# Parallelism of half-intercept and delay-time curves

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Series Geophysical References Series Problems in Exploration Seismology and their Solutions Lloyd P. Geldart and Robert E. Sheriff 11 415 - 468 http://dx.doi.org/10.1190/1.9781560801733 ISBN 9781560801153 SEG Online Store

## Problem 11.10

Prove that a half-intercept curve is parallel to the curve of the total delay time ${\displaystyle \delta }$ (see Figure 11.10a).

### Solution

Referring to Figure 11.10a, we can write

{\displaystyle {\begin{aligned}t_{i}/2=h\left(\cos \theta _{c}/V_{1}\right)\end{aligned}}}

Figure 11.10a.  Delay-time and half-intercept curves.

[see equations (11.9b)]. Thus ${\displaystyle t_{i}/2}$ is a linear function of ${\displaystyle h}$ with slope ${\displaystyle \left(\cos \theta _{c}/V_{1}\right)}$. The total delay time is

{\displaystyle {\begin{aligned}\delta =\delta _{s}+\delta _{g},\end{aligned}}}

${\displaystyle \delta _{s}}$ being constant. If we substitute ${\displaystyle N=h\tan \theta _{c}}$ (see Figure 11.9a) in equation (11.8a), we obtain the result

{\displaystyle {\begin{aligned}\delta _{g}=h\left(\cos \theta _{c}/V_{1}\right),\\{\hbox{so}}\quad \quad \delta =\delta _{s}+h\left(\cos \theta _{c}/V_{1}\right).\end{aligned}}}

Thus the total delay-time curve is parallel to the half-intercept time curve and lies above it the distance ${\displaystyle \delta _{s}}$.