Diving waves

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	4
Pages	79 - 140
DOI	http://dx.doi.org/10.1190/1.9781560801733
ISBN	ISBN 9781560801153
Store	SEG Online Store

Problem 4.20a

When the velocity is a linear function of depth only, as in equation (4.17a), show that the wave will return to the surface again at offset $x$ and with traveltime $t$ given by equations (4.20a,b), and that the maximum depth of penetration $h_{m}$ is given by equation (4.20c).

${\begin{aligned}x&=(2V_{0}/a)\sin h(at/2),\end{aligned}}$

(4.20a)

${\begin{aligned}t&=(2/a)\ln[\cot(i_{0}/2)],\end{aligned}}$

(4.20b)

${\begin{aligned}h_{m}&=(V_{0}/a)[\cos \mathrm {h} (at/2)-1].\end{aligned}}$

(4.20c)

Background

When the velocity increases continuously with depth, a wave will eventually return to the surface (see Figures 4.20a,c): such waves are known as diving waves. When the velocity layers are horizontal, as in Figure 4.20a, or spherically symmetrical as in Figure 4.20c, the raypath is symmetrical about the midpoint and the maximum depth occurs at this point.

Figure 4.20a.  Raypaths where velocity increases linearly with depth.

Solution

Because the raypath is symmetrical about the midpoint, we can find $x$ and $t$ for the midpoint and double the values to get $x$ and $t$ for the entire path. At the midpoint the ray is traveling horizontally, hence $i=90^{\circ }$. Substituting in equations (4.17b,c), we get

${\begin{aligned}x&=(2/pa)\cos i_{0}=(2V_{0}/a)\cot i_{o},\\t&=(2/a)\ln(\cot i_{0}/2),\quad \cot i_{0}/2=e^{at/2}.\end{aligned}}$

(4.20d)

Using the identity $\cot \,2\phi ={\frac {1}{2}}(\cos \phi -\tan \phi )$, equation (4.20d) becomes

${\begin{aligned}x=(2V_{0}/a)(e^{at/2}-e^{-at/2})/2=(2V_{0}/a)\sinh at/2.\end{aligned}}$

The maximum depth $h_{m}$ is the value of $z$ when $i=90^{\circ }$, so from equation (4.17f) we find that

${\begin{aligned}h_{m}&=(1/pa)(\sin i-\sin i_{0})=(1/pa)(1-\sin i_{0})\\&=(V/a)(1/\sin i_{0}-1).\end{aligned}}$

Using the identity: $\sin x=2\tan(x/2)/[1+\tan ^{2}(x/2)]$, we can write

${\begin{aligned}1/\sin i_{o}=\cot(i_{o}/2)[1+\tan ^{2}(i_{o}/2)]/2\\={\frac {1}{2}}[\cot(i_{o}/2)+\tan(i_{o}/2)].\end{aligned}}$

Equation (4.20d) now gives

${\begin{aligned}h_{m}&=(V/a)[(e^{at/2}+e^{-at/2})/2-1]\\&=(V/a)(\cosh at/2-1).\end{aligned}}$

Figure 4.20b.  Raypath parameter for concentric spherical layering.

Problem 4.20b

Show that when the constant velocity layers are concentric spherical shells, the raypath parameter $p$ (see problem 3.1a) becomes

${\begin{aligned}p^{\prime }=r_{n}\sin i_{n}/V_{n}.\end{aligned}}$

(4.20e)

Solution

Equation (4.20e) is a modification of equation (3.1a) to take into account concentric spherical shells instead of plane parallel interfaces. The angle between a ray and the radii changes as the wave travels downward (see Figure 4.20b) so that the angle of entry into a layer does not equal the incident angle at the base of the layer, that is, $i_{2}\neq i_{2}^{*}$. But $OP=r_{3}\sin i_{2}^{*}=r_{2}\sin i_{2}$, and Snell’s law becomes $\sin i_{2}^{*}/V_{2}=\sin i_{3}/V_{3}$, or

${\begin{aligned}{\frac {r_{2}\sin i_{2}}{V_{2}}}={\frac {r_{3}\sin i_{3}}{V_{3}}}=p^{\prime }.\end{aligned}}$

Problem 4.20c

For concentric spherical layering in the Earth, show that diving waves will return to the surface at the time $t$ at the angular distance $\Delta$, where $\Delta$ is the angle subtended at the center of the Earth by the ray $SE$, and $R_{e}$ is the radius of the Earth; (see Sheriff and Geldart, 1995, 99);

Figure 4.20c.  Spherical layering raypaths.

