User:Ageary/Chapter 12

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

12.1 Spatial sampling restrictions

12.1a Show that the maximum spatial sampling $\Delta _{x}$ can be written

${\begin{aligned}\Delta _{x}=V/(2f_{\max }\sin \alpha _{\max }),\end{aligned}}$

(12.1a)

where $f_{\max }$ is the maximum frequency of interest, and $\alpha _{\max }$ is the maximum angle of approach.

Background

A wave is a function of time and space, e.g., $g(x,t)$ (see problem 2.5); therefore it can be sampled in time at a fixed location (problem 9.4) or in space at a fixed time (see Sheriff and Geldart, 1995, section 8.3.10). In both cases the sampling theorem (see problem 9.4) states that the wave can be sampled at fixed intervals $\Delta$ and be recovered exactly from the sampled data provided all frequencies are less than the Nyquist frequency $f_{N}$, that is, less than half the sampling frequency:

${\begin{aligned}f<f_{\rm {N}}=1/2\Delta .\end{aligned}}$

(12.1b)

For spatial sampling, $1/\lambda _{a}$ gives the number of waves per unit length and, hence, corresponds to frequency in the time domain. Therefore, for spatial sampling at intervals $\Delta _{x}$, the equivalent of equation (12.1b) is

${\begin{aligned}1/\lambda _{a}<1/\lambda _{\rm {N}}=1/2\Delta _{x},\end{aligned}}$

(12.1c)

so $\lambda _{a}>\lambda _{\rm {N}}=V_{a}/f_{\rm {N}}=2\Delta _{x},$

where $\lambda _{a}={\mbox{apparent wavelength}}=V_{a}/f=V/f\sin \alpha$, $f_{\rm {N}}=V/\lambda _{\rm {N}}=V/2\Delta _{x}$.

Solution

The maximum sampling interval $\Delta _{x}$ is associated with the minimum apparent wavelength. From equation (12.1c) we have

${\begin{aligned}(\Delta _{x})_{\max }={\frac {1}{2}}(\lambda _{a})_{\min }={\frac {1}{2}}(V_{a}/f)_{\min }=(V/2f_{\max }\sin \alpha _{\max }).\end{aligned}}$

(12.1d)

12.1b Show that the maximum group spacing $D_{\max }$ is

${\begin{aligned}D_{\max }<1000/[2f_{\max }(\Delta t/\Delta x)_{\max }],\end{aligned}}$

(12.1e)

where the dip moveout $(\Delta t/\Delta x)_{\max }$ is in milliseconds/unit distance.

Solution

Assuming $\alpha \approx \zeta$, we replace $\sin \alpha$ in equation (12.1d) with $\sin \zeta =(V/2)(\Delta t/\Delta x)$ [see equation (4.2b)]:

${\begin{aligned}D_{\max }=(\lambda _{a})_{\min }/2=(V_{a}/2f_{\max })=V/2f_{\max }\sin \zeta _{\max }\\=1000/f_{\max }(\Delta t/\Delta x)_{\max },\end{aligned}}$

where $\Delta t={\rm {time}}$ time difference in milliseconds between two geophones separated by a distance $2\Delta x$.

12.2 Bin size in marine work

12.2a What spacings are required to achieve a minimum bin size of $25\times 25\ {\rm {m}}$ for a marine operation towing two sources and three streamers? What mix of offsets and azimuths will result? How far apart should parallel ship traverses be to provide continuous coverage?

Background

Traces whose common midpoints fall within small unit areas (called bins) forming a rectangular grid are assumed to lie at the center of the bin and are stacked together for processing, especially for migrating. The signal that the source generates in the earth is the same as if the entire system were at rest, and the effect on the hydrophone groups at any time is almost the same as if they were at rest. Hence, the fact that the entire recording systems is in motion does not change the dimensions of a bin; the fact that the bin is moving slightly with record time makes so little difference that it is generally ignored.

Solution

For uniform sampling, lines of common midpoints need to be 25 m apart. We assume a symmetrical array with the sources between the streamers.

One possible arrangement, shown in Figure 12.2a, uses streamers 100 m apart and sources offset 25 m on opposite sides of the ship’s track.

