# Comparison of refraction interpretation methods

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

The data in Table 11.14a show refraction traveltimes for geophones spaced 400 m a part between sources ${\displaystyle A}$ and ${\displaystyle B}$ which are separated by 12 km. The columns in the table headed ${\displaystyle t_{A}^{*}}$ and ${\displaystyle t_{B}^{*}}$ give second arrivals.

## Problem 11.14a

Interpret the data using the basic refraction equations (4.24a) to (4.24f).

### Solution

The data are plotted in Figure 11.14a and best-fit lines suggest that this is a two-layer problem. Measurements give the following values:

{\displaystyle {\begin{aligned}V_{1}=2.88\ {\hbox{km/s (average value)}},\\V_{d}=4.65\ {\rm {km/s}},\quad V_{u}=5.71\ {\rm {km/s}},\quad t_{iu}=1.21\ {\rm {s}},\quad t_{id}=0.73\ {\rm {s}}.\end{aligned}}}

Equation (4.24d) gives

{\displaystyle {\begin{aligned}V_{d}=4.65=2.38/\sin \left(\theta _{c}+\xi \right),\quad V_{u}=5.71=2.38/\sin \left(\theta _{c}-\xi \right),\\\left(\theta _{c}+\xi \right)=30.8^{\circ },\quad \left(\theta _{c}-\xi \right)=24.6^{\circ },\quad \theta _{c}=27.7^{\circ },\quad \xi =3.1^{\circ }.\end{aligned}}}

Figure 11.14a.  Plot of the time-distance data.

From equation (4.24f), we have ${\displaystyle V_{2}\approx 2(1/V_{d}+1/V_{u})^{-1}\approx 5.13}$ km/s. From equation (4.24b) we get for the slant depths,

{\displaystyle {\begin{aligned}h_{d}=V_{1}t_{id}/2\cos \theta _{c}=0.98\ {\rm {km}},\quad h_{u}=V_{1}t_{iu}/2\cos \theta _{c}=1.63\ {\rm {km}}.\end{aligned}}}

Checking the values of ${\displaystyle \theta _{c}}$ and ${\displaystyle \xi }$, we obtain

{\displaystyle {\begin{aligned}\theta _{c}=\sin ^{-1}\left(2.38/5.13\right)=27.6^{\circ },\quad \xi =\tan ^{-1}\left[\left(1.63-0.98\right)/12.0\right]=3.1^{\circ }.\end{aligned}}}

## Problem 11.14b

Interpret the data using Tarrant’s method.

### Background

Tarrant’s method (Tarrant, 1956) uses delay times (problem 11.8) to locate the point ${\displaystyle Q}$ [see Figure 11.14b(i)] where the refracted energy that arrives at geophone ${\displaystyle R}$ leaves the refractor. The refractor is defined by finding ${\displaystyle Q}$ for a series of geophone positions. Tarrant’s method is based on the properties of the ellipse.

The delay time for the path ${\displaystyle QR}$ in Figure 11.14b(i) is ${\displaystyle \delta _{g}=\rho /V_{1}-\left(\rho \cos \theta \right)/V_{2}}$. Solving for ${\displaystyle \rho }$, we get

 {\displaystyle {\begin{aligned}\rho =V_{1}\delta _{g}/\left(1-\sin \theta _{c}\cos \phi \right).\end{aligned}}} (11.14a)

This is the polar equation of an ellipse. An ellipse is traced out by a point moving so that the ratio of the distance from a straight line (directrix) to that from a fixed point (${\displaystyle R}$ in Figure 11.14b(ii) is a constant ${\displaystyle \varepsilon }$ (eccentricity)).

The standard polar equation of an ellipse is

 {\displaystyle {\begin{aligned}\rho =\varepsilon h/\left(1-\varepsilon \cos \phi \right).\end{aligned}}} (11.14b)
Figure 11.14b.  Illustrating Tarrant’s method. (i) Relation between ${\displaystyle R}$ and ${\displaystyle Q}$; (ii) locus of ${\displaystyle Q}$ is ellipse, focus at ${\displaystyle R}$; (iii) geometry of ellipse through Q.

