User:Ageary/Chapter 11

Problems in Exploration Seismology and their Solutions

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

11.1 Salt lead time as a function of depth

11.1a The velocity of salt is nearly constant at 4.6 km/s. Calculate the amount of lead time per kilometer of salt diameter as a function of depth assuming the sediments have the Louisiana Gulf Coast velocity distribution shown in Figure 11.1a.

Background

Early seismic prospecting for salt domes involved locating geophones in different directions from the source at roughly the same distance from it. Rays that passed through salt arrived earlier than those that did not, the reduction in traveltime due to the high velocity in salt being the lead time.

Solution

The first two columns of Table 11.1a were obtained from the dashed curve in Figure 11.1a. The third column gives the lead time per kilometer of salt, that is, $\Delta t=(1/V_{i}-1/4.6)$ s/km.

The lead time decreases rapidly with depth to the top of the dome because compaction causes the sediment velocity to increase.

11.1b Early refraction work searching for salt domes in the Gulf Coast considered a significant “lead” to be 0.25 s. Assuming a range of 5.6 km and normal sediment velocity at salt-dome depth of 2.7 km/s, how much salt would this indicate?

Solution

Let $x$ be the path length in the salt. The lead time is the difference in traveltime for a salt path length of $x$. Thus,

${\begin{aligned}0.25=x\left(1/2.7-1/4.6\right);\quad x=1.6\ {\rm {km}}.\end{aligned}}$

Figure 11.1a.  Gulf Coast interval velocity.

Table 11.1a. Calculation of lead time $\Delta t$.

$z$ (km)	$V_{i}$ (km/s)	$\Delta t$ (ms/km)
0.25	1.70	371
0.50	1.92	303
0.75	2.11	257
1.00	2.30	217
1.25	2.46	189
1.50	2.63	163
1.75	2.80	140
2.00	2.93	124

11.2 Effect of assumptions on refraction interpretation

The interpretation of refraction measurements necessarily involves a number of assumptions. How do these affect the interpretation?

Background

Refraction measurements are of apparent velocities (the inverses of the slopes) and intercept times observed from time-distance plots. Refraction events are generally defined by several points that approximately line up to define a straight line, which is drawn through the points. Refractor apparent velocity is then determined from the slope of the line and depth from the intercept with the time axis. If the refraction event from shooting in the opposite direction is also observed, the dip and refractor velocity can be determined. Events from shooting in opposite directions must be correlated correctly.

The basic refraction equations generally assume the following properties:

1. Homogeneous isotropic layers of constant velocity,
2. Each layer’s velocity is larger than that of any shallower layer,
3. Planar interfaces,
4. The profile is perpendicular to the strike.

Solution

Uncertainties in the data and correlations and differences between the real situation and the foregoing assumptions affect the interpretation. Where more than one head-wave event is present, the differences in slope must be large enough to distinguish them as separate events (see problem 11.3). Head waves where offsets are large often show shingling, an en-echelon pattern which may make traveltime picks a cycle late.

In the following we assume that the apparent velocities and intercepts are all measured correctly. Assumption (1) of homogeneous constant-velocity overburden is usually not valid and the velocity in the horizontal direction often exceeds that in the vertical direction. One result is errors in calculating refractor depths. Gradual changes in velocity with depth cause raypaths to be bent or curved, changing calculations as to where a critical raypath reaches the refractor (that is, changing the critical distance) and the distance the head wave travels in the refractor. This is generally the most serious violation of the assumptions. The values for velocity above a refractor generally should be obtained from independent data rather than from the refraction data alone.

Failure of assumption (2) that velocity increases monotonically creates depth errors (see problem 11.3). Layers that have smaller velocity than an overlying layer constitute one type of hidden-layer problem. Layers whose thicknesses are so small that their head waves do not become separate distinct events constitute another type of hidden-layer problem. Changes in overburden velocity in the horizontal direction create similar effects, and they also affect the critical angle at the refractor.

Assumption (3), that the refractor is planar, contrasts with the usual objective of mapping the relief on the refracting interface.

A refraction profile not perpendicular to the strike [assumption (4)] simply results in measuring only a component of the dip rather than the entire dip, and probably does not introduce a large error unless the dip is large.

As will be shown in subsequent problems, refraction mapping often uses more complicated methods than simply applying the refraction equations.

11.3 Effect of a hidden layer

Assume that you wish to map the 5.75 and 6.40 km/s formations in the Illinois basin. Given the velocity information shown in Figure 11.3a, what difficulties would you expect to encounter? The shale at 420–620 m and the lower velocity at 790–960 m form “hidden layers”; how much error will neglect of the hidden layers involve?

Solution

The velocity-depth data are summarized in Table 11.3a. Each of the three high-velocity layers will produce a head wave whose apparent velocity is that of the layers if the layering is all horizontal (which we assume, knowing that dips are generally gentle). We calculate the intercept times in order to plot the time-distance curves.

Because the layers are assumed to be horizontal, equation (3.1a) gives the angles of incidence for the ray that produces the head waves. For the 5150 m/s head wave,

${\begin{aligned}\sin \theta _{c1}/2650=1/5150,\qquad \theta _{c1}=31.0^{\circ }.\end{aligned}}$

Figure 11.3a.  Illinois Basin interval velocity.

Table 11.3a. Velocity-depth data.

Depth range	Velocity
0–300 m	2650 km/s
300–420	5150
420-620	3650
620-790	5750
790-960	5000
960-1200	5750
1200-1550	6400

We use equation (4.18d) to calculate the intercept time $t_{i1}$:

${\begin{aligned}t_{i1}=2\times 300\cos 31.0^{\circ }/2650=0.194\ {\rm {s}}.\end{aligned}}$

For the 5750 m/s head wave we have

${\begin{aligned}\left(\sin \theta _{1}\right)/2650=\left(\sin \theta _{2}\right)/5150=\left(\sin \theta _{c2}\right)/3650=1/5750;\\\theta _{1}=27.4^{\circ },\quad \theta _{2}=63.6^{\circ },\quad \theta _{c2}=39.4^{\circ }.\end{aligned}}$

Its intercept time will be

${\begin{aligned}t_{i2}&=2\times \left(300\cos 27.4^{\circ }/2650+120\cos 63.6^{\circ }/5150+200\cos 39.4^{\circ }/3650\right)\\&=0.306\ {\rm {s}}.\end{aligned}}$

To complete the time-distance curve, we have for the 6400 m/s head wave, allowing for 170 m of 5000 m/s layer that interrupts the 5750 medium (note ray direction is the same in both parts at 5750 m/s),

${\begin{aligned}\left(\sin \theta _{1}\right)/2650&=\left(\sin \theta _{2}\right)/5150=\left(\sin \theta _{3}\right)/3650\\&=\left(\sin \theta _{4}\right)/5000=\left(\sin \theta _{c3}\right)/5750=1/6400;\\\theta _{1}=24.5^{\circ },\quad \theta _{2}&=53.6^{\circ },\quad \theta _{3}=34.8^{\circ },\quad \theta _{4}=51.4^{\circ },\quad \theta _{c3}=64.0^{\circ }.\end{aligned}}$

Its intercept time will be

${\begin{aligned}t_{i3}&=2\times [300\cos 24.5^{\circ }/2650+120\cos 53.6^{\circ }/5150+200\cos 34.8^{\circ }/3650\\&\qquad \qquad +170\cos 51.4^{\circ }/5000+\left(170+240\right)\cos 64.0^{\circ }/5750]\\&=2\times \left(0.103+0.014+0.044+0.021+0.031\right)=0.426\ {\rm {s}}.\end{aligned}}$

The crossover between the 5150 and 5750 m/s head waves is given by

${\begin{aligned}0.194+x/5150=0.306+x/5750,\quad {\rm {or}}\quad x=0.112/0.0203=5.52\ {\rm {km}};\end{aligned}}$

Figure 11.3b.  Time-distance plot.

and the crossover between the 5750 and 6400 m/s head waves is given by

${\begin{aligned}0.306+x/5.75=0.426+x/6.40,\quad {\rm {or}}\quad x=0.120/0.0176=6.78\ {\rm {km}}.\end{aligned}}$

The 5750 m/s curve will be responsible for first breaks for only 1.30 km.

