# Interpretation by the plus-minus 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 11.13

Interpret the data in Table 11.13a using the plus-minus method.

### Background

Fermat’s principle (problem 4.13) states that the raypath between two points $A$ and $B$ is such that the traveltime is either a minimum (e.g., direct waves, reflections and head waves) or a maximum. Therefore, the raypath between $A$ and $B$ is unique so that $t_{AB}=t_{BA}=t_{r}=reciprocaltime$ . As a result, when we have reversed profiles, we can locate the refractor by drawing wavefronts from the two sources $A$ and $B$ ; when the sum of the traveltimes for two intersecting wavefronts equals $t_{r}$ , the point of intersection must lie on the refractor (see problem 11.14c). This is the basic concept of the plus-minus method (Hagedoorn, 1959).

Construction of wavefronts is discussed in problem 11.14c.

Based on the recorded data, we draw and label wavefronts at intervals $\Delta$ as in Figure 11.13a. If the dip is zero, they are at the angles $\pm \theta _{c}$ to the refractor and the intersections give diamond-shaped parallelograms. The horizontal diagonal of a parallelogram is $V_{2}\Delta$ and the vertical diagonal is $V_{1}\Delta /\cos \theta _{c}$ . Lines of constant sum of the traveltimes minus $t_{r}$ (plus values) gives the refractor configuration and differences (minus values) give a check on the value of $V_{2}$ . The refractor lies at plus value = 0.

Table 11.13a. Time-distance data for plus-minus interpretation.
$x$ (km) $t_{A}$ (s) $t_{B}$ (s) $x$ (km) $t_{A}$ (s) $t_{B}$ (s)
0.0 0.00 2.30 6.0 1.30 1.32
0.4 0.15 2.23 6.4 1.33 1.28
0.8 0.28 2.15 6.8 1.40 1.24
1.2 0.44 2.09 7.2 1.51 1.18
1.6 0.52 2.04 7.6 1.57 1.10
2.0 0.63 1.98 8.0 1.60 1.04
2.4 0.70 1.92 8.4 1.72 0.96
2.8 0.76 1.85 8.8 1.78 0.90
3.2 0.84 1.80 9.2 1.80 0.83
3.6 0.91 1.72 9.6 1.91 0.76
4.0 0.95 1.64 10.0 1.93 0.66
4.4 1.04 1.60 10.4 2.04 0.52
4.8 1.12 1.55 10.8 2.07 0.39
5.2 1.16 1.47 11.2 2.17 0.25
5.6 1.25 1.40 11.6 2.20 0.12
12.0 2.30 0.00

### Solution

The traveltime curves are shown in Figure 11.13b. From the figure we obtained the following values: $V_{1}=2.90$ km/s, $V_{2}=6.25$ km/s, $t_{r}=2.28$ s, $\theta _{c}=\sin ^{-1}\left(2.90/6.25\right)=27.6^{\circ }$ .

We next draw straight-line wavefronts at $\pm 27.6^{\circ }$ spaced at intervals $\Delta =0.20$ s. Because $t_{r}=2.30$ s and the refraction from source $B$ starts around 0.8 s, we draw wavefronts for source $B$ for $t_{B}=0.80$ , 1.00, 1.20, 1.40, and 1.60 s. For source $A$ we draw wavefronts for $t_{A}=1.48$ , 1.28, 1.08, 0.88, and 0.68 s. We interpolate to find the starting points of these wavefronts.

The horizontal and vertical diagonals of the parallelograms have lengths of 1.24 and 0.66 km, so

{\begin{aligned}V_{2}\Delta =1.24,\quad V_{2}=1.24/0.20=6.20\ {\rm {km/s}},\\V_{1}\Delta /\cos \theta _{c}=0.66,\quad V_{1}=0.66\cos 27.6^{\circ }/0.20=2.92\ {\rm {km/s}}.\end{aligned}} These values agree with the values in Figure 11.13b within the limits of error.

Table 11.14a. Refraction time-distance data.
$x$ (km) $t_{A}$ (s) $t_{B}$ (s) $t_{A}^{*}$ (s) $x$ (km) $t_{A}$ (s) $t_{B}$ (s) $t_{B}^{*}$ (s)
0.00 0.000 3.310 6.40 2.330 2.003
0.40 0.182 3.182 6.80 2.422 1.862
0.80 0.320 3.140 7.20 2.504 1.743
1.20 0.504 3.063 7.60 2.602 1.622
1.60 0.680 2.917 8.00 2.658 1.610
2.00 0.862 2.839 8.40 2.720 1.482 1.561
2.40 0.997 2.714 8.80 2.744 1.329 1.440
2.80 1.170 2.681 1.682 9.20 2.760 1.140 1.288
3.20 1.342 2.570 1.760 9.60 2.855 1.018 1.202
3.60 1.495 2.505 1.858 10.00 2.920 0.863 1.177
4.00 1.677 2.442 1.881 10.40 2.980 0.660 1.082
4.40 1.821 2.380 1.962 10.80 3.065 0.503
4.80 1.942 2.318 2.053 11.20 3.168 0.340
5.20 2.103 2.220 11.60 3.230 0.198
5.60 2.150 2.125 12.00 3.310 0.000
6.00 2.208 2.030

The refractor is indicated in Figure 11.13b by the dashed line. The variation in the spacing of the vertical minus lines is very slight so that we can assume that $V_{2}$ is constant.