User:Ageary/Chapter 8

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

8.1 Effect of too many groups connected to the cable

Land cables are built in sections that are identical. The connections to plugs at each end of the section effectively rotate the system by the number of groups to be connected to the section. Thus, if pins 1, 2, and 3 at one end of the section are connected to takeouts for 3 groups in a section, at the other end of the section they will connect with pins 4, 5, and 6 of the next section. Usually there are more sets of wires than channels being used at any one time, for example, perhaps 96 independent pairs of wires for use with 48 channels. However, occasionally so many sections and geophones are laid out that a distant group of phones is connected to the same channel as a nearer group. Sketch a possible arrangement for the connections in one section with takeouts for three channels and explain how this would appear on the seismic record.

Solution

For simplicity we assume 9 channels only with 3 groups per section, as diagramed in Figure 8.1a. The circles at the left represent the intake to the recording system, the arrows $\rightarrow$ and $\leftarrow$ represent the connectors at each end of the sections. When two sections are connected, the takeouts for the right-hand section are rotated three channels as shown in the figure. Note that the section is symmetrical; the end pin on the left (#1) relates to the closest takeout, as does the end pin on the right (#9). So flipping the section end-for-end makes no difference. If we connect three sections with nine groups as shown, the profile will be recorded correctly. However, if a 4th section is connected at the right, the rotation will have gone full circle and we will have distant geophones connected to the same channels as near geophones, as shown in Figure 8.1a.

Assuming that the source is to the left of the cable, the signals from groups 10−12 will differ from those from groups 1−3 in NMO and will be weaker due to the longer path. The NMO corrections appropriate to groups 1 to 3 are incorrect for groups 10 to 12 so that, after these corrections have been applied, the signals from groups 10−12 will not reinforce those received by groups 1−3 and so will constitute noise on the record. The effect will be most serious in weak portions of the record.

Figure 8.1a.  Relation between groups and channels.

8.2 Reflection-point smear for dipping reflectors

Assume a reflector 2000 m beneath the midpoint and a dip of $20^{\circ }$ with constant over-burden velocity. How much does the reflecting point move between members of the common-midpoint set for offsets of 0, 500, 1000, 1500, and 2000 m?

Background

Equation (4.11e) gives the shift in the reflecting point $\Delta L$ in terms of the dip $\xi$, slant depth $h_{c}$, and offset 2s:

${\begin{aligned}\Delta L=(s^{2}/2h_{c}){\rm {\;sin\;}}2\xi .\end{aligned}}$

(8.2a)

Solution

Because $\sin 2\xi =\sin 40^{\circ }=0.64$, equation (8.2a) becomes

${\begin{aligned}\Delta L=0.16\times 10^{-3}s^{2}{\mbox{m}}.\end{aligned}}$

Thus, we get the following values of $\Delta L$ for the various offsets:

${\begin{aligned}{\hbox{Offset}}(2s)&\rightarrow &0&500&1000&1500&2000m,\\{\hbox{Shift}}(\Delta L)&\rightarrow &0&40&160&360&640m.\end{aligned}}$

8.3 Stacking charts

8.3a Draw surface and subsurface stacking charts for a 24-trace split spread where the source is midway between the centers of geophone groups 12 and 13, which are separated by three times the normal geophone-group spacing. Assume a source interval double the geophone-group interval and that source and geophone locations beyond a location at the right-hand end of the line cannot be occupied, but the source can move as far as this location. Thus spreads near the end of the line will be asymmetric; this technique is called “shooting through the spread.”

Background

In seismic field work, several relatively closely spaced geophones are connected together to constitute a geophone group; the group is thought of as equivalent to a single large geophone located at the group center. A spread denotes all the geophone groups that record signals from a single source.

A split spread (also called split dip) has the source located at the midpoint of a linear spread. If the source is moved forward half the spread length and the rear half of the cable and geophones are moved forward to become the front half, each subsurface reflecting point will be sampled once, resulting in single-fold continuous subsurface coverage.

In the common midpoint (CMP) method (see problem 5.12), many source and geophone-group locations are used. Stacking charts are used to keep track of locations; stacking charts have geophone-group locations along one axis and source locations along the other (see Figure 8.3a). A trace observed at group location $g$ when the source is at $s$ is plotted at $(g,s)$ in a surface stacking chart, and at $[(g+s)/2,s]$ at in a subsurface stacking chart (see Figures 8.3a,c). Figure 8.3a assumes that the fifth source location is at ${\rm {E}}^{\prime }$ rather than at E. On stacking charts, straight lines locate all of the traces involving a common-geophone, common-source, common-offset, or common-midpoint. In the CMP method, traces having the same midpoint are stacked together. The number of traces stacked together (after statics and NMO corrections) is the multiplicity (see problem 5.12).

Solution

Figure 8.3b shows the end of a surface stacking chart where “shooting through the spread” occurs. Where the pattern is regular the multiplicity is 6, as shown by the common-midpoint lines drawn on the figure at $45^{\circ }$ angles. The irregularity at the end of the line adds additional longer-offset traces and steepens the end-of-line taper. The resulting multiplicity at the end of the line is 6, 6, 6, 7, 7, 8, 8, 9, 9, 9, 8, 8, 7, 7, 6, 6, 4, 4, 2, 2. The corresponding subsurface stacking chart is shown in Figure 8.3c.

Note that where the source occupies a geophone location, pairs of traces on common-midpoint gathers (see problem 9.23) on opposite sides of the source have the same offset distances, whereas if the source occupies a location between geophone-station locations, then offsets in opposite directions from the source (as in this example) do not pair up. The result is more offset distances on the common-midpoint gathers, which is generally preferable.

8.3b Assume that geophones, but not sources, can be located over a distance of geophone-group intervals centered at point $B$ in Figure 8.3d. What is the effect on the multiplicity? If the data deteriorate markedly when multiplicity falls, what can be done to alleviate the deterioration?

Solution

Not being able to locate sources for eight geophone group intervals creates tapers on either side of the zone, as indicated by the common-midpoint lines drawn on Figure 8.3d. Additional source locations (for example, midway between the source locations) on each side of zone $B$ could add to the multiplicity while creating irregularities in the offset distribution.

Figure 8.3a.  Common-midpoint method. (i) Vertical section; (ii) surface stacking chart; (iii) subsurface stacking chart.

Figure 8.3b.  Surface stacking chart.

Figure 8.3c.  Subsurface stacking chart.

Figure 8.3d.  Effect of no-source zone.

8.4 Attenuation of air waves

For an airwave with velocity of 330 m/s and two geophones separated by 5 m, at what frequency is maximum attenuation achieved?

Solution

Maximum attenuation is achieved when the waves recorded at the two geophones differ in phase by one-half wavelength. Therefore $\lambda /2=5\ {\rm {m}}$ and $f=V/\lambda =330/10=33\ {\rm {Hz}}$ Hz.

8.5 Maximum array length for given apparent velocity

Reflections in the zone of interest have apparent velocities around 6300 m/s, whereas the velocity just below a uniform LVL is 2100 m/s. If we wish to avoid severe attenuation below 80 Hz when using an array, what is the maximum inline array length?

Background

The LVL is discussed in problem 4.16. Apparent velocity is defined in problem 4.2d. See problem 8.6 for a discussion of array response. Note that the array length, $n\Delta x$, is larger than the distance between the first and last geophones, $(n-1)\Delta x$.

Solution

To avoid deterioration, the response curve for the array length $n\Delta x$, where n is the number of geophones and $\Delta x$ the geophone interval, must not extend beyond the first null (see Figure 8.6b). Equation (8.6b) shows that the first null occurs when $n\gamma /2=\pi$ or $n=2\pi /\gamma$, where $\gamma$ is the phase difference between adjacent geophones. Problem 8.6a gives $\gamma =2\pi \left(\Delta x/\lambda \right)\sin \alpha$, so

${\begin{aligned}n&=2\pi /2\pi \left(\Delta x/\lambda \right)\sin \alpha ,\\{\mbox{and }}\qquad \qquad n\Delta x&=\lambda /\sin \alpha =\lambda _{a}=V_{a}/f,\end{aligned}}$

(8.5a)

where $V_{a}=$ apparent velocity and $\lambda _{a}=$ apparent wavelength. Thus.

${\begin{aligned}n\Delta x=2100/80{\mbox{ m}}=26{\mbox{ m}}.\end{aligned}}$

The geophone interval must not exceed $\Delta x=26/n$.

8.6 Response of a linear array

8.6a Under what conditions is the response of a linear array of $n$ evenly spaced geophones zero for a wave traveling horizontally (such as ground roll)?

Background

In Figure 8.6a, a plane wave is approaching a linear array of $n$ identical geophones spaced at intervals of $\Delta x$. The wave arrives at the left-hand end of the array at time $t$ giving the output $A\sin \omega t$ (note that the plane of Figure 8.6a is not necessarily vertical).

Figure 8.6a.  Wavefront approaching array.

The path difference for adjacent geophones is $\Delta x\sin \alpha$, the time difference is $\Delta t=(\Delta x\sin \alpha )/V$ and the phase difference is $\gamma =2\pi (\Delta t/T)=\omega \Delta t$ The output of the $r^{\rm {th}}$ geophone is $A\sin(\omega t-r\gamma )$.

