Using a velocity function linear with depth

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Problem 4.17a

When the velocity increases linearly with depth according to the relation


(4.17a)

being constant, show that


(4.17b)


(4.17c)

where velocity, velocity at depth , source-geophone distance, raypath parameter, arrival time, and angle of incidence.

Background

When the velocity is a function of depth only, as in equation (4.17a), expressions for the offset (see problem 4.1) and traveltime can be found by dividing the medium into horizontal layers, each of infinitesimal thickness (Figure 4.17a) and then integrating. We have

Also, using equation (3.1a), we have

In the limit we get the following integrals for and :


(4.17d)

When the velocity increases monotonically with depth, a ray must eventually return to the surface (see Figure 4.20a). For horizontal velocity layering the raypaths are symmetrical about the deepest point.

Figure 4.17a.  Raypaths where .

Solution

In equation (4.17d) we substitute , from equation (4.17a), also . Thus, we get

Problem 4.17b

Show that the angle of incidence and the depth can be written


(4.17e)


(4.17f)

Solution

From equation (4.17c) we get

Solving equation (4.17a) for gives

Problem 4.17c

Given the velocity function km/s, find the depth and offset of the point of reflection and the reflector dip when s and s/km. What interpretation would you give the result?

Solution

First, we note that in the preceding equations is one-way time, so we take s, km/s, . To get , we find the angle of approach [see equation (4.2d)] for s/km:

The dip of the reflector equals , 50.9 in this case, because, in order for the ray to return to the source, it must have been incident on the reflector at right angles.

Because of the unusually steep dip, this event might be from a fault plane or it might represent a steeply dipping bed in an area of highly folded sediments.

Problem 4.17d

If the ray continued without reflection, when and where would it emerge? What moveout would be observed at the recording point? Calculate the maximum depth of penetration.

Solution

[Equations (4.20a,b,c) could be used here instead of the following.]

Because the path is symmetrical about the midpoint, the angle of approach at the receiver equals .

From equations (4.17b,c) we have

Because of symmetry, the moveout on emergence would be s/km. To find the maximum depth of penetration , we use the fact that the ray is traveling horizontally at the depth , that is, . From part (c) we have , so . Equation (4.17f) now gives

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