Resolution of cross-dip

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Problem 4.9a

Sources and are, respectively, 600 m north and 500 m east of source . Traveltimes to zero-offset geophones at , , and for a certain reflection are 1.750, 1.825, and 1.796 s. What are the dip and strike of the horizon, the average velocity being 3.25 km/s?


The dip moveouts are s/km to the north and s/km to the east. We take the -axis east-west (see Figure 4.9a). If we apply equation (4.2b) as in problem 4.2c, we get

Figure 4.9a.  Geometry of dip measures.

Using equations (4.2g,h), we find that

Problem 4.9b

What are the changes in dip and strike if line AC has the bearing N80E?


We use equation (4.2j) and Figure 4.2e to obtain the dip moveout along the east-west -axis when the measured moveout along the -axis with bearing N80E is 0.092 s/km. Equation (4.2j) gives

Figure 4.9b.  Geometry with change of direction.


and the strike angle , that is, N60.0W. Thus, the change in line direction changes the dip about and the strike about .

Problem 4.9c

Solve part (b) graphically.


The graphical construction, shown reduced in Figure 4.9c, yields dip moveout of 0.145 s/km and strike of N61W.

The dip moveout 0.145 gives a dip 13.6. Thus the graphical solution gives almost exactly the same values as in part (b).

Figure 4.9c.  Graphical solution.

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