Chapter 11 Refraction methods
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,
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
be the path length in the salt. The lead time is the difference in traveltime for a salt path length of
. Thus,
Figure 11.1a. Gulf Coast interval velocity.
Table 11.1a. Calculation of lead time
.
(km)
|
(km/s)
|
(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:
- Homogeneous isotropic layers of constant velocity,
- Each layer’s velocity is larger than that of any shallower layer,
- Planar interfaces,
- 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,
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
:
For the 5750 m/s head wave we have
Its intercept time will be
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),
Its intercept time will be
The crossover between the 5150 and 5750 m/s head waves is given by
Figure 11.3b. Time-distance plot.
and the crossover between the 5750 and 6400 m/s head waves is given by
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
as
This gives
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
as
This gives
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:
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
,
for the traveltimes from the sources to a geophone at
and
for the traveltime from
to
, the ABC equation gives the depth
as
|
|
(11.4a)
|
Solution
Assuming that
,
, and
are coplanar and that elevation corrections have been made, we can write
Figure 11.4a. The ABC method.
Thus,
11.5 Adachi’s method
Given the data in Table 11.5a for a reversed refraction profile with sources
and
, 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 (
and
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
and
are angles of incidence relative to the vertical at the
interface for the downgoing rays from sources
and
, respectively (these are angles of approach at the surface for
),
and
are the angles of incidence and refraction for the downgoing ray at interface
,
and
are the same for the upcoming ray,
is the dip of the
interface,
is the vertical thickness of the bed below this interface below the downdip source.
The traveltime
for the refraction along the top of the
layer is given by
|
|
(11.5a)
|
If we set
,
becomes the intercept time
at the downdip source; thus,
|
|
(11.5b)
|
Table 11.5a. Reversed refraction times.
|
0.0
|
0.5
|
1.0
|
1.5
|
2.0
|
2.5
|
3.0
|
3.5
|
4.0
|
4.5
|
5.0
|
(km)
|
|
0.00
|
0.25
|
0.50
|
0.74
|
0.98
|
1.24
|
1.50
|
1.70
|
1.81
|
1.91
|
2.02
|
(s)
|
|
3.00
|
2.90
|
2.80
|
2.68
|
2.52
|
2.41
|
2.31
|
2.20
|
2.07
|
1.91
|
1.80
|
(s)
|
|
|
5.5
|
6.0
|
6.5
|
7.0
|
7.5
|
8.0
|
8.5
|
9.0
|
9.5
|
10.0
|
|
(km)
|
|
2.16
|
2.28
|
2.38
|
2.44
|
2.56
|
2.64
|
2.72
|
2.80
|
2.89
|
3.00
|
|
(s)
|
|
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:
|
|
(11.5c)
|
Snell’s law [equation (3. 1a)] gives
|
|
(11.5d)
|
For the refraction along the
interface,
|
|
(11.5e)
|
The initial interpretation stage is plotting the data and determining
and the apparent velocities
and
, and intercept times
for each of the refraction events. The angles
and
are given by equation (4.2d). Next we use problem 4.24b to get
and
from
,
, and
. The depth
is now found using equation (11.5b).
For the next interface we find new values of
and
using the next pair of apparent velocities. Since
is now known, we use equation (11.5c) to get new values of
and
, after which equation (11.5d) gives
,
and equation (5.11c) gives
,
. We can now find
,
,
, and
.
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
is 2.02 km/s. Two refraction events are observed with the apparent velocities and intercept times listed below.
First we calculate
and
:
Equation (11.5c) gives
,
. Since this interface is the refractor, equation (11.5e) gives
{
}
We find
using equation (11.5b): so
For the second refractor, we calculate new angles of approach:
Then equation (11.5c) gives
Using equation (11.5d), we get
From equation (11.5c) we now get
From equation (5.11e) we have
{
}
Finally, we get the depth from equation (11.5b):
Total vertical depth at
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
km/s and
(To avoid triple subscripts, we denote intercept times at downdip and updip source locations by
and
.)
We start by using equations (4.24f) to get
:
Equations (4.24b,d) can be written
hence
,
. These are the same as those in problem 11.5.
Next we calculate the distances perpendicular to the first refractor at
and
(Figure 11.6a). We use equation (4.24b) to get
and
:
These results are identical with those in problem 11.5. We verify the dip using these depths:
The first step in stripping is to plot the shallow refractor; we do this by swinging arcs with centers
and
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
and
for sources
and
. 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 (
and
). We obtain the required distances by measuring the paths in Figure 11.6a. Calculation of the stripped times is given below. Path lengths:
,
,
,
km.
