# Variation of reflectivity with angle (AVA)

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 3.12a

The values in Table 3.12a illustrate the differences that the interstitial fluid can produce. Calculate the reflectivity for shale-brine sand and shale/gas sand interfaces at incident angles of $0^{\circ }$ , $10^{\circ }$ , $20^{\circ }$ , $30^{\circ }$ , and $40^{\circ }$ .

### Background

It is difficult to tell from the Zoeppritz equations how the variation of amplitude with angle of incidence is affected by changes in the various parameters involved. Shuey (1985) simplified the equations by assuming that the changes in physical properties at an interface are small, so that the raypath bending is small, resulting in Shuey’s equation:

Table 3.12a. Values for AVA/AVO calculations.
Medium $V_{p}$ (m/s) $V_{s}$ (m/s) $p$ (g/cm$^{3}$ ) $V_{p}/V_{s}$ Shale 2742 1394 2.062 1.967
Brine sand 2833 1470 2.078 1.927
Gas sand 2371 1473 2.044 1.610

 {\begin{aligned}R\left(\theta \right)=R_{0}+[PR_{0}+\Delta \sigma /\left(1-\sigma )^{2}\right]\sin ^{2}\theta +\left(\Delta V_{P}/2V_{P}\right)\left(\tan ^{2}\theta -\sin ^{2}\theta \right),\end{aligned}} (3.12a)

where $R_{0}=\left(Z_{2}-Z_{-1}\right)/\left(Z_{2}+Z_{1}\right)\approx \Delta \left(\rho V\right)/2\rho V=\left(1/2\right)\left(\Delta V/V+\Delta \rho /\rho \right),$ {\begin{aligned}\quad P=Q-2\left(1+Q\right)\left(1-2\sigma \right)/\left(1-\sigma \right)+\Delta \sigma /R_{0}(1-\sigma )^{2}\end{aligned}} (3.12b)

 {\begin{aligned}Q=\left(\Delta V_{P}/V_{P}\right)/\left[\left(\Delta V_{P}/V_{P}\right)+\left(\Delta \rho /\rho \right)\right]\end{aligned}} (3.12c)

and $\sigma$ is Poisson’s ratio.

Hilterman (1989) introduced additional approximations resulting in

 {\begin{aligned}R=R_{0}\cos ^{2}\theta +2.25\Delta \sigma \sin ^{2}\theta =R_{0}\cos ^{2}\theta +2.25\Delta \sigma (1-\cos ^{2}\theta )\\=R_{0}\left(1-2.25\Delta \sigma \right)\cos ^{2}\theta +2.25\Delta \sigma .\end{aligned}} (3.12d)

### Solution

Note that 4 significant figures are required to illustrate the effect. We first calculate $\sigma$ for the three beds using equation (10,2) in Table 2.2a:

{\begin{aligned}\sigma _{\rm {shale}}=\left(1.967^{2}-2\right)/2\left(1.967^{2}-1\right)=1.869/5.738=0.326,\\\sigma _{\rm {brine}}=\left(1.927^{2}-2\right)/2\left(1.927^{2}-1\right)=1.713/5.427=0.316,\\\sigma _{\rm {gas}}=\left(1.610^{2}-2\right)/2\left(1.610^{2}-1\right)=0.592/3.184=0.186.\end{aligned}} We take the following average values and increments : $V_{P}=2788$ , $\sigma =0.321$ , $\rho =2.070$ , $\Delta V_{P}=91,\Delta \sigma =-0.010,\Delta \rho =0.016.$ Using these values for the Shuey equation for the shale/brine-sand interface, equations (3.12a,b,c) give