Summing the outputs of the $n$ geophones gives

${\begin{aligned}h(t)=\mathop {\sum } \limits _{r=0}^{n-1}A\sin(\omega t-r\gamma )\end{aligned}}$

(8.6a)

${\begin{aligned}=A\left({\frac {\sin(n\gamma /2)}{\sin(\gamma /2)}}\right)\sin \left(\omega t-{\frac {1}{2}}(n-1)\gamma \right)\end{aligned}}$

(8.6b)

(see Sheriff and Geldart, 1995, problem 15.12c). The amplitude is $A$ times the expression in the first bracket. If the $n$ geophones were located at one point, the amplitude would be $nA$. Dividing the amplitude of $h(t)$ by $nA$ gives the array response $F$:

${\begin{aligned}F={\frac {\sin(n\gamma /2)}{n\sin(\gamma /2)}}.\end{aligned}}$

(8.6c)

But $\gamma =\omega \Delta t=(\omega /V)\Delta x\sin \alpha =2\pi (\Delta x/\lambda )\sin \alpha$, so equation (8.6c) can be written as

${\begin{aligned}F={\frac {\sin[n\pi (\Delta x/\lambda )\sin \alpha ]}{n\sin[\pi (\Delta x/\lambda )\sin \alpha ]}}\end{aligned}}$

(8.6d)

The graph of $|F|$ is shown in Figure 8.6b(i) for five uniformly spaced geophones and in Figure 8.6b(ii) for nine geophones. Note that the abscissa can be expressed in a number of ways; apparent wavelength $=\lambda _{a}=V_{a}/f=V/f\sin \alpha =\lambda /\sin \alpha$, where $\alpha$ is the angle of approach in Figure 8.6a. Figure 8.6b(iii) shows the response for a tapered array.

Ground roll is a Rayleigh wave (see problem 2.14) and usually has low velocity and low frequency. It is nondispersive (see problem 8.11) if the medium is uniform, but it is dispersive on an inhomogeneous earth where velocity and other parameters change with depth.

Solution

For a horizontally traveling wave, the plane of Figure 8.6a is horizontal and equation (8.6d) is unchanged. The only variable at our disposal in equation (8.6d) is $\alpha$ (the other quantities all being fixed by the nature of the array), and the obvious values to check are $\alpha =0^{\circ }$ and $\alpha =90^{\circ }$ For $\alpha =0^{\circ }$, the geophone outputs add up [see part (c)], so we try $\alpha =90^{\circ }$ Equation (8.6d) now reduces to

${\begin{aligned}F={\frac {\sin[n\pi (\Delta x/\lambda )]}{n\sin(\pi \Delta x/\lambda )}}.\end{aligned}}$

For $F$ to be zero, the numerator must vanish and the denominator must be nonzero [see part (b) for the case where both numerator and denominator are zero]. Thus the arguments of the sines must equal $m\pi$ in the numerator and not equal $n\pi$ in the denominator, $m$ and $n$ being integers. Therefore $(\Delta x/\lambda )=m/n$ where $m$ is not a multiple of $n$, i.e., $m=\ldots ,(n-1),(n+1)\ldots$,

8.6b If $n$ geophones are distributed uniformly over one wavelength, show that the response is $F=1/n$

Solution

The effective array length is $(n-1)\Delta x$, If the wavelength is $n\Delta x$, samples in the peak and trough of the wave can be paired up yielding a response of zero. If the wavelength $\lambda =(n-1)\Delta x$, i.e., if the distance between the first and last geophones in the group is $\lambda$, then the outputs of $n-1$ geophones will cancel and the output will be that of the remaining geophone $n$.

${\begin{aligned}F={\frac {\sin[n\pi /(n-1)]}{n\sin[\pi /(n-1)]}}={\frac {\sin[\pi +\pi (n-1)]}{n\sin[\pi /(n-1)]}}=-1/n.\end{aligned}}$

8.6c What is the response of the array in part (a) when the waves arrive perpendicular to the line of geophones?

Figure 8.6b.  Array response to 30-Hz signal. Alternative scales are shown in (i). (i) Five geophones spaced 10 m apart; (ii) nine phones spaced 5.5 m; (iii) five locations as (i) but weighted 1, 2, 3, 2, 1.

Solution

Since the wave is a plane wave, it will arrive at all geophones at the same time, which will give the same result as if all $n$ geophones were together; therefore, $F=1$.

If we set $\alpha =0^{\circ }$ in equation (8.6d), we get $F=0/0$. However, using the simpler equation (8.6b) we have

${\begin{aligned}F=\mathop {\lim } \limits _{\alpha \to 0}\left[{\frac {\sin \left({\frac {1}{2}}n\gamma \right)}{n\sin \left({\frac {1}{2}}\gamma \right)}}\right]=\mathop {\lim } \limits _{\alpha \to 0}\left({\frac {1}{2}}n\gamma /n\times {\frac {1}{2}}\gamma \right)=1,\end{aligned}}$

where we have replaced the sines by their arguments.

8.6d What is the response of the array in part (a) when the waves arrive at $45^{\circ }$ to the line and $n=8$? Repeat for $n=16$.

Solution

We have $\alpha =45^{\circ }$, $n=8$, so equation (8.6d) becomes

${\begin{aligned}F={\frac {\sin \left[8\pi \left(\Delta x/\lambda \right)\sin 45^{\circ }\right]}{8\sin \left[\left(\pi \Delta x/\lambda \right)\sin 45^{\circ }\right]}}={\frac {\sin \left[\left(8\pi /{\sqrt {2}}\right)\left(\Delta x/\lambda \right)\right]}{8\sin \left[\left(\pi /{\sqrt {2}}\right)\left(\Delta x/\lambda \right)\right]}}.\end{aligned}}$

Replacing 8 with 16 in the above equation gives

${\begin{aligned}F={\frac {\sin \left[\left(16\pi /{\sqrt {2}}\right)\left(\Delta x/\lambda \right)\right]}{16\sin \left[\left(\pi /{\sqrt {2}}\right)\left(\Delta x/\lambda \right)\right]}}.\end{aligned}}$

8.6e Repeat part (d) when $\Delta x/\lambda =1/8$. Compare the results when $n=8,16,32$.

Solution

When $\Delta x/\lambda =1/8$ in part (d) and $n=8$, the result is

${\begin{aligned}F=\sin(\pi /{\sqrt {2}})/8\sin(\pi /8{\sqrt {2}})=0.796/(8\times 0.274)=0.363.\end{aligned}}$

For $n=16$, we get

${\begin{aligned}F=\sin(2\pi /{\sqrt {2}})/16\sin(\pi /8{\sqrt {2}})=-0.961/16\times 0.274=-0.220.\end{aligned}}$

When $n=32$, $\Delta x/\lambda =1/8$ the result is

${\begin{aligned}F=\sin(4\pi /{\sqrt {2}})/32\sin(\pi /8{\sqrt {2}})=0.859/32\times 0.274=0.059.\end{aligned}}$

The ratios are 1 : 0.61 : 0.16.

8.7 Directivities of linear arrays and linear sources

Show that the directivity equations (8.6d) and (7.5d) are consistent.

Solution

Since equation (8.6d) applies to a discontinuous array of geophones whereas equation (7.5d) applies to a continuous source, we find the limit of equation (8.6d) as the number of geophones becomes infinite. We require that $n\to \infty$ while $\Delta x\to 0$ in such a way that $n\Delta x\to a\lambda$, $a$ being the same constant as in problem 7.5, thus keeping the array length equal to the source length. In the limit the numerator of equation (8.6d) becomes $\sin(\pi a\sin \alpha )$. To get the limit of the denominator, we replace the sine by its argument (because $\Delta x\to 0$) and get $(n\pi \Delta x/\lambda )\sin \alpha \to \pi a\sin \alpha$. Substituting these values in equation (8.6d), we obtain

${\begin{aligned}F=\sin(\pi a\sin \alpha )/\pi a\sin \alpha =\sin c(\pi a\sin \alpha ).\end{aligned}}$

Because the angles $\alpha$ and $\theta _{0}$ are equivalent, equations (8.6d) and (7.5d) are equivalents.

8.8 Tapered arrays

Array tapering can be achieved by

1. increasing the number of elements at some locations,
2. weighting equally spaced elements either
   1. within the elements, or
   2. in a mixing box, or
3. using unequal spacing of elements.

What arguments are there for/against each of these approaches?

Background

In Figure 8.6b the main lobe is the lobe centered on the origin while its replica to the right is the alias lobe. The region between these lobes is the reject region, so-called because of the high attenuation in this region.

Tapered arrays are used to broaden the main and alias lobes while decreasing the response in the reject region. Compare parts (i) and (iii) of Figure 8.6b.

Solution

1) Increasing the number of elements at some locations improves geophone coupling to the ground and does not require special instrumentation. However, this requires more geophones and more connections, and increases the possibility of adding the extra geophones at the wrong locations or connecting some geophones incorrectly.

2,a) Weighting equally spaced elements within the elements requires fewer geophones. However, it requires specially designed geophones and increases the possiblity of using the wrong geophone at a location.

2,b) Weighting equally spaced elements in a mixing box easily achieves tapering and permits the tapering to be varied easily. However, it involves the cost of the mixing box and possible errors in connecting the geophones to the box or adjusting the weights.

3) Spacing elements unequally does not require special equipment. However, it is inconvenient and expensive to determine accurately the individual geophone locations.

Method 1 is the one almost always used.

8.9 Directivity of marine arrays

Figure 8.9a shows the directivity effect of group length typical of end-on marine shooting. How will the curves change (a) with arrival time, (b) as offset increases, and (c) for greater stacking velocity? Is it better to tow the recording system in the updip or the downdip direction?

Solution

Aside from the effects of the number and spacing of the elements, angles of approach, and wavelength, the array response is affected by NMO and dip moveout. These add in the downdip direction and subtract in the updip direction.

Figure 8.9a.  Array response to apparent dip. Array length 50 m, offset 500 m, time 1.0 s.

1. As $t_{0}$ increases, the NMO decreases and hence has less effect, so the curves in Figure 8.9a become more symmetrical.
2. As the offset increases, the NMO increases and the curves become more asymmetrical.
3. For greater stacking velocity, NMO decreases and the curves become more symmetrical.

Because NMO and dip moveout effects add in the downdip direction, it is better to shoot in the updip direction.

8.10 Response of a triangular array

8.10a The tapered array [1, 2, 3, 3, 2, 1] =[1, 1, 1, 1] * [1, 1, 1] is called a triangular array. Use this fact to sketch the array response.

Background

We use the notation $f(x)$ to represent a continuous function of the variable $x$ while the notation $f_{x}$ denotes a digital function, that is, the result of sampling a continuous function at a fixed sampling interval $\Delta$ (see problem 9.4).

