Depth and dip calculations using velocity functions

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 4.15a

Repeat the calculations of problem 4.14 assuming horizontal velocity layering and dip moveout of 104 ms/km, and find the dips.

Background

While velocity generally follows the layering, especially in structurally deformed areas, isovelocity surfaces may not parallel interfaces. Where the section has not been uplifted significantly, isovelocity surfaces are apt to be nearly horizontal in spite of structural relief.

Solution

The depths will be the same as those calculated in problem 4.14 except that $z$ is now slant depth. Vertical depths are $z_{v}=z\cos \xi$, where the dip $\xi$ is given by $\sin ^{-1}(V\Delta t_{d}/2\Delta x)=\sin ^{-1}(0.052\ V)$, $V$ usually being either ${\bar {V}}$ or $V_{\rm {rms}}$. Using the values of $t_{0}$, ${\bar {V}}$ and $V_{\rm {rms}}$ from Table 4.14a, we obtain the results in Table 4.15a.

Table 4.15a. Calculated depths and dips.

Time	Velocity	Slant	Dip	Vert. depth
$t_{0}$	$V$	depth	$\xi$	$z\cos \xi$
i) Assuming average velocities:
1.00 s	2.00 km/s	1000 m	5.97$^{\circ }$	990 m
2.00	2.50	2500	7.47	2480
2.10	2.67	2800	7.98	2770
3.10	3.10	4800	9.28	4740
ii) Assuming rms velocities:
1.00 s	2.99 km/s	1000 m	5.97$^{\circ }$	990 m
2.00	2.55	2550	7.62	2530
2.10	2.81	2950	8.40	2920
3.10	3.24	5020	9.70	4950
iii) Assuming best-fit depth function:
1.00 s	2.02 km/s	1010 m	6.03$^{\circ }$	1000 m
2.00	2.54	2540	7.59	2520
2.10	2.59	2720	7.74	2700
3.10	3.12	4840	9.34	4780
iv) Assuming best-fit traveltime function:
1.00 s	2.03 km/s	1020 m	6.00$^{\circ }$	1010 m
2.00	2.62	2620	7.83	2800
2.10	2.68	2810	8.01	2780
3.10	3.27	5270	9.79	5000

Problem 4.15b

Trace rays assuming (i) the velocity layering given in Figure 4.13a, and (ii) that the velocity is constant at the values of ${\bar {V}}$ and $V_{\rm {rms}}$ listed in Figure 4.13b. Find the arrival times and reflecting points of reflections at each of the interfaces.

Solution

i) We first calculate the angle of approach $\alpha$ [using equation (4.2d)] and then use Snell’s law to find the other angles:

${\begin{aligned}\alpha =\theta _{1}&=\sin ^{-1}\left(2.00\times {\frac {0.104}{2}}\right)=6.0^{\circ },\\\sin \theta _{2}&=(V_{2}/V_{1})\sin \theta _{1},\\\theta _{2}&=\sin ^{-1}[(3.00/2.00)\sin 6.0^{\circ }]=9.0^{\circ },\\\theta _{3}&=\sin ^{-1}[(6.00/2.00)\sin 6.0^{\circ }]=18.3^{\circ },\\\theta _{4}&=\sin ^{-1}[(4.00/2.00)\sin 6.0^{\circ }]=12.1^{\circ },\\t_{A}&=2z/(V_{1}\cos \theta _{1})=2.000/(2.00\cos 6.0^{\circ })\\&=1.006\ \mathrm {s} ,\\t_{B}&=1.006+2\times 1.500/3.00\cos 9.0^{\circ }\\&=2.018\ \mathrm {s} ,\\t_{C}&=2.018+2\times 0.300/6.00\cos 18.3^{\circ }\\&=2.123\ \mathrm {s} ,\\t_{D}&=2.123+2\times 2.000/4.00\cos 12.1^{\circ }\\&=3.146\ \mathrm {s} .\end{aligned}}$

Next we find $x$-coordinates of intersections of rays and interfaces:

${\begin{aligned}x_{A}&=z\sin 6.0^{\circ }=0.105\ \mathrm {km} ,\\x_{B}&=0.105+0.156=0.261\ \mathrm {km} ,\\x_{C}&=0.261+0.031=0.292\ \mathrm {km} ,\\x_{D}&=0.292+0.209=0.501\ \mathrm {km} .\end{aligned}}$

Figure 4.15a.  Raypath.