Figure 11.6a. Stripping for refraction interpretation. The numbers below the zero-time line are distances from

.
Stripping off the first refractor in effect moves sources
and
down to
and
and geophones at
and
down to
and
, so the stripped times are plotted above these shifted points, the new traveltimes curves being
and
. Measurements on these stripped curves give the following:
Table 11.6a. Comparison of results of Adachi’s and stripping method.
Item
|
Problem 11.5
|
Problem 11.6
|
Difference
|
|
4.92
|
4.88
|
0.8%
|
|
|
|
1%
|
|
|
|
2%
|
|
1.32
|
|
10%
|
*Vertical depth measured at source A.
|
We now get
This dip is relative to
, so the total dip is
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
for all values of
and
.
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
and
,
and
are angles of incidence, those for the deepest interface being critical angles,
is the dip of the interface at the top of the
layer. To get the traveltime from
to
,
, we consider a plane wavefront
that passes through
at time
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
at time
and
at time
where
The same wavefront will travel upward from
to
in time
Since
is the critical angle,
Generalizing, we get for
layers
But
; for
layers, we get
where
|
|
(11.7a)
|
all differences in dip being small, that is,
. 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
where the differences in dip are all small, that is,
. 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
and expand:
Since all differences in dip are small, we expand the right-hand side and set the products of the sines equal to zero. Thus,
Next we treat the right-hand cosine in the same way, writing it as
. We now expand the factor and drop the sine term. Continuing in this way we eventually arrive at the result
We now take
and the result is equation (11.7a).
Figure 11.8a. Illustrating delay time.
11.8 Delay time
Show that
in Figure 11.8a is given by
|
|
(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
to
at the velocity
.
is the projection of the path SMNG onto the refractor), Writing
for the total delay time, we have
|
|
(11.8b)
|
|
|
(11.8c)
|
|
|
(11.8d)
|
|
|
(11.8e)
|
Solution
Referring to Figure 11.8a, we have, by definition,
11.9 Barry’s delay-time refraction interpretation method
Source
is 2 km east of source
. The data in Table 11.9a were obtained with cables extending eastward from
(
is the distance measured from
) with geophones at 200 m intervals. Interpret the data using Barry’s method (Barry, 1967);
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.
and
are sources,
and
are geophones,
being the critical distance (problem 4.18) for source
. We write
and
for the source delay times,
and
for the geophone delay times,
,
, 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
,
, is due to travel along
, so
|
|
(11.9a)
|
|
|
(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)
For zero dip,
, so we can write
|
|
(11.9c)
|
|
|
(11.9d)
|
Table 11.9a. Time-distance data.
(km)
|
(s)
|
(s)
|
(km)
|
(s)
|
(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
, preferably by expressing
in terms of delay times. From Figure 11.9a and equation (11.9b) we get
|
|
(11.9e)
|
Interpretation involves the following steps:
- The traveltimes are corrected for weathering and elevation (problem 8.18)
- Total delay times are calculated and plotted at the geophone positions
- The distance
in Figure 11.9a is calculated for each geophone using equation (11.8a), and the total delay times shifted the distances
toward
- The curves in (b) and (c) should be parallel; if not,
is adjusted until the curves are sufficiently close to being parallel
- The total delay times in (b) are separated into source and geophone delay times and then plotted above points
,
, and
. 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
and intercept times
s,
s,
s. The critical angle is
,
,
. Using equation (11.9e), we have
Figure 11.9b. Plot of data.
Thus,
is located at
km. Also, we need
:
Figure 11.9b shows that we observe refraction data from both sources only for
km. We show the calculations in Table 11.9b. Column 1 is the offset measured from
, columns 2 and 6 are traveltime for sources
and
, columns 3 and 7 are the source-to-geophone distances divided by
, 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
between geophones at
and
, column 9 is
[see equation (11.9c)], column 10 is
, column 11 is column 1 minus column 10 = location of
in Figure 11.9a. Depth values can be obtained by mutliplying
in column 9 by
[see equation (11.9a)].
We used a new value of
. Comparing the new and old values of
for
and
, we see that rounding errors are not responsible for the anomalies. The anomaly at
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
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
|
|
|
|
|
|
|
|
|
|
|
|
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.
.
Table 11.9c. Part of Table 11.9b with increased precision.
1
|
2
|
3
|
4
|
5
|
6
|
7
|
8
|
9
|
|
|
|
|
|
|
|
|
|
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
(see Figure 11.10a).