The triangular array is also used to approximate a cosine array where successive elements are weighted as equally spaced samples of the first half-cycle of a cosine function.

Figure 8.10a.  Response of tapered array.

The notation $g_{t}*h_{t}$ denotes the convolution of $g_{t}$ and $h_{t}$ (see problem 9.2). The convolution is given by the summation in equation (9.2b), namely

${\begin{aligned}g_{t}*h_{t}=\mathop {\sum } \limits _{k}^{}g_{k}h_{t-k}=\mathop {\sum } \limits _{k}^{}h_{k}g_{t-k}.\end{aligned}}$

(8.10a)

Solution

We get for the convolution:

${\begin{aligned}[1,1,1,1]*[1,1,1]=[1\times 1,1\times 1+1\times 1,1\times 1+1\times 1\\\quad +1\times 1,1\times 1+1\times 1+1\times 1,1\times 1\\\quad +1\times 1,1\times 1\left]=\right[1,2,3,3,2,1].\end{aligned}}$

The response of this array to a harmonic signal is shown in Figure 8.10a.

8.10b How could three strings of geophones, each having four equally spaced elements, be laid out to yield a triangular array?

Solution

Number six geophone locations 1 through 6 and lay the first string (see problem 8.13) of four geophones from position 1 to position 4, the second string from 2 to 5, and the third string from 3 to 6. This give the array 1, 2, 3, 3, 2, 1.

8.10c How could a smoother tapered array be approximated?

Solution

We could achieve a smoother array by spacing the geophones unequally such that each represents an equal portion of the area under the desired array response curve, as illustrated in Figure 8.10c.

Figure 8.10c.  Approximating smooth array with unequal geophone spacing.

8.11 Noise tests

8.11a The noise test shown in Figure 8.11a used 36 geophones spaced 10 m apart and six sources spaced 360 m apart. The event $A_{1}-A_{2}$ indicates ground roll. What are the velocites, dominant frequencies, and wavelengths of the noise trains? What length of a geophone group will attenuate them?

Background

A noise test (noise profile) uses single geophones that are closely spaced (as little as 1–3 m apart) and recorded individually. The profile is studied to identify the characteristics (especially apparent velocities) of noise wavetrains so that arrays can be designed to attenuate them.

A pulse is composed of many frequency components, each traveling with a phase velocity that at times varies with the frequency; in this case the pulse changes shape and travels with the group velocity (see problem 2.7c), an effect called dispersion.

Ground roll is discussed in problems 2.14 and 8.6.

Solution

Measurements on Figure 8.11a are very crude. The ground-roll wavetrain $A_{2}-A_{1}$ loses its early cycles with distance because it is dispersive. Its apparent velocity is about 90 m/s. At $A_{2}$ the peak-to-peak period is about 70 ms or 15 Hz frequency, so its wavelength is about 6 m. The dominant frequency near $A_{1}$ is about 5 Hz so the wavelength is about 18 m and a linear geophone array of an integral number of wavelengths would attenuate it. The wavetrain that arrives at long offsets at about 1.8 s has a velocity of about 170 m/s and frequency of about 10 Hz, or wavelength about 17 m. An array about 18 m long would attenuate both wavetrains.

Figure 8.11a.  Walkaway noise test.

8.11b Explain the alignment $B_{1}=B_{2}$

Solution

This event is probably a result of spatial aliasing (problem 9.25) although it might be a backscattered surface wave, scattered from a source not necessarily inline.

8.12 Selecting optimum field methods

Assume that you wish to map objectives 3 to 5 km deep in an area with topography ranging from flat to gentle hills (gradients usually less than 3%). Dips at objective depths may be up to $30^{\circ }$. The velocity at the base of the low-velocity layer is 2000 m/s, that at objective depths is 4000 m/s, and at the basement (8 km) probably about 6000 m/s. Five surface-source units and 96-channel recording equipment are available. Both ground roll $(V_{R}\approx 800{\mbox{ m/s}})$ and air waves $(V\approx 330{\mbox{ m/s}}$ may be problems, but the low-velocity layer (about 10 m thick with a velocity about 600 m/s) is probably fairly uniform. The area is fairly noisy and moderate effort will probably be required to achieve adequate data quality. Propose field methods and explain the bases for your proposals.

Background

The following rules are based to some extent upon theory, but also, to a considerable degree, on experience.

1. The maximum offset should roughly equal the shallowest depth of interest; this usually provides large enough NMO that primaries can be distinguished from multiples while avoiding problems such as large changes in the reflection coefficients and errors arising from approximations in the NMO equation (4.1c) on which CMP corrections depend.
2. The minimum offset should not exceed the depth of the shallowest zone of interest for the reasons given in (1). However, noise produced by the source may dictate a larger value that this.
3. The maximum array length must not exceed the minimum apparent wavelength [see equation (8.5a)]. The minimum apparent velocity $V_{a}$ usually occurs at the maximum offset, and the minimum $\lambda _{a}$ for $V_{a}$ should be within the main lobe in Figure 8.6b(i).
4. The minimum inline geophone spacing within a group should be less than the minimum $\lambda _{a}$ of the noise, which is usually that of the lowest-velocity noise.
5. Provided that rule (3) is not violated thereby, the geophone group interval should not be more than double the desired horizontal resolution (see problem 6.2) at the depth of interest.

Solution

Since the velocity varies from 2 km/s near surface to 4 km/s at depth, we shall base estimates on an average velocity of 3 km/s. Thus the reflection traveltimes at the objective depths range from 2 to 3 s. A basement reflection may occur at about 4 s.

To map at depths of 3–5 km, we would like to use a maximum offset of about 5 km. If we were to use a split-spread configuration to cover the distance, the group interval would have to be about 100 m unless we sacrifice some short-offset data, and 100-m group intervals might be too large for the expected 30-degree dips. Using an end-on configuration, the group interval would need to be only about 50 m, a much more conservative arrangement.

The maximum frequency for the deeper reflections will probably be no more than 40 Hz, in which case the minimum apparent wavelength [see (8.5a)] at the surface will be $2000/(40\sin 30^{\circ })=100{\mbox{ m}}$. To avoid spatial aliasing we must sample at least twice per cycle (see problem 9.25), so 50 m is the maximum group interval.

Because the area is expected to be moderately difficult in terms of signal/noise ratio, we shall probably require high multiplicity (problems 5.12 and 8.3), probably 24 fold, perhaps 48 fold. To start, we will probably use a 100 m source-point interval, twice the geophone-group interval, changing our recording pattern depending on the initial results.

Assuming the ground roll has a broad spectrum from 10 to 40 Hz, the corresponding wavelengths will be 20 to 80 m. We will want the inline geophone group length to equal the geophone-group interval so that the gathers effectively represent a group having the spread length which will permit maximum attenuation in stacking (this is the stack-array concept). Thus we should space geophones within the group at intervals no larger than 10 m, probably 3–4 m, which will require 10–12 geophones per group.

The ground roll with a velocity of 800 m/s will arrive at 3000 m offset at about 3.8 s and at 5000 m offset at 6.2 s, so we should be able to record reflections on most traces before the ground roll arrives.

We might use a split spread with a near-offset of 1200 m and a far-offset of 5000 m, 80-m geophone-group intervals and 80-m groups. An end-on spread from 0 to 5000 m with 50-m geophone-group interval is more conservative, and a split-spread does not offer much advantage over an end-on spread.

8.13 Optimizing field layouts

Seismic field work is usually carried out in a uniform manner with group intervals everywhere the same and the spread either symmetrical about, or on one side of, the source point. Layouts tend to be determined by the equipment at hand or by habit rather than by the problem to be solved; for example, the length of geophone strings (several geophones for a single group permanently wired together) may dictate the geophone interval and the available equipment the number of channels, and thus, the effective spread length. Hybrid spread arrangements are sometimes used to make fuller use of equipment.

Assume that you have more channels available than given by the rules stated in problem 8.12; what circumstances might lead you to use the extra channels

1. to extend the spread to longer offsets than given by rule (1);
2. to use shorter minimum offsets than given by rule (2);
3. to interleaf additional groups somewhere in the spread;
4. to lay out a partial spread on the other side of the source where an end-on arrangement is being used;
5. to lay out a short cross arm? If you have almost but not quite enough channels to use a split arrangement compared to an end-on, what are the advantages and disadvantages of using a split spread with
6. longer group intervals than given by rule (5);
7. shortening the maximum offset; or
8. increasing the minimum offset?

Background

Commonly used spread types are shown in Figure 8.13a. Spreads (ii), (iii), and (v) are used when source noise is a problem (spread (v) is also used to obtain larger offsets). A hybrid spread is one in which the group spacing is different for some groups, usually larger for long-offset groups.

Figure 8.13a.  Typical spreads using 24 groups. Geophone group and source locations are represented by x and o, respectively. (i) Split spread; (ii) split spread with offset source; (iii) gapped split spread; (iv) end-on spread; (v) inline offset spread; (vi) Cross-spread.

Coherence refers to the similarity of an event as seen on successive traces (see also problem 6.1); it is the most important factor in recognizing a reflection (or any event).

The manner in which amplitude varies with offset (or angle of incidence; see problem 3.12), called AVO (or AVA), depends on Poisson’s ratio (or $V_{P}/V_{S}$), which is sensitive to changes in lithology and the fluid contained in pore spaces.

Horizontal resolution is discussed in problem 6.2.