The data are plotted in Figure 11.3b. Interpretation of this time-distance plot will be difficult because the slopes of the three head-wave curves are nearly the same. The ratios of the successive head-wave velocities in this situation are only 1.12 and 1.11; generally ratios should be 1.25 or larger to be interpreted unambiguously.

Failure to recognize a hidden layer means that the time spent in that layer will be interpreted as spent in a layer with higher velocity, which will make the depth appear too large. The shallow refraction event should be interpreted correctly because there are no hidden layers, but the depth calculated for the deeper interfaces will be too great because of the hidden layers.

If we recognize only the 5150 m/s and 6400 m/s head waves (the most probable situation unless additional information is available), that is, the 5750 m/s layer is a hidden layer, then we would calculate the thickness of the 5150 m/s layer $h_{2}$ as

${\begin{aligned}0.426=2\left(300\cos 24.5^{\circ }/2650+h_{2}\cos 53.6^{\circ }/5150\right)=2[0.103+h_{2}(0.000115)].\end{aligned}}$

This gives $h_{2}=960$ m, which, when added to the 300 m thickness of the top layer, gives the depth of the 6400 layer as 1257 m. Comparing with the correct value of 1200 m, the error is 60 m or 5%.

If we should recognize the 5750 m/s head wave, but are not aware of the 3650 m/s layer, we would calculate the thickness of the 5150 m/s layer $h_{2}$ as

${\begin{aligned}0.306=2\left(300\cos 27.4^{\circ }/2650+h_{2}\cos 63.6^{\circ }/5150\right).\end{aligned}}$

This gives $h_{2}=600$ m, which, when added to the 300-m thickness of the top layer, gives a depth of 900 m. Comparing with the correct value of 620 m, the error is 280 m or 45%.

The travel through the 170 m thick 5000 m/s layer, if it is not recognized, would probably be assumed to be at the velocity of 5750 m/s, producing a time error of only 4 ms:

${\begin{aligned}170\left(1/5000-1/5750\right)=170\left(0.000200-0.000174\right)=4\ {\rm {ms}}.\end{aligned}}$

The error is small because the difference in assumed velocities is small.

11.4 Proof of the ABC refraction equation

Prove the ABC refraction equation [equation (11.4a)].

Background

The ABC equation is often used to calculate the weathering thickness. Assuming reversed profiles as shown in Figure 11.4a and writing $t_{AC}$, $t_{BC}$ for the traveltimes from the sources to a geophone at $C$ and $t_{AB}$ for the traveltime from $A$ to $B$, the ABC equation gives the depth $h_{C}$ as

${\begin{aligned}h_{C}={\frac {1}{2}}\left(t_{AC}+t_{BC}-t_{AB}\right)\left[V_{1}V_{2}/\left(V_{2}^{2}-V_{1}^{2}\right)^{1/2}\right].\end{aligned}}$

(11.4a)

Solution

Assuming that $A$, $B$, and $C$ are coplanar and that elevation corrections have been made, we can write

${\begin{aligned}V_{1}\left(t_{AC}+t_{BC}-t_{AB}\right)&=V_{1}\left(t_{MC}+t_{NC}-t_{MN}\right)=2V_{1}t_{MC}-V_{1}t_{MN}\\&=2h_{C}/\cos \theta _{c}-MN\left(V_{1}/V_{2}\right)\\&=2h_{C}/\cos \theta _{c}-\left(2h_{C}\tan \theta _{c}\right)\sin \theta _{c}\\&=2h_{C}/\cos \theta _{c}-\left(2h_{C}\sin ^{2}\theta _{c}/\cos \theta _{c}\right)\\&=\left(2h_{C}/\cos \theta _{c}\right)(1-\sin ^{2}\theta _{c})=2h_{C}\cos \theta _{c}.\end{aligned}}$

Figure 11.4a.  The ABC method.

Thus,

${\begin{aligned}h_{C}&={\frac {1}{2}}\left(t_{AC}+t_{BC}-t_{AB}\right)V_{1}/\cos \theta _{c}\\&={\frac {1}{2}}\left(t_{AC}+t_{BC}-t_{AB}\right)V_{1}/[1-(V_{1}/V_{2})^{2}]^{1/2}\\&={\frac {1}{2}}\left(t_{AC}+t_{BC}-t_{AB}\right)V_{1}V_{2}/[V_{2}^{2}-V_{1}^{2}]^{1/2}.\end{aligned}}$

11.5 Adachi’s method

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)

Table 11.5a. Reversed refraction times.

$x\to$	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)

Figure 11.5a.  Notation used in Adachi’s equations.

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 $h_{2}$.

Figure 11.5b.  Plot of the data in Table 11.5a.

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.

11.6 Refraction interpretation by stripping

11.6a Solve problem 11.5 by stripping off the shallow layer.

Background

Stripping is a method of interpreting refraction data by removing the effect of upper layers, the removal being accomplished by reducing the traveltimes and distances so that in effect the source and geophones are located on the interface at the base of the “stripped” layer. Stripping can be accomplished by calculation or graphically, or by a combination.

Solution

We wish to compare our results with those of problem 11.5, so we use the same measurements, namely $V_{1}=2.02$ km/s and

${\begin{aligned}V_{d2}=3.73\ {\rm {km/s}},\quad V_{u2}=4.51\ {\rm {km/s}},\quad t_{u1}=0.92\ {\rm {s}},\quad t_{d1}=0.46\ {\rm {s}};\\V_{d3}=4.29\ {\rm {km/s}},\quad V_{u3}=5.81\ {\rm {km/s}},\quad t_{u2}=1.28\ {\rm {s}},\quad t_{d2}=0.66\ {\rm {s}}.\end{aligned}}$

(To avoid triple subscripts, we denote intercept times at downdip and updip source locations by $d$ and $u$.)

We start by using equations (4.24f) to get $V_{2}$:

${\begin{aligned}1/V_{2}=\left(1/V_{d2}+1/V_{u2}\right)/2,\quad V_{2}=4.08\ {\rm {km/s}}.\end{aligned}}$

Equations (4.24b,d) can be written

${\begin{aligned}\sin \left(\theta _{c1}+\xi _{2}\right)=V_{1}/V_{d2},\qquad \sin(\theta _{c1}-\xi _{2})=V_{1}/V_{u2};\\{\hbox{so}}\quad \quad \sin(\theta _{c1}+\xi _{2})=2.02/3.73,\qquad (\theta _{c1}+\xi _{2})=32.8^{\circ };\\\sin(\theta _{c1}-\xi _{2})=2.02/4.51,\qquad (\theta _{c1}-\xi _{2})=26.6^{\circ };\end{aligned}}$

hence $\theta _{c1}=29.7^{\circ }$, $\xi _{2}=3.1^{\circ }$. These are the same as those in problem 11.5.

