Two-layer refraction problem

Problems in Exploration Seismology and their Solutions

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

Problem 4.24a

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

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

(4.24a)

where

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

(4.24b)

$t_{d}$ and $t_{u}$ being the traveltimes when shooting downdip and updip, respectively (see Figure 4.24a), and $h_{d}$, $h_{u}$, and $t_{id}$, $t_{iu}$, 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$^{\prime }$ as the receiver. Following the procedure used in problem 4.18, we have

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

(4.24c)

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

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

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

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

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

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

(4.24d)

Problem 4.24b

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

Solution

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

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

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

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

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

(4.24e)

Problem 4.24c

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

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

(4.24f)

Solution

Expanding equation (4.24d) we have

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

Adding the two equations, we get

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

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

Returning to equation (4.24d), we write

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

Table 4.24a. Data from refraction profile.

$x_{A}(m)$	$t_{A}(ms)$	$t_{B}(ms)$	$x_{B}(m)$	$x_{A}(m)$	$t_{A}(ms)$	$t_{B}(ms)$	$x_{B}(m)$
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 $\cos \xi =1$ and expanding by the binomial theorem [see equation (4.1b)], we obtain the result

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

Following the same procedure for $V_{u}$ 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 $\cos \xi =1$.

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 $t$-axes give $t_{1d}$ and $t_{1u}$. We ignore the value of $V_{1}$ obtained on the downdip profile because it is poorly defined. The measured velocities and intercepts are now

Figure 4.24b.  Two-layer arrival times.

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

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

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

the first being more accurate.

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

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

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Also in this chapter

• Accuracy of normal-moveout calculations
• Dip, cross-dip, and angle of approach
• Relationship for a dipping bed
• Reflector dip in terms of traveltimes squared
• Second approximation for dip moveout
• Calculation of reflector depths and dips
• Plotting raypaths for primary and multiple reflections
• Effect of migration on plotted reflector locations
• Resolution of cross-dip
• Cross-dip
• Variation of reflection point with offset
• Functional fits for velocity-depth data
• Relation between average and rms velocities
• Vertical depth calculations using velocity functions
• Depth and dip calculations using velocity functions
• Weathering corrections and dip/depth calculations
• Using a velocity function linear with depth
• Head waves (refractions) and effect of hidden layer
• Interpretation of sonobuoy data
• Diving waves
• Linear increase in velocity above a refractor
• Time-distance curves for various situations
• Locating the bottom of a borehole
• Two-layer refraction problem

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