Reflector dip in terms of traveltimes squared

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 4.4a

Using the dip-moveout equation (4.2b) and the result of problem 4.3, verify that

${\begin{aligned}\tan \xi =t/(t_{SR}^{2}-t_{0}^{2})^{1/2},\end{aligned}}$

where $\xi ={\rm {dip}}$, $t=t_{SM}-t_{MR}$, $t_{SR}=$ traveltime for path ${\rm {SR}}'{\rm {R}}={\rm {S}}'{\rm {R}}'{\rm {R}}$, $t_{0}=2h_{c}/V=$ traveltime to receiver at $M$ (Figure 4.4a).

Solution

Since $t$ in problem 4.3 equals $t_{SR}$ here, we have

${\begin{aligned}(Vt_{SR})^{2}=(2x\cos \xi )^{2}+4h_{c}^{2}=(2x\cos \xi )^{2}+(Vt_{0})^{2},\end{aligned}}$

${\begin{aligned}{\hbox{or}}\qquad \qquad \cos \xi =\left(V/2x\right)(t_{SR}^{2}-t_{0}^{2})^{1/2}.\end{aligned}}$

From equation (4.2b) we get

${\begin{aligned}\sin \xi =\left(V/2\right)\left(\Delta t_{d}/\Delta x\right)=\left(V/2x\right)\left(t_{SM}-t_{MR}\right),\end{aligned}}$

${\begin{aligned}{\hbox{or}}\qquad \qquad \tan \xi =\left(t_{SM}-t_{MR}\right)/(t_{SR}^{2}-t_{0}^{2})^{1/2}=t/(t_{SR}^{2}-t_{0}^{2})^{1/2}.\end{aligned}}$

Figure 4.4a.  Geometry for dipping bed.

Problem 4.4b

Using equation (4.2a), show that

${\begin{aligned}\sin \xi =V^{2}\left(t_{SM}^{2}-t_{MR}^{2}\right)/8h_{c}x.\end{aligned}}$

Solution

Equation (4.2a) gives

${\begin{aligned}(Vt_{SM})^{2}=x^{2}+4h_{c}^{2}+4h_{c}x\sin \xi ,\\(Vt_{MR})^{2}=x^{2}+4h_{c}^{2}+4h_{c}x\sin \xi ,\\{\hbox{thus}}\qquad \qquad \sin \xi =V^{2}\left(t_{SM}^{2}-t_{MR}^{2}\right)/8h_{c}x.\end{aligned}}$

Problem 4.4c

Under what circumstances is the result for part (b) the same as equation (4.2b) and also consistent with part (a)?

Solution

The result in part (b) can be written

${\begin{aligned}\sin \xi =\{[V(t_{MS}-t_{MR})]/2x\}\{[V(t_{MS}+t_{MR})]/4h_{c}\}.\end{aligned}}$

The expression in the first curly brackets is the same as the right-hand side of equation (4.2b). Hence, for the above to be the same as equation (4.2b), we must have $V(t_{MS}+t_{MR})/4h_{c}=1$ that is,

${\begin{aligned}V(t_{MS}+t_{MR})=4h_{c}=2Vt_{0},\end{aligned}}$

${\begin{aligned}{\hbox{so}}\qquad \qquad \qquad \qquad t_{0}={\frac {1}{2}}(t_{MS}+t_{MR}).\end{aligned}}$

In part (a) we got $\tan \xi$ by finding $\sin \xi$ and $\cos \xi$. The expression for $\cos \xi$ involves no approximation, so the only approximation is that used in the derivation of equation (4.2b) to get $\sin \xi$. Thus, results in (a) and (c) are consistent provided we take $t_{0}={\frac {1}{2}}(t_{MS}+t_{MR})$.

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Relationship for a dipping bed	Second approximation for dip moveout
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Partitioning at an interface	Seismic velocity
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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