Next we calculate the distances perpendicular to the first refractor at $A$ and $B$ (Figure 11.6a). We use equation (4.24b) to get $h_{d}$ and $h_{u}$:

${\begin{aligned}h_{d}=\left(V_{1}t_{d1}\right)/2\cos \theta _{c1}=2.02\times 0.46/2\times \cos 29.7^{\circ }=0.53\ {\rm {km}};\\h_{u}=\left(V_{1}t_{u1}\right)/2\cos \theta _{c1}=2.02\times 0.92/2\times \cos 29.7^{\circ }=1.07\ {\rm {km}}.\end{aligned}}$

These results are identical with those in problem 11.5. We verify the dip using these depths:

${\begin{aligned}\xi _{2}=\tan ^{-1}\left[\left(1.07-0.53\right)/10.0\right]=3.1^{\circ }.\end{aligned}}$

The first step in stripping is to plot the shallow refractor; we do this by swinging arcs with centers $A$ and $B$ and radii 1.07 and 0.53 km, the refractor being tangent to the two arcs. To get the “stripped” time values, we subtract the times down to and up from the first refractor, i.e., traveltimes along $AA'$ and $BB'$ for sources $A$ and $B$. Although maximum accuracy would be achieved by stripping times for all geophones, the curves for the shallow refraction are so nearly linear that we calculate the stripped times only for each source and one intermediate point on each profile ($Q$ and $N$). We obtain the required distances by measuring the paths in Figure 11.6a. Calculation of the stripped times is given below. Path lengths: $AA'\approx 1.30$, $MN\approx 0.83$, $BB'\approx 0.65$, $PQ\approx 1.08$ km.

Figure 11.6a.  Stripping for refraction interpretation. The numbers below the zero-time line are distances from $A'$.

${\begin{aligned}{\hbox{Time along path:}}\quad \quad \ AA'\approx 0.64,\ MN\approx 0.41,\ BB'\approx 0.32,\ PQ\approx 0.53\ {\rm {s}}\\{\hbox{Time along path:}}\quad \quad \ AA'MN=2.50\ {\rm {s}}\\{\hbox{Stripped time for}}\quad \quad \ AA'MN\approx 2.50-\left(0.64+0.41\right)\approx 1.45\ {\rm {s}}\ ({\rm {point}}\ E)\\{\hbox{Time along path}}\quad \quad \ AA'B'B=3.00\ {\rm {s}}\\{\hbox{Stripped time for}}\quad \quad \ AA'B'B\approx 3.00-\left(0.64+0.32\right)\approx 2.04\ {\rm {s}}\ ({\rm {points}}\ F,H)\\{\hbox{Time along path}}\quad \quad \ BB'PQ=2.30\ {\rm {s}}\\{\hbox{Stripped time for}}\quad \quad \ BB'PQ\approx 2.30-\left(0.32+0.53\right)\approx 1.45\ {\rm {s}}\ ({\rm {point}}\ G)\end{aligned}}$

Stripping off the first refractor in effect moves sources $A$ and $B$ down to $A'$ and $B'$ and geophones at $Q$ and $N$ down to $P$ and $M$, so the stripped times are plotted above these shifted points, the new traveltimes curves being $EF$ and $GH$. Measurements on these stripped curves give the following:

${\begin{aligned}V_{d3}=4.65\ {\rm {km/s}},\quad V_{u3}=5.13\ {\rm {km/s}},\quad t_{d3}=0.19\ {\rm {s}},\quad t_{u3}=0.30\ {\rm {s}}.\end{aligned}}$

Table 11.6a. Comparison of results of Adachi’s and stripping method.

Item	Problem 11.5	Problem 11.6	Difference
$V_{3}({\hbox{km/s}})$	4.92	4.88	0.8%
$\theta _{c2}$	$56.0^{\circ }$	$56.7^{\circ }$	1%
$\xi _{3}$	$5.8^{\circ }$	$5.7^{\circ }$	2%
$h_{2}^{*}({\hbox{km}})$	1.32	$1.20^{*}$	10%

*Vertical depth measured at source A.

We now get

${\begin{aligned}V_{3}&=[\left(1/4.65+1/5.13\right)/2]^{-1}=4.88\ {\rm {km/s}},\\\theta _{c2}&=\sin ^{-1}\left(4.08/4.88\right)=56.7^{\circ },\\h_{u2}&=V_{2}t_{u3}/2\cos \theta _{c2}=4.08\times 0.30/2\cos 56.7^{\circ }=1.11\ {\rm {km}},\\h_{d2}&=4.08\times 0.19/2\cos 56.7^{\circ }=0.71\ {\rm {km}},\\\xi _{3}&=\tan ^{-1}\left[\left(h_{u2}-h_{d2}\right)/A'B'\right]=\tan ^{-1}\left[\left(1.11-0.71\right)/8.90\right]=2.6^{\circ }.\end{aligned}}$

This dip is relative to $A'B'$, so the total dip is $\xi _{2}+\xi _{3}=\left(3.1^{\circ }+2.6^{\circ }\right)=5.7^{\circ }$

11.6b Compare the solutions by stripping with those using Adachi’s method (problem 11.5).

Solution

To compare depths, we measured vertical depths below A. Results for the first layer are the same for both methods, those for the next layer are given in Table 11.6a.

11.6c What are some of the advantages and disadvantages of stripping?

Solution

Advantages of stripping:

Easy to understand

Straightforward in application

Can be used with beds of different dips if the strike is the same

As rapid as other methods when done graphically

Can be used to interpret irregular or curved surfaces

Disadvantages:

Very sensitive to velocity errors

Like most methods, assumes the same strike for all refractors

Difficult to apply when dips are steep

Figure 11.7a.  Refractors with the same strike but different dips.

11.7 Proof of a generalized reciprocal method relation

Prove equation (11.7a), assuming that $\left(\xi _{i}-\xi _{j}\right)\approx 0$ for all values of $i$ and $j$.

Background

The generalized reciprocal method (GRM) can be used with beds of different dips provided all have the same strike. Figure 11.7a shows a series of such beds. Depths normal to the beds are denoted by $z_{Ai}$ and $z_{Bi}$, $\alpha _{i}$ and $\beta _{i}$ are angles of incidence, those for the deepest interface being critical angles, $\xi _{i}$ is the dip of the interface at the top of the $i^{\rm {th}}$ layer. To get the traveltime from $A$ to $B$, $t_{AB}$, we consider a plane wavefront $PQ$ that passes through $A$ at time $t=0$ in a direction such that it will be totally refracted at one of the interfaces, the third in the case of Figure 11.7a. The wavefront reaches $C$ at time $t_{AC}$ and $R$ at time $t_{AR}$ where

${\begin{aligned}t_{AC}=\left(z_{A1}\cos \alpha _{1}/V_{1}\right),\quad t_{AR}=\sum \limits _{i=1}^{3}z_{Ai}\cos \alpha _{i}/V_{i}.\end{aligned}}$

The same wavefront will travel upward from $T$ to $B$ in time

${\begin{aligned}t_{BT}=\sum \limits _{i=1}^{3}z_{Bi}\cos \beta _{i}/V_{i}.\end{aligned}}$

Since $\alpha _{3}$ is the critical angle,

${\begin{aligned}t_{AB}=\sum \limits _{i=1}^{3}(z_{Ai}+z_{Bi})/V_{i}+RV/V_{4}.\end{aligned}}$

Generalizing, we get for $n$ layers

${\begin{aligned}t_{AB}=\sum \limits _{i=1}^{n-1}(z_{Ai}+z_{Bi})/V_{i}+RV/V_{n}.\end{aligned}}$

But $RV=YJ=EJ\cos \left(\xi _{3}-\xi _{2}\right)=AB\cos \xi _{1}\cos \left(\xi _{2}-\xi _{1}\right)\cos \left(\xi _{3}-\xi _{2}\right)$; for $n$ layers, we get

${\begin{aligned}t_{AB}=\sum \limits _{i=1}^{n-1}(z_{Ai}+z_{Bi})/V_{i}+AB\left(S_{n}/V_{n}\right),\end{aligned}}$

where

${\begin{aligned}S_{n}=\cos \xi _{1}\cos \left(\xi _{2}-\xi _{1}\right)\ldots \cos \left(\xi _{n-1}-\xi _{n-2}\right)\approx \cos \xi _{n-1},\end{aligned}}$

(11.7a)

all differences in dip being small, that is, $\left(\xi _{i}-\xi _{j}\right)\approx 0$. We shall not carry the derivation of the GRM formulas beyond this point; those who are interested should consult Sheriff and Geldart, 1995, Section 11.3.3, or Palmer (1980).

