User:Ageary/Chapter 5
Problems-in-Exploration-Seismology-and-their-Solutions.jpg | |
Series | Geophysical References Series |
---|---|
Title | Problems in Exploration Seismology and their Solutions |
Author | Lloyd P. Geldart and Robert E. Sheriff |
DOI | http://dx.doi.org/10.1190/1.9781560801733 |
ISBN | ISBN 9781560801153 |
Store | SEG Online Store |
Contents
- 1 Chapter 5 Seismic velocity
- 2 5.1 Maximum porosity versus depth
- 3 5.2 Relation between lithology and seismic velocities
- 4 5.3 Porosities, velocities, and densities of rocks
- 5 5.4 Velocities in limestone and sandstone
- 6 5.5 Dependence of velocity-depth curves on geology
- 7 5.6 Effect of burial history on velocity
- 8 5.7 Determining lithology from well-velocity surveys
- 9 5.8 Reflectivity versus water saturation
- 10 5.9 Effect of overpressure
- 11 5.10 Effects of weathered layer (LVL) and permafrost
- 12 5.11 Horizontal component of head waves
- 13 5.12 Stacking velocity versus rms and average velocities
- 14 5.13 “Quick-look” velocity analysis; Effects of errors
- 15 5.14 Well-velocity survey
- 16 5.15 Interval velocities from analyses
- 17 5.16 Finding velocity by the method
- 18 5.17 Effect of timing errors on stacking velocity, depth, and dip
- 19 5.18 Estimating lithology from stacking velocity
- 20 5.19 Velocity versus depth from sonobuoy data
- 21 5.20 Influence of direction on velocity analyses
- 22 5.21 Effect of time picks, NMO stretch, and datum choice on stacking velocity
Chapter 5 Seismic velocity
5.1 Maximum porosity versus depth
What physical fact determines the “limit-of-porosity” line in Figure 5.1a? What is implied for measurements to the right of this line?
Background
The porosity of a rock is the pore volume per unit volume. The pore space is generally filled with salt water except near the surface, where air may be present, and in petroleum deposits, where the pore spaces contain oil and/or gas.
Pore spaces are usually sufficiently interconnected so that the fluid pressure approximates that of a fluid column extending to the surface; this is called normal pressure. The weight of the rock column exerts an overburden pressure. The differential, effective, or net pressure on the rock matrix is the overburden pressure less the interstitial fluid pressure. However, if there is no communication between the pore spaces and the surface, the interstitial fluid pressure may be greater than the normal pressure causing the differential pressure (and velocity) to be lower than usual for a given depth—a situation known as overpressure.
Solution
The line marking the limit of porosity is fixed by the strength of the rock matrix that is presumed to be carrying the major part of the weight of the overburden. The overburden weight is partially supported by the interstitial fluid; in overpressured formations the effective stress on the rock’s matrix structure will be smaller than normal, equal to the stress that would be experienced at a shallower depth. The porosity and velocity will thus be that at the shallower depth and such points may fall to the right of the line.
5.2 Relation between lithology and seismic velocities
Figures 5.2a and 5.2b are based on different experimental data and plotted differently. Show the compatibility of these figures.
Solution
Dashed lines have been drawn on Figure 5.2a to approximate the data for the three lithologies and these have been transferred to Figure 5.2b. The limestone line lies within its domain, while the dolomite and sand lines lie at the edges of their domains. The sandstone data in Figure 5.2a show the greatest scatter. The concept that provides a basis for distinguishing these three lithologies is generally supported.
5.3 Porosities, velocities, and densities of rocks
5.3a Assume that sandstone is composed only of grains of quartz, limestone only of grains of calcite, and shale of equal quantities of kaolinite and muscovite. For sandstone, limestone, and shale saturated with salt water , what porosities are implied by the densities shown in Figure 5.3a? (Mineral densities are: ; ; ; , all in g/cm.)