Solution

1. Extending the spread to offsets larger than the depth of the deepest zone of interest could help in (i) mapping deeper; (ii) increasing the amount of NMO to get better velocity information; (iii) increasing the NMO differences between primaries and multiples to better attenuate multiples; (iv) decreasing source-generated noise; (v) getting better AVO data.
2. Using minimum-offsets less than the depth of the shallowest zone of interest might yield useful shallow information, for example, it might yield data less confused by source-generated wavetrains, such as ground roll.
3. Interleafing additional groups in the middle of the spread might (i) improve coherence and thus increase the detectability of weak and steeply dipping events; (ii) decrease the possibility of aliasing.
4. Laying out a partial spread on the other side of the source where an end-on arrangement is being used would increase the amount of data and, hence, improve noise cancellation, especially shallow noise, and yield better measurement of dip.
5. Laying out a short cross arm would be valuable to (i) measure the cross-dip or check that the line has in fact been laid out in the direction of dip; (ii) check against the possibility of noise arriving from the cross-spread direction.
6. If we have almost but not quite enough channels to use a split arrangement compared to an end-on, a split spread with group intervals more than double the desired horizontal resolution may hurt coherence, permit aliasing of steeply dipping data, and degrade horizontal resolution. The split would add redundancy and perhaps attenuate noise.
7. Without enough channels to use a split arrangement, shortening the maximum offset will give poorer moveout measurements with consequent poorer stacking-velocity values and less multiple attenuation. It will also discriminate against the deeper data. Closer sampling might define events better.
8. Without enough channels to use a split arrangement, increasing the minimum offset may cause deterioration of the shallow data, but traces near the source may be noisy anyway. The advantages are closer sampling and increased redundancy.

8.14 Determining vibroseis parameters

8.14a Signal and noise characteristics determined from a previous dynamite survey are shown in Figure 8.14a. The principal objective is at 3000 m with a stacking velocity of 3000 m/s. What frequencies should be covered by a linear sweep?

Figure 8.14a.  Signal and noise spectra (from Evans, 1997).

Background

The vibroseis method employs a vibrator to impart to the ground a long train of harmonic signals of varying frequency. The vibrator consists of a piston pressing against a steel plate which is held against the ground by the weight of the vehicle. For the usual linear sweep, the vibrator, which is actuated hydraulically, exerts a pressure ${\mathcal {P}}(t)$ against the plate of the form

${\begin{aligned}{\mathcal {P}}(t)=A(t)\sin\{2\pi t[f_{0}+({\rm {d}}f/{\rm {d}}t)t]\},\end{aligned}}$

(8.14a)

where $f_{0}$ is the starting frequency and $({\rm {d}}f/{\rm {d}}t)$ is either positive (for an upsweep) or negative (downsweep). The amplitude $A(t)$ is constant (except for about 0.2 s at the beginning and end of the sweep, when it increases from or decreases to zero. The frequency varies between about 12 and 60 Hz and the duration of the sweep is usually 7 to 35 s.

Each sweep generates a signal train which is reflected at each of the reflectors; since reflections are much more closely spaced than the length of the sweep, the recorded signal is a complex superposition of many reflected wave trains. To interpret a vibroseis record, the input sweep is recorded and crosscorrelated (see problem 9.8) with the record; this compresses the reflected wavetrains into short wavelets, thereby removing much of the overlap of the lengthy reflected wavetrains to produce a more-or-less normal seismic record.

The response of the ground is not an exact reproduction of the motion of the piston and distortion introduces harmonics into the ground, principally the second harmonic. The second harmonic produces reflected wavetrains like the primary reflected wavetrains, but these correlate with the sweep signal to indicate different arrival times. This effect, called correlation ghosts, adds a spurious set of reflection events to the record (Sheriff and Geldart, 1995, 208). The arrival time of this ghost $t_{g}$ is

${\begin{aligned}t_{g}=f_{L}T/(f_{i}-f_{f}),\end{aligned}}$

(8.14a)

where $f_{L}$ is the lowest sweep frequency, $T$ the sweep time, $f_{i}$ the initial sweep frequency, and $f_{f}$ the final sweep frequency. For an upsweep, the ghosts arrive before $t=0$, so the ghost is no problem; to avoid this problem with downsweeps, we can use long sweeps so that the ghost is delayed until after the zone of interest is recorded.

If we have $n$ records on which the signal shape is essentially constant and the noise is random, stacking builds up the signal strength whereas random noise tends to cancel and the signal-to-noise ratio (S/N) varies as $n^{1/2}$ (Sheriff and Geldart, 1995, 184).

When multiple source units are used, the vibrators are actuated synchronously at locations a few meters apart. Generally, several sweeps at closely spaced locations constitute a single vibrator point.

Solution

The signal and noise spectra in Figure 8.14a show that the signal spectrum is strong between 15 to 55 Hz while the noise spectrum is confined largely to the narrow peak centered around 15 Hz. A passband of 20–60 Hz would include most of the signal and exclude much of the noise. This is a bandwidth of roughly 1.5 octaves.

8.14b If a downsweep of 8 s is used, at what time will the correlation ghost appear for a 9-s 60–15 Hz linear sweep? Will it interfere with the objective?

Solution

Assume a downsweep that lasts 8 s and goes from 60 Hz to 15 Hz, that is, 5.6 Hz/s. Fundamental frequencies may generate second harmonic correlation ghosts which fall within the passband. The ghosts of 30–15 Hz may interfere with desired reflections arriving after

${\begin{aligned}t=15\times 8/\left(60-15\right)=2.667{\mbox{ s}},\end{aligned}}$

according to equation (8.14b).

8.14c If a single vibrator sweep of 8 s yields an S/N of 0.2 at the objective depth for a 15 to 60 Hz sweep, how many sweeps will have to be stacked to give ${\mbox{S/N}}=2.0$?

Solution

Assuming the noise is random, the S/N varies as $n^{1/2}$. If we need to increase S/N by a factor of 10, we require $10^{2}=100$ sweeps. Note that the improvement of S/N by the factor $n^{1/2}$ does not apply to coherent noise.

8.14d Assume that recording continues for an additional time of 6 s (listen time) beyond the sweep time and that it takes 10 s to move the vibrators between sweep points; how long will be required for four vibrators to record one vibrator point?

Solution

The sweep time plus the listening time is 14 s; adding the time to move the vibrators, we get a total of 24 s per sweep. To obtain 100 sweeps as required by part (c), we must move the four vibrators 25 times, taking $25\times 24=600{\mbox{ s}}=10$ minutes. However, the vibrators probably build up the signal more than they build up the noise and much of the noise is probably not random, so the number of required sweeps should be cut at least in half.

8.15 Selecting survey parameters

8.15a Assume that you wish to survey a $70\times 70$ km area where Eocene and Cretaceous anticlinal structures with their long axes north-south are expected, the minimum size of an economically viable structure being 2 km across. The maximum dip expected is $15^{\circ }$ and reflectors are listed in Table 8.15a “The Recent” reflection will be useful in making static corrections. A noise test gave a prestack ${\mbox{S/N}}=0.5$. Propose the line spacing and orientation of a reconnaissance survey.

Background

Static corrections are corrections that are independent of traveltime; these include corrections for variations in the surface elevation and the weathered layer (see problem 8.18).

A noise test is discussed in problem 8.11.

Table 8.15a. Reflection data.

Age	Depth (m)	${\bar {V}}(m/s)$	$t_{0}(s)$	$V_{i}(m/s)$
Recent	300	2000	0.300	2000
Eocene	3000	3000	2.000	3180
Cretaceous	5000	3370	2.970	4140

Solution

A reconnaissance will mainly use east-west lines plus a few north-south lines to tie the survey together. Any prior knowledge of the area will help in locating the lines. This includes examination of the land surface to see if surface features may relate to deeper structure. We do not expect to locate all possible structures on the first reconnaissance so east-west lines will be spaced 10 km or more apart and the north-south lines about double this. We shall plan on about seven east-west and three to four north-south lines and we should run the east-west lines first so that we can use their interpretation to locate the north-south lines. We will then select a couple of portions of the area for more detailed surveying where we can infill with east-west lines 2 to 3 km apart plus additional north-south lines to tie the area together. We must keep an open mind about the prior knowledge that the anticlines are oriented north-south and we may alter line orientations as interpretation unfolds. We may wish to shoot additional lines more-or-less perpendicular to the strike of faults that may affect structures.

8.15b What multiplicity is required to give ${\mbox{S/N}}=3.0\$$

Solution

From problem 8.14a we know that S/N varies as $n^{1/2}$ for random noise; so to increase S/N from 0.5 to 3.0 requires a multiplicity of $(3.0/0.5)^{2}=36$.

8.15c What spread geometry should be used, that is, what are the required near- and far-offsets, group spacing to avoid aliasing for 15 to 40 Hz, and minimum number of channels?

Solution

Because the objectives are Eocene to Cretaceous, we will want maximum offsets of 5000 m and, since we expect to use the Recent reflection to make static corrections, we will also need short offset data. Hence, end-on spreads should extend from near the source to 5000 m. We should use 96 channels and 50-m group intervals although 48 channels and 100-m group intervals might suffice. We probably should use geophone spacing within a group no larger than 5 m and have the same group length and group interval.

8.15d How long will the survey require, assuming production of 270 km/month can be achieved for 24-fold multiplicity?

Solution

The reconnaissance of seven east-west and three to four north-south lines, each being 75 km long, amounts to about 800 km and therefore will take about 3 months.

8.15e Answer part (d) assuming 210 km/month production for 48-fold multiplicity.

Solution

Increasing the multiplicity from 24 to 48 will increase the survey cost but will be worthwhile considering the signal/noise improvement expected. The reduction in production rate will increase the time by about 3 weeks. Shortening the group interval might achieve the same improvement in S/N.

8.16 Effect of signal/noise ratio on event picking

8.16a Take a wavelet that has amplitudes at successive 4-ms intervals of $0,\ldots ,0,8,7,-8,-6,0,4,2,0,\ldots ,0$ (with 10 zeros at each end) and add random noise (use Table 6.22a) in the range from $+10$ to $-10$, giving ${\mbox{S/N}}\approx 1.0$. Do this five times for different noise values and plot the results to illustrate how coherence helps in detecting the wavelet.

Background

Variable-area recording is usually achieved by blacking in the peaks of ordinary wiggly-trace recording.

In sign-bit recording the only data recorded are the algebraic signs of the signal at each sampling instant. Band-pass signals have roughly equal probability of being either positive or negative. The superposition of signal on random noise biases this probability in favor of the signal and this bias increases with signal strength. As more traces are added, the stacking tends to reduce the random noise so that the sum is more likely to be that of the signal. By stacking large numbers of traces, a record is obtained that is comparable to a normal record.

