# AVO versus AVA and effect of velocity gradient

Series Geophysical References Series Problems in Exploration Seismology and their Solutions Lloyd P. Geldart and Robert E. Sheriff 3 47 - 77 http://dx.doi.org/10.1190/1.9781560801733 ISBN 9781560801153 SEG Online Store

## Problem 3.11a

How would you recalibrate the scale to change a plot showing amplitude variation with offset (AVO) into a plot of amplitude variation with angle (AVA)? What will be the effect if velocity increases with depth?

### Solution

Assuming constant velocity above a horizontal reflector, the angle of incidence $\theta =$ $\tan ^{-1}\left(x/2h\right)$ , where $x$ is the offset and $h$ the depth. We can calculate $\theta$ for a series of $\left(x,\;t\right)$ values and then stretch the scale to make it linear in $\theta$ . The amount of stretch increases as $x$ and $\theta$ increase.

If the velocity increases with depth, straight raypaths become curved as shown by the solid lines in Figure 3.11a. The effect is to increase $\theta$ as compared with the constant velocity case. Thus angles of incidence are often larger than assumed.

## Problem 3.11b

Calculate the angles of incidence on a reflector at a depth of 2.00 km at offsets of 2.00 and 3.00 km where the velocity increases linearly with depth from 2.20 km/s at the surface to 3.10 km/s at the reflector depth. Assume straightray travel at the average velocity.

### Solution

The average velocity is ${\bar {V}}=2z/t_{0}$ , where $z$ is the depth and $t_{0}$ is the traveltime at the source. Since $V=V_{0}+az$ , $a=\left(3.10-2.20\right)/2.00=0.450\ s^{-1}$ . Then,

{\begin{aligned}t_{0}=2\mathop {\int } \nolimits _{0}^{z=2.00}{\frac {\rm {dz}}{2.20+0.450{\rm {z}}}}=\left(2/0.450\right){\rm {ln}}\ (2.20+0.450z)|_{0}^{2.00}\\=4.44\ln \left({\frac {2.20+0.450\times 2.00}{2.20}}\right)=4.44\ {\rm {ln}}\ \left(1.41\right)=4.44\times 0.344\\=1.53\ {\rm {s}},\end{aligned}} so

 {\begin{aligned}{\bar {V}}=2.00\times 2.00/1.53=2.61\ {\rm {km/s}}.\end{aligned}} (2.1c)

For $x=2.00$ km, the angle of incidence $\theta =\tan ^{-1}\left(x/2z\right)=\tan ^{-1}$ (2.00/4.00) $=26.6^{\circ }$ . Similarly, for $x=3.0$ km, $\theta =36.9^{\circ }$ Table 3.11a Values for curved raypaths.
$\theta _{0}$ $\theta$ $2{\textit {x}}(km)$ 20$^{\circ }$ 28.8$^{\circ }$ 1.82
25$^{\circ }$ 36.5$^{\circ }$ 2.37
30$^{\circ }$ 44.7$^{\circ }$ 3.04

## Problem 3.11c

Repeat part (b) for curved raypaths.

### Solution

We are given $V_{0}$ , $a$ , $z$ (which is equivalent to $V$ ) and must find the values of $\theta$ for the different offsets. Solving equations $\left(4.17b,f\right)$ to determine both $\theta$ and $\theta _{0}$ is not practical, so we assume values of $\theta _{0}$ , then use equation (4.17f) to obtain $\theta$ , and equation (4.17b) to find the offset $2x$ , and then by trial we determine $\theta _{0}$ . The results of part (b) suggest that we start with $\theta _{0}=20^{\circ }$ . Then, $p=\sin \theta _{0}/V_{0}=\sin 20^{\circ }/2.20=0.155$ , $pa=0.155\times 0.450=0.0698$ . From equation (4.17f) we have

{\begin{aligned}2.00=\left(\sin \theta -\sin 20^{\circ }\right)/0.0698,\;\sin \theta =0.140+0.342,\;\theta =28.8^{\circ }\end{aligned}} The offset is $2x=\left(2/pa\right)\left(\cos \theta _{0}-\cos \theta \right)=\left(2/0.0698\right)\left(\cos 20^{\circ }-\cos 28.8^{\circ }\right)$ $\qquad \qquad =1.82\ {\rm {km}}.$ We next take $\theta _{0}=25^{0}$ , so $pa=\left(\sin 25^{\circ }/2.20\right)\times 0.450=0.0864$ .

Then $\sin \theta =paz+\sin \theta _{0}=0.0864\times 2.00+0.423=0.595,\\\theta =36.5^{\circ },$ $2x=\left(2/0.0864\right)\left(\cos 25^{\circ }-\cos 36.5^{\circ }\right)=2.37\ {\rm {km}}.$ For $\theta _{0}=30^{\circ }$ , we get $\theta =44.7^{\circ },2x=3.04\ {\rm {km}}.$ We could calculate offsets for intermediate values of $\theta _{0}$ , e.g., for $22^{\circ }$ , to get more accuracy. However, instead we shall interpolate between these pairs of values. We tabulate the values in Table 3.11a.

Interpretation gives, for the angles of incidence $\theta$ corresponding to $2x=2.00$ and 3.00 km, the values 31.3$^{\circ }$ and 44.2$^{\circ }$ respectively. Comparing these with the results in part (b) of 26.6$^{\circ }$ and 36.9$^{\circ }$ . We see that the angles of incidence for the curved raypaths are 18% and 20% greater than those for the straight-line paths, the difference increasing with offset.