# Vertical seismic profiling

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
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 ${\displaystyle z}$-axis positive downward, the image point ${\displaystyle I}$ (see problem 4.1) in Figure 13.4a is at (0, 4000) and the geophone coordinates are (1000, ${\displaystyle z}$); hence,

{\displaystyle {\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 ${\displaystyle t=1.118\ {\rm {s}}}$; for 1200 m, ${\displaystyle t=0.991\ {\rm {s}}}$; for 1600 m, ${\displaystyle t=0.867\ {\rm {s}}}$.

Figure 13.4b.  Raypaths for vertical profiling and a dipping bed.

## Problem 13.4b

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

### Solution

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

{\displaystyle {\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 ${\displaystyle -7^{\circ }}$ down toward source ${\displaystyle S}$

{\displaystyle {\begin{aligned}a=(2000-1000\tan 7^{\circ })=2120\ {\rm {m}},b=2100\ {\rm {m}},\end{aligned}}}

Coordinates of image are (${\displaystyle +510}$, 4170),

{\displaystyle {\begin{aligned}{\hbox{time}}=[(1000+510)^{2}+(4170-z)^{2}]^{1/2}/3000.\end{aligned}}}

Table 13.4a shows calculated traveltimes for ${\displaystyle z=800}$, 1200, and 1600 m.

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

## Problem 13.4c

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

### Solution

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

{\displaystyle {\begin{aligned}x_{d}=(1000-z\sin 3^{\circ }),\quad z_{d}=z\cos 3^{\circ },\end{aligned}}}

Figure 13.4c.  Vertical profiling in a deviated well.

where subscript ${\displaystyle d}$ denotes values for the deviated well. The image point remains at (0, 4000), so the traveltime ${\displaystyle t_{d}}$ is

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

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