# Barry’s delay-time refraction interpretation method

Series Geophysical References Series Problems in Exploration Seismology and their Solutions Lloyd P. Geldart and Robert E. Sheriff 11 415 - 468 http://dx.doi.org/10.1190/1.9781560801733 ISBN 9781560801153 SEG Online Store

## Problem 11.9

Source $B$ is 2 km east of source $A$ . The data in Table 11.9a were obtained with cables extending eastward from $A$ ($x$ is the distance measured from $A$ ) with geophones at 200 m intervals. Interpret the data using Barry’s method (Barry, 1967); $V_{1}=2.50$ km/s. Assume that the delay-time curve for the reverse profile is sufficiently parallel to yours that step (d) below can be omitted.

### Background

Barry’s method requires that the total delay time be separated into source and geophone delay times. Two sources on the same side of the geophone are used to achieve this. In Figure 11.9a. $A$ and $B$ are sources, $Q$ and $R$ are geophones, $BQ$ being the critical distance (problem 4.18) for source $B$ . We write $\delta _{A}$ and $\delta _{B}$ for the source delay times, $\delta _{Q}$ and $\delta _{R}$ for the geophone delay times, $\delta _{AR}$ , $\delta _{BR}$ , etc., for the total delay times. We get the source delay times from the intercept times if we assume zero dip [see equation (11.9a)]. The delay time at source $B$ , $\delta _{B}$ , is due to travel along $BN$ , so

 {\begin{aligned}\delta _{B}&=BN/V_{1}-N'N/V_{2}=h_{N}/V_{1}\cos \theta _{c}-\left(h_{N}\tan \theta _{c}\right)/V_{2}\\&=\left(h_{N}/V_{1}\cos \theta _{c}\right)\left(1-\sin ^{2}\theta _{c}\right)=\left(h_{N}\cos \theta _{c}\right)/V_{1}={\frac {1}{2}}t_{iB};\end{aligned}} (11.9a)

 {\begin{aligned}{\hbox{and so,}}\quad \quad h_{N}=V_{1}\delta _{B}/\cos \theta _{c}={\frac {1}{2}}V_{1}t_{iB}/\cos \theta _{c}.\end{aligned}} (11.9b)

Note that equations (11.9a,b) apply at any point on the profile where the dip is very small, not merely at souce points.

To find the geophone delay times we have from equation (11.8c)

{\begin{aligned}\delta _{AQ}&=\delta _{A}+\delta _{Q},\\\delta _{AR}&=\delta _{A}+\delta _{R},\\\Delta \delta &=\delta _{AQ}-\delta _{AR}=\delta _{q}-\delta _{R}=differential\ delay\ time.\end{aligned}} For zero dip, $\delta _{B}=\delta _{Q}$ , so we can write

 {\begin{aligned}{\frac {1}{2}}\left(\delta _{BR}+\Delta \delta \right)={\frac {1}{2}}\left(\delta _{Q}+\delta _{R}+\delta _{Q}-\delta _{R}\right)\delta _{Q}-\delta _{R}=\delta _{Q},\end{aligned}} (11.9c)

 {\begin{aligned}{\hbox{hence}}\quad \quad {\frac {1}{2}}\left(\delta _{BR}-\Delta \delta \right)=\delta _{R}.\end{aligned}} (11.9d)
Table 11.9a. Time-distance data.
$x$ (km) $t_{A}$ (s) $t_{B}$ (s) $x$ (km) $t_{A}$ (s) $t_{B}$ (s)
2.6 1.02 0.25 5.4 1.62 1.28
2.8 1.05 0.34 5.6 1.66 1.31
3.0 1.10 0.43 5.8 1.72 1.36
3.2 1.24 0.52 6.0 1.75 1.42
3.4 1.18 0.61 6.2 1.80 1.47
3.6 1.20 0.70 6.4 1.85 1.53
3.8 1.26 0.78 6.6 1.91 1.56
4.0 1.32 0.87 6.8 1.97 1.59
4.2 1.35 0.96 7.0 2.00 1.63
4.4 1.39 1.05 7.2 2.02 1.67
4.6 1.45 1.10 7.4 2.05 1.70
4.8 1.50 1.14 7.6 2.10 1.73
5.0 1.56 1.20 7.8 2.13 1.78
5.2 1.59 1.22 8.0 2.16 1.81

To use these equations we must find the point $Q$ , preferably by expressing $BQ$ in terms of delay times. From Figure 11.9a and equation (11.9b) we get

 {\begin{aligned}BQ=2h_{N}\tan \theta _{C}=2\left(V_{1}\delta _{B}/\cos \theta _{c}\right)\tan \theta _{c}=2V_{2}\ \delta _{B}\ \tan ^{2}\theta _{c}.\end{aligned}} (11.9e)

Interpretation involves the following steps:

1. The traveltimes are corrected for weathering and elevation (problem 8.18)
2. Total delay times are calculated and plotted at the geophone positions
3. The distance $PP'$ in Figure 11.9a is calculated for each geophone using equation (11.8a), and the total delay times shifted the distances $PP'$ toward $A$ 4. The curves in (b) and (c) should be parallel; if not, $V_{2}$ is adjusted until the curves are sufficiently close to being parallel
5. The total delay times in (b) are separated into source and geophone delay times and then plotted above points $M$ , $N$ , and $P$ . Delays times can be converted into depths using equation (11.9b)

