# User:Ageary/Chapter 13

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

## 13.1 S-wave conversion in marine surveys

13.1a In a marine survey, the water depth is 100 m and a reflector is 3 km below the seafloor. Use Figure 13.1a to determine the optimum range of offsets for S-wave generation. Take Poisson’s ratio just below the seafloor as 0.35 and the P-wave velocity as 2.8 km/s. The velocity in sea water is 1.5 km/s.

Solution

Referring to Figure 13.1a(i), we see that the offset ${\displaystyle x}$ for the 3-km reflector is

 {\displaystyle {\begin{aligned}x=2(0.1\times \tan \theta _{1}+3\times \tan \delta _{2})=0.2\tan \theta _{1}+6\tan \delta _{2},\end{aligned}}} (13.1a)

where ${\displaystyle \theta _{1}}$ is the angle of incidence of the P-wave at the bottom of the water layer, and ${\displaystyle \delta _{2}}$ is the angle of refraction of the converted S-wave.

To find ${\displaystyle \delta _{2}}$ for a given ${\displaystyle \theta _{1}}$, we need the S-wave velocity ${\displaystyle \beta _{2}}$. Using equation (1,8) in Table 2.2a, we find that ${\displaystyle \beta /\alpha =0.48}$ for ${\displaystyle \sigma =0.35}$, so ${\displaystyle \beta _{2}=0.48\alpha _{2}=0.48\times 2.8=1.3\ {\rm {km/s}}}$. To get the angles of incidence ${\displaystyle \theta _{1}}$ and of refraction ${\displaystyle \delta _{2}}$, we have from equation (3.1a),

{\displaystyle {\begin{aligned}\sin \theta _{1}/1.5=\sin \delta _{2}/1.3\;,\;\delta _{2}={\rm {sin}}^{-1}\left(0.87\sin \theta _{1}\right).\end{aligned}}}

Figure 13.1a.  PSSP reflections generated by conversion at the sea floor (after Tatham and Stoffa, 1976). (i) Geometry; (ii) the conversion coefficient versus angle of incidence in the water; the curves are labeled with the seafloor P-wave velocity in km/s. [Note: Conversion coefficient = (amplitude of converted wave/amplitude of incident wave.)

The optimum angles of incidence ${\displaystyle \theta _{1}}$ for S-wave conversion for ${\displaystyle \alpha =2.8\ {\rm {km/s}}}$ are obtained by interpolating between the curves for ${\displaystyle \alpha =2.5}$ and 3.0 in Figure 13.1a(ii). This gives a range of about ${\displaystyle 38^{\circ }}$ to ${\displaystyle 80^{\circ }}$ for ${\displaystyle \theta _{1}}$. The corresponding values of ${\displaystyle \delta _{2}}$ are ${\displaystyle 32^{\circ }}$ and ${\displaystyle 59^{\circ }}$. Equation (13.1a) now gives offsets of 4 and 11 km for ${\displaystyle \theta _{1}=38^{\circ }}$ and ${\displaystyle 80^{\circ }}$.

13.1b Most marine S-wave surveys that wish to deal with S-waves utilize conversion at the reflector and recording with three-component geophones laid on the seafloor (OBC, ocean-bottom cables), so that only one mode conversion is involved. Assume a ray leaving an airgun source at ${\displaystyle 10^{\circ }}$ to the vertical in water 100 m deep, an increase in P-wave velocities from 1.5 km/s at the seafloor to 3.0 km/s at a reflector 3 km below the sea floor (average velocity in the sediments of 2.25 km/s). Take ${\displaystyle \sigma }$ in the sediments as 0.3. What will be the source-geophone offset?

Solution

The P-wave gives an incident angle at the reflector of ${\displaystyle \theta =\sin ^{-1}[(3.0/1.5)\sin 10^{\circ }]=20.3^{\circ }}$. Using equation 1,8 in Table 2.2a, the S-wave velocity here is about 0.53 ${\displaystyle \alpha }$ or 1.59 km/s. Hence the angle of reflection is ${\displaystyle \delta =\sin ^{-1}(0.53\sin 20.3^{\circ })=10.6^{\circ }}$. The average direction of the P-wave ray in the sediments is ${\displaystyle \sin ^{-1}[(2.25/1.5)\sin 10^{\circ }]=15.1^{\circ }}$. With constant ${\displaystyle \sigma }$ in the sediments, ${\displaystyle \beta /\alpha }$ is constant and, hence, ${\displaystyle \sin \delta /\sin \theta =0.53}$. Hence, the average direction of the S-wave is ${\displaystyle \sin ^{-1}(0.53\sin 15.1^{\circ }=7.9^{\circ })}$. The offset is thus

{\displaystyle {\begin{aligned}100\tan 10^{\circ }+3000\tan 15.1^{\circ }+3000\tan 7.9^{\circ }=1240\ {\rm {m}}.\end{aligned}}}

This is a very reasonable offset.

