# Effect of horizontal velocity gradient

Series Geophysical References Series Problems in Exploration Seismology and their Solutions Lloyd P. Geldart and Robert E. Sheriff 10 367 - 414 http://dx.doi.org/10.1190/1.9781560801733 ISBN 9781560801153 SEG Online Store

## Problem

In Figure 10.14a, $V_{1}=2.00\ {\rm {km/s}}$ , $V_{2}=4.00\ {\rm {km/s}}$ , the horizon dips $20^{\circ }$ , and the diffracting point $P$ is at a vertical depth of 2.00 km, the dipping horizon being midway between $P$ and $P'$ (at the surface). Compare the diffraction curve with what would have been observed on a CMP section if $V_{1}=V_{2}=3.00\ {\rm {km/s}}$ .

### Solution

The curve for the diffracting point below the dipping interface can be solved analytically or graphically by ray tracing. The crest of the diffraction is located beneath where the angle of approach to the surface is $90^{\circ }$ ; such a raypath is called an image ray. If we denote the angle between a ray at $P$ and the vertical as $\theta$ , then Snell’s law at the interface gives for the image ray the equation

{\begin{aligned}\sin \left(\theta +20^{\circ }\right)=\left(V_{2}/V_{1}\right)\sin 20^{\circ }=2\sin 20^{\circ }=0.684,\\{\rm {or}}\qquad \qquad \qquad \qquad \theta ={\sin }^{-1}0.684-20^{0}=23^{\circ }.\end{aligned}} The fact that an image ray approaches the surface vertically is employed in depth migration (see problem 10.16) to accommodate horizontal changes in velocity.

The diffraction curve (solid curve, Figure 10.14c) is not symmetrical and the crest is displaced about 500 m updip from $P$ . The right limb is nearly flat because the increased traveltimes at the high velocity is almost compensated for shorter travel distance at the low velocity. Figure 10.14b.  Ray tracing for the diffracting point $P$ . Figure 10.14c.  Diffraction curves. Solid curve is 2-layer case, dashed is constant velocity case.

The diffraction curve for the constant velocity case (dashed curve, Figure 10.14c) can be calculated using the equations

{\begin{aligned}\tan \theta =x/z=x/2.00,\\t=2z/(V\cos \theta )=4.00/(3.00\cos \theta )=1.33/\cos \theta ,\end{aligned}} where $\theta$ is the angle of approach at the surface.