In Figure 11.14b(ii) ${\displaystyle Q}$ moves so that the ratio ${\displaystyle QR/QM=\varepsilon =\rho /\left(\rho \cos \phi +h\right)<1}$. The major axis 2a of the ellipse is

 {\displaystyle {\begin{aligned}2a=\rho _{\phi =0}+\rho _{\phi =\pi }\\=\varepsilon h/\left(1-\varepsilon \right)+\varepsilon h/\left(1+\varepsilon \right)\\=2\varepsilon h/\left(1-\varepsilon ^{2}\right).\end{aligned}}} (11.14c)

To get the minor axis, we set the first derivative of ${\displaystyle 2b=2\rho \sin \phi }$ equal to zero. Using equation (11.14b), we find that

 {\displaystyle {\begin{aligned}2b=2\varepsilon h/(1-\varepsilon ^{2})^{1/2},\end{aligned}}} (11.14d)

The distance from the focal point ${\displaystyle R}$ to the center ${\displaystyle O}$ is

{\displaystyle {\begin{aligned}OR&=\left(\rho \phi =0-a\right)\\&=\varepsilon h/\left(1-\varepsilon \right)-\varepsilon h/\left(1-\varepsilon ^{2}\right)\varepsilon a.\end{aligned}}}

If we substitute ${\displaystyle \varepsilon =\sin \theta _{c}}$, and ${\displaystyle h=V_{2}\delta _{g}}$ in equation (11.14b), we get equation (11.14a). Also these values give the following results.

 {\displaystyle {\begin{aligned}a=V_{1}\delta _{g}/{\rm {cos}}^{2}\theta _{c}\\&=V_{2}\delta _{g}\tan \theta _{c}/\cos \theta _{c}\end{aligned}}} (11.14e)

 {\displaystyle {\begin{aligned}&=OQ=V_{2}\delta _{g}\sin \theta _{c}/\cos \theta _{c}=V_{2}\delta _{g}\tan \theta _{c},\end{aligned}}} (11.14f)

 {\displaystyle {\begin{aligned}OR&=\varepsilon a=V_{2}\delta _{g}\tan ^{2}\theta _{c},\end{aligned}}} (11.14g)

 {\displaystyle {\begin{aligned}OQR&=\tan ^{-1}\left(OR/b\right)=\tan ^{-1}\left(V_{2}\delta _{g}\tan ^{2}\theta _{c}/V_{2}\delta _{g}\tan \theta _{c}\right)=\theta _{c}\;,\end{aligned}}} (11.14h)

 {\displaystyle {\begin{aligned}OC&=OR\;\tan =V_{2}\delta _{g}\tan ^{3}\theta _{c}.\end{aligned}}} (11.14i)

To approximate the ellipse in the vicinity of ${\displaystyle Q}$ with a circle, we need to find the center of curvature of the ellipse. The general equation of an ellipse is

 {\displaystyle {\begin{aligned}(x/a)^{2}+(y/b)^{1}=1.\end{aligned}}} (11.14j)

The equation for the radius of curvature of a function ${\displaystyle y\left(x\right)}$ is

{\displaystyle {\begin{aligned}r=[1+\left(y)^{2}\right]/y''.\end{aligned}}}

Differentiating, we obtain

{\displaystyle {\begin{aligned}y'=-(b/a)^{2}\left(x/y\right);\quad y''=-(b/a)^{2}\left[1/y-\left(x/y^{2}\right)y'\right].\end{aligned}}}

{\displaystyle {\begin{aligned}{\hbox{At Q,}}\quad \quad x=0,y=-b,\ {\rm {so}}\ y'=0,y''=\left(b/a^{2}\right),r=a^{2}/b=V_{2}\delta _{g}\tan \theta _{c}/\cos ^{2}\theta _{c}.\end{aligned}}}

The center of curvature is a distance ${\displaystyle r}$ above ${\displaystyle Q}$, so the ${\displaystyle y}$-coordinate of ${\displaystyle C}$ [Figure 11.14b(iii)] is ${\displaystyle \left(r-b\right)=\left(V_{2}\delta _{g}\tan \theta _{c}/\cos ^{2}\theta _{c}-V_{2}\delta _{g}\tan \theta _{c}\right)=V_{2}\delta _{g}\tan ^{3}\theta _{c}}$. A circle with center ${\displaystyle C}$ and radius ${\displaystyle r}$ will approximate the ellipse in the vicinity of ${\displaystyle Q}$.

### Solution

We need the total delay time at source ${\displaystyle A}$, ${\displaystyle \delta _{SA}}$, and the delay times ${\displaystyle \delta _{g}}$ at the geophones where the head wave is observed.

