# Effect of velocity change on VSP traveltime

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 13.5a

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 $h$ .

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}} 