Background
Gardner et al. (1974) plotted the log of velocity against the log of density for sedimentary rocks and obtained the empirical relation known as Gardner’s rule:
( )
Porosity | ||||||
---|---|---|---|---|---|---|
Rock | Density range | Density av. | Mineral density | Max | Av. | Min |
Ss | 2.00–2.60 g/cm | 2.35 g/cm | 2.68 g/cm | 41% | 20% | 5% |
Ls | 2.20–2.75 | 2.55 | 2.71 | 30 | 10 | 0 |
Sh | 1.90–2.70 | 2.40 | 2.72 | 48 | 19 | 0 |
being in g/cm and or 0.23 when is in m/s or ft/s, respectively. The rule is valid for the major sedimentary rock types, but not for evaporites or carbonaceous rocks (coal, lignite).
When a porous rock is saturated with a fluid, its density is given by the equation
( )
being the porosity, and the densities of the fluid and rock matrix, respectively.
Solution
The density ranges in Table 5.3a were obtained from Figure 5.3a. The mineral densities are for , the values for shale being averages for kaolinite and muscovite.
We solve equation (5.3b) for , obtaining
( )
The histogram in Figure 5.3a does not encompass the complete range of samples and the range limits have been picked somewhat arbitrarily. Porosity in rocks ranges from about 50% to 0%. The upper limits of the density range sometimes exceed the mineral densities, hence heavier minerals must be present in the rocks; in these cases we assume that . We take as the fluid density.
5.3b What velocities would be expected for the density values in Table 5.3a according to Gardner’s rule? Plot these on Figure 5.3b.
Solution
We solve equation (5.3a) for the velocity , obtaining
The velocities in Table 5.3b are plotted as triangles on Figure 5.3b.
Rock | |||
---|---|---|---|
Ss | 1.8 (41%) | 3.4 (20%) | 6.8 (0%) |
Ls | 2.6 (30%) | 4.6 (10%) | 6.8 (0%) |
Sh | 1.4 (48%) | 3.4 (19%) | 6.8 (0%) |
Note: The values in parentheses are the porosities. |
5.3c From Figure 5.3c, what densities would you expect at 7500 ft and how do these compare with Figures 5.3d and 5.3e from offshore Louisiana?
Solution
Using and from Figure 5.3c, equation (5.3c) gives , which is in accord with Figure 5.3d. Using , equation (5.3c) gives , which is slightly lower than most values in Figure 5.3e.
5.4 Velocities in limestone and sandstone
Assume that the velocity in calcite is 6.86 km/s and in quartz 5.85 km/s. What velocities should be expected for 10, 20, and 30% porosities in (a) limestone composed only of calcite; (b) sandstone composed only of quartz? Where do these values plot on lithology versus velocity curves (Figure 5.4a) and porosity versus velocity curves (Figure 5.4b)? Assume that the pore fluid is water with velocity 1.55 km/s.
Background
The velocity in a porous rock depends primarily on the matrix velocity, the porosity, and the nature of and velocity in the pore fluid. The empirical time-average equation [of the same form as equation (5.3b)] relates the specific transit time (reciprocal of the velocity) or slowness to the volume fraction of pore space and the remaining volume :
( )
where , , and are the velocities in the saturated rock, the fluid, and the rock matrix, respectively.
Solution
Substituting in equation (5.4a) yields . For values of 10, 20, and 30%, we obtain the velocities in Table 5.4a.
Rock | ||||
---|---|---|---|---|
Limestone | 6.86 km/s | 5.11 km/s | 4.07 km/s | 3.38 km/s |
Sandstone | 5.85 | 4.58 | 3.76 | 3.19 |
5.5 Dependence of velocity-depth curves on geology
5.5a Why do the velocity-depth curves for various areas shown in Figures 5.5a and 5.5b depart from each other? Incorporate your knowledge of the geology of the various areas in your answer.