Solution

We are asked to prove that

${\begin{aligned}\cos \xi _{1}\cos \left(\xi _{2}-\xi _{1}\right)\cos \left(\xi _{3}-\xi _{2}\right)\ldots \cos \left(\xi _{n-1}-\xi _{n-2}\right)\approx \cos \xi _{n-1},\end{aligned}}$

where the differences in dip are all small, that is, $\xi _{j}-\xi _{j-1}\approx 0$. We start with the single cosine on the right-hand side of the equation and try to express it as a product of cosines. We write it as $\cos \left(\xi _{n-1}-\xi _{m}\right)$ and expand:

${\begin{aligned}\cos \left(\xi _{n-1}-\xi _{m}\right)=\cos[\left(\xi _{n-1}-\xi _{n-2}\right)+(\xi _{n-2}-\xi _{m})].\end{aligned}}$

Since all differences in dip are small, we expand the right-hand side and set the products of the sines equal to zero. Thus,

${\begin{aligned}\cos \left(\xi _{n-1}-\xi _{m}\right)\approx \cos \left(\xi _{n-1}-\xi _{n-2}\right)\cos \left(\xi _{n-2}-\xi _{m}\right).\end{aligned}}$

Next we treat the right-hand cosine in the same way, writing it as $\cos[\left(\xi _{n-2}-\xi _{n-3}\right)+(\xi _{n-3}-\xi _{m})]$. We now expand the factor and drop the sine term. Continuing in this way we eventually arrive at the result

${\begin{aligned}\cos \left(\xi _{n-1}-\xi _{m}\right)&\approx \cos \left(\xi _{n-1}-\xi _{n-2}\right)\cos \left(\xi _{n-2}-\xi _{n-3}\right)\;\ldots \\&\qquad \times \cos \left(\xi _{2}-\xi _{1}\right)\cos \left(\xi _{1}-\xi _{m}\right).\end{aligned}}$

We now take $\xi _{m}=0$ and the result is equation (11.7a).

Figure 11.8a.  Illustrating delay time.

11.8 Delay time

Show that $NQ$ in Figure 11.8a is given by

${\begin{aligned}NQ=V_{2}\delta _{NQ}\tan ^{2}\theta _{c},\\{\hbox{i.e.,}}\quad \quad \delta _{g}=\delta _{NQ}=NQ/V_{2}\tan ^{2}\theta _{c}.\end{aligned}}$

(11.8a)

Background

The concept of delay time has found wide application in refraction interpretation (see problems 11.9 and 11.11). We define the delay time associated with the refraction path SMNG in Figure 11.8a as the observed traveltime minus the time required to travel from $P$ to $Q$ at the velocity $V_{2}$. $PQ$ is the projection of the path SMNG onto the refractor), Writing $\delta$ for the total delay time, we have

${\begin{aligned}\delta &=t_{SG}-PQ/V_{2}\end{aligned}}$

(11.8b)

${\begin{aligned}&=\left({\frac {SM+NG}{V_{1}}}+{\frac {MN}{V_{2}}}\right)-{\frac {PQ}{V_{2}}}=\left({\frac {SM}{V_{1}}}-{\frac {PM}{V_{2}}}\right)+\left({\frac {NG}{V_{1}}}-{\frac {NQ}{V_{2}}}\right)\end{aligned}}$

${\begin{aligned}&=\delta _{s}+\delta _{g},\end{aligned}}$

(11.8c)

${\begin{aligned}{\hbox{where}}\quad \quad \delta _{s}={\hbox{source delay time}}\ =\left({\frac {SM}{V_{1}}}-{\frac {PM}{V_{2}}}\right)\end{aligned}}$

(11.8d)

${\begin{aligned}{\hbox{and}}\quad \quad \delta _{g}={\hbox{geophone delay time}}\ =\left({\frac {NG}{V_{1}}}-{\frac {NQ}{V_{2}}}\right).\end{aligned}}$

(11.8e)

Solution

Referring to Figure 11.8a, we have, by definition,

${\begin{aligned}\delta _{g}&=NG/V_{1}-NQ/V_{2}\\&=NQ\left({\frac {1}{V_{1}\sin \theta _{c}}}-{\frac {1}{V_{2}}}\right)={\frac {NQ}{V_{2}}}\left({\frac {1}{\sin ^{2}\theta _{c}}}-1\right)\\&=NQ/V_{2}\tan ^{2}\theta _{c}.\end{aligned}}$

11.9 Barry’s delay-time refraction interpretation method

Source $B$ is 2 km east of source $A$. The data in Table 11.9a were obtained with cables extending eastward from $A$ ($x$ is the distance measured from $A$) with geophones at 200 m intervals. Interpret the data using Barry’s method (Barry, 1967); $V_{1}=2.50$ km/s. Assume that the delay-time curve for the reverse profile is sufficiently parallel to yours that step (d) below can be omitted.

Background

Barry’s method requires that the total delay time be separated into source and geophone delay times. Two sources on the same side of the geophone are used to achieve this. In Figure 11.9a. $A$ and $B$ are sources, $Q$ and $R$ are geophones, $BQ$ being the critical distance (problem 4.18) for source $B$. We write $\delta _{A}$ and $\delta _{B}$ for the source delay times, $\delta _{Q}$ and $\delta _{R}$ for the geophone delay times, $\delta _{AR}$, $\delta _{BR}$, etc., for the total delay times. We get the source delay times from the intercept times if we assume zero dip [see equation (11.9a)]. The delay time at source $B$, $\delta _{B}$, is due to travel along $BN$, so

${\begin{aligned}\delta _{B}&=BN/V_{1}-N'N/V_{2}=h_{N}/V_{1}\cos \theta _{c}-\left(h_{N}\tan \theta _{c}\right)/V_{2}\\&=\left(h_{N}/V_{1}\cos \theta _{c}\right)\left(1-\sin ^{2}\theta _{c}\right)=\left(h_{N}\cos \theta _{c}\right)/V_{1}={\frac {1}{2}}t_{iB};\end{aligned}}$

(11.9a)

${\begin{aligned}{\hbox{and so,}}\quad \quad h_{N}=V_{1}\delta _{B}/\cos \theta _{c}={\frac {1}{2}}V_{1}t_{iB}/\cos \theta _{c}.\end{aligned}}$

(11.9b)

Figure 11.9a.  Determining delay times.

Note that equations (11.9a,b) apply at any point on the profile where the dip is very small, not merely at souce points.

To find the geophone delay times we have from equation (11.8c)

${\begin{aligned}\delta _{AQ}&=\delta _{A}+\delta _{Q},\\\delta _{AR}&=\delta _{A}+\delta _{R},\\\Delta \delta &=\delta _{AQ}-\delta _{AR}=\delta _{q}-\delta _{R}=differential\ delay\ time.\end{aligned}}$

For zero dip, $\delta _{B}=\delta _{Q}$, so we can write

${\begin{aligned}{\frac {1}{2}}\left(\delta _{BR}+\Delta \delta \right)={\frac {1}{2}}\left(\delta _{Q}+\delta _{R}+\delta _{Q}-\delta _{R}\right)\delta _{Q}-\delta _{R}=\delta _{Q},\end{aligned}}$

(11.9c)

${\begin{aligned}{\hbox{hence}}\quad \quad {\frac {1}{2}}\left(\delta _{BR}-\Delta \delta \right)=\delta _{R}.\end{aligned}}$

(11.9d)

Table 11.9a. Time-distance data.