## 13.2 Equally inclined orthogonal geophones

Determine the inclination angles for three orthogonal geophones that are equally inclined to the vertical.

Solution

The three orthogonal geophones define a rectangular coordinate system with the ${\displaystyle x}$-, ${\displaystyle y}$-, and ${\displaystyle z}$-axes along the geophones axes. We take a straight line through the origin of this system such that it is equally inclined to each axis, and rotate the coordinate system so that the straight line is vertical. The direction cosines of the vertical line are ${\displaystyle (l,m,n)}$ and ${\displaystyle l=m=n}$. Because ${\displaystyle l^{2}+m^{2}+n^{2}=1}$ [Sheriff and Geldart, problem (15.9a)],

{\displaystyle {\begin{aligned}3l^{2}=1,\quad l=1{\sqrt {3}},\cos ^{-1}(1/{\sqrt {3}})=54.74^{\circ }.\end{aligned}}}

Thus the geophones must be inclined ${\displaystyle 54.74^{\circ }}$ to the vertical.

## 13.3 Guided (Channel) waves; Normal-mode propagation

13.3a In Figure 13.3a the first arrival (${\displaystyle f_{o}=40\ {\rm {Hz}}}$) has traveled at the velocity 2.7 km/s; find the water depth.

Background

A wave guide is a layer in which a wave can propagate with little loss of energy. In a water layer nearly total reflection occurs at both boundaries, at the surface because of the very large impedance contrast and at the bottom reflection beyond the critical angle. The phase is inverted at the surface because the reflectivity is nearly ${\displaystyle -1}$, but not at the seafloor (beyond the critical angle) until the angle becomes very large.

Figure 13.3b(i) shows waves bouncing back and forth in a wave guide. For certain incident angles ${\displaystyle \theta }$ and frequencies ${\displaystyle f}$, they interfere constructively. In Figure 13.3b(ii), ${\displaystyle AC}$ is a wavefront traveling upward at the angle ${\displaystyle \theta }$. The previous cycle of a parallel wavefront that passed through the position earlier followed paths such as EFGH and BDA and thus coincides with ${\displaystyle AC}$. Clearly ${\displaystyle (EF+FG+GH)=BD+DA}$. ${\displaystyle DA=h/\cos \theta }$, ${\displaystyle BD=DA\cos 2\theta }$, so ${\displaystyle (BD+DA)=2h/\cos \theta }$. Taking into account the phase reversal at the surface the condition for constructive interference is

{\displaystyle {\begin{aligned}(BD+DA)=2h\cos \theta =(2n+1)\lambda /2,\end{aligned}}}

 {\displaystyle {\begin{aligned}{\mbox{so}}\qquad \qquad h=(2n+1)\lambda /4\cos \theta .\end{aligned}}} (13.3a)

Writing ${\displaystyle V}$ for the phase velocity, the frequencies that are reinforced are

 {\displaystyle {\begin{aligned}f_{n}=V/\lambda =(2n+1)V/4h\cos \theta ,\quad n=0,1,2,\ldots .\end{aligned}}} (13.3b)

In addition to the upgoing waves parallel to ${\displaystyle AC}$ and ${\displaystyle A^{\prime }C^{\prime }}$ in Figure 13.3b(iii), a second downgoing set ${\displaystyle PQ}$ and ${\displaystyle P^{\prime }Q^{\prime }}$ will combine with the set ${\displaystyle AC}$ to build up the energy traveling in the direction ${\displaystyle RR^{\prime }}$ along the wave guide. Energy travels from ${\displaystyle R}$ to ${\displaystyle R^{\prime }}$ in the time that wavefront ${\displaystyle AC}$ moves to ${\displaystyle A^{\prime }C^{\prime }}$, so the phase velocity of the energy traveling along ${\displaystyle RR^{\prime }}$, ${\displaystyle V_{g}}$, is

 {\displaystyle {\begin{aligned}V_{g}=V/\sin \theta .\end{aligned}}} (13.3c)
Figure 13.3a.  Wave-layer channel wave, source at 4 km (after Clay and Medwin, 1977).
Figure 13.3b.  (i) Raypaths in a wave guide; (ii) showing reinforcement of reflection wavefronts; (iii) relation between phase and group velocities.

