# Properties of a coincident-time curve

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 11.12a

A coincident-time curve connects points where waves traveling by different paths arrive at the same time. In Figure 11.12a, the curve $AC$ is where the head wave and direct wave arrive simultaneously. On a vertical section through the source with constant-velocity above a refractor, head-wave wavefronts are parallel straight lines. In Figure 11.12b, show that the virtual wavefront $DE$ for $t=0$ is at a slant depth $SD=2h=2z\cos \theta _{c}$ .

### Background

Figure 11.12a shows first-arrival wavefronts at intervals of 0.1 s generated by the source $S$ for a three-layer situation where the velocities are in the ratio 2:3:4. The critical angle at the first interface is reached at $A$ , so head waves are generated to the right of this point, the wavefronts in the upper layer being straight lines that join with the direct wavefronts having the same traveltimes. The locus of the junction point where the first-arrival wavefronts abruptly change direction is a coincident time curve. $ABC$ is a coincident-time curve. In general a coincident-time curve (for example, DEFG) is the locus of the junction points where two wavefronts having the same traveltimes but have traveled different paths.

A curve that is equidistant from a fixed point and a fixed straight line is a parabola.

### Solution

In Figure 11.12a, the wave generated at $S$ at time $t=0$ arrives at $A$ at time $t_{A}=SA/V_{1}$ , the angle of incidence being the critical angle $\theta _{c}$ . Head waves traveling upwards at the critical angle are generated to the right of $A$ . We assume that a fictitious source generates plane wavefronts traveling parallel to the head-wave wavefronts with velocity $V_{1}$ , $DE$ being their position at $t=0$ . This wavefront arrives at $A$ at time $t_{A}$ so that $CD=SA$ . Hence,

{\begin{aligned}SD&=SC+CD=SC+SA\\&=SA\left(1+\cos 2\theta _{c}\right)\\&=2SA{\rm {cos}}^{2}\theta _{c}=2z\cos \theta _{c}.\end{aligned}} ## Problem 11.12b

Show that after $DE$ reaches $A$ , wavefronts such as $BF$ coincde with the head-wave wavefronts.

### Solution

If the wavefront $CA$ arrives at $FB$ at time $t_{A}+\Delta t$ , then $AF=V_{1}\Delta t$ . During the time $\Delta t$ , the headwave travels from $A$ to $B$ at velocity $V_{2}$ , that is, $AB=V_{2}\Delta t$ . Therefore $AF/AB=V_{1}/V_{2}=\sin \theta _{c}$ , so $BF$ parallels the refracted wavefronts.

## Problem 11.12c

Show that the coincident-time curve is a parabola.

### Solution

At any point on the coincident-time curve, the traveltime of the direct wave equals that of a wavefront coming from $DE$ . Since both wavefronts travel with the velocity $V_{1}$ , the point on the curve is equidistant from $S$ and from the straight line $DE$ , hence the curve is a parabola.

## Problem 11.12d

Show that, taking $DE$ and $DS$ as the $x$ and $y$ -axes, the equation of $AH$ is $4hy=x^{2}+4h^{2}$ , where $DS=2h$ .

### Solution

We take $H$ as $H\left(x,\;y\right)$ and $S\left(0,2h\right)$ . We know from part (c) that $SH=$ distance from $H$ to the line $DE$ . The squares of these distances are also equal, so

{\begin{aligned}SH^{2}=HE^{2},\ {\hbox{that is}},x^{2}+(y-2h)^{2}=y^{2},\\{\hbox{and}}\quad \quad x^{2}+4h^{2}=4hy.\end{aligned}} ## Problem 11.12e

Show that the coincident-time curve is tangent to the refractor at $A$ .

### Solution

We must show that the coincident-time curve passes through $A$ with the same slope as the refractor. Obviously the curve starts at $A$ because the head wave starts at the instant the direct wave reaches $A$ . We use the equation of the curve in part (d) to get the slope and then substitute the coordinates of $A$ . Thus,

{\begin{aligned}x^{2}+4h^{2}=4hy,\qquad dy/dx=x/2h.\end{aligned}} The $x$ -coordinate of $A$ is

{\begin{aligned}x_{A}&=AS\sin 2\theta _{c}=\left(z/\cos \theta _{c}\right)\sin 2\theta _{c}=2z\sin \theta _{c}\\&=\left(2h/\cos \theta _{c}\right)\sin \theta _{c}=2h\tan \theta _{c},\end{aligned}} where we used the result in (a) in the last step. Substitution in $dy/dx$ gives the slope $\tan \theta _{c}$ which is the same as the refractor slope. Therefore, the coincident-time curve is tangent to the refractor at $A$ .