Solution
Referring to Figure 11.10a, we can write
Figure 11.10a. Delay-time and half-intercept curves.
[see equations (11.9b)]. Thus
is a linear function of
with slope
. The total delay time is
being constant. If we substitute
(see Figure 11.9a) in equation (11.8a), we obtain the result
Thus the total delay-time curve is parallel to the half-intercept time curve and lies above it the distance
.
11.11 Wyrobek’s refraction interpretation method
Sources
,
,
,
, and
in Figure 11.11a are 5 km a part. The data in Table 11.11a are for three profiles
,
, and
with sources at
,
, and
, no data being recorded for offsets less than 3 km. For profiles from
and
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.
- The traveltimes are measured, corrected, and plotted, and apparent velocities and intercepts are measured. If
cannot be measured,
is calculated from an assumed value.
- The total delay times
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.
- 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,
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
.
Table 11.11a. Time-offset data for three refraction profiles.
(km)
|
(s)
|
(s)
|
(s)
|
(km)
|
(s)
|
(s)
|
(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
and
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,
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
,
, and
.
1
|
2
|
3
|
4
|
5
|
6
|
7
|
8
|
9
|
|
|
5.22
|
|
|
6.25
|
|
7.7
|
5.6
|
(km)
|
(s)
|
(s)
|
(s)
|
(s)
|
(s)
|
(s)
|
(s)
|
(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
s,
s,
s, and we are also given
s,
s. Obviously the refractor is dipping down from
towards
and
above is in fact
. However, initially we shall ignore dip and use
km/s.
The calculated delay times are listed in Table 11.11b;
is the offset distance from the sources for profiles
,
, and
, while
,
and
are total delay times. These were obtained in the same way as
and
in Table 11.9b using the value
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
,
, and
with slopes
(the lines
,
, and
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
,
, and
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
curve up and the
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
.
The delay-time curve is not parallel to the half-intercept line and, to achieve parallelism, we must change
to increase the delay times at large values of
relative to those at small values. For profile
we need to change
so that
moves downward about 0.2 s more than
; this gives the curve
with slope equal to
km/s, the other two curves becoming
and
. We recalculate the delay times using
km/s; the new values are given in columns 5, 6, and 7 of Table 11.11b and plotted as
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
and
upward by 0.2 s and 0.3 s, respectively, to join the
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
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
-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
could be 6.25 at the two ends but higher than 6.25 in the range
km and lower than 6.25 in the range
km, (ii) velocity changes due to dip (the intercepts show an overall dip down from
to
, so
is the apparent velocity
. 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
so that the difference between the values of
at
and
km increases by 0.1 s. Letting
be the required velocity and using equation (11.8b), we get
We also need a new velocity that will increase
about 0.1 s more at
than at
. Thus
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 (
) from
to
and by crosses
from
to 20.0. The values can be changed to depths by multiplying the half-intercept times by
[see equation (11.9b)].
We now get approximate dip
by finding depths at
and
using equation (11.9b); then we use
and
to calculate
,
, and
which give a more accurate depth factor
. Thus, we have
Using these values, the depths become
Using
km/s, we solve equation (4.24d) for
, giving
Thus, the refractor is nearly flat over the region where we used
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
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
for
is at a slant depth
.
Background
Figure 11.12a shows first-arrival wavefronts at intervals of 0.1 s generated by the source
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
, 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.
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
at time
arrives at
at time
, the angle of incidence being the critical angle
. Head waves traveling upwards at the critical angle are generated to the right of
. We assume that a fictitious source generates plane wavefronts traveling parallel to the head-wave wavefronts with velocity
,
being their position at
. This wavefront arrives at
at time
so that
. Hence,
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
reaches
, wavefronts such as
coincde with the head-wave wavefronts.
Solution
If the wavefront
arrives at
at time
, then
. During the time
, the headwave travels from
to
at velocity
, that is,
. Therefore
, so
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
. Since both wavefronts travel with the velocity
, the point on the curve is equidistant from
and from the straight line
, hence the curve is a parabola.
11.12d Show that, taking
and
as the
and
-axes, the equation of
is
, where
.
Solution
We take
as
and
. We know from part (c) that
distance from
to the line
. The squares of these distances are also equal, so
11.12e Show that the coincident-time curve is tangent to the refractor at
.
