# Diffraction traveltime curves

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
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 ${\displaystyle S_{2}}$ in Figure 6.5a(i) approaches ${\displaystyle \pm 1/V}$ for large ${\displaystyle x}$.

### Solution

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

{\displaystyle {\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 ${\displaystyle x\gg h}$, the equation of the curve becomes

{\displaystyle {\begin{aligned}t_{d}\approx x/V+h/V\end{aligned}}}

which is a straight line with slope ${\displaystyle +1/V}$ for ${\displaystyle x>0}$,${\displaystyle -1/V}$ for ${\displaystyle x<0}$. The traveltime approaches these asymptotes as ${\displaystyle x\to \pm \infty .}$

Figure 6.5a.  Diffraction traveltime curves.

### Alternative solution

The slope of the traveltime curve is

{\displaystyle {\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 ${\displaystyle |x|\gg h}$, the slope is ${\displaystyle \pm 1/V}$ 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

{\displaystyle {\begin{aligned}t_{d}=(2/V)(x^{2}+h^{2})^{1/2}=(\pm 2x/V)[1+(h/x)^{2}]^{1/2}\end{aligned}}}

{\displaystyle {\begin{aligned}\approx (\pm \ 2x/V)[1+{\frac {1}{2}}(h/x_{1})^{2}].\end{aligned}}}

As ${\displaystyle |x|}$ increases, ${\displaystyle t_{d}\to \pm 2x/V}$. The asymptote has the equation ${\displaystyle t_{d}=\pm 2x/V}$, which is a straight line with slope ${\displaystyle \pm 2/V.}$