AVO versus AVA and effect of velocity gradient

Problems in Exploration Seismology and their Solutions

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

Figure 3.11a.  Curved raypaths.

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.

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Table of Contents (book)

Also in this chapter

• General form of Snell’s law
• Reflection/refraction at a solid/solid interface and displacement of a free surface
• Reflection/refraction at a liquid/solid interface
• Zoeppritz’s equations for incident SV- and SH-waves
• Reinforcement depth in marine recording
• Complex coefficient of reflection
• Reflection and transmission coefficients
• Amplitude/energy of reflections and multiples
• Reflection/transmission coefficients at small angles and magnitude
• Magnitude
• Variation of reflectivity with angle (AVA)

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