Since both ${\displaystyle V_{g}}$ and ${\displaystyle f}$ are functions of ${\displaystyle \theta }$, ${\displaystyle V_{g}}$ is dispersive with a group velocity ${\displaystyle U}$ given by equation (2.7a):

 {\displaystyle {\begin{aligned}U=V_{g}+f{\frac {{\rm {d}}V_{g}}{{\rm {d}}f}}.\end{aligned}}} (13.3d)

The derivative is always negative for a water channel wave (Figure 13.3c), so ${\displaystyle U.

If the wave-guide effect had been due to reflection beyond the critical angle at both boundaries, as with a low-velocity coal seam, the phase would not have been reversed at either boundary and equations (13.3a,b) would reduce to

 {\displaystyle {\begin{aligned}h=n\lambda /2\cos \theta ,\quad f_{n}=nV/2h\cos \theta .\end{aligned}}} (13.3e)

Solution

Equation (13.3c) gives

{\displaystyle {\begin{aligned}\sin \theta =V/V_{g}=1.5/2.7,\quad \theta =33.7^{\circ },\end{aligned}}}

and we use equations (13.3b,e) with ${\displaystyle n=0}$ to get ${\displaystyle h}$:

{\displaystyle {\begin{aligned}h=V/4f_{0}\cos \theta =1500/4\times 40\cos 33.7^{\circ }=11.3\ {\rm {m}}.\end{aligned}}}

Figure 13.3c.  Phase and group velocities ${\displaystyle (V,U)}$ versus normalized frequency where ${\displaystyle \alpha _{2}/\alpha _{1}=2/3}$, ${\displaystyle \sigma _{1}=0.5}$, ${\displaystyle \sigma _{2}=0.25}$, and ${\displaystyle \rho _{2}/\rho _{1}=2.5}$ (from Ewing, Jardetsky, and Press, 1957).

13.3b What frequency is reinforced when ${\displaystyle \theta =40^{\circ }}$?

Solution

We use equation (13.3b) with ${\displaystyle n=0}$, ${\displaystyle h=11.3\ {\rm {m}}}$, ${\displaystyle \theta =40^{\circ }}$:

{\displaystyle {\begin{aligned}f=V/4h\cos \theta =1500/4\times 11.3\cos 40^{\circ }=43.3\ {\rm {Hz}}.\end{aligned}}}

## 13.4 Vertical seismic profiling

13.4a (a) A source is offset 1000 m from a vertical well in which a geophone is suspended, and a horizontal reflector is present at a depth of 2000 m. Calculate the traveltimes for the reflection when the geophone is 800, 1200, and 1600 m deep if the velocity =3000 m/s

Solution

Using a coordinate system with origin at the source and the ${\displaystyle z}$-axis positive downward, the image point ${\displaystyle I}$ (see problem 4.1) in Figure 13.4a is at (0, 4000) and the geophone coordinates are (1000, ${\displaystyle z}$); hence,

{\displaystyle {\begin{aligned}t_{c}=[1000^{2}+(4000-z)^{2}]^{1/2}/3000.\end{aligned}}}

Figure 13.4a.  Raypaths for vertical profiling and a horizontal bed.

For 800 m depth ${\displaystyle t=1.118\ {\rm {s}}}$; for 1200 m, ${\displaystyle t=0.991\ {\rm {s}}}$; for 1600 m, ${\displaystyle t=0.867\ {\rm {s}}}$.

Figure 13.4b.  Raypaths for vertical profiling and a dipping bed.

13.4b Repeat for reflectors dipping ${\displaystyle \pm 7^{\circ }}$, where the reflector intersects the well at the same point as in part (a) (see Figure 13.4b).

