# Two-layer refraction problem

Series Geophysical References Series Problems in Exploration Seismology and their Solutions Lloyd P. Geldart and Robert E. Sheriff 4 79 - 140 http://dx.doi.org/10.1190/1.9781560801733 ISBN 9781560801153 SEG Online Store

## Problem 4.24a

Show that time-distance curves for dipping refractors take the form

 {\displaystyle {\begin{aligned}t_{d}&=(x/V_{1})\sin(\theta _{c}+\xi )+t_{1d},\\t_{u}&=(x/V_{1})\sin(\theta _{c}-\xi )+t_{1u},\end{aligned}}} (4.24a)

where

 {\displaystyle {\begin{aligned}t_{1d}&=(2h_{d}/V_{1})\cos \theta _{c},\\t_{1u}&=(2h_{u}/V_{1})\cos \theta _{c},\end{aligned}}} (4.24b)

${\displaystyle t_{d}}$ and ${\displaystyle t_{u}}$ being the traveltimes when shooting downdip and updip, respectively (see Figure 4.24a), and ${\displaystyle h_{d}}$, ${\displaystyle h_{u}}$, and ${\displaystyle t_{id}}$, ${\displaystyle t_{iu}}$, the corresponding slant depths and sourcepoint intercepts times.

### Background

Traveltime curves for horizontal refractors are discussed in problem 4.18.

Figure 4.24a.  Traveltime curves and raypaths for dipping refractor.

### Solution

For the downdip case, we take O as the sourcepoint and O${\displaystyle ^{\prime }}$ as the receiver. Following the procedure used in problem 4.18, we have

 {\displaystyle {\begin{aligned}t_{d}&=(OM+PO^{\prime })/V_{1}+MP/V_{2}\\&={\frac {h_{d}+h_{u}}{V_{1}\cos \theta _{c}}}+{\frac {OO^{\prime }-(h_{d}+h_{u})\tan \theta _{c}}{V_{2}}}\\&={\frac {x}{V_{2}}}+\left({\frac {h_{d}+h_{u}}{V_{1}\cos \theta _{1}}}\right)\left[1-{\frac {V_{1}}{V_{2}}}\sin \theta _{c}\right]\\&={\frac {x}{V_{2}}}+(h_{d}+h_{u})\cos \theta _{c}/V_{1}.\end{aligned}}} (4.24c)

But ${\displaystyle h_{u}=h_{d}+x\sin \xi }$, so we can express equation (4.24c) in terms of ${\displaystyle h_{d}}$:

{\displaystyle {\begin{aligned}t_{d}&=(x/V_{1})(\sin \theta _{c}\cos \xi +\cos \theta _{c}\sin \xi )+(2h_{d}/V_{1})\cos \theta _{c}\\&=(x/V_{1})\sin(\theta _{c}+\xi )+t_{id}.\end{aligned}}}

Expressing equation (4.24c) in terms of ${\displaystyle h_{u}}$, we obtain

{\displaystyle {\begin{aligned}t_{u}=(x/V_{1})\sin(\theta _{c}-\xi )+t_{1u}.\end{aligned}}}

The slopes of the two traveltime curves are sin (${\displaystyle \theta _{c}\pm \xi )/V_{1}}$, the reciprocals being the apparent velocities, ${\displaystyle V_{d}}$ and ${\displaystyle V_{u}}$ (see problem 4.2d), where

 {\displaystyle {\begin{aligned}V_{d}=V_{1}/\sin(\theta _{c}+\xi ),\quad V_{u}=V_{1}/\sin(\theta _{c}-\xi ).\end{aligned}}} (4.24d)

## Problem 4.24b

Show how to find ${\displaystyle V_{2}}$ and ${\displaystyle \xi }$ from the observed data.

### Solution

We obtain ${\displaystyle V_{1}}$, ${\displaystyle V_{d}}$, and ${\displaystyle V_{u}}$ from the slopes of the time-distance curves. From equation (4.24d), we get

{\displaystyle {\begin{aligned}\theta _{c}+\xi =\sin ^{-1}(V_{1}/V_{d}),\\\theta _{c}-\xi =\sin ^{-1}(V_{1}/V_{u}).\end{aligned}}}

Adding and subtracting the two equations gives ${\displaystyle \theta _{c}}$ and ${\displaystyle \xi }$. Since sin ${\displaystyle \theta _{c}=(V_{1}/V_{2})}$ and ${\displaystyle V_{1}}$ is known, we can find ${\displaystyle V_{2}}$

The dip ${\displaystyle \xi }$ can also be found (usually more accurately) from the relation

 {\displaystyle {\begin{aligned}\tan \xi =(h_{d}-h_{u})/x.\end{aligned}}} (4.24e)

## Problem 4.24c

Show that ${\displaystyle V_{2}}$ is given approximately by either of the following equations, the latter being less accurate:

 {\displaystyle {\begin{aligned}{\frac {1}{V_{2}}}\approx {\frac {1}{2}}\left({\frac {1}{V_{d}}}+{\frac {1}{V_{u}}}\right),\quad V_{2}\approx {\frac {1}{2}}(V_{d}+V_{u}).\end{aligned}}} (4.24f)

