# Diffraction traveltime curves

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 6.5a

Show that the slope of the diffraction curve with source $S_{2}$ in Figure 6.5a(i) approaches $\pm 1/V$ for large $x$ .

### Solution

The diffraction path is $S_{2}AG_{2}$ in Figure 6.5a(i), so the traveltime curve is

{\begin{aligned}t_{d}=h/V+(x^{2}+h^{2})^{1/2}/V=h/V+(x/V)[1+(h/x)^{2}]^{1/2}.\end{aligned}} For $x\gg h$ , the equation of the curve becomes

{\begin{aligned}t_{d}\approx x/V+h/V\end{aligned}} which is a straight line with slope $+1/V$ for $x>0$ ,$-1/V$ for $x<0$ . The traveltime approaches these asymptotes as $x\to \pm \infty .$ ### Alternative solution

The slope of the traveltime curve is

{\begin{aligned}{\frac {{\rm {d}}t_{d}}{{\rm {d}}x}}={\frac {x}{V(x^{2}+h^{2})^{1/2}}}=\pm {\frac {1}{V}}[1+(h/x)^{2}]^{-1/2}.\end{aligned}} For $|x|\gg h$ , the slope is $\pm 1/V$ as before.

## Problem 6.5b

What is the asymptote slope for a coincident source-receiver?

### Solution

The traveltime curve for Figure 6.5a(ii) is given by

{\begin{aligned}t_{d}=(2/V)(x^{2}+h^{2})^{1/2}=(\pm 2x/V)[1+(h/x)^{2}]^{1/2}\end{aligned}} {\begin{aligned}\approx (\pm \ 2x/V)[1+{\frac {1}{2}}(h/x_{1})^{2}].\end{aligned}} As $|x|$ increases, $t_{d}\to \pm 2x/V$ . The asymptote has the equation $t_{d}=\pm 2x/V$ , which is a straight line with slope $\pm 2/V.$ 