Resolution of cross-dip
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
Using equations (4.2g,h), we find that
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
and the strike angle , that is, N60.0W. Thus, the change in line direction changes the dip about and the strike about .
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).
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Also in this chapter
- Accuracy of normal-moveout calculations
- Dip, cross-dip, and angle of approach
- Relationship for a dipping bed
- Reflector dip in terms of traveltimes squared
- Second approximation for dip moveout
- Calculation of reflector depths and dips
- Plotting raypaths for primary and multiple reflections
- Effect of migration on plotted reflector locations
- Resolution of cross-dip
- Variation of reflection point with offset
- Functional fits for velocity-depth data
- Relation between average and rms velocities
- Vertical depth calculations using velocity functions
- Depth and dip calculations using velocity functions
- Weathering corrections and dip/depth calculations
- Using a velocity function linear with depth
- Head waves (refractions) and effect of hidden layer
- Interpretation of sonobuoy data
- Diving waves
- Linear increase in velocity above a refractor
- Time-distance curves for various situations
- Locating the bottom of a borehole
- Two-layer refraction problem