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 11.5

Given the data in Table 11.5a for a reversed refraction profile with sources $A$ and $B$ , use Adachi’s method to find velocities, depths, and dips.

### Background

Adachi (1954; see also Johnson, 1976) derived equations for reversed refraction profiles similar to equations (4.18b,d) but with two important differences: he used angles of incidence measured relative to the vertical ($\alpha _{i}$ and $\beta _{i}$ in Figure 11.5a) and vertical depths. The equations are valid for a series of refractors of different dips but with the same strike. Derivation of his equations is lengthy but not difficult (see Sheriff and Geldart, 1995, Section 11.3.2); we quote the final results without proof.

The notation is illustrated in Figure 11.5a where $\alpha _{i}$ and $\beta _{i}$ are angles of incidence relative to the vertical at the $i^{\rm {th}}$ interface for the downgoing rays from sources $A$ and $B$ , respectively (these are angles of approach at the surface for $i=1$ ), $a_{i}$ and $a_{i}^{'}$ are the angles of incidence and refraction for the downgoing ray at interface $i$ , $b_{i}$ and $b_{i}^{'}$ are the same for the upcoming ray, $\xi _{i+1}$ is the dip of the $i^{\rm {th}}$ interface, $h_{i}$ is the vertical thickness of the bed below this interface below the downdip source.

The traveltime $t_{n}$ for the refraction along the top of the $n^{\rm {th}}$ layer is given by

 {\begin{aligned}t_{n}={\frac {x\sin \beta _{1}}{V_{1}}}+\sum \limits _{i=1}^{n-1}{\frac {h_{i}}{V_{i}}}\left(\cos \alpha _{i}+\cos \beta _{i}\right).\end{aligned}} (11.5a)

If we set $x=0$ , $t_{n}$ becomes the intercept time $t_{in}$ at the downdip source; thus,

 {\begin{aligned}t_{in}=\sum \limits _{i=1}^{n-1}{\frac {h_{i}}{V_{i}}}\left(\cos \alpha _{i}+\cos \beta _{i}\right).\end{aligned}} (11.5b)
 $x\to$ 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 (km) $t_{A}\to$ 0.00 0.25 0.50 0.74 0.98 1.24 1.50 1.70 1.81 1.91 2.02 (s) $t_{B}\to$ 3.00 2.90 2.80 2.68 2.52 2.41 2.31 2.20 2.07 1.91 1.80 (s) $x\to$ 5.5 6.0 6.5 7.0 7.5 8.0 8.5 9.0 9.5 10.0 (km) $t_{A}\to$ 2.16 2.28 2.38 2.44 2.56 2.64 2.72 2.80 2.89 3.00 (s) $t_{B}\to$ 1.65 1.50 1.40 1.25 1.12 1.00 0.75 0.49 0.23 0.00 (s)

The angles are related as follows:

 {\begin{aligned}\left.{\begin{array}{l}\alpha _{i+1}=a_{i}^{'}+\xi _{i+1}=\alpha _{i+1}+\xi _{i+2},\\\beta _{i+1}=b_{i}^{'}-\xi _{i+1}=b_{i+1}-\xi _{i+2}.\end{array}}\right\}\end{aligned}} (11.5c)

Snell’s law [equation (3. 1a)] gives

 {\begin{aligned}\sin \alpha _{i}'=\left(V_{i+1}/V_{i}\right)\sin a_{i},\quad \sin b_{i}'=\left(V_{i+1}/V_{i}\right)\sin b_{i}.\end{aligned}} (11.5d)

For the refraction along the $n^{\rm {th}}$ interface,

 {\begin{aligned}a_{n}=b_{n}=\theta _{cn}=\left(\alpha _{n}+\beta _{n}\right)/2,\qquad \xi _{n+1}=\left(\alpha _{n}-\beta _{n}\right)/2.\end{aligned}} (11.5e)

The initial interpretation stage is plotting the data and determining $V_{1}$ and the apparent velocities $V_{un}$ and $V_{dn}$ , and intercept times $t_{in}$ for each of the refraction events. The angles $\alpha _{1}$ and $\beta _{1}$ are given by equation (4.2d). Next we use problem 4.24b to get $\theta _{c1}$ and $\xi _{2}$ from $V_{1}$ , $\alpha _{1}$ , and $\beta _{1}$ . The depth $h_{1}$ is now found using equation (11.5b).

For the next interface we find new values of $\alpha _{1}$ and $\beta _{1}$ using the next pair of apparent velocities. Since $\xi _{2}$ is now known, we use equation (11.5c) to get new values of $a_{1}$ and $b_{1}$ , after which equation (11.5d) gives $a_{1'}$ , $b_{1'}$ and equation (5.11c) gives $\alpha _{3}$ , $\beta _{3}$ . We can now find $\theta _{c3}$ , $\xi _{3}$ , $V_{3}$ , and $h_{2}$ .

### Solution

Figure 11.5b shows the plotted data and the measured slopes and time intercepts. The average value of the near-surface velocity $V_{1}$ is 2.02 km/s. Two refraction events are observed with the apparent velocities and intercept times listed below.

