# User:Ageary/Chapter 11

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 |

## Contents

- 1 Chapter 11 Refraction methods
- 2 11.1 Salt lead time as a function of depth
- 3 11.2 Effect of assumptions on refraction interpretation
- 4 11.3 Effect of a hidden layer
- 5 11.4 Proof of the ABC refraction equation
- 6 11.5 Adachi’s method
- 7 11.6 Refraction interpretation by stripping
- 8 11.7 Proof of a generalized reciprocal method relation
- 9 11.8 Delay time
- 10 11.9 Barry’s delay-time refraction interpretation method
- 11 11.10 Parallelism of half-intercept and delay-time curves
- 12 11.11 Wyrobek’s refraction interpretation method
- 13 11.12 Properties of a coincident-time curve
- 14 11.13 Interpretation by the plus-minus method
- 15 11.14 Comparison of refraction interpretation methods
- 16 11.15 Feasibility of mapping a horizon using head waves
- 17 11.16 Refraction blind spot
- 18 11.17 Interpreting marine refraction data.

## 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,

(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.

**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,

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

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

**(**)

**Solution**

Assuming that , , and are coplanar and that elevation corrections have been made, we can write

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

**(**)

If we set , becomes the intercept time at the downdip source; thus,

**(**)

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) |

The angles are related as follows:

**(**)

Snell’s law [equation (3. 1a)] gives

**(**)

For the refraction along the interface,

**(**)

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 .

**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.

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:

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

## 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

**(**)

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).

## 11.8 Delay time

**Show that in Figure 11.8a is given by**

**(**)

**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

**(**)

**(**)

**(**)

**(**)

**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

**(**)

**(**)

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

**(**)

**(**)

(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

**(**)

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

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.

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. .

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

[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:

- 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 .

(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,

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,

**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.

(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 |

**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.

(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

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

**(**)

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

**(**)

In Figure 11.14b(ii) moves so that the ratio . The major axis 2a of the ellipse is

**(**)

To get the minor axis, we set the first derivative of equal to zero. Using equation (11.14b), we find that

**(**)

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.

**(**)

**(**)

**(**)

**(**)

**(**)

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

**(**)

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.

(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

**(**)