In the inline direction hydrophone group spacing should be no more than 50 m, but this is not a limitation since streamer groups are usually spaced appreciably closer than this. Allowing 15 s between shots implies about 100 m source spacing when using two sources with the ship traveling at about 6 knots (problem 12.4a.) Shots from the two sources will be about 50 m apart in the inline direction so adjacent bins in the crossline direction involve different mixes of offsets. The CMP bins for this arrangement are shown in Figure 12.2a. The numbers within the bins show, respectively, the shortest inline and shortest crossline offsets of gathers (see problem 9.23) within the bins, assuming that the survey is regular and that the shortest inline offset is 100 m.

Figure 12.2a.  Configuration for two sources and three streamers. Numbers in bins indicate minimum inline and crossline offsets.

The azimuths for most of the traces in a gather will be nearly zero, that is, they are mostly inline. The largest azimuth would be for the bin where the crossline offset is 62.5 m and inline offset 100 m and where the angle with the inline direction is $\tan ^{-1}(62.5/100)=32^{\circ }$. This type of marine surveying is narrow-azimuth. The ship’s track for the next swath to extend coverage in the crossline direction should be moved 150 m assuming regular operations.

12.2b Answer the questions for two sources and four streamers.

Solution

One possible arrangement with four streamers is shown in Figure 12.2b. The streamers are separated by 100 m and the sources are 25 m on opposite sides of the ship’s track.

Some of the CMP bins for this arrangement are shown in Figure 12.2b. The numbers within the bins show, respectively, the shortest inline and crossline offsets of the gathers within the bins, assuming that the survey is regular and that the shortest inline offset is 100 m.

Successive ship tracks for continuous coverage should be separated by 200 m assuming regular operations.

Figure 12.2b.  Configuration for two sources and four streamers.

12.3 Effect of crosscurrents

Assuming that 96 hydrophone group centers in Figure 12.3a are 50 m apart and that the ship speed is 6 knots, calculate the crosscurrent at two locations. Locations are plotted for every 25th source. The active streamer length is 5000 m.

Solution

The effects of streamer feathering were discussed in problem 7.10.

Toward the left the angle that the streamer makes with the ship’s traverse is about $38^{\circ }$ decreasing to about $20^{\circ }$ to the right. The tangents of these angles equal the ratio of the crosscurrent velocity to the ship’s velocity. Hence if the ship’s speed is 6 knots, the crosscurrents are, respectively, 4.7 and 2.2 knots.

Relying on feathering to obtain offsets perpendicular to the ship’s track produces unequal distributions of offsets and azimuths. The areal coverage that feathering produces is capricious as currents change. The strip of subsurface coverage shown in Figure 12.3a varies from about 1800 m at the left side of the figure to about 700 m at the right.

Figure 12.3a.  Sideways drift of streamers because of crosscurrent.

12.4 Number of seismic sources

12.4a Whereas seismic ships sometimes tow several streamers, only rarely do they use more than two source arrays. Why?

Solution

The ship must move at a steady rate to keep the towed equipment under control. Also it must travel an appreciable distance between air-gun releases to avoid overlap of the recordings from successive shots. A speed of 6 knots means about 3 m/s. The data received by the streamer ordinarily lasts for about 15 s following a source release, so that the source firing interval must be at least this large if overlap of data is to be avoided. During 15 s, the ship travels about 45 m, about the maximum without excessive smearing. With two source arrays firing alternately, minimum source spacing is thus about 90 m. The use of more than two sources would result in unacceptable source spacing and gaps in the continuous coverage.

12.4b Why do some surveys use more detector locations than source locations whereas others do the opposite?

Solution

The cost of a seismic survey depends primarily on the time required. Depending on local conditions, laying out detectors may be either fairly rapid or slow, and similar considerations apply to movement of the source. Hence the optimum combination of numbers and disposition of detectors and sources in order to achieve a given density of coverage varies from situation to situation.

12.5 Circle shooting

What are the advantages and disadvantages of marine circle recording as opposed to acquiring data along parallel lines?

Background

Circle shooting is sometimes used to acquire 3-D data over salt domes and other features. One type involves the ship encircling the prospect in circles of ever-increasing radius [Figure 12.5a(i)], and another type involves circles of the same radius where the centers of the circles progress along a straight line (Figure 12.5a(ii)).

Solution

Where ship traverses are parallel, the acquisition assembly has to turn around to acquire the next traverse and the ship has to turn very slowly when towing a lot of equipment to keep the towed equipment under control. Typically, it takes two or more hours to turn around, that is, the cost of turning is about the same as that of acquiring about 15 km of data. With circle shooting, no time is lost turning around.

