Vertical seismic profiling

Problems in Exploration Seismology and their Solutions

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

Table 13.4a. Traveltimes for dipping reflectors.

	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	1.00	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}} $

Figure 13.4c.  Vertical profiling in a deviated well.

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).

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Table of Contents (book)

Also in this chapter

• S-wave conversion in marine surveys
• Equally inclined orthogonal geophones
• Guided (channel) waves and normal-mode propagation
• Vertical seismic profiling
• Effect of velocity change on VSP traveltime
• Mapping the vertical flank of a salt dome
• Poission’s ratio from P- and S-wave traveltimes

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