Solution
Velocities are increased not only by compaction with depth of burial but also by cementation and other factors attendant upon age. They are also affected by lithology and other factors. The U.S. Gulf Coast and offshore Venezuela sediments are predominantly young siliciclastics that have never been buried deeper than they are now. Hence their velocities generally relate to the maximum porosities such as shown in Figure 5.1a. Similar reasoning applies to the shallower portions of the offshore U.S. East Coast and the Gulf of Alaska curves of Figure 5.5b. The high-velocity values in the upper part of the Gulf of Alaska curve probably indicate limestone or volcanics. The Texas Gulf Coast-1 curve penetrated Cretaceous rocks containing limestone at fairly shallow depths whereas the Gulf Coast-2 well did not encounter this section until a depth of 4.25 km. The Illinois Basin and Permian Basin wells contain much older, higher-velocity rocks including limestone.
5.5b Plot the shale and limestone values from Figure 5.5c for depths of 1000 and 2000 m on the velocity-depth Figure 5.5a. How do they compare?
Solution
The values read from Figure 5.5c are:
shale at 1000 m | 2.2 km/s |
shale at 2000 m | 2.8 |
limestone at 1000 | 4.2 |
limestone at 2000 | 5.4 |
These points are plotted as triangles on Figure 5.5a for regions where the sections are sand-shale. The shale values fit nicely, but the limestone values are much too large, as one would expect. The limestone values are also plotted on Figure 5.5b where they fit in nicely with data for regions that are mostly carbonate.
5.6 Effect of burial history on velocity
5.6a Assume a subsiding area where there has been no uplift and a shale that is normally pressured until it reaches a depth of burial of 1.40 km, at which point fluid communication is cut off, that is, interstitial fluid can no longer escape. If it is at a depth of 2.00 km, what velocity and fluid pressure would you expect? If at a depth of 3.00 km?
Background
The normal fluid pressure at depth is , where is the fluid density. The fluid pressure gradient is MPa/km, whereas the pressure gradient due to the rock overburden is about 22.5 MPa/km. The differential pressure gradient is therefore approximately 12 MPa/km.
Solution
At a depth of 1.40 km, the velocity is about 2.5 km/s (see Figure 5.6a) and the fluid pressure is about 10 MPa/km 1.4 km = 14 MPa, while the pressure on the matrix , the differential pressure being 18 MPa. The porosity will be about 30% (see Figure 5.1a). If the shale is cut off at a depth of 1.4 km and the depth is then increased to 2.0 km, the differential pressure and velocity will not change greatly.
At 2.0 km, the overburden pressure , the fluid pressure will be about . At 3.0 km, MPa and .
5.6b Assume a shale buried to 3.00 km and then uplifted to 2.00 km, being normally pressured all the time. What velocity and fluid pressure would you expect?
Solution
The velocity at 3.00 km is about 3.2 km/s from Figure 5.6a and the fluid pressure about 30 MPa. Burial to this depth will have reduced the porosity to about 25% (see Figure 5.1a) and very little of the porosity will be recovered upon uplift. Hence the velocity will be larger than for a situation where the rock had not been buried deeper.
5.6c Assume the shale in part (a) is buried to 3.00 km and then uplifted to 2.00 km without fluid communication being established. What velocity and fluid pressure would you expect? What if the shale is uplifted to 1.00 km?
Solution
Since the differential pressure was not changed, we expect about the same result at 2.00 km as in part (a). At 2.00 km, . When the shale is then uplifted to 1.40 km, its pressure will again be normal. If further uplifted to 1.00 km without fluid communication being established, the porosity will remain about 32%. The velocity and differential pressure will be about the same as they were at 1.40 km, but will now be (instead of the usual value of 10 Mpa); the shale will be under-pressured and will have approximately the same velocity as at 1.40 km.
5.7 Determining lithology from well-velocity surveys
By comparing the velocity-depth relations in Figures 5.6a and 5.7a, what can you deduce about the nature of the rocks in the well for Figure 5.7a?