{\begin{aligned}V_{d2}=3.73\ {\rm {km/s}},\quad V_{u2}=4.51\ {\rm {km/s}},\quad t_{i1}=0.92\ {\rm {s}};\\V_{d3}=4.29\ {\rm {km/s}},\quad V_{u3}=5.81\ {\rm {km/s}},\quad t_{i2}=1.28\ {\rm {s}}.\end{aligned}} First we calculate $\alpha _{1}$ and $\beta _{1}$ :

{\begin{aligned}\alpha _{1}=\sin ^{-1}\left(V_{1}/V_{d2}\right)=32.8^{\circ },\quad \beta _{1}=\sin ^{-1}\left(V_{1}/V_{u2}\right)=26.6^{\circ }.\end{aligned}} Equation (11.5c) gives $\alpha _{1}=a_{1}+\xi _{2}$ , $\beta _{1}=b_{1}-\xi _{2}$ . Since this interface is the refractor, equation (11.5e) gives

{\begin{aligned}a_{1}&=b_{1}=\theta _{c1}=\left(\alpha _{1}+\beta _{1}\right)/2=29.7^{\circ },\\\xi _{2}&=\left(\alpha _{1}-\beta _{1}\right)/2=3.1^{\circ },\\V_{2}&=V_{1}/\sin \theta _{c1}=2.02/\sin 29.7^{\circ }=4.08\ {\rm {km/s}}.\end{aligned}} {{\begin{aligned}{\hbox{Checking:}}\quad \quad V_{2}&=[(1/V_{d2}+1/V_{u3})^{2}/2]^{-1}=4.08\ {\rm {km/s}}.\end{aligned}} }

We find $h_{1}$ using equation (11.5b): so

{\begin{aligned}h_{1}&=V_{1}t_{i1}/\left(\cos \alpha _{1}+\cos \beta _{1}\right)\\&=2.02\times 0.92/\left(\cos 32.8^{\circ }+\cos 26.6^{\circ }\right)=1.07\ {\rm {km}}.\end{aligned}} For the second refractor, we calculate new angles of approach:

{\begin{aligned}\alpha _{1}=\sin ^{-1}\left(V_{1}/V_{d3}\right)=\sin ^{-1}\left(2.02/4.29\right)=28.1^{\circ },\\\beta _{1}=\sin ^{-1}\left(V_{1}/V_{u3}\right)=\sin ^{-1}\left(2.02/5.81\right)=20.3^{\circ }.\end{aligned}} Then equation (11.5c) gives

{\begin{aligned}a_{1}=\alpha _{1}-\xi _{2}=28.1^{\circ }-3.1^{\circ }=25.0^{\circ },\\b_{1}=\beta _{1}+\xi _{2}=20.3^{\circ }+3.1^{\circ }=23.4^{\circ }.\end{aligned}} Using equation (11.5d), we get

{\begin{aligned}a_{1^{'}}=\sin ^{-1}\left[\left(V_{2}/V_{1}\right)\sin a_{1}\right]=\sin ^{-1}[(4.08/2.02)\sin 25.0^{\circ }]=58.6^{\circ },\\b_{1^{'}}=\sin ^{-1}\left[\left(V_{2}/V_{1}\right)\sin b_{1}\right]=\sin ^{-1}[(4.08/2.02)\sin 23.4^{\circ }]=53.3^{\circ }.\end{aligned}} From equation (11.5c) we now get

{\begin{aligned}\alpha _{2}=a_{1^{'}}+\xi _{2}=58.6^{\circ }+3.1^{\circ }=61.7^{\circ },\\\beta _{2}=b_{1^{'}}-\xi _{2}=53.3^{\circ }-3.1^{\circ }=50.2^{\circ }.\end{aligned}} From equation (5.11e) we have

{\begin{aligned}a_{2}&=b_{2}=\theta _{c2}=\left(\alpha _{2}+\beta _{2}\right)/2=56.0^{\circ },\\\xi _{3}&=\left(\alpha _{2}-\beta _{2}\right)/2=5.8^{\circ },\\V_{3}&=V_{2}/\sin \theta _{c2}=4.08/\sin 56.0^{\circ }=4.92\ {\rm {km/s}}.\end{aligned}} {{\begin{aligned}{\hbox{Checking:}}\quad \quad V_{3}&=[\left(1/V_{d3}+1/V_{u3}\right)/2]^{-1}=4.94\ {\rm {km/s}}.\end{aligned}} }

Finally, we get the depth from equation (11.5b):

{\begin{aligned}t_{i2}&=\left(h_{1}/V_{1}\right)\left(\cos \alpha _{1}+\cos \beta _{1}\right)+\left(h_{2}/V_{2}\right)\left(\cos \alpha _{2}+\cos \beta _{2}\right)\\&=t_{u1}+\left(h_{2}/V_{2}\right)\left(\cos \alpha _{2}+\cos \beta _{2}\right),\\h_{2}&=\left(1.28-0.92\right)\times 4.08/\left(\cos 61.7^{\circ }+\cos 50.2^{\circ }\right)=1.32\ {\rm {km}}.\end{aligned}} Total vertical depth at $A=h_{1}+h_{2}=1.07+1.32=2.39$ km.