Table 8.16a. Sequences of signal plus random noise; ${\mbox{S/N}}\approx 1.0.$

S	${\rm {N}}_{1}$	$\sum _{1}$	${\rm {N}}_{2}$	$\sum _{2}$	${\rm {N}}_{3}$	$\sum _{3}$	${\rm {N}}_{4}$	$\sum _{4}$	${\rm {N}}_{5}$	$\sum _{5}$	C	D
0	2	2	0	0	3	3	5	5	3	3	13	5
0	0	0	1	1	1	1	3	3	8	8	13	3
0	8	8	–3	–3	5	5	–10	–10	–3	–3	–3	–1
0	9	9	8	8	–2	–2	4	4	–2	–2	17	1
0	–7	–7	0	0	–2	–2	0	0	–4	–4	–13	–3
0	–1	–1	–7	–7	–5	–5	–8	–8	–8	–8	–29	–5
0	3	3	9	9	–9	–9	4	4	5	5	12	3
0	0	0	–5	–5	–2	–2	1	1	–10	–10	–16	–1
0	10	10	10	10	–8	–8	–1	–1	6	6	17	1
0	–7	–7	–8	–8	2	2	–7	–7	0	0	–20	–3
8	–9	–1	–1	7	1	9	–5	3	–6	2	20	3
7	5	12	–2	5	6	13	0	7	–1	6	43	5
–8	2	–6	–7	–15	8	0	3	–5	–2	–10	–36	–3
–6	–1	–7	–2	–8	–5	–11	9	3	–2	–8	–31	–3
0	–7	–7	–1	–1	–6	–6	1	1	–9	–9	–22	–3
4	–10	–6	–3	1	3	7	7	11	10	14	27	3
2	–9	–7	4	6	–8	–6	3	5	2	4	2	1
0	–2	–2	8	8	6	6	7	7	–10	–10	9	1
0	–2	–2	–10	–10	5	5	0	0	7	7	0	–1
0	1	1	6	6	5	5	–2	–2	3	3	13	3
0	–1	–1	–6	–6	–3	–3	–1	–1	–7	–	–18	–5
0	–5	–5	0	0	1	1	–7	–7	0	0	–11	–1
0	–4	–4	–6	–6	–8	–8	–5	–5	–4	–4	–27	–5
0	–3	–3	0	0	6	6	6	6	9	9	18	3
0	3	3	5	5	2	2	–8	–8	8	8	10	3
0	9	9	9	9	8	8	1	1	3	3	30	5
0	4	4	–7	–7	–4	–4	4	4	2	2	–1	–1

S = signal; ${\rm {N}}_{i}=i^{\rm {th}}$ set of random noises, $i=1,2,3,4,5$; $\Sigma _{i}=\Sigma ({\rm {S}}+{\rm {N}}_{i})$,

${\rm {C}}=\Sigma _{i}$ for part (c), D = sign-bit sum for part (d)

Figure 8.16a.  Vartiable-area display of signal plus random noise; ${\mbox{S/N}}\approx 1.0.$

Solution

Although we have few data, we calculated S/N for each of the five sets (rather than use an average value). Using absolute values (since S/N does not depend upon the polarity) we get for the averages ${\rm {S}}=5$, ${\rm {S/N}}_{1}=0.8$, ${\rm {S/N}}_{2}=1.7$, ${\rm {S/N}}_{3}=1.0$, ${\rm {S/N}}_{4}=1.2$, ${\rm {S/N}}_{5}=1.0$. The average of these five, 1.1, is close to 1.

Table 8.16a lists five random noise sequences and ${\hbox{S}}+{\hbox{N}}$ for each noise sequence. The sums are then plotted in Figure 8.16a with positive values shaded in. The first peak and trough of the signal are coherent enough on the five traces that they probably would be picked, but the second signal peak is lost in the background noise.

Figure 8.16b.  Variable-area display of signal plus random noise; ${\rm {S/N}}=0.5$

Table 8.16b. Sequences of signal plus random noise; ${\rm {S/N}}\approx 0.5.$

S	${\rm {N}}_{1}$	$\sum _{1}$	${\rm {N}}_{2}$	$\sum _{2}$	${\rm {N}}_{3}$	$\sum _{3}$	${\rm {N}}_{4}$	$\sum _{4}$	${\rm {N}}_{5}$	$\sum _{5}$	C	D
0	2	2	0	0	3	3	5	5	3	3	13	5
0	0	0	1	1	1	1	3	3	8	8	13	3
0	8	8	–3	–3	5	5	–10	–10	–3	–3	–3	–1
0	9	9	8	8	–2	–2	4	4	–2	–2	17	1
0	–7	–7	0	0	–2	–2	0	0	–4	–4	–13	–3
0	–1	–1	–7	–7	–5	–5	–8	–8	–8	–8	–29	–5
0	3	3	9	9	–9	–9	4	4	5	5	12	3
0	0	0	–5	–5	–2	–2	1	1	–10	–10	–16	–1
0	10	10	10	10	–8	–8	–1	–1	6	6	17	1
0	–7	–7	–8	–8	2	2	–7	–7	0	0	–20	–3
8	–9	–1	–1	7	1	9	–5	3	–6	2	20	3
7	5	12	–2	5	6	13	0	7	–1	6	43	5
–8	2	–6	–7	–15	8	0	3	–5	–2	–10	–36	–3
–6	–1	–7	–2	–8	–5	–11	9	3	–2	–8	–31	–3
0	–7	–7	–1	–1	–6	–6	1	1	–9	–9	–22	–3
4	–10	–6	–3	1	3	7	7	11	10	14	27	3
2	–9	–7	4	6	–8	–6	3	5	2	4	2	1
0	–2	–2	8	8	6	6	7	7	–10	–10	9	1
0	–2	–2	–10	–10	5	5	0	0	7	7	0	–1
0	1	1	6	6	5	5	–2	–2	3	3	13	3
0	–1	–1	–6	–6	–3	–3	–1	–1	–7	–	–18	–5
0	–5	–5	0	0	1	1	–7	–7	0	0	–11	–1
0	–4	–4	–6	–6	–8	–8	–5	–5	–4	–4	–27	–5
0	–3	–3	0	0	6	6	6	6	9	9	18	3
0	3	3	5	5	2	2	–8	–8	8	8	10	3
0	9	9	9	9	8	8	1	1	3	3	30	5
0	4	4	–7	–7	–4	–4	4	4	2	2	–1	–1

S = signal; ${\rm {N}}_{i}=i^{\rm {th}}$ set of random noises; $i=1,2,3,4,5$; $\Sigma _{i}=\Sigma ({\rm {S}}+{\rm {N}}_{i})$; ${\rm {C}}=\Sigma _{i}$ for part (c);

D = sign-bit sum for part (d).

Figure 8.16c.  Display of sums of five samples of signal plus random noise. Top, signal S; center, $S/N\approx 0.8$; bottom, ${\rm {S/N}}\approx 0.4$.

8.16b Repeat for noise ranging from $\pm 20$, giving ${\rm {S/N}}\approx 0.5.$

Solution

The data are shown in Table 8.16b and the results graphed in Figure 8.16b. The signal appears to be completely lost at this S/N. The value ${\rm {S/N}}\approx 1$ is roughly the limit of our ability to extract a signal visually from a noisy trace.

8.16c Sum the five waveforms in parts (a) and (b) to show how stacking enhances the signal.

Solution

The columns headed C in Tables 8.16a and 8.16b are the sums of the five values of $\Sigma _{i}$; these are plotted in Figure 8.16c. The signal is clearly evident in the curve for part (a) where ${\rm {S/N}}\approx 1$, but not in the curve for part (b) where ${\rm {S/N}}\approx 0.5$. However, even with ${\rm {S/N}}\approx 1$ false signals can be seen (such as a negative wavelet at the right-hand end).

Figure 8.16d.  Display of sign-bit sums of five samples of signal plus random noise. Top, signal; center, ${\hbox{S/N}}\approx 0.8$; bottom, ${\hbox{S/N}}\approx 0.4$.

8.16d Replace the elements in the wavelets in parts (a) and (b) with $+1$ or $-1$ as the value is positive or negative (sign-bit expression) and repeat part (c).

Solution

We have replaced each value of the sums of S + N in Tables 8.16a and 8.16b with +1 or –1 as the value was positive or negative (zero values are alternately made + and –) and summed the five values in the columns headed D. The results, plotted in Figure 8.16d, are similar to Figure 8.16c, because the same sequences are used. The signal is becoming evident although noise can lead to false picks. More than five sequences need to be summed to make the signal stand out clearly.

8.17 Interpreting uphole surveys

Uphole surveys in five different (unrelated) areas give the uphole-time versus depth information in Table 8.17a. Explain the possible velocity layering for each case. How reliably are velocities and depths of weathering defined?

Background

An uphole geophone is a geophone placed at or near a borehole for the purpose of recording the uphole time, that is, the time for a wave generated by a subsurface source to reach the surface.

Solution

Because the uphole geophone is very close to the borehole, the raypath is essentially vertical.

Figures 8.17a, b, c, d, e are plots of the uphole times in Table 8.17a. The plotted data were approximated by series of straight lines and the velocities determined from the slopes of these lines. The thickness of the LVL is determined by the abrupt change in velocity at the base of the layer.

Table 8.17a. Uphole times versus depth in five areas.

Depth (m)	Area $A\left(s\right)$	Area $B\left(s\right)$	Area $C\left(s\right)$	Area $D\left(s\right)$	Area $E\left(s\right)$
5	0.012	0.011		0.012	0.008
8			0.020	0.010
10	0.025	0.023	0.024	0.018
12		0.024	0.027		0.020
15	0.030		0.031	0.022
18		0.028	0.034	0.030	0.030
21	0.034		0.036	0.033	0.031
25	0.036	0.032		0.035	0.032
30	0.039	0.035	0.039
35		0.037		0.039	0.036
40	0.046		0.044	0.044	0.042
50	0.051	0.044	0.048	0.047	0.043

Figure 8.17a-e.  Uphole time-depth plots.