With the method of figure 12.5a(i), all azimuths from the center are acquired in the same way so that acquisition differences should be minimized and comparisons of data at different azimuths should be freer from acquisition bias. Multiplicity should be nearly constant because end-of-line tapers are almost eliminated. However, processing almost always requires a rectangular grid so a polar acquisition pattern has to be converted to rectangular bins for processing. This type of acquisition is usually not suitable for structures that are long in one direction. Difficulties with locating acquisition points are not a problem with today’s navigation systems, although they were earlier.

Figure 12.5a.  Two types of marine circle shooting.

With the method of Figure 12.5b(ii), multiplicity varies considerably, and much of the time that would be lost in turning the ship is lost in oversampling and undersampling different portions of the area. This technique is rarely used.

For structures with some radial symmetry, such as salt domes and their associated faulting, the technique shown in Figure 12.5a(i) works well.

12.6 Ocean-bottom cable surveys

12.6a A marine survey (Figure 12.6a) used four parallel ocean-bottom cables (OBC), each having 48 geophone/hydrophone groups spaced at 50-m intervals, the receiver lines being 400 m apart. A source boat towing an air-gun source traversed 19 lines spaced 250 m apart perpendicular to the receiver lines, each of the source lines being 2000 m long with air-gun pops every 50 m, thus covering nearly double the area occupied by the receiver lines. What minimum bin size can be used, and what multiplicity will be achieved?

Figure 12.6a.  Configuration used in ocean-bottom survey (only the left half of the area is shown, i.e., 10 of 19 source lines).

Background

Ocean-bottom recording permits recording among obstacles such as platforms where ships towing long streamers cannot operate. The use of geophone/hydrophone combinations permits excellent attenuation of surface multiples (see Sheriff and Geldart, 1995, section 7.5.5), usually the strongest multiples encountered. Ocean-bottom cables are also used to record three-component data so that S-waves (converted waves) can be studied, a technique called 4-C recording (three orthogonal geophones plus a hydrophone). Laying the cables is time consuming and not very precise, so that the receiver groups are not positioned as regularly as on land. The actual locations of the receiver groups can be determined from the arrivals of waves traveling directly from the various sources.

A reflection recorded by a detector is the sum of the waves reflected from all points of the reflecting surface, the major portion coming from a small circular area called the Fresnel zone (see problem 6.2). To achieve the correct amplitude for any point on the reflector, in migration we must sum all of the traces to which the point contributes, that is, over an area of the surface equal to the Fresnel-zone area.

Solution

For orthogonal source and receiver lines, the minimum bin size in the direction of source movement is half the source spacing and in the receiver direction, half the geophone group interval. With 50-m spacing for both receiver groups and source pops, the minimum bin size is $25\times 25\ {\rm {m}}$.

Since data from all source locations are recorded at all receiver locations, the multiplicity for the first row of midpoints increases from five ones starting at the corners of the covered area, then five twos, five threes, etc., until it reaches nineteens. This multiplicity will be repeated for the first eight inlines of bins, then the multiplicity will double as receivers on the second cable begin to contribute, and for an area in the center the multiplicity will triple. Then it will decrease symmetrically toward the other edge of the survey. A portion of a corner of the covered area is shown in Figure 12.6b. However, if the cables are then moved forward and the pattern repeated, the coverage to the left of the cables will compensate for the taper and produce a more uniform coverage. Irregularities in the cable layout will produce minor variation in the uniformity of coverage. The long dashes outline the area over which CMP subsurface coverage is obtained.

12.6b Assume that a deep objective horizon is a nearly horizontal erosional surface and that the trapping is stratigraphic, so that amplitudes must be mapped accurately. How large an area can be mapped with confidence?

Figure 12.6b.  Corner of Figure 12.6a showing bins and multiplicity in some of them.

Solution

Three factors affect the size of the areas that can be mapped with confidence.

1. A region around the periphery of a survey area involves a taper zone where multiplicity decreases, and often this zone is about half the length of the spreads employed in acquiring the data; thus the useful area of full multiplicity is smaller than the acquisition area by the amount of this taper zone.
2. Data usually have to be migrated to position features correctly and this introduces another peripheral zone whose dimensions depend on the depth and the angles that need to be incorporated in the migration, typically up to $30^{\circ }$ or $45^{\circ }$. Migration is needed even for horizontal reflectors to sharpen fault evidences.
3. A given point in the subsurface affects all detectors within an area equivalent to the Fresnel zone area so that, if amplitudes near the survey edge are to be compared to those in the central area, data must be available to be stacked. This area increases with the depth of the reflector. It involves a distance of the order of $(V/4)(t/f)$, where $V$ is the average velocity, $t$ the traveltime, and $f$ the frequency.

