# Resolution of cross-dip

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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 4.9a

Sources ${\displaystyle B}$ and ${\displaystyle C}$ are, respectively, 600 m north and 500 m east of source ${\displaystyle A}$. Traveltimes to zero-offset geophones at ${\displaystyle A}$, ${\displaystyle B}$, and ${\displaystyle C}$ 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 ${\displaystyle 0.075/0.600=0.125}$ s/km to the north and ${\displaystyle 0.046/0.500=0.092}$ s/km to the east. We take the ${\displaystyle x}$-axis east-west (see Figure 4.9a). If we apply equation (4.2b) as in problem 4.2c, we get

{\displaystyle {\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}}}

Figure 4.9a.  Geometry of dip measures.

Using equations (4.2g,h), we find that

{\displaystyle {\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${\displaystyle ^{\circ }}$E?

### Solution

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

{\displaystyle {\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}}}

Figure 4.9b.  Geometry with change of direction.

thus,

{\displaystyle {\begin{aligned}\ell =0.150/\cos 10^{\circ }-0.203\tan 10^{\circ }=0.117.\end{aligned}}}

{\displaystyle {\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 ${\displaystyle \Xi =30.0^{\circ }}$, that is, N60.0${\displaystyle ^{\circ }}$W. Thus, the ${\displaystyle 10^{\circ }}$ change in line direction changes the dip about ${\displaystyle 1.0^{\circ }}$ and the strike about ${\displaystyle 6^{\circ }}$.

## 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${\displaystyle ^{\circ }}$W.

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

Figure 4.9c.  Graphical solution.