The results for each area are given below.

Area A (Figure 8.17a):

$V_{1}=400\ {\rm {m/s}}$, $V_{2}=1690\ {\rm {m/s}}$, $D_{W}=11\ {\rm {m}}$. All three values are accurately determined.

Area B (Figure 8.17b):

This is a three-layer situation; $V_{1}=440{\rm {m/s}}$, $V_{2}=1660{\rm {m/s}}$, $V_{3}=2200{\rm {m/s}}$, $D_{W}=10{\rm {m}}$. To get the thickness of the second layer, we note that the bases of the LVL and second layer correspond to uphole times of 23 and 35 ms. Since the path is vertical, we can determine the depth to the base of the second layer, $D_{2}$, to be $D_{2}=10+\left(0.035-0.024\right)\times 1660=30\ {\hbox{m}}$. Values are reasonably well determined.

Using the simpler two-layer solution, the depth of the 440 m/s layer would be picked as 12 m rather than 10 m, a 20% error, but the statics correction would have an error of only 2 ms.

Area C (Figure 8.17c)

$V_{1}=410\ \mathrm {m/s} ,V_{2}=750\ \mathrm {m/s} ,V_{3}=2500\ {\hbox{m/s}},D_{1}=10\ {\hbox{m}},D_{2}=10+(0.035-0.024)\times 750=18\ {\hbox{m}}$. Values are moderately well determined.

As in the case of Area B, this area could be interpreted as a two-layer situation but with larger errors.

Area D (Figure 8.17d):

The time-depth curve can be interpreted either in terms of three or four layers. The three-layer solution assumes that the measurement at 15 m is in error and the four-layer solution honors the data more closely.

The three-layer solution is given by the dashed lines. Measured values are: $V_{1}=420\ {\hbox{m/s}},V_{2}=740\ {\hbox{m/s}},V_{3}=2060\ {\hbox{m/s}},D_{1}=6\ {\hbox{m}},D_{2}=6+\left(0.033-0.015\right)\times 740=19\ {\hbox{m}}$.

The four-layer case is shown by the solid lines. Measured values are: $V_{1}=420\ {\hbox{m/s}}$, $V_{2}=1070\ {\hbox{m/s}},V_{3}=390\ {\hbox{m/s}},V_{4}=1890\ {\hbox{m/s}},D_{1}=7\ {\hbox{m}},D_{2}=7+\left(0.022-0.015\right)\times 1070=14\ {\hbox{m}}$, $D_{3}=14+\left(0.030-0.022\right)\times 390=17\,{\rm {m}}$. This interpretation postulates a high-velocity layer $\left(V_{2}\right)$ within the LVL.

In both of these interpretations velocities and depths are questionable except for the velocity of the deepest layer.

Area E (Figure 8.17e):

Two-layer and three-layer solutions are possible. Assuming two layers (dashed-line), measured values are: $V_{1}=610\,{\rm {m/s}}$, $V_{2}=2360\,{\rm {m/s}}$, $D_{W}=17\,{\rm {m}}$.

The three-layer solution (solid line) gives: $V_{1}=610\ {\hbox{m/s}}$, $V_{2}=2910\ {\hbox{ms}}$, $V_{3}=2140\ {\hbox{m/s}},D_{W}=18m,D_{2}=18+\left(0.035-0.031\right)\times 2910=30\ {\hbox{m}}$.

8.18 Weathering and elevation (near-surface) corrections

8.18a Show that when the source is below the LVL, weathering and elevation corrections for a geophone at the source are given by

${\begin{aligned}\Delta t_{0}=2\left(E_{S}-D_{S}-E_{D}\right)/V_{H}+t_{uh},\end{aligned}}$

Figure 8.18a.  Calculation of weathering corrections.

where $E_{S}$ and $E_{D}$ are the elevations of the source point and datum, $D_{S}$ is the depth of the source, and $t_{uh}$ is the uphole time.

Background

Corrections are necessary to eliminate the effect of changes in the elevation of the surface and in the thickness of and velocity in the LVL. The corrections in effect reduce the traveltimes to those that would be observed if the source and geophones were located on a reference datum, usually a horizontal plane below the base of the low-velocity layer. These corrections are called static corrections because they are the same for all reflections regardless of their arrival times.

Solution

In Figure 8.18a, the correction to the traveltime for a wave going from the source at $A$ down to the datum is

${\begin{aligned}\Delta t_{S}=\left(E_{S}-D_{S}-E_{D}\right)/V_{H}.\end{aligned}}$

(8.18a)

The correction for travel from the datum up to a geophone at the source point is

${\begin{aligned}\Delta t_{g}=\Delta t_{S}+t_{uh},\end{aligned}}$

(8.18b)

so the total correction for the traveltime for a geophone at the source point is

${\begin{aligned}\Delta t_{0}=2\left(E_{S}-D_{S}-E_{D}\right)/V_{H}+t_{uh}.\end{aligned}}$

(8.18c)

8.18b If the split spread $ACB$ in Figure 8.18a is used to find the dip, what correction must be applied to the dip moveout?

Solution

The dip moveout is obtained by subtracting traveltimes at sources $A$ and $B$ in Figure 8.18b. If traveltimes have not been corrected for weathering and elevation, the dip moveout must be corrected; this is the differential weathering correction $\Delta t_{d}$. Using equation (8.18b) we have

${\begin{aligned}\Delta t_{d}=(\Delta t_{g})_{B}-(\Delta t_{g})_{A}=(\Delta t_{S}+t_{uh})_{B}-(\Delta t_{S}+t_{uh})_{A}.\end{aligned}}$

(8.18d)

Figure 8.18b.  Differential weathering correction.

8.18c Derive an expression to correct the traveltime for a geophone at $C$ in Figure 8.18b $A$ and $B$ being source points.

Solution

We use the first-break traveltimes $t_{AC}$ and $t_{BC}$ that correspond to the paths $A'C''C$ and $B'C'C$. The velocity contrast at the base of the LVL is usually large enough that the paths $C'C$ and $C''C$ are so close to vertical that the distance $C'C''$ is very small. Hence the sum $\left(t_{AC}+t_{BC}\right)$ is given by

${\begin{aligned}\left(t_{AC}+t_{BC}\right)\approx \left(A'B'\right)/V_{H}+2t_{w},\\{\mbox{so}}\qquad \qquad t_{w}\approx {\frac {1}{2}}[\left(t_{AC}+t_{BC}\right)-(AB/V_{H})],\end{aligned}}$

(8.18e)

where $t_{w}$ is the traveltime through the LVL at $C$, and $D_{W}=V_{W}t_{w}$. Therefore the correction that effectively places the geophone at $C$ on the datum is

${\begin{aligned}\Delta t_{C}=t_{w}+\left(E_{C}-D_{W}-E_{D}\right)/V_{H}.\end{aligned}}$

(8.18f)

We must add to this the correction that locates the source on the datum, namely $\Delta t_{S}$, given by equation (8.18a). Thus the total correction for a traveltime recorded at $C$ is

${\begin{aligned}\Delta t_{C}=t_{w}+\left(E_{C}-D_{W}-E_{D}\right)/V_{H}+\Delta t_{S}.\end{aligned}}$

(8.18g)

8.18d The weathering and elevation corrections given by equations (8.18a) to (8.18g) assume that the source is below the base of the LVL. What changes are required if the source is within the LVL?

Solution

When the source is within the LVL, the wave traveling down to the datum is in the LVL for the distance $(D_{W}-D_{S}$, hence equation (8.18a) is changed to

${\begin{aligned}\Delta t_{S}&=\left(D_{W}-D_{S}\right)/V_{W}+\left(E_{S}-D_{W}-E_{D}\right)/V_{H}\\&=\left(E_{S}-D_{W}-E_{D}\right)/V_{H}+\Gamma _{S},\end{aligned}}$

(8.18h)

where $\Gamma _{S}=\left(D_{W}-D_{S}\right)/V_{W}$. Equation (8.18b) is unchanged provided we use the value of $\Delta t_{S}$ in equation (8.18h). Equation (8.18c) becomes

${\begin{aligned}\Delta t_{0}&=\left(D_{W}-D_{S}\right)/V_{W}+2\left(E_{S}-D_{W}-E_{D}\right)/V_{H}+\left[\left(D_{W}-D_{S}\right)/V_{W}+t_{uh}\right]\\&=2\Gamma _{S}+2\left(E_{S}-D_{W}-E_{D}\right)/V_{H}]+t_{uh}.\end{aligned}}$

(8.18i)

Equation (8.18d) is unchanged provided we use equation (8.18h) for $\Delta t_{S}$. Equation (8.18e) becomes

${\begin{aligned}\left(t_{AG}+t_{BG}\right)=AB/V_{H}+2t_{w}+\left(\Gamma _{A}+\Gamma _{B}\right),\end{aligned}}$

where $\Gamma _{A}=\Gamma _{SA}$, $\Gamma _{B}=\Gamma _{SB}$. Thus, equation (8.18e) becomes

${\begin{aligned}t_{w}={\frac {1}{2}}[\left(t_{AG}+t_{BG}\right)-AB/V_{H}-(\Gamma _{A}+\Gamma _{B})].\end{aligned}}$

(8.18j)

Equations (8.18f,g) are unchanged except that we must use the values of $t_{w}$ and $\Delta t_{S}$ from equations (8.18j) and (8.18h).

8.19 Determining static corrections from first breaks

8.19a Figure 8.19a shows the first arrivals (first breaks) at geophone stations 100 m apart from sources 25 m deep at each end of the spread. The geophone group at each end is not recorded because of hole noise. The uphole time is on the third trace from the right. Elevations for each group are given at the top. Weathering velocity is 500 m/s. The valley midway between the sources produces a change in the firstbreak slope, as if two refractors were involved, which is not the case. How can we be sure?

Figure 8.19a.  First breaks on reflection record.

Table 8.19a. Values observed from Figure 8.19a.

