S-wave conversion in marine surveys

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Series Geophysical References Series Problems in Exploration Seismology and their Solutions Lloyd P. Geldart and Robert E. Sheriff 13 485 - 496 http://dx.doi.org/10.1190/1.9781560801733 ISBN 9781560801153 SEG Online Store

Problem 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 ${\displaystyle 80^{\circ }}$.

Problem 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.