# Vertical seismic profiling

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Series Geophysical References Series Problems in Exploration Seismology and their Solutions Lloyd P. Geldart and Robert E. Sheriff 13 485 - 496 http://dx.doi.org/10.1190/1.9781560801733 ISBN 9781560801153 SEG Online Store

## Problem 13.4a

A source is offset 1000 m from a vertical well in which a geophone is suspended, and a horizontal reflector is present at a depth of 2000 m. Calculate the traveltimes for the reflection when the geophone is 800, 1200, and 1600 m deep if the velocity =3000 m/s

### Solution

Using a coordinate system with origin at the source and the $z$ -axis positive downward, the image point $I$ (see problem 4.1) in Figure 13.4a is at (0, 4000) and the geophone coordinates are (1000, $z$ ); hence,

{\begin{aligned}t_{c}=[1000^{2}+(4000-z)^{2}]^{1/2}/3000.\end{aligned}}  Figure 13.4a.  Raypaths for vertical profiling and a horizontal bed.

For 800 m depth $t=1.118\ {\rm {s}}$ ; for 1200 m, $t=0.991\ {\rm {s}}$ ; for 1600 m, $t=0.867\ {\rm {s}}$ . Figure 13.4b.  Raypaths for vertical profiling and a dipping bed.

## Problem 13.4b

Repeat for reflectors dipping $\pm 7^{\circ }$ , where the reflector intersects the well at the same point as in part (a) (see Figure 13.4b).

### Solution

i) Dip $+7^{\circ }$ down toward the well

{\begin{aligned}a&={\hbox{vertical depth of reflector at source S}}\\&=(2000-1000\tan 7^{\circ })=1880\ {\rm {m}},\\b&={\hbox{slant depth of reflector}}=a\cos 7^{\circ }\\&=1870\ {\rm {m}},\\{\hbox{Coordiantes of image}}&=(-2{\hbox{b}}\sin 7^{\circ },2{\hbox{b}}\cos 7^{\circ })\\&=(-460,3710),\\{\hbox{time}}_{+}&=[(1000+460)^{2}+(3710-Z)^{2}]^{1/2}/3000.\end{aligned}} ii) Dip $-7^{\circ }$ down toward source $S$ {\begin{aligned}a=(2000-1000\tan 7^{\circ })=2120\ {\rm {m}},b=2100\ {\rm {m}},\end{aligned}} Coordinates of image are ($+510$ , 4170),

{\begin{aligned}{\hbox{time}}=[(1000+510)^{2}+(4170-z)^{2}]^{1/2}/3000.\end{aligned}} Table 13.4a shows calculated traveltimes for $z=800$ , 1200, and 1600 m.

 Depth(m) $\to$ 800 1200 1600 ${\rm {Dip}}=+7^{\circ }$ time (s) $\to$ 1.09 0.97 0.86 ${\rm {Dip}}=-7^{\circ }$ time (s) $\to$ 1.14 1 0.87
Table 13.4b. Effect of well deviation on $t_{a}$ .
$z(m)$ $x_{d}(m)$ $Z_{d}(m)$ $t_{d}(s)$ $\Delta t(s)$ 800 958 799 1.114 $-0.004$ 1200 937 1198 0.985 $-0.006$ 1600 916 1598 0.857 $-0.010$ ## Problem 13.4c

By how much do the values in part (a) change if the well deviates by $3^{\circ }$ towards the source (see Figure 13.4c)?

### Solution

Well inclination changes the geophone coordiantes from (1000, $z$ ) where the in-hole depth $z=800$ , 1200, 1600, to $(x_{d},z_{d})$ where

{\begin{aligned}x_{d}=(1000-z\sin 3^{\circ }),\quad z_{d}=z\cos 3^{\circ },\end{aligned}} where subscript $d$ denotes values for the deviated well. The image point remains at (0, 4000), so the traveltime $t_{d}$ is

 {\begin{aligned}t_{d}=[x_{\rm {d}}^{2}+(4000-z_{\rm {d}})^{2}]^{1/2}/3000.\end{aligned}} (13.4a)

Substituting the values of $z$ , we get the results in Table 13.4b. $\Delta t$ is the difference in traveltime from that calculated in part (a).