# Calculation of reflector depths and dips

Series Geophysical References Series Problems in Exploration Seismology and their Solutions Lloyd P. Geldart and Robert E. Sheriff 4 79 - 140 http://dx.doi.org/10.1190/1.9781560801733 ISBN 9781560801153 SEG Online Store

## Problem 4.6

In Figure 4.6a assume that the depth to 1.0 s is 1500 m and that the interval velocity between 1.0 and 1.4 s is 3300 m/s, and the trace spacing is 100 m. Calculate the depths and dips of the four picked reflectors.

### Solution

There are 126 trace intervals corresponding to 12.6 km between the end traces. We have timed four reflections (on an enlargement) at the end traces to the nearest millisecond, giving the depths:

{\begin{aligned}A&=1.075\ \mathrm {s} \to 1500+(3300/2)\times 0.075=1624\ \mathrm {m} ;\quad A'=1.045\ \mathrm {s} \to 1574\ \mathrm {m} ;\\B&=1.200\ \mathrm {s} \to 1830\ \mathrm {m} ;\quad B'=1.184\ \mathrm {s} \to 1804\ \mathrm {m} ;\\C&=1.308\ \mathrm {s} \to 2008\ \mathrm {m} ;\quad C'=1.302\ \mathrm {s} \to 1998\ \mathrm {m} ;\\D&=1.347\ \mathrm {s} \to 2073\ \mathrm {m} ;\quad D'=1.347\ \mathrm {s} \to 2074\ \mathrm {m} .\end{aligned}} The dips of the four reflections are given by tan $\xi =\Delta z/\Delta x$ :

{\begin{aligned}\tan \xi _{\rm {A}}&=(1624-1574)/12.60=3.97\ \mathrm {m/km} ,\qquad \xi _{\rm {A}}=0^{\circ }14';\\\tan \xi _{\rm {B}}&=2.06\ \mathrm {m/km} ,\qquad \xi _{\rm {B}}=0^{\circ }7';\\\tan \xi _{\rm {C}}&=0.79\ \mathrm {m/km} ,\qquad \xi _{\rm {C}}=0^{\circ }3';\\\tan \xi _{\rm {D}}&=-0.08\ \mathrm {m/km} ,\qquad \xi _{\rm {D}}=0^{\circ }0'.\end{aligned}} The section is thickening to the right. Although the seismic reflections are nearly flat, the small dip can be measured.

Note that all four reflections have slight bending and changes in character about one-third of the way from the left end, suggesting that something unresolvable is happening here, perhaps a very small fault.