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 3.11

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 ${\displaystyle \theta =}$ ${\displaystyle \tan ^{-1}\left(x/2h\right)}$, where ${\displaystyle x}$ is the offset and ${\displaystyle h}$ the depth. We can calculate ${\displaystyle \theta }$ for a series of ${\displaystyle \left(x,\;t\right)}$ values and then stretch the scale to make it linear in ${\displaystyle \theta }$. The amount of stretch increases as ${\displaystyle x}$ and ${\displaystyle \theta }$ 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 ${\displaystyle \theta }$ as compared with the constant velocity case. Thus angles of incidence are often larger than assumed.

Figure 3.11a.  Curved raypaths.

Problem 3.11b

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

{\displaystyle {\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

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

For ${\displaystyle x=2.00}$ km, the angle of incidence ${\displaystyle \theta =\tan ^{-1}\left(x/2z\right)=\tan ^{-1}}$(2.00/4.00) ${\displaystyle =26.6^{\circ }}$. Similarly, for ${\displaystyle x=3.0}$ km, ${\displaystyle \theta =36.9^{\circ }}$

Table 3.11a Values for curved raypaths.
${\displaystyle \theta _{0}}$ ${\displaystyle \theta }$ ${\displaystyle 2{\textit {x}}(km)}$
20${\displaystyle ^{\circ }}$ 28.8${\displaystyle ^{\circ }}$ 1.82
25${\displaystyle ^{\circ }}$ 36.5${\displaystyle ^{\circ }}$ 2.37
30${\displaystyle ^{\circ }}$ 44.7${\displaystyle ^{\circ }}$ 3.04

Problem 3.11c

3.11c Repeat part (b) for curved raypaths.

Solution

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

{\displaystyle {\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 ${\displaystyle 2x=\left(2/pa\right)\left(\cos \theta _{0}-\cos \theta \right)=\left(2/0.0698\right)\left(\cos 20^{\circ }-\cos 28.8^{\circ }\right)}$

${\displaystyle \qquad \qquad =1.82\ {\rm {km}}.}$

We next take ${\displaystyle \theta _{0}=25^{0}}$, so ${\displaystyle pa=\left(\sin 25^{\circ }/2.20\right)\times 0.450=0.0864}$.

Then ${\displaystyle \sin \theta =paz+\sin \theta _{0}=0.0864\times 2.00+0.423=0.595,\\\theta =36.5^{\circ },}$

${\displaystyle 2x=\left(2/0.0864\right)\left(\cos 25^{\circ }-\cos 36.5^{\circ }\right)=2.37\ {\rm {km}}.}$

For ${\displaystyle \theta _{0}=30^{\circ }}$, we get ${\displaystyle \theta =44.7^{\circ },2x=3.04\ {\rm {km}}.}$

We could calculate offsets for intermediate values of ${\displaystyle \theta _{0}}$, e.g., for ${\displaystyle 22^{\circ }}$, 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 ${\displaystyle \theta }$ corresponding to ${\displaystyle 2x=2.00}$ and 3.00 km, the values 31.3${\displaystyle ^{\circ }}$ and 44.2${\displaystyle ^{\circ }}$ respectively. Comparing these with the results in part (b) of 26.6${\displaystyle ^{\circ }}$ and 36.9${\displaystyle ^{\circ }}$. 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.