Solution

i) Dip ${\displaystyle +7^{\circ }}$ down toward the well

{\displaystyle {\begin{aligned}a&={\hbox{vertical depth of reflector at source S}}\\&=(2000-1000\tan 7^{\circ })=1880\ {\rm {m}},\\b&={\hbox{slant depth of reflector}}=a\cos 7^{\circ }\\&=1870\ {\rm {m}},\\{\hbox{Coordiantes of image}}&=(-2{\hbox{b}}\sin 7^{\circ },2{\hbox{b}}\cos 7^{\circ })\\&=(-460,3710),\\{\hbox{time}}_{+}&=[(1000+460)^{2}+(3710-Z)^{2}]^{1/2}/3000.\end{aligned}}}

ii) Dip ${\displaystyle -7^{\circ }}$ down toward source ${\displaystyle S}$

{\displaystyle {\begin{aligned}a=(2000-1000\tan 7^{\circ })=2120\ {\rm {m}},b=2100\ {\rm {m}},\end{aligned}}}

Coordinates of image are (${\displaystyle +510}$, 4170),

{\displaystyle {\begin{aligned}{\hbox{time}}=[(1000+510)^{2}+(4170-z)^{2}]^{1/2}/3000.\end{aligned}}}

Table 13.4a shows calculated traveltimes for ${\displaystyle z=800}$, 1200, and 1600 m.

 Depth(m) ${\displaystyle \to }$ 800 1200 1600 ${\displaystyle {\rm {Dip}}=+7^{\circ }}$ time (s) ${\displaystyle \to }$ 1.09 0.97 0.86 ${\displaystyle {\rm {Dip}}=-7^{\circ }}$ time (s) ${\displaystyle \to }$ 1.14 1 0.87
Table 13.4b. Effect of well deviation on ${\displaystyle t_{a}}$.
${\displaystyle z(m)}$ ${\displaystyle x_{d}(m)}$ ${\displaystyle Z_{d}(m)}$ ${\displaystyle t_{d}(s)}$ ${\displaystyle \Delta t(s)}$
800 958 799 1.114 ${\displaystyle -0.004}$
1200 937 1198 0.985 ${\displaystyle -0.006}$
1600 916 1598 0.857 ${\displaystyle -0.010}$

13.4c By how much do the values in part (a) change if the well deviates by ${\displaystyle 3^{\circ }}$ towards the source (see Figure 13.4c)?

Solution

Well inclination changes the geophone coordiantes from (1000, ${\displaystyle z}$) where the in-hole depth ${\displaystyle z=800}$, 1200, 1600, to ${\displaystyle (x_{d},z_{d})}$ where

{\displaystyle {\begin{aligned}x_{d}=(1000-z\sin 3^{\circ }),\quad z_{d}=z\cos 3^{\circ },\end{aligned}}}

Figure 13.4c.  Vertical profiling in a deviated well.

where subscript ${\displaystyle d}$ denotes values for the deviated well. The image point remains at (0, 4000), so the traveltime ${\displaystyle t_{d}}$ is

 {\displaystyle {\begin{aligned}t_{d}=[x_{\rm {d}}^{2}+(4000-z_{\rm {d}})^{2}]^{1/2}/3000.\end{aligned}}} (13.4a)

Substituting the values of ${\displaystyle z}$, we get the results in Table 13.4b. ${\displaystyle \Delta t}$ is the difference in traveltime from that calculated in part (a).

## 13.5 Effect of velocity change on VSP traveltime

13.5a A source is offset 3.00 km west of a vertical well in which a geophone is suspended. There is a vertical north-south fault 0.80 km west of the well with velocities ${\displaystyle V=4.00}$ and 3.00 km/s west and east of the fault, respectively. A horizontal reflector is present west of the fault at a depth of 2.00 km. Find the reflection traveltime for geophone depths of 0.60 and 1.20 km.

Solution

This problem must be solved by trial-and-error methods, either graphical or numerical. We have chosen the latter, because it is easier, quicker, and more accurate than graphical methods.

Figure 13.5a shows the geometry. The horizontal reflector at a depth of 2 km has an image point 4.00 km below the source. A typical ray is shown going from the image point to a geophone located in the well at a depth ${\displaystyle h}$.

Figure 13.5a.  Geometry of the problem.

The angle of incidence ${\displaystyle \theta _{1}}$ and of refraction ${\displaystyle \theta _{2}}$ where the ray passes through the fault are related by Snell’s law: that is, ${\displaystyle \sin \theta _{2}=(3.00/4.00)\sin \theta _{1}=0.75\sin \theta _{1}}$. We have a further constraint in that the vertical distance between the image and the geophone is ${\displaystyle (4.00-h)}$. These relations fix ${\displaystyle h}$:

1. ${\displaystyle \sin \theta _{2}=0.75\sin \theta _{1}}$;
2. ${\displaystyle 2.20\tan \theta _{1}+0.80\tan \theta _{2}=(4.00-h)}$.

