velocity being 3.25 km/s?

### Solution

The dip moveouts are $0.075/0.600=0.125$ s/km to the north and $0.046/0.500=0.092$ s/km to the east. We take the $x$ -axis east-west (see Figure 4.9a). If we apply equation (4.2b) as in problem 4.2c, we get

{\begin{aligned}\sin \xi _{y}&=(3.25/2)(0.075/0.60)=0.203\\&=m;\qquad \xi _{y}=11.7^{\circ };\\\sin \xi _{x}&=(3.25/2)(0.046/0.50)=0.150\\&=\ell ;\qquad \xi =8.6^{\circ }.\end{aligned}} Using equations (4.2g,h), we find that

{\begin{aligned}\sin \xi &=(\ell ^{2}+m^{2})^{1/2}=(0.150^{2}+0.203^{3})^{1/2}=0.252,\quad \xi =14.6^{\circ };\\\mathrm {strike} \ \Xi &=\tan ^{-1}(\ell /m)=\tan ^{-1}(0.150/0.203)\\&=36.4^{\circ }\ {\hbox{with respect to the}}\ x{\hbox{-axis}}{}={\hbox{N}}53.6^{\circ }{\hbox{W}}.\end{aligned}} ## Problem 4.9b

What are the changes in dip and strike if line AC has the bearing N80$^{\circ }$ E?

### Solution

We use equation (4.2j) and Figure 4.2e to obtain the dip moveout along the east-west $x$ -axis when the measured moveout along the $x'$ -axis with bearing N80$^{\circ }$ E is 0.092 s/km. Equation (4.2j) gives

{\begin{aligned}{\frac {3.25}{2}}\left({\frac {0.046}{0.500}}\right)&=0.150\\&=\ell \cos 10^{\circ }+0.203\sin 10^{\circ };\end{aligned}} thus,

{\begin{aligned}\ell =0.150/\cos 10^{\circ }-0.203\tan 10^{\circ }=0.117.\end{aligned}} {\begin{aligned}{\hbox{So}}\qquad \qquad \sin \xi =(0.117^{2}+0.203^{2})^{1/2}=0.234,\quad \xi =13.6^{\circ },\\\tan \Xi =0.117/0.203=0.576,\end{aligned}} and the strike angle $\Xi =30.0^{\circ }$ , that is, N60.0$^{\circ }$ W. Thus, the $10^{\circ }$ change in line direction changes the dip about $1.0^{\circ }$ and the strike about $6^{\circ }$ .

## Problem 4.9c

Solve part (b) graphically.

### Solution

The graphical construction, shown reduced in Figure 4.9c, yields dip moveout of 0.145 s/km and strike of N61$^{\circ }$ W.

The dip moveout 0.145 gives a dip 13.6$^{\circ }$ . Thus the graphical solution gives almost exactly the same values as in part (b).