Effect of velocity change on VSP traveltime

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 13.5a

A source is offset 3.00 km west of a vertical well in which a geophone is suspended. There is a vertical north-south fault 0.80 km west of the well with velocities $V=4.00$ and 3.00 km/s west and east of the fault, respectively. A horizontal reflector is present west of the fault at a depth of 2.00 km. Find the reflection traveltime for geophone depths of 0.60 and 1.20 km.

Solution

This problem must be solved by trial-and-error methods, either graphical or numerical. We have chosen the latter, because it is easier, quicker, and more accurate than graphical methods.

Figure 13.5a shows the geometry. The horizontal reflector at a depth of 2 km has an image point 4.00 km below the source. A typical ray is shown going from the image point to a geophone located in the well at a depth $h$.

Figure 13.5a.  Geometry of the problem.

The angle of incidence $\theta _{1}$ and of refraction $\theta _{2}$ where the ray passes through the fault are related by Snell’s law: that is, $\sin \theta _{2}=(3.00/4.00)\sin \theta _{1}=0.75\sin \theta _{1}$. We have a further constraint in that the vertical distance between the image and the geophone is $(4.00-h)$. These relations fix $h$:

1. $\sin \theta _{2}=0.75\sin \theta _{1}$;
2. $2.20\tan \theta _{1}+0.80\tan \theta _{2}=(4.00-h)$.

We denote the left-hand side of (b) by $L$. For the geophone at depth 0.60 km, the bending of the ray will have a small effect because of the shallowness and proximity to the fault, so we take as our first approximation the straight path,

${\begin{aligned}\theta _{1}\approx \tan ^{-1}(4.00-0.60)/3.00\approx 49^{\circ }.\end{aligned}}$

To allow for some bending, we try $\theta _{1}=51^{\circ }$ and get $\theta _{2}=35.7^{\circ }$. Substituting theses values in relation (b) we get

${\begin{aligned}L=2.20\tan 51^{\circ }+0.80\tan 35.7^{\circ }=3.29\ {\rm {km}}.\end{aligned}}$

This should be $(4.00-0.60)=3.40$, so we must increase $\theta _{1}$ by a small amount; taking $\theta _{1}=52^{\circ }$ gives $\theta _{2}=36.2^{\circ }$ and $L=3.40\ {\rm {m}}$. We now find the traveltime:

${\begin{aligned}{\rm {time}}=(2.20/4.00\cos \theta _{1})+(0.800/3.00\cos \theta _{2})=1.224\ {\rm {s}}.\end{aligned}}$

For $h=1.20\ {\rm {km}}$, $L=2.80\ {\rm {km}}$. We know that $\theta _{1}$ will be less than for the shallower geophone, so we take $\theta _{1}=45^{\circ }$ and get $\theta _{2}=32.0^{\circ }$, $L=2.70$. Taking $\theta _{1}=47.0^{\circ }$, $\theta _{2}=33.3^{\circ }$, $L=2.88$. The change of $2^{\circ }$ increased $L$ by 0.18 m; we need an increase of 0.10, so we take $\theta _{1}=45^{\circ }+2^{\circ }\times (0.10/0.18)=46.1^{\circ }$, $\theta _{2}=32.7^{\circ }$, and $L=2.80\ {\rm {km}}$. The traveltime is

${\begin{aligned}{\rm {time}}=(2.20/4.00\cos 46.1^{\circ })+(0.800/3.00\cos 32.7^{\circ })=1.110\ {\rm {s}}.\end{aligned}}$

Problem 13.5b

What is the deepest geophone location for which the reflection can be recorded?

Solution

The deepest geophone position is that for which the ray passes through the intersection of the fault and reflector. For this point,

${\begin{aligned}\theta _{1}=\tan ^{-1}(2.00/2.20)=42.3^{\circ },\quad \theta _{2}=30.3^{\circ },\\L=2.20\tan 42.3^{\circ }+0.80\tan 30.3^{\circ }=2.47\ {\rm {km}},\\(400-h)=2.47,\quad h=1.53\ {\rm {km}}.\end{aligned}}$

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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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