We have from part (a): ${\displaystyle V_{1}=2.38}$ km/s, ${\displaystyle V_{2}=5.13}$ km/s, ${\displaystyle \theta _{c}=27.7^{\circ }}$; from equation(11.9b), we get ${\displaystyle \delta _{SA}=t_{iA}/2=0.60}$ s, ${\displaystyle \delta _{SB}=0.36}$ s. For a geophone ${\displaystyle R}$ at a distance ${\displaystyle x}$ from ${\displaystyle A}$, equation (11.8b) gives for source ${\displaystyle A}$,

{\displaystyle {\begin{aligned}\delta _{g}=t_{R}-x/V_{2}-\delta _{SA}=t_{R}-\left(x/5.13+0.60\right),\end{aligned}}}

and for source ${\displaystyle B}$,

{\displaystyle {\begin{aligned}\delta _{g}=t_{R}-x/V_{2}-\delta _{SB}=t_{R}-\left(x/5.13+0.36\right)\end{aligned}}}

(note that ${\displaystyle x}$ is measured from ${\displaystyle A}$ for ${\displaystyle \delta _{SA}}$ and from ${\displaystyle B}$ for ${\displaystyle \delta _{SB}}$).

We can obtain values of ${\displaystyle \delta _{g}}$ either by using the above equations or graphically by drawing straight lines with slope ${\displaystyle 1/V_{2}}$ starting at the half-intercept values (there by subtracting it); the vertical distances between these lines and the traveltime curves give ${\displaystyle \delta _{g}}$. The values of ${\displaystyle \delta _{g}}$ in Tables 11.14b,c were calculated.

Table 11.14b. Calculations of ${\displaystyle OC}$ and ${\displaystyle r}$ for source ${\displaystyle A}$.
${\displaystyle x_{A}}$(km) ${\displaystyle T}$ (s) ${\displaystyle \delta _{g}}$ (s) ${\displaystyle OQ}$ (km) ${\displaystyle OR}$ (km) ${\displaystyle OC}$ (km) ${\displaystyle r}$ (km)
2.80 1.15 0.53 1.43 0.75 0.39 1.82
3.20 1.22 0.54 1.45 0.76 0.40 1.85
3.60 1.30 0.56 1.51 0.79 0.41 1.92
4.00 1.38 0.50 1.34 0.70 0.37 1.71
4.40 1.46 0.50 1.34 0.70 0.37 1.71
4.80 1.54 0.51 1.37 0.72 0.38 1.75
5.20 1.62 0.48 1.29 0.68 0.36 1.65
5.60 1.69 0.46 1.24 0.65 0.34 1.58
6.00 1.77 0.44 1.18 0.62 0.33 1.51
6.40 1.85 0.48 1.29 0.68 0.36 1.65
6.80 1.93 0.49 1.32 0.69 0.36 1.68
7.20 2.01 0.49 1.32 0.69 0.36 1.68
7.60 2.08 0.52 1.40 0.73 0.39 1.79
8.00 2.16 0.50 1.34 0.70 0.37 1.71
8.40 2.24 0.48 1.29 0.68 0.36 1.65
8.80 2.32 0.42 1.13 0.59 0.31 1.44
9.20 2.40 0.36 0.97 0.51 0.27 1.24
9.60 2.48 0.38 1.02 0.54 0.28 1.30
10.00 2.55 0.37 1.00 0.52 0.27 1.27
10.40 2.63 0.35 0.94 0.49 0.26 1.20
10.80 2.71 0.36 0.97 0.51 0.27 1.24
11.20 2.79 0.38 1.02 0.54 0.28 1.30
11.60 2.87 0.36 0.97 0.51 0.27 1.24
12.00 2.94 0.37 1.00 0.52 0.27 1.27

The last step is to find the center of curvature ${\displaystyle C}$ in Figure 11.14b(iii) and to draw an arc with radius ${\displaystyle r=CQ}$. We first calculate ${\displaystyle OQ}$ and then find ${\displaystyle C}$ by calculating ${\displaystyle OC}$ or ${\displaystyle OR}$ and drawing a line normal to ${\displaystyle RQ}$. The first method was used to get Tables 11.14b,c (although ${\displaystyle OR}$ is given in the tables, it was not used). We repeat equations (11.14f,g,h) and get

 {\displaystyle {\begin{aligned}OQ&=V_{2}\delta _{g}\tan \theta _{c}=2.69\delta _{g},\end{aligned}}} (11.14l)

 {\displaystyle {\begin{aligned}OR&=V_{2}\delta _{g}\tan ^{2}\theta _{c}=1.41\delta _{g},\end{aligned}}} (11.14m)