${\begin{aligned}t=2\int _{R_{e}-h_{m}}^{R_{e}}{\frac {\mathrm {d} r}{V(r)\{1-[p^{\prime }V(r)/r]^{2}\}^{1/2}}},\end{aligned}}$

(4.20f)

${\begin{aligned}\Delta =2\int _{R_{e}-h_{m}}^{R_{e}}{\frac {[p^{\prime }V(r)/r]\mathrm {d} r}{r\{1-[p^{\prime }V(r)/r]^{2}\}^{1/2}}}.\end{aligned}}$

(4.20g)

Solution

Using equation (4.20e) and $\Delta ABC$ in Figure 4.20c, we get

${\begin{aligned}p^{\prime }={\frac {r\sin i}{V}}=\left({\frac {r}{V}}\right)\left({\frac {r\delta \Delta }{V\delta t}}\right)=\left({\frac {r}{v}}\right)^{2}\left({\frac {\delta \Delta }{\delta t}}\right).\end{aligned}}$

(4.20h)

$\Delta ABC$ also gives

${\begin{aligned}(V\delta t)^{2}=(\delta r)^{2}+(r\delta \Delta )^{2}.\end{aligned}}$

(4.20i)

Eliminating $\delta \Delta$ between equations (4.20h,i) gives

${\begin{aligned}&(r\delta \Delta )^{2}=(V\delta t)^{2}-(\delta r)^{2}=(p^{\prime }V^{2}\delta t/r)^{2},\\&(\delta t)^{2}[V^{2}-(p^{\prime }V^{2}/r)^{2}]=(\delta r)^{2},\\\mathrm {so} \qquad &\delta t=(\delta r/V)[1-(p^{\prime }V/r)^{2}]^{-1/2}.\end{aligned}}$

Integrating from $(R_{e}-h_{m})$ to $R_{e}$ and multiplying by 2 to allow for the return path gives equation (4.20f).

Eliminating $\delta t$ between equations (4.20h,i), we have

${\begin{aligned}&(\delta t)^{2}=(1/V^{2})[(\delta r)^{2}+(r\delta \Delta )^{2}]=(r/V)^{4}(\delta \Delta /p^{\prime })^{2},\\&(\delta r/r)^{2}+(\delta \Delta )^{2}=(r/p^{\prime }V)^{2}(\delta \Delta )^{2}.\end{aligned}}$

Again we multiply by 2 and integrate, obtaining equation (4.20g).

Alternative solution

Equation (4.20f) is a relation between $t$ and $r$, so we use $\Delta ABC$ in Figure 4.20c to get a relation between $\delta t$ and $\delta r$, then integrate to get $t$. From $\Delta ABC$ and equation (4.20e), we have

${\begin{aligned}\cos i=(\delta r/V\delta t),\quad \delta t=(\delta r/V)[1-(p^{\prime }/r)^{2}]^{-1/2},\end{aligned}}$

since $\sin i=p^{\prime }V/r$. Integrating this expression for $\delta t$ gives equation (4.20f).

Equation (4.20g) expresses $\Delta$ in terms of $r$, so we use $\Delta ABC$ to get the relation $\tan i=r\delta \Delta /\delta r$. Using equation (4.20e), this becomes

${\begin{aligned}\delta \Delta =(\delta r/r)(p^{\prime }V/r)[1-(p^{\prime }/r)^{2}]^{1/2}.\end{aligned}}$

Multiplying by 2 and integrating gives equation (4.20g).

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Table of Contents (book)

Also in this chapter

• Accuracy of normal-moveout calculations
• Dip, cross-dip, and angle of approach
• Relationship for a dipping bed
• Reflector dip in terms of traveltimes squared
• Second approximation for dip moveout
• Calculation of reflector depths and dips
• Plotting raypaths for primary and multiple reflections
• Effect of migration on plotted reflector locations
• Resolution of cross-dip
• Cross-dip
• Variation of reflection point with offset
• Functional fits for velocity-depth data
• Relation between average and rms velocities
• Vertical depth calculations using velocity functions
• Depth and dip calculations using velocity functions
• Weathering corrections and dip/depth calculations
• Using a velocity function linear with depth
• Head waves (refractions) and effect of hidden layer
• Interpretation of sonobuoy data
• Diving waves
• Linear increase in velocity above a refractor
• Time-distance curves for various situations
• Locating the bottom of a borehole
• Two-layer refraction problem

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