The result of these factors is that the acquisition area has to be larger than the area to be mapped with confidence. The effective peripheral fringe zone is at least as large as the largest of the individual fringe zones, but smaller than the sum of the three individual fringe zones. Since they will overlap. It should be considered in planning a survey.

12.6c Assume that the objective formations dip away from one edge of the area, how does this affect the area that can be mapped confidently?

Solution

Dip offsets the area of confident coverage in the updip direction, but, if the dip is uniform, it does not significantly shrink the area of coverage. Dip of $20^{\circ }$ foreshortens coverage by only $\cos 20^{\circ }=0.94$ or 6%. The distance that dip moves the subsurface coverage depends on the raypath curvature, which, in turn, depends on the velocity gradient; it is apt to be about $(z/2)\tan \zeta$, where $z$ is the depth and $\zeta$ is the dip angle.

12.7 Vibroseis land survey

A land survey layout is shown in Figure 12.7a. A single swath used 112 geophone groups spaced 110 ft apart on each of six east-west lines (solid lines) with vibrators traversing north-south lines (dashed) with vibrator points spaced 220 ft apart. For the next swath to the north, the three southernmost lines of geophones were moved to new lines north of swath #1. The entire area of 8.2 square miles was thus mapped in four swaths, but the layout pattern had to be modified in the north because of permit restrictions.

Figure 12.7a.  Layout for a vibroseis survey shot in four swaths. Layouts for the first and last of the four swaths are shown (after Hardage, 1993).

Figure 12.2b.  Zones of multiplicity for $110\times 110\ {\rm {ft}}$ bins for half the swath. Note that each zone is eight bins wide by 12 bins tall. Bins in the top three rows get additional coverage from the swath to the north. The pattern uses six lines of geophones spaced 1320 ft apart and 15 lines of sources spaced 880 ft apart.

In the southern two-thirds of the area where spacing was regular, what is the smallest bin size that should be used? What is the best multiplicity achieved? How wide is the multiplicity taper area? What is the smallest bin size if square bins are desired, and, for the best multiplicity bins, what are the offset and azimuth ranges? How will this change if four of the smallest square bins are combined to give larger square bins?

Solution

Midpoints in the cable direction will be spaced at half the group interval or 55 ft; in the source direction at half the vibrator-point spacing, or 110 ft. Thus the minimum bin size is $55\times 110\ {\rm {ft}}$ as any smaller size will leave many empty bins. If square bins are desired, $110\times 110\ {\rm {ft}}$ bins would have a multiplicity of two in the corners of the survey area and build up to 28 along the southern edge (see Figure 12.7b). This figure shows the multiplicity of bins in each rectangle (each rectangle contains $8\times 12$ bins, each $110\times 110\ {\rm {ft}}$ bins). The buildup of multiplicity along the east and west edges of the survey area will increase by two up to a maximum of eight. Azimuths for the center bins cover all directions. If $220\times 220\ {\rm {ft}}$ bins are used, the multiplicity values will simply be the sums of the four bins that are combined. The second swath will overlap the two northernmost sets of rectangles, resulting in zones of coverage along the east and west edges of 2, 4, 6, 8, 6, 6, 6, 6, 8, and then repeats in descending order, and in the north-south direction through the center of the area, 28, 56, 84, 112, 112, 84, 84, 84, 84, 112, 112, etc.

Access restrictions in the northwest portion of the area limited the vibrator lines, which results in irregularities in the multiplicity but not to a major degree. The multiplicity degradation will not be very severe since no change in geophone layout is indicated and only a few vibrator lines are missing completely.

12.8 Loop layout for a 3D survey

In one 3D technique, source points ($\times$) and geophones ($\circ$) are laid out around a loop such as the square shown in Figure 12.8a and all of the geophones are recorded for each source point. This example employs 48 geophone stations spaced 50 m apart and 24 source points spaced 100 m apart. Locate the midpoints and determine their multiplicity.

Solution

The minimum bin size has sides half the geophone spacing; the lines in Figure 12.8b indicate the bin centers.