 {\displaystyle {\begin{aligned}OC&=V_{2}\delta _{g}\tan ^{3}\theta _{c}=0.74\delta _{g}.\end{aligned}}} (11.14n)

The calculations are shown in Tables 11.14b,c. The quantity ${\displaystyle T=\left(x_{A}/5.12+0.60\right)}$ in Table 11.14b and ${\displaystyle \left(x_{B}/5.12+0.36\right)}$ in Table 11.14c. Because complete reversed profiles were obtained, there is considerable duplication of the values of ${\displaystyle OC}$ and ${\displaystyle r}$. Rather than plot all of the arcs, we used the average values of ${\displaystyle OC}$ and ${\displaystyle r}$ (calculated in Table 11.14d). The results are shown in Figure 11.14c.

## Problem 11.14c

Interpret the data in Table 11.14a using the wavefront method illustrated in Figure 11.14d

### Background

In Figure 11.14d(i) ${\displaystyle MCD}$ and ${\displaystyle PCE}$ are two wavefronts generated at sources ${\displaystyle A}$ and ${\displaystyle B}$, respectively, and meeting at C. Clearly, ${\displaystyle t_{AC}+t_{BC}=t_{AB}=t_{r}}$. If wavefronts ${\displaystyle MC}$ and ${\displaystyle CP}$ continue upward to the surface at velocity ${\displaystyle V_{1}}$ and are recorded, we can project them backwards using the method shown in Figure 11.14d(ii). Point ${\displaystyle C}$ where they meet locates a point on the refractor. This is the basis of the wavefront method.

Table 11.14c. Calculations of ${\displaystyle OC}$ and ${\displaystyle r}$ for source ${\displaystyle B}$.
${\displaystyle x_{A}}$(km) ${\displaystyle T}$(s) ${\displaystyle \delta _{g}}$(s) ${\displaystyle OQ}$(km) ${\displaystyle OR}$(km) ${\displaystyle OC}$(km) ${\displaystyle r}$(s)
1.60 0.67 0.41 1.10 0.58 0.30 1.40
2.00 0.75 0.43 1.16 0.61 0.32 1.48
2.40 0.83 0.37 1.00 0.52 0.27 1.27
2.80 0.91 0.38 1.02 0.54 0.28 1.30
3.20 0.98 0.46 1.24 0.65 0.34 1.58
3.60 1.06 0.50 1.34 0.70 0.37 1.71
4.00 1.14 0.47 1.26 0.66 0.35 1.61
4.40 1.22 0.40 1.08 0.56 0.30 1.38
4.80 1.30 0.44 1.18 0.62 0.33 1.51
5.20 1.38 0.48 1.29 0.68 0.36 1.65
5.60 1.45 0.55 1.48 0.78 0.41 1.89
6.00 1.53 0.50 1.34 0.70 0.37 1.71
6.40 1.61 0.51 1.37 0.72 0.38 1.75
6.80 1.69 0.53 1.43 0.75 0.39 1.82
7.20 1.77 0.55 1.48 0.78 0.41 1.89
7.60 1.84 0.54 1.45 0.76 0.40 1.85
8.00 1.92 0.52 1.40 0.73 0.39 1.79
8.40 2.00 0.50 1.34 0.70 0.37 1.71
8.80 2.08 0.49 1.32 0.69 0.36 1.68
9.20 2.16 0.52 1.40 0.73 0.39 1.79
9.60 2.24 0.47 1.26 0.66 0.35 1.67
10.00 2.31 0.53 1.45 0.75 0.39 1.82
10.40 2.39 0.53 1.43 0.75 0.39 1.82
10.80 2.47 0.59 1.59 0.83 0.44 2.03
11.20 2.55 0.59 1.59 0.83 0.44 2.03
11.60 2.63 0.55 1.48 0.78 0.41 1.89
12.00 2.70 0.61 1.64 0.86 0.45 2.09
Table 11.14d. Calculating average values of ${\displaystyle OC}$ and ${\displaystyle r.}$
${\displaystyle A}$ ${\displaystyle A}$ ${\displaystyle B}$ ${\displaystyle B}$ ${\displaystyle Av}$ ${\displaystyle Av}$
${\displaystyle x_{A}}$ ${\displaystyle OC}$ ${\displaystyle r}$ ${\displaystyle OC}$ ${\displaystyle r}$ ${\displaystyle OC}$ ${\displaystyle r}$
0.00 0.00 0.00 0.45 2.09 0.45 2.09
0.40 0.41 1.89 0.41 1.89
0.80 0.44 2.03 0.44 2.03
1.20 0.44 2.03 0.44 2.03
1.60 0.39 1.82 0.39 1.82
2.00 0.39 1.82 0.39 1.82
2.40 0.35 1.61 0.35 1.61
2.80 0.39 1.82 0.39 1.79 0.39 1.80
3.20 0.40 1.85 0.36 1.68 0.38 1.76
3.60 0.41 2.92 0.37 1.71 0.39 1.82
4.00 0.37 1.71 0.39 1.79 0.38 1.75
4.40 0.37 1.71 0.40 1.85 0.38 1.78
4.80 0.38 1.75 0.41 1.89 0.39 1.82
5.20 0.36 1.65 0.39 1.62 0.38 1.64
5.60 0.34 1.58 0.38 1.75 0.36 1.66
6.00 0.33 1.51 0.37 1.71 0.35 1.61
6.40 0.36 1.65 0.41 1.89 0.38 1.77
6.80 0.35 1.68 0.36 1.65 0.36 1.66
7.20 0.36 1.68 0.33 1.51 0.34 1.60
7.60 0.39 1.79 0.30 1.38 0.34 1.59
8.00 0.37 1.71 0.35 1.61 0.36 1.66
8.40 0.36 1.65 0.37 1.71 0.36 1.68
8.80 0.39 1.44 0.34 1.58 0.36 1.51
9.20 0.27 1.24 0.28 1.30 0.28 1.27
9.60 0.28 1.30 0.27 1.27 0.28 1.28
10.00 0.27 1.27 0.32 1.48 0.30 1.38
10.40 0.26 1.20 0.30 1.40 0.28 1.30
10.80 0.27 1.24 0.27 1.24
11.20 0.28 1.30 0.28 1.30
11.60 0.27 1.24 0.27 1.24
12.00 0.27 1.27 0.27 1.27
Figure 11.14c.  Solution by Tarrant’s method; the small circles at the top are centers of the arcs. The wavefront solution is shown by small squares, the formula solution by the dashed line.
Figure 11.14d.  Illustrating the wave-front method. (i) Two wavefronts where ${\displaystyle t_{AC}+t_{BC}=t_{r}}$; (ii) reconstructing wavefronts.