$x$ (km)	$t_{A}$ (s)	$t_{B}$ (s)	$x$ (km)	$t_{A}$ (s)	$t_{B}$ (s)
2.6	1.02	0.25	5.4	1.62	1.28
2.8	1.05	0.34	5.6	1.66	1.31
3.0	1.10	0.43	5.8	1.72	1.36
3.2	1.24	0.52	6.0	1.75	1.42
3.4	1.18	0.61	6.2	1.80	1.47
3.6	1.20	0.70	6.4	1.85	1.53
3.8	1.26	0.78	6.6	1.91	1.56
4.0	1.32	0.87	6.8	1.97	1.59
4.2	1.35	0.96	7.0	2.00	1.63
4.4	1.39	1.05	7.2	2.02	1.67
4.6	1.45	1.10	7.4	2.05	1.70
4.8	1.50	1.14	7.6	2.10	1.73
5.0	1.56	1.20	7.8	2.13	1.78
5.2	1.59	1.22	8.0	2.16	1.81

To use these equations we must find the point $Q$, preferably by expressing $BQ$ in terms of delay times. From Figure 11.9a and equation (11.9b) we get

${\begin{aligned}BQ=2h_{N}\tan \theta _{C}=2\left(V_{1}\delta _{B}/\cos \theta _{c}\right)\tan \theta _{c}=2V_{2}\ \delta _{B}\ \tan ^{2}\theta _{c}.\end{aligned}}$

(11.9e)

Interpretation involves the following steps:

1. The traveltimes are corrected for weathering and elevation (problem 8.18)
2. Total delay times are calculated and plotted at the geophone positions
3. The distance $PP'$ in Figure 11.9a is calculated for each geophone using equation (11.8a), and the total delay times shifted the distances $PP'$ toward $A$
4. The curves in (b) and (c) should be parallel; if not, $V_{2}$ is adjusted until the curves are sufficiently close to being parallel
5. The total delay times in (b) are separated into source and geophone delay times and then plotted above points $M$, $N$, and $P$. Delays times can be converted into depths using equation (11.9b)

Solution

The data are plotted in Figure 11.9b. Measurements give an average value of 4.60 km/s for $V_{2}$ and intercept times $t_{iA}=0.45$ s, $t_{iB}=0.55$ s, $\delta _{B}=0.55/2=0.28$ s. The critical angle is $\theta _{c}=\sin ^{-1}\left(2.50/4.60\right)=32.9^{\circ }$, $\cos \theta _{c}=0.840$, $\tan \theta _{c}=0.647$. Using equation (11.9e), we have

${\begin{aligned}BQ=2V_{2}\delta _{B}\tan ^{2}\theta _{c}=2\times 4.60\times 0.28\times 0.647^{2}=1.1\ {\rm {km}}.\end{aligned}}$

Figure 11.9b.  Plot of data.

Thus, $Q$ is located at $x=3.1$ km. Also, we need $\delta _{AQ}$:

${\begin{aligned}\delta _{AQ}=t_{AQ}-x_{AQ}/V_{2}=1.12-3.1/4.60=0.45\ {\rm {s}}.\end{aligned}}$

Figure 11.9b shows that we observe refraction data from both sources only for $x\geq 4.6$ km. We show the calculations in Table 11.9b. Column 1 is the offset measured from $A$, columns 2 and 6 are traveltime for sources $A$ and $B$, columns 3 and 7 are the source-to-geophone distances divided by $V_{2}$, columns 4 and 8 are the total delay times [the differences between columns 2 and 3, 6 and 7, respectively—see equation (11.8b)], column 5 is the differential delay time $\Delta \delta$ between geophones at $Q$ and $R=\delta _{AQ}-\delta _{AR}=\left(0.45-\delta _{AR}\right)$, column 9 is $\delta _{R}=\left(\delta _{BR}-\Delta \delta \right)/2$ [see equation (11.9c)], column 10 is $PP'$, column 11 is column 1 minus column 10 = location of $P'$ in Figure 11.9a. Depth values can be obtained by mutliplying $\delta _{PR}$ in column 9 by $V_{1}/\cos \theta _{c}=2.98$ [see equation (11.9a)].

We used a new value of $\delta _{AQ}=1.123-3.10/4.60=0.449s$. Comparing the new and old values of $\delta _{PR}$ for $x=5.0$ and $x=5.6$, we see that rounding errors are not responsible for the anomalies. The anomaly at $x=5.6$ km is 0.01 s whereas the original data are also given to the nearest 0.01 s, so this anomaly could be the result of rounding off of the original time values; however, the anomaly at $x=5.0$ is too large to be due to this.

Table 11.9b. Delay-time calculations.

1	2	3	4	5	6	7	7	8	10	11
$x$	$t_{AR}$	$x/V_{2}$	$\delta _{AR}$	$\Delta \delta$	$t_{BR}$	$x'/V_{2}$	$\delta _{BR}$	$\delta _{R}$	$PP'$	$x-PP'$
4.6	1.45	1.00	0.45	0.00	1.10	0.57	0.53	0.27	0.52	4.08
4.8	1.50	1.04	0.46	–0.01	1.14	0.61	0.53	0.27	0.52	4.28
5.0	1.56	1.09	0.47	–0.02	1.20	0.65	0.55	0.29	0.56	4.44
5.2	1.59	1.13	0.46	–0.01	1.22	0.70	0.52	0.27	0.52	4.68
5.4	1.62	1.17	0.45	0.00	1.28	0.74	0.54	0.27	0.52	4.88
5.6	1.66	1.22	0.44	0.01	1.31	0.78	0.53	0.26	0.50	5.10
5.8	1.72	1.26	0.46	–0.01	1.36	0.83	0.53	0.27	0.52	5.28
6.0	1.73	1.30	0.43	0.02	1.42	0.87	0.55	0.27	0.52	5.48
6.2	1.80	1.35	0.45	0.00	1.47	0.91	0.56	0.28	0.54	5.66
6.4	1.85	1.39	0.46	–0.01	1.53	0.96	0.57	0.29	0.56	5.84
6.6	1.91	1.43	0.48	–0.03	1.56	1.00	0.56	0.30	0.58	6.02
6.8	1.97	1.48	0.49	–0.04	1.59	1.04	0.55	0.30	0.58	6.22
7.0	2.00	1.52	0.48	–0.03	1.63	1.09	0.54	0.29	0.56	6.44
7.2	2.02	1.57	0.45	0.00	1.67	1.13	0.54	0.27	0.52	6.68
7.4	2.05	1.61	0.44	0.01	1.70	1.17	0.53	0.26	0.50	6.90
7.6	2.10	1.65	0.45	0.00	1.73	1.22	0.51	0.26	0.50	7.10
7.8	2.13	1.70	0.43	0.02	1.78	1.26	0.52	0.25	0.48	7.32
8.0	2.16	1.74	0.42	0.03	1.81	1.30	0.51	0.24	0.46	7.54

Note. $x'=x-2$.

Table 11.9c. Part of Table 11.9b with increased precision.

