Effect of velocity change on VSP traveltime

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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 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 .

Figure 13.5a.  Geometry of the problem.

The angle of incidence and of refraction where the ray passes through the fault are related by Snell’s law: that is, . We have a further constraint in that the vertical distance between the image and the geophone is . These relations fix :

  1. ;
  2. .

We denote the left-hand side of (b) by . 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,

To allow for some bending, we try and get . Substituting theses values in relation (b) we get

This should be , so we must increase by a small amount; taking gives and . We now find the traveltime:

For , . We know that will be less than for the shallower geophone, so we take and get , . Taking , , . The change of increased by 0.18 m; we need an increase of 0.10, so we take , , and . The traveltime is

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,

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