### Solution

The earliest refracted wavefront from ${\displaystyle A}$ that we can reconstruct is at 1.60 s (using the best-fit line to get ${\displaystyle x}$ for ${\displaystyle t=1.60}$) and the last is about 3.00 s; the corresponding limits for source ${\displaystyle B}$ are 1.10 and 3.00 s. We take ${\displaystyle \Delta =0.20}$ s and draw wavefronts such that ${\displaystyle \left(t_{A}+t_{B}\right)=t_{r}=3.31\ {\hbox{s}}}$. We reconstruct the four wavefront pairs ${\displaystyle (t_{A},t_{B})=(1.60,1.71)}$, (1.80, 1.51), (2.00, 1.31), (2.20, 1.11) using ${\displaystyle V_{1}=2.38}$ km/s [from part (a)].

We swing arcs from points on the ${\displaystyle x}$-axis as in Figure 11.14d(ii). Because we are interested only in the portions near the points of intersection, we determine the refractor depth. Using the intercept times at the sources in part (a) to obtain slant depths, we get ${\displaystyle h_{A}\approx 1.6}$ km, ${\displaystyle h_{B}\approx 1.0}$ km; multiplying by ${\displaystyle \cos \theta _{c}\approx 0.9}$ to get vertical depths; the maximum vertical depth is about 1.40 km.

The radius of the arcs is given by ${\displaystyle R=V_{1}\Delta t=2.38\Delta t}$. The maximum value of ${\displaystyle \Delta t}$ is about ${\displaystyle (1.40/2.38)\approx 0.6\ {\hbox{s}}}$ but, to be on the safe side, we calculate ${\displaystyle R}$ for values of ${\displaystyle \Delta t}$ ranging from 0.45 to 0.75 s for the pair (1.60, 1.71), then adjust the range as necessary to achieve wavefront intersections for the other three pairs.