We denote the left-hand side of (b) by ${\displaystyle L}$. For the geophone at depth 0.60 km, the bending of the ray will have a small effect because of the shallowness and proximity to the fault, so we take as our first approximation the straight path,

{\displaystyle {\begin{aligned}\theta _{1}\approx \tan ^{-1}(4.00-0.60)/3.00\approx 49^{\circ }.\end{aligned}}}

To allow for some bending, we try ${\displaystyle \theta _{1}=51^{\circ }}$ and get ${\displaystyle \theta _{2}=35.7^{\circ }}$. Substituting theses values in relation (b) we get

{\displaystyle {\begin{aligned}L=2.20\tan 51^{\circ }+0.80\tan 35.7^{\circ }=3.29\ {\rm {km}}.\end{aligned}}}

This should be ${\displaystyle (4.00-0.60)=3.40}$, so we must increase ${\displaystyle \theta _{1}}$ by a small amount; taking ${\displaystyle \theta _{1}=52^{\circ }}$ gives ${\displaystyle \theta _{2}=36.2^{\circ }}$ and ${\displaystyle L=3.40\ {\rm {m}}}$. We now find the traveltime:

{\displaystyle {\begin{aligned}{\rm {time}}=(2.20/4.00\cos \theta _{1})+(0.800/3.00\cos \theta _{2})=1.224\ {\rm {s}}.\end{aligned}}}

For ${\displaystyle h=1.20\ {\rm {km}}}$, ${\displaystyle L=2.80\ {\rm {km}}}$. We know that ${\displaystyle \theta _{1}}$ will be less than for the shallower geophone, so we take ${\displaystyle \theta _{1}=45^{\circ }}$ and get ${\displaystyle \theta _{2}=32.0^{\circ }}$, ${\displaystyle L=2.70}$. Taking ${\displaystyle \theta _{1}=47.0^{\circ }}$, ${\displaystyle \theta _{2}=33.3^{\circ }}$, ${\displaystyle L=2.88}$. The change of ${\displaystyle 2^{\circ }}$ increased ${\displaystyle L}$ by 0.18 m; we need an increase of 0.10, so we take ${\displaystyle \theta _{1}=45^{\circ }+2^{\circ }\times (0.10/0.18)=46.1^{\circ }}$, ${\displaystyle \theta _{2}=32.7^{\circ }}$, and ${\displaystyle L=2.80\ {\rm {km}}}$. The traveltime is

{\displaystyle {\begin{aligned}{\rm {time}}=(2.20/4.00\cos 46.1^{\circ })+(0.800/3.00\cos 32.7^{\circ })=1.110\ {\rm {s}}.\end{aligned}}}

13.5b What is the deepest geophone location for which the reflection can be recorded?

Solution

The deepest geophone position is that for which the ray passes through the intersection of the fault and reflector. For this point,

{\displaystyle {\begin{aligned}\theta _{1}=\tan ^{-1}(2.00/2.20)=42.3^{\circ },\quad \theta _{2}=30.3^{\circ },\\L=2.20\tan 42.3^{\circ }+0.80\tan 30.3^{\circ }=2.47\ {\rm {km}},\\(400-h)=2.47,\quad h=1.53\ {\rm {km}}.\end{aligned}}}

## 13.6 Mapping the vertical flank of a salt dome

13.6a Two media of velocities ${\displaystyle V_{1}}$ and ${\displaystyle V_{2}}$ are separated by a vertical plane interface. A source is located at ${\displaystyle S}$ on the surface of the high-velocity medium and a geophone at point ${\displaystyle G}$ in a well in the other medium. If ${\displaystyle t_{A}}$ is the traveltime of a wave from ${\displaystyle S}$ to ${\displaystyle G}$ along the path SAG where ${\displaystyle A}$ is on the salt/sediment interface, discuss the locus of points located at the intersection of arcs centered at ${\displaystyle G}$ and ${\displaystyle S}$ having radii ${\displaystyle V_{1}\Delta t_{1}}$ and ${\displaystyle V_{2}\Delta t_{2}}$, respectively, where ${\displaystyle \Delta t_{1}+\Delta t_{2}=t_{A}}$.

Solution

Any path where the interface is tangent to the locus curve satisfies the observed traveltime. To draw the locus we try a series of points ${\displaystyle {\rm {B}}_{i}}$ such that ${\displaystyle ({\rm {SB}}_{i}/V_{2}+{\rm {GB}}_{i}/V_{1})=t_{\hbox{A}}}$. If a number of loci can be drawn for various geophone (or source) loctions, then their common tangent must locate the interface.

