# Traveltime curves for various events

Series Geophysical References Series Problems in Exploration Seismology and their Solutions Lloyd P. Geldart and Robert E. Sheriff 6 181 - 220 http://dx.doi.org/10.1190/1.9781560801733 ISBN 9781560801153 SEG Online Store

## Problem 6.14

Draw arrival-time curves for the five events in Figure 6.14a.

### Solution

We have for the depth to the mesa, 1900 m; height of mesa, 900 m. The traveltime curves were obtained graphically. We let $R$ stand for receiver locations.

For the reflected diffraction from $S_{1}$ (diffracted at A), the virtual source (see problem 4.1) for the event is $I_{1}$ in Figure 6.14b(i) (note that traveltime increases upward), so that

{\begin{aligned}t=(S_{1}A+I_{1}R)/V_{1}=(2.20+I_{1}R)/2.00.\end{aligned}} For the reflection from $S_{2}$ , we use the virtual source $I_{2}$ . We will also have a diffraction from the $S_{2}$ source (paths not shown).

For the reflected refraction from $S_{3}$ (reflected at C), we find two traveltimes and then draw a straight line through them.

{\begin{aligned}{\hbox{At}}\ S_{3},t=2(2.20/2.00+2.20/3.64)=2.80{\hbox{s}}.\\{\hbox{At}}\ S_{4},t=2(2.20/2.00)+1.10/3.64=2.50{\hbox{s}}.\end{aligned}} For the diffraction at $C$ from $S_{4}$ ,

{\begin{aligned}t=(S_{4}C+CR)/2.00=(2.20+CR)/2.00.\end{aligned}} For the diffracted reflection from $S_{5}$ (diffracted at C), we use the image point of $S_{5}$ (not shown) so that

{\begin{aligned}t=(I_{5}C+CR)/2.00,\end{aligned}} which gives the same curve as for the diffraction from $S_{4}$ except that it is displaced towards increased time by the difference in traveltimes for $S_{4}C$ and $I_{5}C$ .