Reflector dip in terms of traveltimes squared

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Problem 4.4a

Using the dip-moveout equation (4.2b) and the result of problem 4.3, verify that

where , , traveltime for path , traveltime to receiver at (Figure 4.4a).

Solution

Since in problem 4.3 equals here, we have

From equation (4.2b) we get

Figure 4.4a.  Geometry for dipping bed.

Problem 4.4b

Using equation (4.2a), show that

Solution

Equation (4.2a) gives

Problem 4.4c

Under what circumstances is the result for part (b) the same as equation (4.2b) and also consistent with part (a)?

Solution

The result in part (b) can be written

The expression in the first curly brackets is the same as the right-hand side of equation (4.2b). Hence, for the above to be the same as equation (4.2b), we must have that is,

In part (a) we got by finding and . The expression for involves no approximation, so the only approximation is that used in the derivation of equation (4.2b) to get . Thus, results in (a) and (c) are consistent provided we take .

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