$x$	$E$	$\Delta _{t}$	$t_{A}$	$t_{B}$	$t_{A}^{*}$	$t_{B}^{*}$
0	1125	—	—	433	—	433
100	1130	—10	66	398	56	388
200	1127	—4	106	357	102	353
300	1128	—6	151	317	145	311
400	1125	0	184	270	184	270
500	1120	+10	218	220	228	230
600	1120	+10	257	172	267	182
700	1125	0	313	143	313	143
800	1133	—16	367	118	351	102
900	1138	—26	418	83	392	57
1000	1140	—30	466	—	436	—

Background

The source instant $\left(t=0\right)$ or time break is the sharp deflection on the third trace from the right. Each time division is 10 ms.

Hole noise is caused by reverberations within the shothole and by material ejected from the shothole and falling back to the earth when an explosive charge is detonated.

Solution

We correct first-break readings for elevation by taking as the reference datum the elevation of the left-hand source point $A$ and adding or subtracting 2 ms (=1 m/500 m/s) for each meter below or above the datum. The corrected times $t_{A}^{*}$ and $t_{B}^{*}$ are given in Table 8.19a where $t_{A}$ and $t_{B}$ are first-break times for sources at $A$ and $B$ at offset $x$ from source $A$, All times are in milliseconds, distances in meters.

The plots of the corrected times in Figure 8.19b give straight lines whose slopes have an average value of 2390 m/s for $V_{H}$ whereas the plots of the uncorrected times suggest a 3-layer situation with a low-velocity layer in between two higher-velocity layers. The corrected times fit a straight line in each case which is strong evidence that there is only one high-velocity layer.

8.19b Determine the weathering thickness at the two sourcepoints from the uphole times.

Figure 8.19b.  Plots of first breaks in Figure 8.19a.

Solution

To find $D_{W}$ at $A$ and $B$ from $t_{uh}$, we have

${\begin{aligned}t_{uh}=D_{W}/V_{W}+\left(D_{S}-D_{W}\right)/V_{H}\\{\mbox{or}}\qquad \qquad D_{W}=\left(t_{uh}-D_{S}/V_{H}\right)/\left(1/V_{W}-1/V_{H}\right).\end{aligned}}$

The uphole times are 0.025 s and 0.049 s at $A$ and $B$, $D_{S}=25$ m at both $A$ and $B$, $V_{W}=500$ m/s $V_{H}=2390$ m/s; thus $D_{W}=9.2$ m at $A$ and 24 m at $B$.

8.19c What correction $\Delta t_{0}$ should be applied to reflection times at the two source-points for a datum of 1125 m?

Solution

Applying equation (8.18c), we have

${\begin{aligned}{\mbox{at A,}}\qquad \Delta t_{0}=2(1125-25-1125/2390+0,025=0.004\quad {\rm {s}},\\{\mbox{and at B,}}\qquad \Delta t_{0}=2\left(1140-25-1125\right)/2390+0.049=0.041\quad {\rm {s}}.\end{aligned}}$

Table 8.19b. Calculating $D_{W}$ and $\Delta t_{g}$ for each geophone.

$x_{A}$	$t_{AC}+tBC$	$t_{w}$	$D_{W}$	$E_{g}$	$(\Delta t_{g})_{A}$	$(\Delta t_{g})_{B}$
0	—	—	10	1125	—	32
100	464	23	12	1130	31	37
200	463	22	11	1127	29	35
300	468	25	12	1128	32	38
400	454	18	9	1125	25	31
500	438	10	5	1120	16	22
600	429	6	3	1120	13	19
700	456	19	10	1125	25	31
800	485	34	17	1133	41	47
900	501	42	21	1138	49	55
1000	—	—	24	1140	55	—

Times are in milliseconds and distances in meters.

8.19d Calculate the weathering thickness and the time correction for each geophone station.

Solution

In Table 8.19b the second column is the sum of the uncorrected times at geophone $C$ from Table 8.19a. To get $t_{w}$ we require the quantity $\left(AB/V_{H}\right)=\left(1000/2390\right)=0.418\ {\hbox{s}}$; we now get $t_{w}=\left[\left(t_{AC}+t_{BC}\right)-0.418\right]/2$. Next, $D_{W}=500\ t_{w}$ (except for the source points where $D_{W}$ is obtained from the uphole times [part (b)]. The weathering correction for a geophone group is given by equation (8.18g), that is,

${\begin{aligned}\Delta t_{g}=t_{w}+\left(E_{C}-D_{W}-E_{D}\right)/V_{H},\end{aligned}}$

where $D_{W}=V_{W}t_{w}$, $t_{w}$ being given by equation (8.18e), namely,

${\begin{aligned}t_{w}={\frac {1}{2}}[t_{AC}+t_{BC}-(AB/V_{H}].\end{aligned}}$

This correction is equivalent to placing the geophone group at C on the datum. To locate the sources on the datum also, we must add to $\Delta t_{g}$ the time from the source down to the datum, that is, $\left(E_{S}-D_{S}-E_{D}\right)/V_{H}$; this amounts to 0 and 4 ms at sources $A$ and$B$, respectively. The column headed $(\Delta t_{g})_{A}$ gives the corrections for the arrival times at the given offset for source $A$ while that headed $(\Delta t_{g})_{B}$ gives the corrections for source $B$.

8.19e Plot corrected reflection arrival times in an $X^{2}$-$T^{2}$ plot and determine the depth, dip, and average velocity to the reflector giving the reflection at 0.30 and 0.21 s in Figure 8.19a.

Table 8.19c. $X^{2}$-$T^{2}$ data for profile from A.

$x_{A}$	$x_{A}^{2}$	$t_{A}$	$(\Delta t_{g})_{A}$	$t_{AC}$	$t_{AC}^{2}$
100	$1\times 10^{4}$	0.303	31	0.272	0.0740
200	4	0.305	29	0.276	0.0762
300	9	0.317	32	0.285	0.0812
400	16	0.325	25	0.300	0.0900
500	25	0.334	16	0.318	0.1011
600	36	0.341	13	0.328	0.1076
700	49	0.355	25	0.330	0.1089

Distances are in meters, traveltimes in seconds, corrections in milliseconds. Underlined values are doubtful.

Table 8.19d. $X^{2}$-$T^{2}$ data for profile from B.

$x_{B}$	$x_{B}^{2}$	$t_{B}$	$(\Delta t_{g})_{B}$	$t_{BC}$	$t_{BC}^{2}$
100	$11\times 10^{4}$	0.211	55	0.156	0.0243
200	4	0.224	47	0.177	0.0313
300	9	0.237	31	0.206	0.0424
400	10	0.251	19	0.232	0.0538
500	25	0.272	22	0.258	0.0666

Solution

Tables 8.19c and 8.19d list the offsets and their squares, the uncorrected times and the correction for each (from Table 8.18a), the corrected times $t_{AC}$ and their squares. Table 8.19c is for source $A$ , Table 8.19d is for source $B$. The $X^{2}-T^{2}$ data are plotted in Figure 8.19c. Drawing the best-fit straight lines, we measure the intercepts on the $t^{2}$-axis and the slopes. Assuming that the dip is small so that the factor ${\rm {\;cos\;}}\xi$ in equation (4.3a) can be neglected, the reciprocals of the slopes give the velocities squared. The measured results are:

${\begin{aligned}V_{A}=2.36\,\mathrm {km/s} ,\;t_{i}=0.15\,\mathrm {s} ,\;h_{A}=180\,\mathrm {m} ;\\V_{B}=2.98\,\mathrm {km/s} ,\;t_{i}=0.27\,\mathrm {s} ,\;h_{B}=400\,\mathrm {m} ;\\\mathrm {dip} \ \xi ={\tan }^{-1}\left[\left(400-180\right)/1000\right]=12^{\circ }.\end{aligned}}$

Figure 8.19c.  $X^{2}$-$T^{2}$ graph.

8.20 Determining reflector location

8.20a The arrival time of a reflection at the source point is 1.200 s, near-surface corrections having been applied. Determine the reflector depth and horizontal location with respect to the source point, assuming zero dip and that the average velocity associated with a vertical traveltime is $V=2630\,\mathrm {m/s} .$

Background

After records have been picked, that is, after reflections have been identified and $t_{0}$ and $\Delta t_{d}$ measured, the next stage is to prepare a section displaying the reflection events in two-dimensions. Such a section can be prepared in several ways, one of which is by using a wavefront chart such as that shown in Figure 8.20a. A wavefront chart is a two-dimensional graph showing approximate wavefronts and raypaths for a given distribution of constant-velocity layers, Raypaths are found by starting with rays leaving the source at different angles and tracing them downward. Traveltimes to various points on the raypaths are calculated and contoured to show wavefronts. This assumes that waves started from, and returned to, the source, hence they must have been reflected by a bed perpendicular to the ray (parallel to a wavefront); thus the dip as well as the location of the reflector is determined. The reflection event denoted by the symbol -o- in Figure 8.20a corresponds to $t_{0}=2.350\,{\rm {s}},$ $\Delta t_{d}/\Delta x=$ 110 ms/km.

Figure 8.20a.  Wavefront chart.

Solution

We are given $t_{0}=1.200\,\mathrm {s} ,$ $V=2630\,{\rm {m/s}}$; hence $h=Vt_{0}/2=1580\,\mathrm {m}$ vertical depth, and the location is directly below the source.

8.20b Determine the reflector depth, dip, and horizontal location assuming that the dip moveout is 0.150 s/km and that the line is normal to strike.

Solution

As in (a) $h=1580\ {\hbox{m}}$, but now it is slant depth;

${\begin{aligned}\mathrm {dip\ moveout} &=\Delta t_{d}/\Delta x=0.150/1000=0.000150\,\mathrm {s/m} ,\\\xi &={\sin }^{-1}\left({\frac {1}{2}}\times 2630\times 0.000150\right)=11.4^{\circ };\\\mathrm {vertical\ depth} &=1580\,\cos 11.4^{\circ }=1550\,\mathrm {m} ;\\\mathrm {horizontal\ up-dip\ displacement} &=1580\sin \,11.4^{\circ }=310\,\mathrm {m} .\end{aligned}}$

8.20c Determine the depth, dip, and horizontal location assuming straight-line travel at the angle of approach and $V_{H}=1830\,\mathrm {m/s}$.