Ignoring the sources outside the square for the moment, the sources and geophones along the top of the square will produce inline multiplicity along this line. The source and geophone in the upper left-hand corner will produce a zero-offset trace at that midpoint. This source and geophone #2 will give a midpoint trace between geophones 1 and 2. This source and geophone #3 will give a midpoint trace at geophone #2 as will the source at geophone 3 into geophone #1, giving 2-fold data here. However, both of these involve the same travelpath and so do not produce independent information and so only one counts as increasing the multiplicity. The midpoint at geophone #3 will have multiplicity of 2, once for source-geophone locations at 1 and 5 and once for coincident source and geophone at location 3. The sources along the west side of the area and geophones along the north side will give single-fold coverage over much of the interior of the square, but will leave many bins empty because there are only half as many source locations as geophone locations. Sources along the west side and geophone locations along the east side will provide multiplicity along the north-south bisector of the square. Most of these will not duplicate the raypaths from sources on the east and geophones on the west, and hence, the multiplicity down this bisector will be larger than that along the edges of the square. The multiplicity achieved is shown in Figure 12.8b.

Figure 12.8a.  Loop of geophones ($\circ$), sources ($\times$).

Figure 12.2b.  Multiplicity achieved in one quadrant ignoring sources outside the square.

The sources outside the square will expand the coverage area and also increase the multiplicity along the edges of the square, as indicated in Figure 12.8c.

If adjacent squares are shot by repeating locations along an edge of the square, the midpoints outside one square will fall inside another square. But, in general, this will not increase the multiplicity because reciprocal raypaths will be involved.

Figure 12.8c.  Multiplicity achieved in one quadrant from all sources.

If the loop had been irregular rather than square, the irregularity would have produced irregularities in the distribution of the midpoints and changed the multiplicity somewhat.

Working out the multiplicity pattern longhand is not only tedious but also very subject to errors. Ordinarily a computer is used to make maps showing the multiplicity. The mixture of offsets involved is often at least as important as the multiplicity, and so, maps are also made showing the offset distribution. Likewise maps are often made showing the distribution of azimuths.

12.9 Fault interpretation using time slices

12.9a The series of time slices shown in Figure 12.9a may be faulted. Where are possible faults and what can you infer about them? Assume a velocity of 3000 m/s.

Figure 12.9a.  Sequence of time slices (after Brown, 1983).

Figure 12.9b.  Interpretation of Figure 12.9a showing postulated fault. The contour map (viii) is made by tracing the inner black area in (i) and its changes on (ii) to (vii).

Background

Time slices show the strike of features. Faults are suggested where reflections terminate systematically. A horizon slice that cuts through the 3-D volume along a picked horizon is often the best way to see stratigraphic features.

Solution

Assuming that the closure is due to an anticline, the only offset of the pattern that might suggest faulting is seen on slices (iii) to (v) in Figure 12.9a. This suggests an east-west down-to-the-south fault dipping to the north, i.e., a reverse fault, as suggested in Figure 12.9b. However, as drawn, the fault dip is only slightly greater than the bedding dip, which is geologically unreasonable. An alternative and more probable explanation is that it represents a re-entrant (valley) cutting into the structure roughly along the same alignment. A horizon slice should be created to corroborate a channel interpretation, since channels show up best on horizon slices.

12.9b Locate possible faults on Figure 12.9c.

Solution

The very obvious offset of the contours (A on Figure 12.9d) almost certainly indicates a fault. The abrupt change of dip ($B$) probably indicates another fault. The location of these features becomes less clear toward the southern edge of this time slice. Study of other time slices or vertical sections would probably clarify the matter.

12.10 Acquisition direction for marine 3D surveys

Does it make any difference whether a 3D marine survey is recorded in the dip or strike direction?

Solution

The common assumption is that reciprocity applies, i.e., that one gets the identical result when source and receiver are interchanged. Both source and receiver groups have linear dimensions that are orthogonal to each other, and, hence, the smearing effects of group lengths will be in different directions. However, the dimensions are so small that this aspect is probably not important. Differences in midpoint locations within a bin will make minor differences when stacking, but this also is probably not important. Hence, it is probably unimportant that the entire system is moving during the recording. While the offsets will be slightly different for different streamers where multiple streamers are used, such effects can be accommodated in processing. We generally assume that the earth is isotropic, and this assumption probably does not create significant differences. Thus, we generally do not expect the shooting direction to introduce differences in the survey results.

Figure 12.9c.  A time slice.

Figure 12.9d.  Interpretation of Figure 12.9c showing possible faults.

However, where there are appreciable lateral changes in velocity, the raypaths for common reflecting points in different directions may differ appreciably, and this can produce quite different effects, especially if they are not allowed for before stacking. Poststack migration is especially apt to produce different results.