# Finding velocity

Series Geophysical References Series Problems in Exploration Seismology and their Solutions Lloyd P. Geldart and Robert E. Sheriff 5 141 - 180 http://dx.doi.org/10.1190/1.9781560801733 ISBN 9781560801153 SEG Online Store

## Problem 5.16

Determine the velocity by the ${\displaystyle X^{2}-T^{2}}$ method using the data given in Table 5.16a, ${\displaystyle t_{A}}$ being for a horizontal reflector and ${\displaystyle t_{B}}$ for a reflector dipping ${\displaystyle 10^{\circ }}$ toward the source. (The two reflections are observed on different records.)

### Background

Using equation (4.3a), we have for a dipping horizon,

{\displaystyle {\begin{aligned}(Vt)^{2}=(x\cos \xi )^{2}+(2h_{c})^{2},\end{aligned}}}

where ${\displaystyle x}$ is the offset, ${\displaystyle \xi }$ is the dip, and ${\displaystyle h_{c}}$ the slant depth at the midpoint between source and receiver.

Table 5.16a. ${\displaystyle X^{2}-T^{2}}$ data.
${\displaystyle x}$ (km) ${\displaystyle t_{A}}$ (s) ${\displaystyle t_{B}}$ (s) ${\displaystyle x}$ (km) ${\displaystyle t_{A}}$ (s) ${\displaystyle t_{B}}$ (s) ${\displaystyle x}$ (km) ${\displaystyle t_{A}}$ (s) ${\displaystyle t_{B}}$ (s)
0.0 0.855 0.906 1.4 1.005 0.977 2.8 1.330 1.202
0.1 0.856 0.902 1.5 1.017 0.991 2.9 1.360 1.234
0.2 0.858 0.898 1.6 1.037 1.004 3.0 1.404 1.253
0.3 0.864 0.898 1.7 1.068 1.019 3.1 1.432 1.272
0.4 0.868 0.899 1.8 1.081 1.037 3.2 1.457 1.296
0.5 0.874 0.902 1.9 1.105 1.058 3.3 1.487 1.304
0.6 0.882 0.903 2.0 1.118 1.066 3.4 1.513 1.334
0.7 0.892 0.909 2.1 1.151 1.083 3.5 1.548 1.356
0.8 0.900 0.916 2.2 1.166 1.102 3.6 1.580 1.377
0.9 0.906 0.922 2.3 1.203 1.121 3.7 1.610 1.407
1.0 0.930 0.932 2.4 1.237 1.127 3.8 1.649 1.415
1.1 0.945 0.943 2.5 1.255 1.158 3.9 1.674 1.438
1.2 0.950 0.950 2.6 1.283 1.177 4.0 1.708 1.459
1.3 0.979 0.965 2.7 1.304 1.195
Table 5.16b. Data used in ${\displaystyle X^{2}-T^{2}}$ calculations.
${\displaystyle x^{2}}$ ${\displaystyle t_{A}^{2}}$ (${\displaystyle x\cos \xi )^{2}}$ ${\displaystyle t_{B}^{2}}$ ${\displaystyle x^{2}}$ ${\displaystyle t_{A}^{2}}$ (${\displaystyle x\cos \xi )^{2}}$ ${\displaystyle t_{B}^{2}}$
0.00 0.731 0.00 0.821 4.41 1.325 4.28 1.173
0.01 0.733 0.01 0.814 4.84 1.360 4.69 1.214
0.04 0.736 0.04 0.806 5.29 1.447 5.13 1.257
0.09 0.746 0.09 0.806 5.76 1.530 5.59 1.270
0.16 0.753 0.16 0.808 6.25 1.575 6.06 1.341
0.25 0.764 0.24 0.814 6.76 1.646 6.56 1.385
0.36 0.778 0.35 0.815 7.29 1.700 7.07 1.426
0.49 0.796 0.48 0.826 7.84 1.769 7.60 1.445
0.64 0.817 0.62 0.839 8.41 1.850 8.16 1.523
0.81 0.821 0.79 0.850 9.00 1.971 8.73 1.570
1.00 0.865 0.97 0.869 9.61 2.051 9.32 1.618
1.21 0.893 1.17 0.889 10.24 2.123 9.93 1.680
1.44 0.902 1.40 0.902 10.89 2.211 10.56 1.700
1.69 0.968 1.64 0.931 11.56 2.289 11.21 1.780
1.96 1.010 1.90 0.965 12.25 2.396 11.88 1.839
2.25 1.034 2.18 0.962 12.96 2.496 12.57 1.896
2.56 1.075 2.48 1.008 13.69 2.592 13.28 1.980
2.89 1.141 2.80 1.030 14.43 2.719 14.00 2.002
3.24 1.169 3.14 1.075 15.21 2.802 14.75 2.068
3.61 1.221 3.50 1.119 16.00 2.917 15.52 2.129
3.61 1.250 3.88 1.136
Figure 5.16a.  ${\displaystyle X^{2}-T^{2}}$ plot.

### Solution

The values of ${\displaystyle (x^{2},\;t_{A}^{2})}$ and ${\displaystyle [(x\cos \xi )^{2},\;t_{B}^{2}]}$ are tabulated in Table 5.16b and plotted in Figure 5.16a. The best-fit lines in Figure 5.16a determined by eye give

{\displaystyle {\begin{aligned}V_{A}=2.71\ {\rm {km/s}},h_{A}=1.16\ {\rm {km}};V_{B}=3.35\ {\rm {km/s}},h_{B}=1.49\ {\rm {km}}.\end{aligned}}}

Using the least-squares method (see problem 9.33), we get

{\displaystyle {\begin{aligned}V_{A}=2.71\ {\rm {km/s}},h_{A}=1.15\ {\rm {km}};V_{B}=3.38\ {\rm {km/s}},h_{B}=1.51\ {\rm {km}}.\end{aligned}}}