Solution

Because the path is a straight line, the velocity must be constant and we assume it is the starting velocity of 1830 m/s. Then,

${\begin{aligned}h&={\frac {1}{2}}\times 1830\times 1.200=1100\,\mathrm {m} =\mathrm {slantdepth} ;\\\xi &={\sin }^{-1}\left({\frac {1}{2}}\times 1830\times 0.150/1000\right)=7.9^{\circ };\\\mathrm {vertical\ depth} &=1100\,\cos \,7.9^{\circ }=1090\,\mathrm {m} ;\\\mathrm {horizontal\ displacement} &=1100\sin 7.9^{\circ }=150\,\mathrm {m} .\end{aligned}}$

8.20d Determine the reflector depth, dip, and horizontal location assuming straight-line travel at the local velocity above the reflector, 3840 m/s.

Solution

${\begin{aligned}h&=3840\times 1.200/2=2300\,\mathrm {m} =\mathrm {slantdepth} ;\\\xi &=\sin ^{-1}\left({\frac {1}{2}}\times 3840\times 0.150/1000\right)=16.7^{\circ };\\\mathrm {vertical\ depth} &=2300\cos \,16.7^{\circ }=2200\mathrm {m} ;\\\mathrm {horizon\ displacement} &=2300\sin 16.7^{\circ }=660\mathrm {m} .\end{aligned}}$

8.20e Assume that the migrated position is determined from the wavefront chart in Figure 8.20d.

Solution

Figure 8.20a gives 1200 m vertical depth, 170 m horizontal displacement, and $\xi =12^{\circ }$ dip.

Results are summarized in Table 8.20e, z being the vertical depth and $x$ the horizontal displacement.

Table 8.20e. Summary of results.

	$V\left(m/s\right)$	dip	$z\left(m\right)$	$x\left(m\right)$
a) Vertical path	2630	$0^{\circ }$	1580	0
b) Dip moveout 150 ms/km	2630	$11.4^{\circ }$	1550	310
c) At approach angle	1830	$7.9^{\circ }$	1090	150
d) At local velocity	3840	$16.7^{\circ }$	2200	660
e) Curved raypath		$12^{\circ }$	1200	170

8.21 Blondeau weathering corrections

The Blondeau method of making weathering corrections (Musgrave and Bratton, 1967, 231−246) is useful in areas where, because of appreciable compaction within the low-velocity layer, the velocity is given approximately by the equation

${\begin{aligned}V=az^{1/n};\end{aligned}}$

(8.21a)

$a$ and $n$ being constants, $n>+1.$

The Blondeau method starts with a curve of first breaks versus offset [Figure 8.21a(ii)] plotted on log-log graph paper; the curve is approximately a straight line with slope $B=\left(1-1/n\right)$ (see Musgrave and Bratton, 1967, 244).

To remove the effect of a surface layer of thickness $z_{m}$, we find $F$, a tabulated function of $B$. Then $x=Fz_{m}$. Next we use the $x-t$ plot to find the corresponding $t$. Finally, $t_{v}$, the vertical traveltime to the depth $z_{m}$, is given by $t_{v}=t/F$.

Verify the Blondeau procedure by deriving these relations:

1. $z=z_{m}\sin ^{n}i$, where $i$ is the angle of incidence measured with respect to the vertical at depth $z$;
2. $x=Fz_{m}$, where $F=2n\mathop {\int } \nolimits _{0}^{\pi /2}\sin ^{n}i\ {\hbox{d}}i=$ function of $n$, hence also of $B$;
3. $t=\left(G/a\right)(x/F)^{B}$, where $G=2n\mathop {\int } \limits _{0}^{\pi /2}{\hbox{sin}}^{n-2}i\ {\hbox{d}}i$; [Note that $G$ can be obtained from the table for $F$ by writing $m=n-2$, finding for $F$, and multiplying the value by $\left(m+2\right)/m=n/(n-2$
4. ${\hbox{d}}x/{\hbox{d}}t$, the horizontal component of the apparent velocity at any point $\left(x_{j},\;z_{j}\right)$ of the trajectory, is $V_{m}=x/Bt$;
5. $t_{v}=\mathop {\int } \nolimits _{0}^{t_{m}}{\hbox{d}}z/V=t/F$.

Solution

i) Note that the quantities $x$ and $t$ are the offset and traveltime for the point of emergence. For intermediate points we write $\left(x_{j},\;t_{j},\;z_{j},\;V_{j},\;i_{j}\right)$ except for the deepest point, where we have $x_{m}=x/2,$ $t_{m}=t/2,$ $z_{j}=z_{m}$, $V_{j}=V_{m}$, $i_{m}=\pi /2$.

Solving equation (8.21a) for $z_{j}$ gives

${\begin{aligned}z_{j}=(V_{j}/a)^{n}.\end{aligned}}$

(8.21b)

From Snell’s law we have

${\begin{aligned}\left(\sin t_{j}/V_{j}\right)=1/V_{m}.\end{aligned}}$

(8.21c)

Thus, using equations (8.21b,c) we get

${\begin{aligned}\left(z_{j}/z_{m}\right)=(V_{j}/V_{m})^{n}={\sin }^{n}i_{j}.\end{aligned}}$

(8.21d)

ii) From equations (4.17d), we get

${\begin{aligned}x_{j}-\int _{0}^{z_{j}}\tan i\ \mathrm {d} z=\int _{0}^{z_{i}}(\tan i)[nz_{m}\sin ^{(n-1)}i\cos i]\mathrm {d} i=nz_{m}\int _{0}^{i_{j}}\sin ^{n}i\ \mathrm {d} i,\end{aligned}}$

where we have differentiated equation (8.21d) to replace $z$ with $i$. If we integrate from 0 to $\pi /2$ and multiply by 2, the result is

${\begin{aligned}x=2x_{m}=2nz_{m}\int \limits _{0}^{\pi /2}{\sin }^{n}i\ \mathrm {d} i=Fz_{m},\end{aligned}}$

${\begin{aligned}{\mbox{where}}\qquad \qquad F=2n\int _{0}^{\pi /2}\sin ^{n}i\ \mathrm {d} i.\end{aligned}}$

(8.21e)

iii) From Figure 4.17a we have $\Delta t=\Delta z/V\cos i$, so

${\begin{aligned}t_{j}=\int _{0}^{z_{j}}\mathrm {d} z/V\cos i=\left(nz_{m}\right)\int _{0}^{i_{j}}{\frac {\sin ^{\left(n-1\right)}i}{V}}\mathrm {d} i,\end{aligned}}$

where equation (8.21d) was used to replace $z$ with $i$. Using equation (8.21c) to eliminate $V$, we get

${\begin{aligned}t_{j}=(nz_{m}/V_{m})\int _{0}^{i_{j}}\,\sin ^{(n-2)}i\ \mathrm {d} i.\end{aligned}}$

Integration from 0 to $\pi /2$ gives $t_{m}$, so

${\begin{aligned}t=2t_{m}=\left(2nz_{m}/V_{m}\right)\int _{0}^{\pi /2}\sin ^{\left(n-2\right)}i\ \mathrm {d} i=\left(z_{m}/V_{m}\right)G.\end{aligned}}$

Since $V_{m}=az_{m}^{1/n}$ from equations (8.21a) and $z_{m}=x/F$ from equation (8.21e), this result can be written

${\begin{aligned}t=(G/a)(x/F)^{B}.\end{aligned}}$

(8.21f)

iv) Solving equation (8.21f) for $x$ gives

${\begin{aligned}x=F(at/G)^{1/B};\end{aligned}}$

${\begin{aligned}{\mbox{then}}\qquad \qquad {\frac {\mathrm {d} x}{\mathrm {d} t}}=\left(F/B\right)(a/G)^{1/B}t^{\left(1/B-1\right)}.\end{aligned}}$

(8.2lg)

Equation (8.21f) can be used to eliminate $\left(a/G\right)$, so equation (8.21g) becomes

${\begin{aligned}{\frac {\mathrm {d} x}{\mathrm {d} t}}=\left(F/B\right)t^{-1/B}\left(x/F\right)t^{\left(1/B-1\right)}=x/Bt.\end{aligned}}$

(8.21h)

The angle between the wavefront and a horizontal line equals the angle between a ray and the vertical, both angles being the angle of incidence (see Figure 4.2c), so

${\begin{aligned}\sin i_{j}=\left(V_{j}\Delta t_{j}/\Delta x_{j}\right)\end{aligned}}$

[compare with equation (4.2d)]. Therefore,

${\begin{aligned}\left(\Delta t_{j}/\Delta x_{j}\right)=\sin i_{j}/V_{j}=1/V_{m}\end{aligned}}$

using Snell’s law. Thus the apparent velocity $\left(\Delta x_{j}/\Delta t_{j}\right)=V_{m}$; since this is a constant, it holds at every point of the trajectory, including the point of emergence. Therefore, from equation (8.21h),

${\begin{aligned}V_{m}=\mathrm {d} x/\mathrm {d} t=x/Bt.\end{aligned}}$

(8.21i)

v) We have $t_{v}=$ time to travel vertically from the surface to the depth $z_{m}$, Thus,

${\begin{aligned}t_{v}=\int _{0}^{z_{m}}dz/V=\int _{0}^{z_{m}}\mathrm {d} z/az^{1/n}={\frac {z_{m}^{\left(1-1/n\right)}}{a\left(1-1/n\right)}}={\frac {z_{m}}{aBz_{m}^{1/n}}}={\frac {z_{m}}{BV_{m}}},\end{aligned}}$

where we have used equation (8.21a) in the first and last steps. From equations (8.21e) and (8.21i) we have $z_{m}=x/F$ and $V_{m}=x/Bt$, so

${\begin{aligned}t_{v}=t/F.\end{aligned}}$

(8.21j)