Table 11.14e. Calculation of wavefront radii for sources ${\displaystyle A}$ and ${\displaystyle B}$.
Wavefront: (A-1.600) Wavefront: (B-1.710)
${\displaystyle x_{A}}$ (km) ${\displaystyle t_{A}(s)}$ ${\displaystyle \Delta t(s)}$ ${\displaystyle R}$ (km) ${\displaystyle x_{A}}$ (km) ${\displaystyle t_{A}}$ (s) ${\displaystyle \Delta t}$ (s) ${\displaystyle R}$ (km)
4.80 1.942 0.453 1.08 4.00 2.442 0.732 1.74
5.20 2.103 0.503 1.20 4.40 2.380 0.670 1.59
5.60 2.150 0.550 1.31 4.80 2.318 0.608 1.45
6.00 2.208 0.608 1.45 5.20 2.220 0.510 1.21
6.40 2.330 0.730 1.74 5.60 2.125 0.415 0.99
Wavefront: (A-1.800) Wavefront: (B-1.510)
6.00 2.208 0.408 0.97 4.80 2.318 0.808 1.92
6.40 2.330 0.530 1.26 5.20 2.220 0.710 1.69
6.80 2.422 0.622 1.48 5.60 2.125 0.615 1.46
7.20 2.504 0.704 1.68 6.00 2.030 0.520 1.24
7.60 2.602 0.802 1.91 6.40 2.003 0.493 1.17
Wavefront: (A-2.000) Wavefront: (B-1.310)
6.80 2.422 0.422 1.00 6.00 2.030 0.720 1.71
7.20 2.504 0.504 1.20 6.40 2.003 0.693 1.65
7.60 2.602 0.602 1.43 6.80 1.802 0.552 1.31
8.00 2.658 0.658 1.57 7.20 1.743 0.433 1.03
8.40 2.720 0.720 1.71 7.60 1.622 0.312 0.74
Wavefront: (A-2.200) Wavefront: (B-1.110)
8.40 2.720 0.520 1.24 6.40 2.003 0.893 2.13
8.80 2.744 0.544 1.29 6.80 1.862 0.752 1.79
9.20 2.760 0.560 1.33 7.20 1.743 0.633 1.51
9.60 2.855 0.655 1.56 7.60 1.622 0.512 1.22
10.00 2.920 0.720 1.71

The calculations are shown in Table 11.14e. Columns 1 and 2 in each sub-table come from table 11.14a. Column 3 is the difference between ${\displaystyle t_{A}}$ or ${\displaystyle t_{B}}$ and the wavefront traveltime shown above each subtable; the radius ${\displaystyle R=V_{1}\Delta t=2.38\Delta t}$.

Figure 11.14e shows the result for the wavefront pair (1.60, 1.71). We leave to the reader the construction of the other three pairs of wavefronts and determination of the points of intersection. Our results are shown by the four small squares in Figure 11.14c and, more clearly, in Figure 11.14g.

File:Ch11 fig11-14e.png
Figure 11.14e.  Determining the point of intersection of wavefronts (1.60,1.71).

## Problem 11.14d

Interpret the data in Table 11.14a using Hales’s method

### Background

Hales’s (1958) method is a graphical method based on reversed profiles. It enables us to find ${\displaystyle Q}$ in Figure 11.14f(i) in terms of data recorded at geophones ${\displaystyle R}$ and ${\displaystyle S}$. The circumscribed circle through points ${\displaystyle R}$, ${\displaystyle S}$ and ${\displaystyle Q}$ is shown in Figure 11.14f(iv), all of the angles inside the circle being expressible in terms of ${\displaystyle \theta _{c}}$ and ${\displaystyle \xi }$.

Point ${\displaystyle C}$ is on the perpendicular bisector of ${\displaystyle RS}$; writing ${\displaystyle \rho =CQ}$, we have {\displaystyle {\begin{aligned}\rho \cos \theta _{c}=QN=RQ-RN,\quad {\rm {but}}\quad QN=QG,\end{aligned}}}

{\displaystyle {\begin{aligned}{\hbox{so}}\qquad \qquad \rho \cos \theta _{c}=QG=SQ+SG.\end{aligned}}}

Adding the two expressions, we get

{\displaystyle {\begin{aligned}2\rho \cos \theta _{c}=\left(RQ+SQ\right)+\left(SG-RN\right).\end{aligned}}}

Since ${\displaystyle RN=CN\tan \xi =CG\tan \xi =SG}$,

 {\displaystyle {\begin{aligned}\rho =\left(RQ+SQ\right)/2\cos \theta _{c}.\end{aligned}}} (11.14o)

Also we have

{\displaystyle {\begin{aligned}t_{AS}+t_{BR}=t_{r}+\left(RQ+SQ\right)/V_{1}=t_{r}+t',\end{aligned}}}