### Solution

The data are plotted in Figure 11.9b. Measurements give an average value of 4.60 km/s for $V_{2}$ and intercept times $t_{iA}=0.45$ s, $t_{iB}=0.55$ s, $\delta _{B}=0.55/2=0.28$ s. The critical angle is $\theta _{c}=\sin ^{-1}\left(2.50/4.60\right)=32.9^{\circ }$ , $\cos \theta _{c}=0.840$ , $\tan \theta _{c}=0.647$ . Using equation (11.9e), we have

{\begin{aligned}BQ=2V_{2}\delta _{B}\tan ^{2}\theta _{c}=2\times 4.60\times 0.28\times 0.647^{2}=1.1\ {\rm {km}}.\end{aligned}} Thus, $Q$ is located at $x=3.1$ km. Also, we need $\delta _{AQ}$ :

{\begin{aligned}\delta _{AQ}=t_{AQ}-x_{AQ}/V_{2}=1.12-3.1/4.60=0.45\ {\rm {s}}.\end{aligned}} Figure 11.9b shows that we observe refraction data from both sources only for $x\geq 4.6$ km. We show the calculations in Table 11.9b. Column 1 is the offset measured from $A$ , columns 2 and 6 are traveltime for sources $A$ and $B$ , columns 3 and 7 are the source-to-geophone distances divided by $V_{2}$ , columns 4 and 8 are the total delay times [the differences between columns 2 and 3, 6 and 7, respectively—see equation (11.8b)], column 5 is the differential delay time $\Delta \delta$ between geophones at $Q$ and $R=\delta _{AQ}-\delta _{AR}=\left(0.45-\delta _{AR}\right)$ , column 9 is $\delta _{R}=\left(\delta _{BR}-\Delta \delta \right)/2$ [see equation (11.9c)], column 10 is $PP'$ , column 11 is column 1 minus column 10 = location of $P'$ in Figure 11.9a. Depth values can be obtained by mutliplying $\delta _{PR}$ in column 9 by $V_{1}/\cos \theta _{c}=2.98$ [see equation (11.9a)].

We used a new value of $\delta _{AQ}=1.123-3.10/4.60=0.449s$ . Comparing the new and old values of $\delta _{PR}$ for $x=5.0$ and $x=5.6$ , we see that rounding errors are not responsible for the anomalies. The anomaly at $x=5.6$ km is 0.01 s whereas the original data are also given to the nearest 0.01 s, so this anomaly could be the result of rounding off of the original time values; however, the anomaly at $x=5.0$ is too large to be due to this.

Table 11.9b. Delay-time calculations.
1 2 3 4 5 6 7 7 8 10 11
$x$ $t_{AR}$ $x/V_{2}$ $\delta _{AR}$ $\Delta \delta$ $t_{BR}$ $x'/V_{2}$ $\delta _{BR}$ $\delta _{R}$ $PP'$ $x-PP'$ 4.6 1.45 1.00 0.45 0.00 1.10 0.57 0.53 0.27 0.52 4.08
4.8 1.50 1.04 0.46 –0.01 1.14 0.61 0.53 0.27 0.52 4.28
5.0 1.56 1.09 0.47 –0.02 1.20 0.65 0.55 0.29 0.56 4.44
5.2 1.59 1.13 0.46 –0.01 1.22 0.70 0.52 0.27 0.52 4.68
5.4 1.62 1.17 0.45 0.00 1.28 0.74 0.54 0.27 0.52 4.88
5.6 1.66 1.22 0.44 0.01 1.31 0.78 0.53 0.26 0.50 5.10
5.8 1.72 1.26 0.46 –0.01 1.36 0.83 0.53 0.27 0.52 5.28
6.0 1.73 1.30 0.43 0.02 1.42 0.87 0.55 0.27 0.52 5.48
6.2 1.80 1.35 0.45 0.00 1.47 0.91 0.56 0.28 0.54 5.66
6.4 1.85 1.39 0.46 –0.01 1.53 0.96 0.57 0.29 0.56 5.84
6.6 1.91 1.43 0.48 –0.03 1.56 1.00 0.56 0.30 0.58 6.02
6.8 1.97 1.48 0.49 –0.04 1.59 1.04 0.55 0.30 0.58 6.22
7.0 2.00 1.52 0.48 –0.03 1.63 1.09 0.54 0.29 0.56 6.44
7.2 2.02 1.57 0.45 0.00 1.67 1.13 0.54 0.27 0.52 6.68
7.4 2.05 1.61 0.44 0.01 1.70 1.17 0.53 0.26 0.50 6.90
7.6 2.10 1.65 0.45 0.00 1.73 1.22 0.51 0.26 0.50 7.10
7.8 2.13 1.70 0.43 0.02 1.78 1.26 0.52 0.25 0.48 7.32
8.0 2.16 1.74 0.42 0.03 1.81 1.30 0.51 0.24 0.46 7.54

Note. $x'=x-2$ .

Table 11.9c. Part of Table 11.9b with increased precision.
1 2 3 4 5 6 7 8 9
$x$ $t_{AR}$ $x/V_{2}$ $\delta _{AR}$ $\Delta \delta$ $t_{BR}$ $x'/V_{2}$ $\delta _{BR}$ $\delta _{PR}$ 4.8 1.50 1.043 0.457 –0.008 1.14 0.609 0.531 0.270
5.0 1.56 1.087 0.473 –0.024 1.20 0.652 0.548 0.286
5.2 1.59 1.130 0.460 –0.011 1.22 0.696 0.524 0.268
5.4 1.62 1.174 0.446 0.003 1.28 0.739 0.541 0.269
5.6 1.66 1.217 0.443 0.006 1.31 0.783 0.527 0.261
5.8 1.72 1.261 0.459 –0.010 1.36 0.826 0.534 0.272