# Deducing fault geometry from well data

Series Geophysical References Series Problems in Exploration Seismology and their Solutions Lloyd P. Geldart and Robert E. Sheriff 10 367 - 414 http://dx.doi.org/10.1190/1.9781560801733 ISBN 9781560801153 SEG Online Store

## Problem 10.2a

Well B is 500 m due east of well A and well C is 600 m due north of A. A fault cuts A, B, and C at depths of 800, 1200, and 600 m, respectively. Assuming that these wells are vertical and the fault is planar and extends to the surface, find the surface trace and strike of the fault.

### Background

It is shown in Sheriff and Geldart, 1995, problem 15.9a that the direction cosines ${\displaystyle \left(l,\;m,\;n\right)}$ of a straight line satisfy the equation

 {\displaystyle {\begin{aligned}l^{2}+m^{2}+n^{2}=1,\end{aligned}}} (10.2a)

and Sheriff and Geldart, 1995, problem 15.9b gives the equation of a plane whose perpendicular from the origin has length h and direction cosines ${\displaystyle \left(l,\;m,\;n\right)}$ as

 {\displaystyle {\begin{aligned}lx+my+nz=h.\end{aligned}}} (10.2b)
Figure 10.1b.  Conventional display of seismic line shown in Figure 10.1a.

### Solution

We take the origin at well A, the x- and y-axes being positive towards the east and north, respectively, and the z-axis positive vertically downward. The coordinates of the points of intersection of the fault plane with wells A, B, and C are, respectively (0,0,800), (500, 0, 1200), and (0, 600, 600) and these three points all lie on the fault plane. Hence, equation (10.2b) shows that

{\displaystyle {\begin{aligned}800n=h\;,\ h;500l+1200n=h\;,\;600m+600n=h.\end{aligned}}}

In addition to these equations we have equation (10.2a) so that we can solve for the four unknowns. Thus ${\displaystyle n/h=1/800=1.25\times 10^{-3}}$, ${\displaystyle l/h+2.40n/h=1/500=2.00\times 10^{-3}}$, ${\displaystyle m/h+n/h=1/600=1.67\times 10^{-3}}$, Solving these equations we get ${\displaystyle l/h=-1.00\times 10^{-3}}$, ${\displaystyle m/h=0.42\times 10^{-3}}$, ${\displaystyle n/h=1.25\times 10^{-3}}$. Using equation (10.2a) we have

{\displaystyle {\begin{aligned}l^{2}+m^{2}+n^{2}=1=h^{2}\times 10^{-6}\left(1.00^{2}+0.42^{2}+1.25^{2}\right)\;,\;h=604\ {\rm {m}}.\end{aligned}}}

We now get ${\displaystyle l=-0.604,}$, ${\displaystyle m=0.254}$, ${\displaystyle n=0.755}$, and the equation of the fault plane is

 {\displaystyle {\begin{aligned}-0.604x+0.254y=0.755z=604.\end{aligned}}} (10.2c)

The surface trace is obtained by setting ${\displaystyle z=0}$ in equation (10.2c), giving

{\displaystyle {\begin{aligned}-0.604x+0.254y=604.\end{aligned}}}

The trace intersects the ${\displaystyle x}$-axis at ${\displaystyle x=-1000\ {\rm {m}}}$, that is, west of A, and cuts the ${\displaystyle y}$-axis at ${\displaystyle y=2380\ {\rm {m}}}$ north of A. The strike is ${\displaystyle \tan ^{-1}(1000/2380)={N}22.8^{\circ }{\rm {E}}}$.

## Problem 10.2b

At what depth would you look for this fault in well D located 500 m ${\displaystyle {\hbox{N}}30^{\circ }{\hbox{W}}}$. of well C?

### Solution

The coordinates of the wellhead at D are

{\displaystyle {\begin{aligned}x=-500\sin 30^{\circ }=-250\ {\rm {m}},\;y=600+500\cos 30^{\circ }=1030\ {\rm {m}}.\end{aligned}}}

Substituting these values in equation (10.2c) gives

{\displaystyle {\begin{aligned}-0.604\times \left(-250\right)+0.254\times 1039+0.755z=60,\\{\rm {so}}\qquad \qquad \qquad \qquad z=\left(604-0.604\times 250-0.254\times 1039\right)/0.755=253\ {\rm {m}}.\end{aligned}}}

## Problem 10.2c

Another fault known to strike N20${\displaystyle ^{\circ }}$W cuts wells A and C at depths of 1300 and 1000 m, respectively. Where should it cut well B?

### Solution

Let the equation of the fault plane be ${\displaystyle l'x+m'y+n'z=h'}$. With four unknows (${\displaystyle l'}$, ${\displaystyle m'}$, ${\displaystyle n'}$ and ${\displaystyle h'}$) we need four equations. The fault intersection in well A gives the equation ${\displaystyle 1300n{'}=h'}$ and the intersection in well C gives ${\displaystyle 600m{'}+1000n{'}=h'}$. The strike is N20${\displaystyle ^{\circ }}$W, so the slope of the strike line relative to the ${\displaystyle x}$-axis is ${\displaystyle \tan(-20^{\circ })={\rm {d}}y/{\rm {d}}x=-m'/l'}$ [see equation (4.2)], so ${\displaystyle m'/l'=0.364}$. We use these three equations to find ${\displaystyle l'}$, ${\displaystyle m'}$ and ${\displaystyle n'}$ in terms of ${\displaystyle h'}$, and then the sum ${\displaystyle l^{'2}+m^{'2}+n^{'2}=1}$ to get ${\displaystyle h'}$.

Solving the first three equations gives ${\displaystyle n'/h'=7.69\times 10^{-4}}$,

{\displaystyle {\begin{aligned}m'/h'=\left(1-0.769\right)/600=\left(0.231/600\right)=3.85\times 10^{-4},\\l'/h'=m'/0.364h'=10.58\times 10^{-4}.\end{aligned}}}

Then ${\displaystyle \left(10.58^{2}+3.85^{2}+7.69^{2}\right)\times 10^{-8}h^{'2}=1}$, ${\displaystyle h'=733\ {\rm {m}}}$,

{\displaystyle {\begin{aligned}l'=0.776,\;m'=0.282,\;n'=0.564.\end{aligned}}}

The equation of the fault plane is thus

{\displaystyle {\begin{aligned}0.776x+0.282y+0.564z=733.\end{aligned}}}

Substituting the coordinates of well B, we get

{\displaystyle {\begin{aligned}500\times 0.776+0.564z=733\;,\;z=612\ {\rm {m}}.\end{aligned}}}