# Variation of reflection point with offset

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 4.11a

Equation (4.3a) for an offset geophone can be written

 {\begin{aligned}(Vt)^{2}=(2h_{c})^{2}+(2s\cos \xi )^{2},\end{aligned}} (4.11a)

where $2s$ is the offset and $h_{c}$ is the slant depth at the midpoint between the source $S$ and receiver $G$ (see Figure 4.11a). The point of reflection $R(x_{1},z_{1})$ is displaced updip the distance $\Delta L$ from the zero-dip position $P(x_{0},z_{0})$ . Show that the coordinates of a point $x,z$ on the line $IG$ must satisfy the relation

 {\begin{aligned}(2s-x)/(s+h\ell )=z/hn=k,\end{aligned}} (4.11b)

where $I$ is the image point, $\ell ,n$ are the direction cosines of $SI$ , $h$ is the slant depth at the source, and $k$ is a parameter fixing the location of a point on $IG$ .

### Solution

Referring to Figure 4.11a, $(\ell ,n)$ are the direction cosines of $SI$ where

{\begin{aligned}\ell =\sin \xi ,\quad n=\cos \xi ,\quad \xi =\tan ^{-1}(\ell /n).\end{aligned}} To get the coordinates of $A(x,z)$ , a point on $IG$ , we draw $AB$ and $IC$ perpendicular to $TG$ . Then, using the similar triangles $ABG$ and $ICG$ , we have $AB/BG=IC/CG$ , that is,

 {\begin{aligned}z/(2s-x)=2hn/2(s+h\ell ),\qquad \mathrm {so} \ z/hn=(2s-x)/(s+h\ell ),\end{aligned}} (4.11c)

$x$ being the horizontal distance from $S$ . If we write

{\begin{aligned}k=z/hn=(2s-x)/(s+h\ell ),\end{aligned}}  Figure 4.11a.  Displacement of reflection point for offset geophone.

we can vary $k$ to get different points on $IG$ .

## Problem 4.11b

Verify the following relations:

 {\begin{aligned}x_{1}=x_{0}-s^{2}\ell n^{2}/h_{c},\quad z_{1}=z_{0}-s^{2}\ell ^{2}n/h_{c},\end{aligned}} (4.11d)

 {\begin{aligned}{\hbox{and}}\qquad \qquad \Delta L=RP=-(s^{2}/2h_{c})\sin 2\xi .\end{aligned}} (4.11e)

### Solution

To get $R(x_{1},z_{1})$ , the point of intersection of $IG$ and $PT$ , we first find the equation of $PT$ ; the line $PT$ has slope $\tan \xi$ and passes through $T(-h/\ell ,0)$ , so the equation is

 {\begin{aligned}z=x\tan \xi +h/n=(\ell /n)x+h/n=(\ell xh)/n.\end{aligned}} (4.11f)

We now solve equations (4.11c) and (4.11f) as simultaneous equations. Eliminating $z$ gives

{\begin{aligned}z=hn((2s-x_{1})/(s+h\ell ))=(\ell x_{1}+h)/n,\end{aligned}} {\begin{aligned}{\hbox{so}}\qquad \qquad hn^{2}(2s-x_{1})=(s+h\ell )(\ell x_{1}+h).\end{aligned}} Using the equations $(\ell ^{2}+n^{2})=1$ , $h_{c}=(h+s\ell )$ ; this reduces to

 {\begin{aligned}x_{1}=h(s-2s\ell ^{2}-h\ell )/h_{c}.\end{aligned}} (4.11g)

 {\begin{aligned}{\hbox{Also}}\qquad \qquad x_{0}=s-h_{c}\ell =s-h-s\ell ^{2}=sn^{2}-h\ell ,\end{aligned}} (4.11h)

{\begin{aligned}{\hbox{so}}\qquad \qquad \Delta x=x_{1}-x_{0}&=[h(s-2s\ell ^{2}-h\ell )-(sn^{2}-h\ell )/h_{c}]/h_{c}\\&=[hs-h\ell (h+2s\ell )+(h+s\ell )(h\ell -sn^{2})]/h_{c}\\&=[hs-h\ell (h+2s\ell )+h^{2}\ell -hsn^{2}+hs\ell ^{2}-s^{2}\ell n^{2}]/h_{c}\\&=-s^{2}\ell n^{2}/h_{c}.\end{aligned}} From equation (4.11f) we get

{\begin{aligned}z_{1}&=(x_{1}\ell +h)/n=h\ell [(s-2s\ell ^{2}-h\ell )+h_{c}h]/h_{c}n\\&=[h\ell (2s-2s\ell ^{2}-h\ell )+h^{2}]/h_{c}n.\end{aligned}} Since $z_{0}=h_{c}n$ ,

{\begin{aligned}\Delta z&=z_{1}-z_{0}=\left[h\ell (2s-2s\ell ^{2}-h\ell )+h^{2}-h_{c}^{2}n^{2}\right]/h_{c}n\\&=[h\ell (2s-2s\ell -h\ell )+h^{2}-n^{2}(h^{2}+2hs\ell +s^{2}\ell ^{2})]/h_{c}n\\&=[2hs\ell -2hs\ell ^{3}+h^{2}(1-\ell ^{2}-n^{2})-2hs\ell n-s^{2}\ell ^{2}n^{2}]/h_{c}n\\&=[2hs\ell (1-\ell ^{2}-n)-s^{2}\ell ^{2}n^{2}]/h_{c}n=-S^{2}\ell ^{2}n/h_{c}.\end{aligned}} We now have

{\begin{aligned}\Delta L=[(\Delta x)^{2}+(\Delta z)^{2}]^{1/2}=-s^{2}\ell n/h_{c}=-(s^{2}/2h_{c})\sin 2\xi .\end{aligned}} 