# Depth and dip calculations using velocity functions

Series Geophysical References Series Problems in Exploration Seismology and their Solutions Lloyd P. Geldart and Robert E. Sheriff 4 79 - 140 http://dx.doi.org/10.1190/1.9781560801733 ISBN 9781560801153 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 ${\displaystyle z}$ is now slant depth. Vertical depths are ${\displaystyle z_{v}=z\cos \xi }$, where the dip ${\displaystyle \xi }$ is given by ${\displaystyle \sin ^{-1}(V\Delta t_{d}/2\Delta x)=\sin ^{-1}(0.052\ V)}$, ${\displaystyle V}$ usually being either ${\displaystyle {\bar {V}}}$ or ${\displaystyle V_{\rm {rms}}}$. Using the values of ${\displaystyle t_{0}}$, ${\displaystyle {\bar {V}}}$ and ${\displaystyle V_{\rm {rms}}}$ 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
${\displaystyle t_{0}}$ ${\displaystyle V}$ depth ${\displaystyle \xi }$ ${\displaystyle z\cos \xi }$
i) Assuming average velocities:
1.00 s 2.00 km/s 1000 m 5.97${\displaystyle ^{\circ }}$ 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${\displaystyle ^{\circ }}$ 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${\displaystyle ^{\circ }}$ 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${\displaystyle ^{\circ }}$ 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 ${\displaystyle {\bar {V}}}$ and ${\displaystyle V_{\rm {rms}}}$ 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 ${\displaystyle \alpha }$ [using equation (4.2d)] and then use Snell’s law to find the other angles:

{\displaystyle {\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 ${\displaystyle x}$-coordinates of intersections of rays and interfaces:

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

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

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

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

{\displaystyle {\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.