{\begin{aligned}R_{0}=\left(1/2\right)\left(91/2788+0.016/2.070\right)=0.0202,\\Q=\left(91/2788\right)/\left[\left(91/2788\right)+\left(0.016/2.07\right)\right]\\=0.0326/\left(0.0326+0.0077\right)=0.0326/0.0403=0.809,\\P=0.809-2\left(1.809\right)\left(0.358/0.679\right)-0.010/0.0202\times 0.679^{2}\\=0.809-1.908-1.074=-2.173,\\R=0.0202+\left(-2.173\times 0.0202-0.010/0.679^{2}\right)\sin ^{2}\theta \\+\left(91/2\times 2788\right)\left(\tan ^{2}\theta -\sin ^{2}\theta \right)\\=0.0202-\left(0.0439+0.0217\right)\sin ^{2}\theta +0.0163\left(\tan ^{2}\theta -\sin ^{2}\theta \right)\\=0.0202-0.0656\sin ^{2}\theta +0.0163\left(\tan ^{2}\theta -\sin ^{2}\theta \right)\\=0.0202-0.0819\sin ^{2}\theta +0.0163\tan ^{2}\theta ,\\R_{10}=0.0202-0.0819\times 0.1736^{2}+0.0163\times 0.1763^{2}=0.0182,\\R_{20}=0.0202-0.0819\times 0.3420^{2}+0.0163\times 0.3640^{2}=0.0128,\\R_{30}=0.0202-0.0819\times 0.5000^{2}+0.0163\times 0.5774^{2}=0.0052,\\R_{40}=0.0202-0.0819\times 0.6428^{2}+0.0163\times 0.8391^{2}=0.0022.\end{aligned}} At the shale-gas sand interface, averages and increments are $V_{P}=2556{\rm {m/s}}$ , $\sigma =0.256$ , $\rho =2.053$ , $\Delta V_{P}=-371{\rm {m/s}}$ , $\Delta \sigma =-0.140$ , $\Delta \rho =-0.018.$ {\begin{aligned}{\text{Then}},R_{0}=-{\frac {1}{2}}\left({\frac {371}{2556}}+{\frac {0.016}{2.053}}\right)=-0.0765,\\Q=\left(-371/2556\right)/\left[\left(-371/2556\right)+\left(-0.018/2.053\right)\right]\\=0.1451/\left(0.1451+0.0088\right)=0.943,\\P=0.943-2\left(1.943\times 0.488/0.744\right)+(-0.140/\left(-0.0765\times 0.744^{2}\right)\\=0.943-2.549+3.306=1.700.\\{\text{Thus}},R=-0.0765+\left(-1.700\times 0.0765-0.140/0.744^{2}\right)\sin ^{2}\theta \\+\left(-371/2\times 2556\right)\left(\tan ^{2}\theta -\sin ^{2}\theta \right)\\=-0.0765-0.3830\sin ^{2}\theta -0.0726\left(\tan ^{2}\theta -\sin ^{2}\theta \right)\\=-0.0765-0.3104\sin ^{2}\theta -0.0726\tan ^{2}\theta .\\{\text{So}},R_{10}=-0.0765-0.0094-0.0023=-0.0882,\\R_{20}=-0.0765-0.0363-0.0096=-0.1221,\\R_{30}=-0.0765-0.0776-0.0242=-0.1783,\\R_{40}=-0.0765-0.1282-0.0511=-0.2558.\end{aligned}} Substituting $R_{0}=0.0202$ , $\Delta \sigma =-0.010$ in equation (3.12d), we get for the Hilterman equation (3.12d) for the shale/brine-interface,

Table 3.12b. Comparison of predictions by Shuey and Hilterman equations.
$0^{\circ }$ $10^{\circ }$ $20^{\circ }$ $30^{\circ }$ $40^{\circ }$ For shale/brine sand
Shuey equation 0.0202 0.0182 0.0128 0.0052 –0.0022
Hilterman equation 0.0202 0.0189 0.0152 0.0095 0.0026
For the shale/gas sand
Shuey equation –0.0765 –0.0881 –0.1221 –0.1782 –0.2558
Hilterman equation –0.0765 –0.0837 –0.1044 –0.1361 –0.1750 Figure 3.12a.  Reflectivity versus angle; solid line, Shuey equation; dashed, Hilterman equation.

{\begin{aligned}R=R_{0}\left(1-2.25\Delta \sigma \right)\cos ^{2}\theta +2.25\Delta \sigma \\=0.0202\left(1-0225\right)\cos ^{2}\theta +0.0225=0.0427\cos ^{2}\theta -0.0225;\\R_{0}=0.0202,\\R_{10}=0.0427\left(0.985^{2}\right)-0.0225=0.0189,\\R_{20}=0.0427\left(0.940^{2}\right)-0.0225=0.0152,\\R_{30}=0.0427\left(0.866^{2}\right)-0.0225=0.0095,\\R_{40}=0.0427\left(0.766^{2}\right)-0.0225=0.0026.\end{aligned}} The Hilterman equation (3.12d) for the shale/gas-sand interface is

{\begin{aligned}R=-0.2385\cos ^{2}\theta -0.315,\\R_{0}=-0.0765,\\R_{10}=-0.2385\left(0.985^{2}\right)-0.315=0.084,\\R_{20}=-0.2385\left(0.940^{2}\right)-0.315=0.105,\\R_{30}=-0.2385\left(0.866^{2}\right)-0.315=0.137,\\R_{40}=-0.2385\left(0.766^{2}\right)-0.315=0.175.\end{aligned}} Table 3.12b compares the values given by the Shuey and Hilterman equations and the results are graphed in Figure 3.12a.

The two equations give essentially the same results for angles up to $20^{\circ }$ . The increase of amplitude with angle (offset) is larger with the Shuey equation. An additional term that becomes important at large angles is sometimes added to these equations.