 {\displaystyle {\begin{aligned}{\hbox{so}}\qquad \qquad t_{AS}=\left(t_{r}-t_{BR}\right)+t{'},\end{aligned}}} (11.14p)

where ${\displaystyle t_{r}=}$ reciprocal time (see problem 11.13). Thus we have

 {\displaystyle {\begin{aligned}t'=\left(RQ+SQ\right)/V_{1},\quad \rho =V_{1}t'/2\cos \theta _{c},\end{aligned}}} (11.14q)
Figure 11.14f.  Hales’s graphical method. (i) Relation between two receivers ${\displaystyle R}$ and ${\displaystyle S}$ having a common emergent point ${\displaystyle Q}$; (ii) geometrical properties of points on the traveltime curves corresponding to sources ${\displaystyle A}$ and ${\displaystyle B}$; (iii) construction for locating ${\displaystyle Q}$; (iv) properties of circumscribed circle through ${\displaystyle R}$, ${\displaystyle S}$, and ${\displaystyle Q}$; (v) offset due to errors in ${\displaystyle x'}$ in part (ii).

using equation (11.14g). Using the law of sines and equation (11.14i) we get

 {\displaystyle {\begin{aligned}x'=RS=RH+HS={\frac {RQ\sin \theta _{c}}{\sin \left(\pi /2+\xi \right)}}+{\frac {SQ\sin \theta _{c}}{\sin \left(\pi /2-\xi \right)}}\\={\frac {\left(RQ+SQ\right)\sin \theta _{c}}{\cos \xi }}=V_{1}t'\sin \theta _{c}/\cos \xi ,\end{aligned}}} (11.14r)

 {\displaystyle {\begin{aligned}{\hbox{so}}\qquad \qquad x'/t'=\tan \alpha =V_{1}\sin \theta _{c}/\cos \xi =V_{1}\sin \theta _{c},\quad {\rm {when}}\ \xi =0.\end{aligned}}} (11.14s)

The traveltime curves are shown in Figure 11.14f(ii). We draw the vertical through point ${\displaystyle S}$ on the ${\displaystyle x}$-axis and locate the point ${\displaystyle K}$ so that ${\displaystyle KS=\left(t_{r}-t_{AS}\right)}$. Assuming ${\displaystyle \xi =0}$, we draw a line at ${\displaystyle K}$ at the angle ${\displaystyle \alpha }$ to meet the other traveltime curve. From equations (11.14h,j,k) we see that the line from ${\displaystyle K}$ at the angle ${\displaystyle \alpha }$ must intersect the traveltime curve for source ${\displaystyle B}$ at time ${\displaystyle t_{BR}}$, which locates the point ${\displaystyle R}$ and gives the values of ${\displaystyle t_{BR},x'}$, and, ${\displaystyle t'}$. At point ${\displaystyle R}$ on the ${\displaystyle x}$-axis we draw a line at angle ${\displaystyle \theta _{c}}$ to the horizontal [see Figure 11.14f(iii)] and locate point ${\displaystyle C}$ vertically above the midpoint of ${\displaystyle RS}$, i.e., at the distance ${\displaystyle x'/2}$ from ${\displaystyle R}$; With center ${\displaystyle C}$ and radius ${\displaystyle \rho =V_{1}t'/2\cos \theta _{c}}$. [see equation (11.14i)], we draw an arc. Repeating this process for a series of points S, we obtain several arcs to define the refracting surface.

When ${\displaystyle \xi \neq 0,x',t'}$, and ${\displaystyle \rho }$ change, but it can be shown (see Sheriff and Geldart, 1995, 444) that the change in ${\displaystyle \rho }$ is negligible and the only effect of dip is to displace the point ${\displaystyle Q}$ updip the distance ${\displaystyle \Delta x'/2}$ [Figure 11.14e(v)], which is usually negligible for moderate dips.

Figure 11.14g.  Illustrating Hales’s solution. (i) Determining values of ${\displaystyle t_{BR},x'}$, and ${\displaystyle t'}$; (ii) locating ${\displaystyle C}$ and drawing arcs of radius ${\displaystyle \rho }$.

### Solution

From part (a) we have: ${\displaystyle V_{1}=2.38}$ km/s, ${\displaystyle V_{2}=5.12}$ km/s, ${\displaystyle \theta _{c}=27.7^{\circ }}$, the reciprocal time ${\displaystyle t_{r}=3.310}$ s.