Solution
We must show that the coincident-time curve passes through
with the same slope as the refractor. Obviously the curve starts at
because the head wave starts at the instant the direct wave reaches
. We use the equation of the curve in part (d) to get the slope and then substitute the coordinates of
. Thus,
The
-coordinate of
is
where we used the result in (a) in the last step. Substitution in
gives the slope
which is the same as the refractor slope. Therefore, the coincident-time curve is tangent to the refractor at
.
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
and
is such that the traveltime is either a minimum (e.g., direct waves, reflections and head waves) or a maximum. Therefore, the raypath between
and
is unique so that
. As a result, when we have reversed profiles, we can locate the refractor by drawing wavefronts from the two sources
and
; when the sum of the traveltimes for two intersecting wavefronts equals
, 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
as in Figure 11.13a. If the dip is zero, they are at the angles
to the refractor and the intersections give diamond-shaped parallelograms. The horizontal diagonal of a parallelogram is
and the vertical diagonal is
. Lines of constant sum of the traveltimes minus
(plus values) gives the refractor configuration and differences (minus values) give a check on the value of
. The refractor lies at plus value = 0.
Table 11.13a. Time-distance data for plus-minus interpretation.
(km)
|
(s)
|
(s)
|
(km)
|
(s)
|
(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:
km/s,
km/s,
s,
.
We next draw straight-line wavefronts at
spaced at intervals
s. Because
s and the refraction from source
starts around 0.8 s, we draw wavefronts for source
for
, 1.00, 1.20, 1.40, and 1.60 s. For source
we draw wavefronts for
, 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
These values agree with the values in Figure 11.13b within the limits of error.
Table 11.14a. Refraction time-distance data.
(km)
|
(s)
|
(s)
|
(s)
|
(km)
|
(s)
|
(s)
|
(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
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
and
which are separated by 12 km. The columns in the table headed
and
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:
Equation (4.24d) gives
Figure 11.14a. Plot of the time-distance data.
From equation (4.24f), we have
km/s. From equation (4.24b) we get for the slant depths,
Checking the values of
and
, we obtain
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
[see Figure 11.14b(i)] where the refracted energy that arrives at geophone
leaves the refractor. The refractor is defined by finding
for a series of geophone positions. Tarrant’s method is based on the properties of the ellipse.
The delay time for the path
in Figure 11.14b(i) is
. Solving for
, we get
|
|
(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 (
in Figure 11.14b(ii) is a constant
(eccentricity)).
The standard polar equation of an ellipse is
|
|
(11.14b)
|
Figure 11.14b. Illustrating Tarrant’s method. (i) Relation between

and

; (ii) locus of

is ellipse, focus at

; (iii) geometry of ellipse through Q.
In Figure 11.14b(ii)
moves so that the ratio
. The major axis 2a of the ellipse is
|
|
(11.14c)
|
To get the minor axis, we set the first derivative of
equal to zero. Using equation (11.14b), we find that
|
|
(11.14d)
|
The distance from the focal point
to the center
is
If we substitute
, and
in equation (11.14b), we get equation (11.14a). Also these values give the following results.
|
|
(11.14e)
|
|
|
(11.14f)
|
|
|
(11.14g)
|
|
|
(11.14h)
|
|
|
(11.14i)
|
To approximate the ellipse in the vicinity of
with a circle, we need to find the center of curvature of the ellipse. The general equation of an ellipse is
|
|
(11.14j)
|
The equation for the radius of curvature of a function
is
Differentiating, we obtain
The center of curvature is a distance
above
, so the
-coordinate of
[Figure 11.14b(iii)] is
. A circle with center
and radius
will approximate the ellipse in the vicinity of
.
Solution
We need the total delay time at source
,
, and the delay times
at the geophones where the head wave is observed.
We have from part (a):
km/s,
km/s,
; from equation(11.9b), we get
s,
s. For a geophone
at a distance
from
, equation (11.8b) gives for source
,
and for source
,
(note that
is measured from
for
and from
for
).
We can obtain values of
either by using the above equations or graphically by drawing straight lines with slope
starting at the half-intercept values (there by subtracting it); the vertical distances between these lines and the traveltime curves give
. The values of
in Tables 11.14b,c were calculated.
Table 11.14b. Calculations of
and
for source
.
(km)
|
(s)
|
(s)
|
(km)
|
(km)
|
(km)
|
(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
in Figure 11.14b(iii) and to draw an arc with radius
. We first calculate
and then find
by calculating
or
and drawing a line normal to
. The first method was used to get Tables 11.14b,c (although
is given in the tables, it was not used). We repeat equations (11.14f,g,h) and get
|
|
(11.14l)
|