1	2	3	4	5	6	7	8	9
$x$	$t_{AR}$	$x/V_{2}$	$\delta _{AR}$	$\Delta \delta$	$t_{BR}$	$x'/V_{2}$	$\delta _{BR}$	$\delta _{PR}$
4.8	1.50	1.043	0.457	–0.008	1.14	0.609	0.531	0.270
5.0	1.56	1.087	0.473	–0.024	1.20	0.652	0.548	0.286
5.2	1.59	1.130	0.460	–0.011	1.22	0.696	0.524	0.268
5.4	1.62	1.174	0.446	0.003	1.28	0.739	0.541	0.269
5.6	1.66	1.217	0.443	0.006	1.31	0.783	0.527	0.261
5.8	1.72	1.261	0.459	–0.010	1.36	0.826	0.534	0.272

11.10 Parallelism of half-intercept and delay-time curves

Prove that a half-intercept curve is parallel to the curve of the total delay time $\delta$ (see Figure 11.10a).

Solution

Referring to Figure 11.10a, we can write

${\begin{aligned}t_{i}/2=h\left(\cos \theta _{c}/V_{1}\right)\end{aligned}}$

Figure 11.10a.  Delay-time and half-intercept curves.

[see equations (11.9b)]. Thus $t_{i}/2$ is a linear function of $h$ with slope $\left(\cos \theta _{c}/V_{1}\right)$. The total delay time is

${\begin{aligned}\delta =\delta _{s}+\delta _{g},\end{aligned}}$

$\delta _{s}$ being constant. If we substitute $N=h\tan \theta _{c}$ (see Figure 11.9a) in equation (11.8a), we obtain the result

${\begin{aligned}\delta _{g}=h\left(\cos \theta _{c}/V_{1}\right),\\{\hbox{so}}\quad \quad \delta =\delta _{s}+h\left(\cos \theta _{c}/V_{1}\right).\end{aligned}}$

Thus the total delay-time curve is parallel to the half-intercept time curve and lies above it the distance $\delta _{s}$.

11.11 Wyrobek’s refraction interpretation method

Sources $C$, $D$, $E$, $F$, and $G$ in Figure 11.11a are 5 km a part. The data in Table 11.11a are for three profiles $CE$, $DF$, and $EG$ with sources at $C$, $D$, and $E$, no data being recorded for offsets less than 3 km. For profiles from $F$ and $G$ the intercepts were 1.52 and 1.60 s, respectively. Use Wyrobek’s method (Wyrobek, 1956) to interpret the data.

Background

Wyrobek’s method is based on a series of unreversed profiles such as those shown in Figure 11.11a. The steps in the interpretation are as follows:

Figure 11.11a.  Unreversed refraction profiles.

1. The traveltimes are measured, corrected, and plotted, and apparent velocities and intercepts are measured. If $V_{1}$ cannot be measured, $\theta _{c}$ is calculated from an assumed value.
2. The total delay times $\delta$ are calculated [see equation (11.8b)] for each geophone location for each profile. The curves for the different profiles are displaced up or down to obtaina composite curve covering the entire range.
3. The half-intercept times are plotted at the source locations and a curve drawn through them. This curve is compared with the composite curve in (d); if the curves are not sufficiently parallel, $V_{2}$ is adjusted to achieve parallelism. The composite delay-time curve is also used to interpolate or extrapolate the half-intercept curve to cover the complete range. Delay times are now converted into depths using equation (11.9a), i.e., by multiplying half-intercept times by $V_{1}/\left(\cos \theta _{c}\right)$.

Table 11.11a. Time-offset data for three refraction profiles.

$x$ (km)	$t_{CE}$ (s)	$t_{DF}$ (s)	$t_{EG}$ (s)	$x$ (km)	$t_{CE}$ (s)	$t_{DF}$ (s)	$t_{EG}$ (s)
3.00	1.18	1.20	1.19	6.60	1.90	2.12	2.49
3.20	1.22	1.29	1.28	6.80	1.94	2.16	2.54
3.40	1.24	1.38	1.35	7.00	1.97	2.20	2.57
3.60	1.28	1.45	1.43	7.20	2.01	2.25	2.60
3.80	1.35	1.54	1.50	7.40	2.06	2.30	2.65
4.00	1.38	1.60	1.58	7.60	2.10	2.33	2.68
4.20	1.41	1.70	1.68	7.80	2.14	2.37	2.71
4.40	1.47	1.74	1.76	8.00	2.17	2.41	2.74
4.60	1.51	1.77	1.82	8.20	2.20	2.45	2.77
4.80	1.53	1.80	1.89	8.40	2.24	2.47	2.82
5.0	1.58	1.82	2.00	8.60	2.30	2.52	2.85
5.20	1.63	1.85	2.06	8.80	2.32	2.55	2.89
5.40	1.65	1.91	2.15	9.00	2.35	2.61	2.93
5.60	1.69	1.95	2.21	9.20	2.38	2.64	2.97
5.80	1.74	1.97	2.29	9.40	2.44	2.68	3.00
6.00	1.78	1.99	2.38	9.60	2.47	2.73	3.04
6.20	1.82	2.03	2.43	9.80	2.50	2.78	3.07
6.40	1.87	2.08	2.46	10.00	2.54	2.82	3.10

Solution

The traveltimes in Table 11.11a are plotted in the upper part of Figure 11.11b. The values of $V_{1}$ and $V_{2}$ have different accuracies since different numbers of points are used for each value, so we obtain weighted averages using as weights the horizontal extent of the data for each value. Thus,

${\begin{aligned}V_{1}&=\left(2.50\times 1+2.52\times 3\right)/4=2.52\ {\rm {km/s}},\\V_{2}&=\left(5.13\times 7+5.08\times 6+5.59\times 4\right)/17=5.22\ {\rm {km/s}},\\{\hbox{so}}\quad \quad \theta _{c}=\sin ^{-1}\left(2.52/5.22\right)=28.9^{\circ }.\end{aligned}}$

Figure 11.11b.  Time-distance plot (top half ) and plot of delay-times and half-intercept times (bottom).

Table 11.11b. Delay times for profiles $CE$, $DF$, and $EG$.

1	2	3	4	5	6	7	8	9
		5.22			6.25		7.7	5.6
$x$(km)	$\delta _{CE}$(s)	$\delta _{DF}$(s)	$\delta _{EG}$(s)	$\delta _{CE}$(s)	$\delta _{DF}$(s)	$\delta _{EG}$(s)	$\delta _{CE}$(s)	$\delta _{DF}$(s)
3.0	0.61			0.70
3.2	0.61			0.71
3.4	0.59			0.70
3.6	0.59			0.70
3.8	0.62			0.74
4.0	0.61			0.74
4.2	0.61	0.90		0.74	1.03
4.4	0.63	0.90		0.77	1.04
4.6	0.63	0.89		0.77	1.03
4.8	0.61	0.88		0.76	1.03
5.0	0.62	0.86		0.78	1.02			0.93
5.2	0.63	0.85		0.80	1.02			0.92
5.4	0.62	0.88		0.79	1.05			0.95
5.6	0.62	0.88		0.79	1.05			0.95
5.8	0.63	0.86		0.81	1.04			0.93
6.0	0.63	0.84		0.82	1.03			0.92
6.2	0.63	0.84	1.24	0.83	1.04	1.44		0.92
6.4	0.64	0.85	1.23	0.85	1.06	1.44		0.94
6.6	0.64	0.86	1.23	0.84	1.06	1.43	1.04	0.94
6.8	0.64	0.86	1.24	0.85	1.07	1.45	1.06	0.95
7.0	0.63	0.86	1.23	0.85	1.08	1.45	1.06	0.95
7.2	0.63	0.87	1.22	0.86	1.10	1.45	1.07	0.96
7.4	0.64	0.88	1.23	0.88	1.12	1.47	1.10	0.98
7.6	0.64	0.87	1.22	0.88	1.11	1.46	1.11	0.97
7.8	0.65	0.88	1.22	0.89	1.12	1.46	1.13	0.98
8.0	0.64	0.88	1.21	0.89	1.13	1.46	1.13	0.98
8.2	0.63	0.88	1.20	0.89	1.14	1.46	1.14	0.99
8.4	0.63	0.86	1.21	0.90	1.13	1.48	1.15	0.97
8.6	0.65	0.87	1.20	0.92	1.14	1.47	1.18	0.98
8.8	0.63	0.86	1.20	0.91	1.14	1.48	1.18	0.98
9.0	0.63	0.89	1.21	0.91	1.17	1.49	1.18	1.00
9.2	0.62	0.88	1.21	0.91	1.17	1.50	1.19	1.00
9.4	0.64	0.88	1.20	0.94	1.18	1.50	1.22	1.00
9.6	0.63	0.89	1.20	0.93	1.19	1.50	1.22	1.02
9.8	0.62	0.90	1.19	0.93	1.21	1.50	1.23	1.03
10.0	0.62	0.90	1.18	0.94	1.22	1.50	1.24	1.03