### Solution

Expanding equation (4.24d) we have

{\displaystyle {\begin{aligned}\sin \theta _{c}\cos \xi +\cos \theta _{c}\sin \xi =V_{1}/V_{d},\\\sin \theta _{c}\cos \xi -\sin \theta _{c}\cos \xi =V_{1}/V_{u}.\end{aligned}}}

Adding the two equations, we get

{\displaystyle {\begin{aligned}2\sin \theta _{c}\cos \xi =(1/V_{d}+1/V_{u}).\end{aligned}}}

Because ${\displaystyle \xi }$ is usually small, we set ${\displaystyle \cos \xi =1}$. Since ${\displaystyle \sin \theta _{c}=V_{1}/V_{2}}$, we get the first result in equation (4.24f).

Returning to equation (4.24d), we write

{\displaystyle {\begin{aligned}V_{d}=V_{1}/\sin(\theta _{c}+\xi )=(V_{1}/\sin \theta _{c})(\cos \xi +\cot \theta _{c}\sin \xi )^{-1}.\end{aligned}}}

Table 4.24a. Data from refraction profile.
${\displaystyle x_{A}(m)}$ ${\displaystyle t_{A}(ms)}$ ${\displaystyle t_{B}(ms)}$ ${\displaystyle x_{B}(m)}$ ${\displaystyle x_{A}(m)}$ ${\displaystyle t_{A}(ms)}$ ${\displaystyle t_{B}(ms)}$ ${\displaystyle x_{B}(m)}$
0 0 98 225 120 70 52 105
15 10 92 210 135 73 46 90
30 21 87 195 150 78 43 75
45 30 81 180 165 81 37 60
60 41 73 165 180 85 31 45
75 50 71 150 195 89 21 30
90 59 63 135 210 94 10 15
105 65 60 120 225 98 0 0

Setting ${\displaystyle \cos \xi =1}$ and expanding by the binomial theorem [see equation (4.1b)], we obtain the result

{\displaystyle {\begin{aligned}V_{d}=V_{2}(1-\cot \theta _{c}\sin \xi ).\end{aligned}}}

Following the same procedure for ${\displaystyle V_{u}}$ and adding the two expansions gives the second result in equation (4.24f). This result is less accurate than the first because we approximated the binomial expansion and also set ${\displaystyle \cos \xi =1}$.

## Problem 4.24d

Sources A and B are located at the ends of a 225-m spread of 16 geophones. Using the data in Table 4.24a, find the velocities, dip, and depth to the refractor.

### Solution

The data in Table 4.24a are plotted in Figure 4.24b and straight-line curves drawn through the data points. The slopes of these lines give the direct-wave velocity and the apparent updip and downdip velocities, and the intercepts with the ${\displaystyle t}$-axes give ${\displaystyle t_{1d}}$ and ${\displaystyle t_{1u}}$. We ignore the value of ${\displaystyle V_{1}}$ obtained on the downdip profile because it is poorly defined. The measured velocities and intercepts are now

Figure 4.24b.  Two-layer arrival times.

{\displaystyle {\begin{aligned}V_{1}&=1.48\,\mathrm {km/s} ,\ V_{d}=2.72\,\mathrm {km/s} ,\ V_{u}=3.75\,\mathrm {km/s} ,\\t_{1d}&=15\,\mathrm {ms} ,\quad t_{1u}=38\,\mathrm {ms} .\end{aligned}}}

From these data, we calculate first ${\displaystyle V_{2}}$, then ${\displaystyle \theta _{c},\xi ,h_{d}}$, and ${\displaystyle h_{u}}$. The two equations (4.24f) give

{\displaystyle {\begin{aligned}{\frac {1}{V_{2}}}&={\frac {1}{2}}\left({\frac {1}{V_{d}}}+{\frac {1}{V_{u}}}\right)={\frac {1}{2}}\left({\frac {1}{2.72}}+{\frac {1}{3.72}}\right),\quad V_{2}=3.15\,\mathrm {km/s} ;\\V_{2}&={\frac {1}{2}}(V_{d}+V_{u})={\frac {1}{2}}(2.72+3.75),\quad V_{2}=3.23\,\mathrm {km/s} ,\end{aligned}}}

the first being more accurate.

Next, ${\displaystyle \theta _{c}=\sin ^{-1}(V_{1}/V_{2})=\sin ^{-1}(1.48/3.15)=28.0^{\circ }}$. Now we find ${\displaystyle h_{d}}$ and ${\displaystyle h_{u}}$, and finally ${\displaystyle \xi }$. From equation (4.24b)

{\displaystyle {\begin{aligned}h_{d}&={\frac {1}{2}}V_{1}t_{id}/\cos \theta _{c}={\frac {1.48\times 0.015}{2\times 0.883}}=13\,\mathrm {m} ,\quad h_{u}={\frac {1.48\times 0.038}{2\times 0.883}}=32\,\mathrm {m} ,\\\xi &=\tan ^{-1}[(32-13)/225]=4.8^{\circ }.\end{aligned}}}