While this problem illustrates the concept, it can also be applied in three dimensions to allow for situations where ${\displaystyle V_{1}\Delta t_{1}}$ and ${\displaystyle V_{2}\Delta t_{2}}$ do not lie in the same plane.

Table 13.6a. Survey to define flank of a salt dome.
${\displaystyle z({\rm {m}})}$ ${\displaystyle t({\rm {s}})}$ ${\displaystyle z({\rm {m}})}$ ${\displaystyle t({\rm {s}})}$ ${\displaystyle z({\rm {m}})}$ ${\displaystyle t({\rm {s}})}$
500 0.44 1250 0.52 2000 0.6
750 0.46 1500 0.56 2250 0.67
1000 0.49 1750 0.60 2500 0.70

13.6b An outcropping salt dome has roughly vertical flanks. A source is located on the salt and a geophone is suspended in a vertical well in the sediments 1600 m from the source point. Determine the outline of the salt dome from the ${\displaystyle t-z}$ data in Table 13.6a. Take the velocities in the salt and adjacent sediments as 5.00 and 3.00 km/s.

Solution

The usual method of resolving this problem is to prepare a vertical section through the well W and source point S, then draw a series of circles centered at S with radii equal to the distances traveled in salt for convenient time intervals such as 0.2 s, 0.4 s, etc., the circles being labeled with the time value. A second set of concentric circles using the sediment velocity is drawn on a transparency; the center of these circles is then placed over a geophone position, and intersections of the two sets of circles are marked wherever the time values of the two radii add up to give the traveltime to the geophone.

An alternative method is to dispense with the circles and use only the radii. The centers of the circles and the extremities of the radii are marked along the edges of two narrow strips of light cardboard (one for salt, one for sediments), the time values being marked as before (the marks and the centers should be on opposite sides of the two strips to permit accurate determinations of the intersections). Map pins can be used to attach the zero points of the two strips to the vertical section, one at the source point and the other at the geophone location in the well. Intersections are found as before and marked directly on the section. After curves have been plotted for each geophone position, the flank is outlined by the curve through the apices of the curves. The results are shown in Figure 13.6a.

Figure 13.6a.  Construction for mapping the flank of a salt dome. Small circles mark the intersections of arcs with centers ${\displaystyle S}$ and ${\displaystyle G_{\hbox{i}}}$.

## 13.7 Poission’s ratio from P- and S-wave traveltimes

Find Poisson’s ratio for the five events in Figure 13.7a.

Solution

Poisson’s ratio ${\displaystyle \sigma }$ can be obtained from the ratio ${\displaystyle V_{\hbox{P}}/V_{\hbox{S}}}$ using equation (10.2) Table 2.2a. Since ${\displaystyle z=V_{\hbox{P}}t_{\hbox{P}}=V_{\hbox{S}}t_{\hbox{S}}}$, ${\displaystyle V_{\hbox{P}}/V_{\hbox{S}}=t_{\hbox{S}}/t_{\hbox{P}}}$, we can get ${\displaystyle (V_{\hbox{P}}/V_{\hbox{S}})}$ from the traveltimes and then get ${\displaystyle \sigma }$ using equation (10,2) (see Table 2.2a), that is,

{\displaystyle {\begin{aligned}\sigma ={\frac {(V_{\hbox{P}}/V_{\hbox{S}})^{2}-2}{2[(V_{\hbox{P}}/V_{\hbox{S}})^{2}-1]}}.\end{aligned}}}

Measurements and calculations are listed in Table 13.7a.

Figure 13.7a.  Comparison of P- and S-wave records (courtesy of CGG.) (i) P-wave record; (ii) S-wave record displayed at double the speed to facilitate comparison.
Table 13.7a. Determination of ${\displaystyle \sigma }$.
Event ${\displaystyle t_{\hbox{P}}}$ ${\displaystyle t_{\hbox{S}}}$ ${\displaystyle \alpha /\beta }$ ${\displaystyle (\alpha /\beta )^{2}}$ ${\displaystyle \sigma }$
1 0.39 0.96 2.46 6.05 0.40
2 0.62 1.50 2.42 5.86 0.40
3 0.74 1.65 2.23 4.97 0.37
4 0.80 1.83 2.29 5.24 0.38
5 0.99 2.16 2.18 4.75 0.37