Solution
We superimpose the sand-shale curve from Figure 5.6a on Figure 5.7a (the dashed line) and note that the data are nearly parallel. It appears that the geologic section represented by Figure 5.7a is normally pressured, mainly clastics, and has not been uplifted significantly (i.e., the velocity is about what is expected for normal compaction).
5.8 Reflectivity versus water saturation
What shale velocities are consistent with the Figure 5.8a oil-sand data?
Solution
We determine velocities and reflectivities for two values of water saturation for each of the three depths, neglecting differences between sand and shale densities. Setting , equation (3.6a) becomes
Depth | Water saturation | |||
---|---|---|---|---|
610 m | 30% | –0.08 | 1.7 km/s | 2.0 km/s |
70 | –0.03 | 1.8 | 1.9 | |
1830 | 30 | –0.10 | 2.3 | 2.8 |
70 | –0.07 | 2.4 | 2.8 | |
3050 | 30 | –0.10 | 2.7 | 3.3 |
70 | –0.07 | 2.8 | 3.2 |
From Figures 5.8a(i,ii) we obtain the data in Table 5.8a for oil sands. These are plotted as triangles on Figure 5.6a.
5.9 Effect of overpressure
5.9a Figure 5.6a shows velocity versus depth for normally pressured shales. How do the Figure 5.9a velocities above and below the top of the abnormal pressure zone compare with the sand and shale velocity-depth curves of Figure 5.6a? What depth corresponds to normal pressure for the top overpressure? What porosity would you expect for the overpressured shale?
Solution
The transit times above and below the “top of the overpressure zone” are roughly and , respectively, that is, velocities of 8.0 and 5.7 kft/s (2.4 and 1.7 km/s). According to Figure 5.6a, the shale velocity at 6000 ft (2.1 km) should be higher (about 8.8 kft/s); the shale is probably somewhat undercompacted. The 5.7 kft/s velocity of the overpressured shale corresponds to a depth of about 1000 ft (300 m) and the porosity of the overpressured shale is probably close to 50%.
5.9b Plot the velocities for 100% water saturated sands from Figure 5.8a(i) on Figure 5.6a. How do they compare?
Solution
The values for 100% water saturated sands from Figure 5.8a(i) are listed below and plotted as triangles on Figure 5.6a. They lie slightly below the Gulf Coast sand curve.
Depth | 0.61 km | 1.83 | 3.05 |
Velocity | 1.9 km/s | 2.5 | 3.0 |
The data from Figure 5.9a are plotted as squares on Figure 5.6a.
5.10 Effects of weathered layer (LVL) and permafrost
5.10a Assume that raypaths have angles of approach of , , , , and in the subweathering where the velocity is 2400 m/s. For a weathered layer 10 m thick with velocity 500 m/s, how do travel-times through the weathered layer compare with that for a vertically traveling ray? What are the horizontal components of the raypaths in the LVL?
Solution
Referring to Figure 5.10a,
Incident | In low-velocity layer | In permafrost | ||||
---|---|---|---|---|---|---|
(m) | (ms) | (m) | (ms) | |||
0.0 | 20.0 | 0 | 27.8 | |||
0.4 | 20.0 | 27 | 28.8 | |||
0.7 | 20.1 | 60 | 32.4 | |||
1.1 | 20.1 | 113 | 42.0 | |||
1.4 | 20.1 | 363 | 104.6 | |||
1.4 | 20.1 | |||||
1.8 | 20.3 |
Substituting the values of , we get the results in Table 5.10a. The traveltimes in the LVL vary by only 0.5% over most of the range of , and, even for , change by only 1.5%.
5.10b For permafrost 100 m thick with a velocity of 3600 m/s, answer the questions in part (a).
Solution
We repeat the calculations of part (a) changing to 3.60 km/s and layer thickness to 100 m. The results are also shown in Table 5.10a. Because rays now have large horizontal components, the changes in and are considerable. This large ray bending makes corrections for permafrost very difficult. If , upcoming waves are totally reflected.