ii) Assuming ${\bar {V}}=0.288z+1.77$ (see Figure 4.13b) and the given depths,

${\begin{aligned}V_{A}=0.288+1.77=2.058\ \mathrm {km/s} ,\ \alpha =6.1^{\circ },\\t_{A}=2z/(V_{1}\cos \theta _{1})=0.977\ \mathrm {s} ,\\x_{A}=2z\sin 6.1^{\circ }=0.107\ \mathrm {km} ;\\V_{B}=0.72+1.77=2.49\ \mathrm {km/s} ,\alpha =7.4^{\circ },\\t_{B}=5.00/2.562\cos 7.4^{\circ }=1.968\ \mathrm {s} ,x_{B}=0.324\ \mathrm {km} ;\\V_{C}=0.806+1.77=2.576\ \mathrm {km/s} ,\alpha =7.7^{\circ },\\t_{C}=5.600/2.660\cos 7.7^{\circ }=2.124\ \mathrm {s} ,x_{C}=0.365\ \mathrm {km} ;\\V_{D}=1.382+1.77=2.940\ \mathrm {km/s} ,\alpha =9.4^{\circ }\\t_{D}=9.600/3.310\cos 9.4^{\circ }=3.146\ \mathrm {s} ,x_{D}=0.787\ \mathrm {km} .\end{aligned}}$

Assuming $V_{\rm {rms}}=0.325z+1.75$ and the given depths,

${\begin{aligned}V_{A}=0.325+1.75=2.075\ \mathrm {km/s} ,\alpha =6.2^{\circ }\\t_{A}=2z(V_{1}\cos \theta _{1})=2.00/(2.075\cos 6.2^{\circ })=0.970\ \mathrm {s} \\x_{A}=2z\sin 6.2^{\circ }=0.108\ \mathrm {km} ;\\V_{B}=0.812+1.75=2.562\ \mathrm {km/s} ,\alpha =7.7^{\circ }\\t_{B}=5.000/2.562\cos 7.7^{\circ }=2.018\ \mathrm {s} ,x_{B}=0.335\ \mathrm {km} ;\\V_{C}=0.910+1.75=2.660\ \mathrm {km/s} ,\alpha =8.0^{\circ },\\t_{C}=5.600/2.660\cos 8.0^{\circ }=2.123\ \mathrm {s} ,x_{C}=0.390\ \mathrm {km} ;\\V_{D}=1.560+1.75=3.310\ \mathrm {km/s} ,\alpha =9.9^{\circ },\\t_{D}=9.600/3.310\cos 9.9^{\circ }=3.146\ \mathrm {s} ,x_{D}=0.825\ \mathrm {km} .\end{aligned}}$

Assuming ${\bar {V}}=0.526t+1.49$ and times in part (i):

${\begin{aligned}t_{A}=1.006\ \mathrm {s} ,{\bar {V}}=2.019\ \mathrm {km/s} ,\alpha =6.0^{\circ },z_{A}=2.042\ \mathrm {km} ,x_{A}=0.213\ \mathrm {km} ,\\t_{B}=2.018\ \mathrm {s} ,{\bar {V}}=2.551\ \mathrm {km/s} ,\alpha =7.6^{\circ },z_{B}=5.194\ \mathrm {km} ,x_{B}=0.683\ \mathrm {km} ,\\t_{C}=2.123\ \mathrm {s} ,{\bar {V}}=2.607\ \mathrm {km/s} ,\alpha =7.8^{\circ },z_{C}=5.586\ \mathrm {km} ,x_{C}=0.751\ \mathrm {km} ,\\t_{D}=3.146\ \mathrm {s} ,{\bar {V}}=3.145\ \mathrm {km/s} ,\alpha =9.6^{\circ },z_{D}=10.029\ \mathrm {km} ,x_{D}=1.618\ \mathrm {km} .\end{aligned}}$

Assuming $V_{\rm {rms}}=0.595\mathrm {t} +1.43$ and times in part (i):

${\begin{aligned}t_{A}=1.006\ \mathrm {s} ,V_{\rm {rms}}=2.029\ \mathrm {km/s} ,\alpha =6.1^{\circ },z_{A}=2.053\ \mathrm {km} ,x_{A}=0.215\ \mathrm {km} ,\\t_{B}=2.018\ \mathrm {s} ,V_{\rm {rms}}=2.631\ \mathrm {km/s} ,\alpha =7.9^{\circ },z_{B}=5.360\ \mathrm {km} ,x_{B}=0.726\ \mathrm {km} ,\\t_{C}=2.123\ \mathrm {s} ,V_{\rm {rms}}=2.693\ \mathrm {km/s} ,\alpha =8.1^{\circ },z_{C}=5.774\ \mathrm {km} ,x_{C}=0.800\ \mathrm {km} ,\\t_{D}=3.146\ \mathrm {s} ,V_{\rm {rms}}=3.302\ \mathrm {km/s} ,\alpha =9.9^{\circ },z_{D}=10.454\ \mathrm {km} ,x_{D}=1.784\ \mathrm {km} .\end{aligned}}$

The use of any functional form involves approximation, so it is not surprising that values of the depths and horizontal displacements depend upon the way in which they are calculated.

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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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