# Variation of reflection point with offset

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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 4.11a

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

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

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

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

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

### Solution

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

{\displaystyle {\begin{aligned}\ell =\sin \xi ,\quad n=\cos \xi ,\quad \xi =\tan ^{-1}(\ell /n).\end{aligned}}}

To get the coordinates of ${\displaystyle A(x,z)}$, a point on ${\displaystyle IG}$, we draw ${\displaystyle AB}$ and ${\displaystyle IC}$ perpendicular to ${\displaystyle TG}$. Then, using the similar triangles ${\displaystyle ABG}$ and ${\displaystyle ICG}$, we have ${\displaystyle AB/BG=IC/CG}$, that is,

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

${\displaystyle x}$ being the horizontal distance from ${\displaystyle S}$. If we write

{\displaystyle {\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 ${\displaystyle k}$ to get different points on ${\displaystyle IG}$.

## Problem 4.11b

Verify the following relations:

 {\displaystyle {\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)

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

### Solution

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

 {\displaystyle {\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 ${\displaystyle z}$ gives

{\displaystyle {\begin{aligned}z=hn((2s-x_{1})/(s+h\ell ))=(\ell x_{1}+h)/n,\end{aligned}}}

{\displaystyle {\begin{aligned}{\hbox{so}}\qquad \qquad hn^{2}(2s-x_{1})=(s+h\ell )(\ell x_{1}+h).\end{aligned}}}

Using the equations ${\displaystyle (\ell ^{2}+n^{2})=1}$, ${\displaystyle h_{c}=(h+s\ell )}$ ; this reduces to

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

 {\displaystyle {\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)

{\displaystyle {\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

{\displaystyle {\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 ${\displaystyle z_{0}=h_{c}n}$,

{\displaystyle {\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

{\displaystyle {\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}}}