Figure 11.14g shows details of the solution. We select points ${\displaystyle S}$ at intervals of 0.80 km, then find ${\displaystyle \left(t_{r}-t_{AS}\right)}$ and the points ${\displaystyle K}$ [shown as small triangles (${\displaystyle \Delta }$) in Figure 11.14g(i)]. Points ${\displaystyle R}$ in Figure 11.14f(i) are found by laying off at ${\displaystyle K}$ the angle ${\displaystyle \alpha =\tan ^{-1}\left(V_{1}\sin \theta _{c}\right)=\tan ^{-1}\left(2.38\times \sin 27.7^{\circ }\right)=\tan ^{-1}1.11}$ [see equation (11.14s)]; to take into account the scale factors, we write ${\displaystyle \alpha =\tan ^{-1}}$(2.22/2.00), then draw a vertical line equal to 2.00 s, then a lineto the left equivalent to 2.22 km [see lines going up at about ${\displaystyle 10^{\circ }-15^{\circ }}$ from the vertical from the triangles in Figure 11.14g(i)]. This gives the locations of points ${\displaystyle R}$ and values of ${\displaystyle t_{BR},x'}$, and ${\displaystyle t'}$; we can now calculate ${\displaystyle \rho }$ from equation (11.14g). The calculations are shownin Table 11.14f.

Finally we locate ${\displaystyle C}$ as in Figure 11.14g(ii), with ${\displaystyle C}$ as center and radius ${\displaystyle \rho }$ we draw arcs which pass through point ${\displaystyle Q}$ in Figure 11.14f(v). The refractor is fairly well defined by the arcs (except for two arcs marked with?).

## Problem 11.14f

On the basis of your results, compare the methods in terms of (1) time involved; (2) effect of refractor curvature; (3) effect of random errors; (4) suitability for routine production; and (5) for special effort where high accuracy is essential.

### Solution

1. Time involved: The formula method is by far the quickest, the wavefront method is next, and Tarrant’s and Hales’s methods are the most time consuming, being more-or-less the same in this respect.
2. Effect of refractor curvature: The formula method does not take curvature into account (except on a broad scale over two or more profiles). The remaining methods all work well for curved refractors. Tarrant’s and Hales’s methods give good results over the commonly observed angle of curvature, while the wavefront method is useable over a smaller range of curvatures.
3. Effect of refractor random errors: These errors affect the measured slopes and intercepts and therefore affect the formula solution; however, the effects are usually minimized by the use of best-fit lines which utilize most of the available data. Tarrant’s and Hales’s methods use ${\displaystyle V_{1}}$ and ${\displaystyle V_{2}}$ and the wavefront method uses ${\displaystyle V_{1}}$, so that errors affect all three methods. In addition Tarrant’s and Hales’s methods use two traveltimes in each calculation so that random errors cause further errors. The wavefront method is less susceptible to this type of error because several time values enter into each wavefront determination.
4. Suitability for routine production: The formula method, being the quickest, is satisfactory where refractor relief is minimal. Tarrant’s and Hales’s methods are slightly less suitable, and the wavefront method is least suitable.
5. Suitability for high accuracy: The formula method is unsuitable, the wavefront method the best, Tarrant’s and Hales’s methods being almost as good.

Table 11.14f. Calculating position of center ${\displaystyle C}$ and radius ${\displaystyle \rho }$ of arcs in Hales’s method.
S S S R R
${\displaystyle x}$ ${\displaystyle t_{AS}}$ ${\displaystyle (t_{r}-t_{RS})}$ ${\displaystyle x_{R}}$ ${\displaystyle t_{SA}}$ ${\displaystyle t'}$ ${\displaystyle x'}$ ${\displaystyle \rho }$ ${\displaystyle x'/2}$
11.20 3.17 0.14 10.10 1.14 1.00 1.10 1.34 0.55
10.40 2.98 0.33 9.40 1.25 0.92 1.00 1.24 0.50
9.60 2.86 0.45 8.38 1.56 1.11 1.22 1.49 0.61
8.80 2.74 0.57 7.65 1.62 1.05 1.15 1.41 0.57
8.00 2.66 0.65 6.57 1.95 1.30 1.43 1.75 0.71
7.20 2.50 0.81 5.78 2.08 1.27 1.42 1.71 0.71
6.40 2.33 0.98 4.95 2.28 1.30 1.45 1.75 0.72
5.60 2.15 1.16 4.22 2.40 1.24 1.38 1.67 0.69
4.80 2.05 1.26 3.37 2.54 1.28 1.43 1.72 0.71
4.00 1.88 1.43 2.57 2.70 1.27 1.43 1.71 0.71
3.20 1.76 1.55 1.70 2.90 1.35 1.50 1.81 0.75