# Difference between revisions of "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. Figure 6.14b.  Arrival times of events. (i) Geometry; (ii) traveltimes.

### Solution

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

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

\displaystyle \begin{align} t=(S_{1} A+I_{1} R)/V_{1} =(2.20+I_{1} R)/2.00. \end{align}

For the reflection from $\displaystyle S_{2}$ , we use the virtual source $\displaystyle I_{2}$ . We will also have a diffraction from the $\displaystyle S_{2}$ source (paths not shown).

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

\displaystyle \begin{align} \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{align}

For the diffraction at $\displaystyle C$ from $\displaystyle S_{4}$ ,

\displaystyle \begin{align} t=(S_{4} C+CR)/2.00=(2.20+CR)/2.00. \end{align}

For the diffracted reflection from $\displaystyle S_{5}$ (diffracted at C), we use the image point of $\displaystyle S_{5}$ (not shown) so that

\displaystyle \begin{align} t=(I_{5} C+CR)/2.00, \end{align}

which gives the same curve as for the diffraction from $\displaystyle S_{4}$ except that it is displaced towards increased time by the difference in traveltimes for $\displaystyle S_{4}C$ and $\displaystyle I_{5}C$ .