# Determining reflector location

Series Geophysical References Series Problems in Exploration Seismology and their Solutions Lloyd P. Geldart and Robert E. Sheriff 8 253 - 294 http://dx.doi.org/10.1190/1.9781560801733 ISBN 9781560801153 SEG Online Store

## Problem 8.20a

The arrival time of a reflection at the source point is 1.200 s, near-surface corrections having been applied. Determine the reflector depth and horizontal location with respect to the source point, assuming zero dip and that the average velocity associated with a vertical traveltime is ${\displaystyle V=2630\,\mathrm {m/s} .}$

### Background

After records have been picked, that is, after reflections have been identified and ${\displaystyle t_{0}}$ and ${\displaystyle \Delta t_{d}}$ measured, the next stage is to prepare a section displaying the reflection events in two-dimensions. Such a section can be prepared in several ways, one of which is by using a wavefront chart such as that shown in Figure 8.20a. A wavefront chart is a two-dimensional graph showing approximate wavefronts and raypaths for a given distribution of constant-velocity layers, Raypaths are found by starting with rays leaving the source at different angles and tracing them downward. Traveltimes to various points on the raypaths are calculated and contoured to show wavefronts. This assumes that waves started from, and returned to, the source, hence they must have been reflected by a bed perpendicular to the ray (parallel to a wavefront); thus the dip as well as the location of the reflector is determined. The reflection event denoted by the symbol -o- in Figure 8.20a corresponds to ${\displaystyle t_{0}=2.350\,{\rm {s}},}$ ${\displaystyle \Delta t_{d}/\Delta x=}$ 110 ms/km.

Figure 8.20a.  Wavefront chart.

### Solution

We are given ${\displaystyle t_{0}=1.200\,\mathrm {s} ,}$ ${\displaystyle V=2630\,{\rm {m/s}}}$; hence ${\displaystyle h=Vt_{0}/2=1580\,\mathrm {m} }$ vertical depth, and the location is directly below the source.

## Problem 8.20b

Determine the reflector depth, dip, and horizontal location assuming that the dip moveout is 0.150 s/km and that the line is normal to strike.

### Solution

As in (a) ${\displaystyle h=1580\ {\hbox{m}}}$, but now it is slant depth;

{\displaystyle {\begin{aligned}\mathrm {dip\ moveout} &=\Delta t_{d}/\Delta x=0.150/1000=0.000150\,\mathrm {s/m} ,\\\xi &={\sin }^{-1}\left({\frac {1}{2}}\times 2630\times 0.000150\right)=11.4^{\circ };\\\mathrm {vertical\ depth} &=1580\,\cos 11.4^{\circ }=1550\,\mathrm {m} ;\\\mathrm {horizontal\ up-dip\ displacement} &=1580\sin \,11.4^{\circ }=310\,\mathrm {m} .\end{aligned}}}

## Problem 8.20c

Determine the depth, dip, and horizontal location assuming straight-line travel at the angle of approach and ${\displaystyle V_{H}=1830\,\mathrm {m/s} }$.

### Solution

Because the path is a straight line, the velocity must be constant and we assume it is the starting velocity of 1830 m/s. Then,

{\displaystyle {\begin{aligned}h&={\frac {1}{2}}\times 1830\times 1.200=1100\,\mathrm {m} =\mathrm {slantdepth} ;\\\xi &={\sin }^{-1}\left({\frac {1}{2}}\times 1830\times 0.150/1000\right)=7.9^{\circ };\\\mathrm {vertical\ depth} &=1100\,\cos \,7.9^{\circ }=1090\,\mathrm {m} ;\\\mathrm {horizontal\ displacement} &=1100\sin 7.9^{\circ }=150\,\mathrm {m} .\end{aligned}}}

## Problem 8.20d

Determine the reflector depth, dip, and horizontal location assuming straight-line travel at the local velocity above the reflector, 3840 m/s.

### Solution

{\displaystyle {\begin{aligned}h&=3840\times 1.200/2=2300\,\mathrm {m} =\mathrm {slantdepth} ;\\\xi &=\sin ^{-1}\left({\frac {1}{2}}\times 3840\times 0.150/1000\right)=16.7^{\circ };\\\mathrm {vertical\ depth} &=2300\cos \,16.7^{\circ }=2200\mathrm {m} ;\\\mathrm {horizon\ displacement} &=2300\sin 16.7^{\circ }=660\mathrm {m} .\end{aligned}}}

## Problem 8.20e

Assume that the migrated position is determined from the wavefront chart in Figure 8.20d.

### Solution

Figure 8.20a gives 1200 m vertical depth, 170 m horizontal displacement, and ${\displaystyle \xi =12^{\circ }}$ dip.

Results are summarized in Table 8.20e, z being the vertical depth and ${\displaystyle x}$ the horizontal displacement.

Table 8.20e. Summary of results.
${\displaystyle V\left(m/s\right)}$ dip ${\displaystyle z\left(m\right)}$ ${\displaystyle x\left(m\right)}$
a) Vertical path 2630 ${\displaystyle 0^{\circ }}$ 1580 0
b) Dip moveout 150 ms/km 2630 ${\displaystyle 11.4^{\circ }}$ 1550 310
c) At approach angle 1830 ${\displaystyle 7.9^{\circ }}$ 1090 150
d) At local velocity 3840 ${\displaystyle 16.7^{\circ }}$ 2200 660
e) Curved raypath ${\displaystyle 12^{\circ }}$ 1200 170