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Source line and receiver line interval

In this case the desired fold for a 10 m x 10 m bin is 24. This fold can be decomposed as, say, 6 inline and 4 cross-line. Since the fold cross-line is half the number of receiver lines in the recording patch, this means the recording patch will have 8 receiver lines. The distance between the receiver lines drl can be computed as  
 \begin{displaymath}
drl=\frac{2*h_{\mbox{xmax}}}{nrl-1}\end{displaymath} (3)
where $h_{\mbox{xmax}}$ is the maximum offset cross-line, 3000 m in this case, and nrl is the number of receiver lines per patch, 8 in this case. This gives drl=400 m.

The source line interval (dsl) can be similarly computed from the expectation that the inline fold be 6.  
 \begin{displaymath}
Fold_x=\frac{nchl\Delta x}{2dsl}\end{displaymath} (4)
Here nchl is the number of channels per receiver line, 300 in this case (maximum inline offset of 3000 m and receiver interval 20 m), $\Delta x$is the receiver interval, 20 m, and Foldx is 6. This gives dsl=500 m. If we wanted to apply the theory of symmetric wavefield sampling we might set dsl=400 m. Here I will not use the symmetric wavefield sampling theory.


next up previous print clean
Next: Salvo Up: STANDARD GEOMETRY Previous: Maximum offset inline and
Stanford Exploration Project
7/8/2003