The intercept times from the data in Table 11.11a are $t_{C}=0.60$ s, $t_{D}=0.82$ s, $t_{E}=1.31$ s, and we are also given $t_{F}=1.52$ s, $t_{G}=1.60$ s. Obviously the refractor is dipping down from $C$ towards $G$ and $V_{2}$ above is in fact $V_{d}$. However, initially we shall ignore dip and use $V_{2}=5.22$ km/s.

The calculated delay times are listed in Table 11.11b; $x$ is the offset distance from the sources for profiles $CE$, $DF$, and $EG$, while $\delta _{CE}$, $\delta _{DF}$ and $\delta _{EG}$ are total delay times. These were obtained in the same way as $\delta _{AR}$ and $\delta _{BR}$ in Table 11.9b using the value $V_{2}=5.22$ km/s to get columns 2, 3, and 4 in Table 11.11b.

The delay times can also be obtained by drawing straight lines through sources $C$, $D$, and $E$ with slopes $1/V_{2}$ (the lines $HJ$, $KL$, and $MN$ in Figure 11.11b) and then measuring the time differences between these lines and the observed times.

The delay times in columns 2, 3, and 4 are plotted in the lower part of Figure 11.11b using small circles (o). The half-intercept times for sources $C$, $D$, and $G$ are also plotted (solid line at top of the lower figure) but using a different scale from that used for delay times.

The next step is to shift the delay-time values to form a continuous composite curve; we achieve this by moving the $CE$ curve up and the $EG$ curve down. Since this is merely a preliminary step we do not move individual values but displace the average straight lines through the points, giving the composite curve $PQ$.

The delay-time curve is not parallel to the half-intercept line and, to achieve parallelism, we must change $V_{2}$ to increase the delay times at large values of $x$ relative to those at small values. For profile $CE$ we need to change $V_{2}$ so that $J$ moves downward about 0.2 s more than $H$; this gives the curve $H'J'$ with slope equal to $1/V_{2}=1/6.25$ km/s, the other two curves becoming $K'L'$ and $M'N'$. We recalculate the delay times using $V_{2}=6.25$ km/s; the new values are given in columns 5, 6, and 7 of Table 11.11b and plotted as $x's$ in Figure 11.11b. The new curves do roughly parallel the half-intercept curve, and we obtain a new composite delay-time curve by moving $\delta _{DF}$ and $\delta _{CE}$ upward by 0.2 s and 0.3 s, respectively, to join the $\delta _{EG}$ values to form a continuous curve. The values agree exactly except for the first and last overlapping values, which differ by 2 ms; we used the average values at these two points.

Comparison of the composite delay-time curve with the half-intercept time curve shows reasonably good agreement at the two ends but significant divergence in the central part. We might assume that the intercept time at source $E$ is in error but the value 1.31 s would have to decrease to about 1.15 s (for a half-intercept time of about 0.58 s) to agree with the delay-time curve. Although the $EG$-curve is short, it is regular so that it is difficult to fit a line having an intercept of 1.15 s. A more likely source of error is variations of velocity; these could be of two kinds: (i) the actual value of $V_{2}$ could be 6.25 at the two ends but higher than 6.25 in the range $7<x_{c}<10$ km and lower than 6.25 in the range $10<x_{c}<15$ km, (ii) velocity changes due to dip (the intercepts show an overall dip down from $C$ to $G$, so $V_{2}$ is the apparent velocity $V_{d}$. While velocity variations due to changes in dip are the more likely explanation, we can proceed with the interpretation without deciding which velocity effect is the cause.

To reduce the gap between the two curves, we change $V_{2}$ so that the difference between the values of $\delta _{CE}$ at $X_{C}=10.0$ and $x_{C}=6.6$ km increases by 0.1 s. Letting $V$ be the required velocity and using equation (11.8b), we get

${\begin{aligned}\left(2.54-10.0/V\right)-\left(1.90-6.6/V\right)=\left(0.94-0.84\right)+0.10,\\V=3.4/\left(0.64-0.20\right)=7.7\ {\rm {km/s}}.\end{aligned}}$

We also need a new velocity that will increase $\delta _{DF}$ about 0.1 s more at $x_{C}=5.0$ than at $x_{C}=10.0$. Thus

${\begin{aligned}\left(1.82-5.0/V\right)-\left(2.82-10.0/V\right)=\left(1.02-1.22\right)+0.10,\\V=5.0/0.90=5.6\ {\rm {km/s}}.\end{aligned}}$

These two velocities were used to calculate revised delay times in columns 8 and 9 of Table 11.11b, and the revised values are plotted in Figure 11.11b (using small squares).

The final interpreted curve is represented by inverted triangles ($\nabla$) from $x_{c}=3.0$ to $x_{c}=15.0$ and by crosses $\left(\times \right)$ from $x_{c}=16.2$ to 20.0. The values can be changed to depths by multiplying the half-intercept times by $V_{1}/\left(\cos \theta _{c}\right)$ [see equation (11.9b)].

We now get approximate dip $\xi$ by finding depths at $C$ and $G$ using equation (11.9b); then we use $\xi$ and $V_{d}$ to calculate $V_{u}$, $V_{2}$, and $\theta _{c}$ which give a more accurate depth factor $V_{1}/\cos \theta _{c}$. Thus, we have

${\begin{aligned}V_{1}=2.52\ {\rm {km/s}},\quad V_{2}=5.22\ {\rm {km/s}},\quad \theta _{c}=28.9^{\circ },\quad \delta _{c}=0.60/2,\quad \delta _{G}=1.60/2.\end{aligned}}$

Using these values, the depths become

${\begin{aligned}h_{C}&=0.30\times 2.52/\cos 28.9^{\circ }=0.86\ {\rm {km}},\\h_{G}&=0.80\times 2.52/\cos 28.9^{\circ }=2.30\ {\rm {km}},\\\xi &=\tan ^{-1}\left[\left(2.30-0.86\right)/20\right]=4.1^{\circ }.\end{aligned}}$

Using $V_{d}=6.25$ km/s, we solve equation (4.24d) for $V_{u}$, giving

${\begin{aligned}\xi =\left(1/2\right)[\sin ^{-1}\left(2.52/6.25\right)-\sin ^{-1}(2.52/V_{u})],\\{\hbox{so}}\quad \quad 4.1^{\circ }&=\left(1/2\right)[23.8^{\circ }-\sin ^{-1}(2.52/V_{u})]\\\sin ^{-1}\left(2.52/V_{u}\right)&=23.8^{\circ }-8.2^{\circ }=15.6^{\circ },\\\left(2.52/V_{u}\right)&=\sin 15.6^{\circ }=0.269,V_{u}=9.37\ {\rm {km/s}},\\V_{2}\approx {\frac {1}{2}}\left(V_{d}+V_{u}\right)&={\frac {1}{2}}\left(6.25+9.37\right)=7.81\ {\rm {km/s}},\\\theta _{c}&\approx \sin ^{-1}\left(2.52/7.81\right)\approx 18.8^{\circ },\;\cos \theta _{c}\approx 0.947,\\{\hbox{depth factor}}&\approx V_{1}/\cos \theta _{c}\approx 2.52/0.947\approx 2.66.\end{aligned}}$

Thus, the refractor is nearly flat over the region where we used $V_{2}=6.25$ km/s, so local dip is mainly in the places where we carried out the second revision using velocities of 7.7 and 5.6 km/s.

We shall not refine our interpretation further because of the limited acccuracy of the data.

11.12 Properties of a coincident-time curve

11.12a A coincident-time curve connects points where waves traveling by different paths arrive at the same time. In Figure 11.12a, the curve $AC$ is where the head wave and direct wave arrive simultaneously. On a vertical section through the source with constant-velocity above a refractor, head-wave wavefronts are parallel straight lines. In Figure 11.12b, show that the virtual wavefront $DE$ for $t=0$ is at a slant depth $SD=2h=2z\cos \theta _{c}$.

Background

Figure 11.12a shows first-arrival wavefronts at intervals of 0.1 s generated by the source $S$ for a three-layer situation where the velocities are in the ratio 2:3:4. The critical angle at the first interface is reached at $A$, so head waves are generated to the right of this point, the wavefronts in the upper layer being straight lines that join with the direct wavefronts having the same traveltimes. The locus of the junction point where the first-arrival wavefronts abruptly change direction is a coincident time curve. $ABC$ is a coincident-time curve. In general a coincident-time curve (for example, DEFG) is the locus of the junction points where two wavefronts having the same traveltimes but have traveled different paths.

A curve that is equidistant from a fixed point and a fixed straight line is a parabola.

Solution

In Figure 11.12a, the wave generated at $S$ at time $t=0$ arrives at $A$ at time $t_{A}=SA/V_{1}$, the angle of incidence being the critical angle $\theta _{c}$. Head waves traveling upwards at the critical angle are generated to the right of $A$. We assume that a fictitious source generates plane wavefronts traveling parallel to the head-wave wavefronts with velocity $V_{1}$, $DE$ being their position at $t=0$. This wavefront arrives at $A$ at time $t_{A}$ so that $CD=SA$. Hence,

${\begin{aligned}SD&=SC+CD=SC+SA\\&=SA\left(1+\cos 2\theta _{c}\right)\\&=2SA{\rm {cos}}^{2}\theta _{c}=2z\cos \theta _{c}.\end{aligned}}$

Figure 11.12a.  First-arrival wavefronts at 0.1-s intervals.

Figure 11.12b.  Deriving properties of a coincident-time curve.

11.12b Show that after $DE$ reaches $A$, wavefronts such as $BF$ coincde with the head-wave wavefronts.

Solution

If the wavefront $CA$ arrives at $FB$ at time $t_{A}+\Delta t$, then $AF=V_{1}\Delta t$. During the time $\Delta t$, the headwave travels from $A$ to $B$ at velocity $V_{2}$, that is, $AB=V_{2}\Delta t$. Therefore $AF/AB=V_{1}/V_{2}=\sin \theta _{c}$, so $BF$ parallels the refracted wavefronts.

11.12c Show that the coincident-time curve is a parabola.

Solution

At any point on the coincident-time curve, the traveltime of the direct wave equals that of a wavefront coming from $DE$. Since both wavefronts travel with the velocity $V_{1}$, the point on the curve is equidistant from $S$ and from the straight line $DE$, hence the curve is a parabola.

11.12d Show that, taking $DE$ and $DS$ as the $x$ and $y$-axes, the equation of $AH$ is $4hy=x^{2}+4h^{2}$, where $DS=2h$.

Solution

We take $H$ as $H\left(x,\;y\right)$ and $S\left(0,2h\right)$. We know from part (c) that $SH=$ distance from $H$ to the line $DE$. The squares of these distances are also equal, so

${\begin{aligned}SH^{2}=HE^{2},\ {\hbox{that is}},x^{2}+(y-2h)^{2}=y^{2},\\{\hbox{and}}\quad \quad x^{2}+4h^{2}=4hy.\end{aligned}}$

11.12e Show that the coincident-time curve is tangent to the refractor at $A$.

Solution

We must show that the coincident-time curve passes through $A$ with the same slope as the refractor. Obviously the curve starts at $A$ because the head wave starts at the instant the direct wave reaches $A$. We use the equation of the curve in part (d) to get the slope and then substitute the coordinates of $A$. Thus,

${\begin{aligned}x^{2}+4h^{2}=4hy,\qquad dy/dx=x/2h.\end{aligned}}$

The $x$-coordinate of $A$ is

${\begin{aligned}x_{A}&=AS\sin 2\theta _{c}=\left(z/\cos \theta _{c}\right)\sin 2\theta _{c}=2z\sin \theta _{c}\\&=\left(2h/\cos \theta _{c}\right)\sin \theta _{c}=2h\tan \theta _{c},\end{aligned}}$

where we used the result in (a) in the last step. Substitution in $dy/dx$ gives the slope $\tan \theta _{c}$ which is the same as the refractor slope. Therefore, the coincident-time curve is tangent to the refractor at $A$.

11.13 Interpretation by the plus-minus method

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

Figure 11.13a.  Illustrating the plus-minus method.

Figure 11.13b.  Solution by the plus-minus method.

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.

11.14 Comparison of refraction interpretation methods

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

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:

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

Equation (4.24d) gives

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

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

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

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

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

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

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 $Q$ [see Figure 11.14b(i)] where the refracted energy that arrives at geophone $R$ leaves the refractor. The refractor is defined by finding $Q$ for a series of geophone positions. Tarrant’s method is based on the properties of the ellipse.

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

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

(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 ($R$ in Figure 11.14b(ii) is a constant $\varepsilon$ (eccentricity)).

The standard polar equation of an ellipse is

${\begin{aligned}\rho =\varepsilon h/\left(1-\varepsilon \cos \phi \right).\end{aligned}}$

(11.14b)

Figure 11.14b.  Illustrating Tarrant’s method. (i) Relation between $R$ and $Q$; (ii) locus of $Q$ is ellipse, focus at $R$; (iii) geometry of ellipse through Q.

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

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

(11.14c)

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

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

(11.14d)

The distance from the focal point $R$ to the center $O$ is

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

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

${\begin{aligned}a=V_{1}\delta _{g}/{\rm {cos}}^{2}\theta _{c}\\&=V_{2}\delta _{g}\tan \theta _{c}/\cos \theta _{c}\end{aligned}}$

(11.14e)

${\begin{aligned}&=OQ=V_{2}\delta _{g}\sin \theta _{c}/\cos \theta _{c}=V_{2}\delta _{g}\tan \theta _{c},\end{aligned}}$

(11.14f)

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

(11.14g)

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

(11.14h)

${\begin{aligned}OC&=OR\;\tan =V_{2}\delta _{g}\tan ^{3}\theta _{c}.\end{aligned}}$

(11.14i)

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

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

(11.14j)

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

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

Differentiating, we obtain

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

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

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

Solution

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

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

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

and for source $B$,

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

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

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

Table 11.14b. Calculations of $OC$ and $r$ for source $A$.

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

${\begin{aligned}OQ&=V_{2}\delta _{g}\tan \theta _{c}=2.69\delta _{g},\end{aligned}}$

(11.14l)

${\begin{aligned}OR&=V_{2}\delta _{g}\tan ^{2}\theta _{c}=1.41\delta _{g},\end{aligned}}$

(11.14m)