5.11 Horizontal component of head waves
In the early days of refraction exploration for salt domes, sketches of expected raypaths indicated that the angle of approach to the surface should have a large horizontal component, but measurements showed very little horizontal component. Controversy arose over whether the travelpaths could be as drawn. Explain the apparent discrepancy.
Solution
The raypath bending at the base of the low-velocity layer made travel through the LVL nearly vertical, as illustrated in problem 5.10a so the horizontal component at the surface was very small.
5.12 Stacking velocity versus rms and average velocities
5.12a Assume six horizontal layers, each 300 m thick and with constant velocity (Figure 5.12a). The successive layers have velocities of 1.5, 1.8, 2.1, 2.4, 2.7, and 3.0 km/s. Ray-trace through the model to determine offset distances and arrival times for rays that make angles of incidence at the base of the 3.0 km/s layer of , , , and . Calculate stacking velocity for each angle and compare with the average velocity and the rms velocity .
Background
Average velocity and rms velocity were discussed in problem 4.13 [see equations (4.13a,b)].
In the common-midpoint (CMP) technique, a number of traces are obtained with different source-geophone distances (offsets, see problem 4.1) but the same midpoint. After correcting for NMO (and for dip if necessary), they are added together (stacked), the number of traces added together being the multiplicity. The velocity used to remove the NMO is the stacking velocity . If we use equation (4.1c) to remove the NMO, that is, if we assume a single horizontal constant-velocity layer, the velocity in equations (4.1a,c) becomes . The plot of equation (4.1a) is a straight line with slope ; thus,
( )
where and are the two-way traveltimes at the origin and at offset while . When the velocity changes with depth, the plot is curved but the curvature is generally small enough that the best-fit straight line gives reasonably accurate results. For horizontal velocity layering and small offsets, .
Solution
We use Snell’s law to calculate the raypath angles in each layer. The two-way time in a layer is and the offset in a layer is . The values in Table 5.12a have been calculated without regard to the number of significant figures to illustrate the sensitivity of the calculations. The average velocities along the respective raypaths have also been calculated for comparisons.
The calculations for the intermediate layer boundaries assume that reflections are generated at each boundary. Traveltime differences, shown in parentheses in Table 5.12b, are very small for most of the situations, and, especially where the differences are less than 20 ms, are not very reliable for calculating . A general rule for calculations, that the offset should be comparable to the depth, is not reached for any of these situations.
layer 1 | layer 2 | layer 3 | layer 4 | layer 5 | layer 6 | |
---|---|---|---|---|---|---|
0.400 | 0.333 | 0.286 | 0.250 | 0.222 | 0.200 | |
0.400 | 0.733 | 1.019 | 1.269 | 1.491 | 1.691 | |
0 | 0 | 0 | 0 | 0 | 0 | |
1500 | 1640 | 1770 | 1890 | 2010 | 2130 | |
1500 | 1640 | 1780 | 1920 | 2060 | 2190 | |
0.1600 | 0.5373 | 1.0384 | 1.6104 | 2.2231 | 2.8595 | |
0.402 | 0.325 | 0.288 | 0.252 | 0.225 | 0.203 | |
0.402 | 0.737 | 1.025 | 1.277 | 1.502 | 1.705 | |
52 | 63 | 73 | 84 | 95 | 106 | |
52 | 105 | 189 | 273 | 368 | 473 | |
0 | 0.077 | 0.111 | 0.143 | 0.181 | 0.218 | |
* | 1400* | 1700* | 1900* | 2029 | 2170 | |
1500 | 1636 | 1767 | 1892 | 2013 | 2131 | |
0.406 | 0.340 | 0.294 | 0.260 | 0.224 | 0.213 | |
0.406 | 0.746 | 1.041 | 1.301 | 1.534 | 1.747 | |
104 | 126 | 146 | 171 | 194 | 218 | |
104 | 230 | 378 | 549 | 743 | 961 | |
0.0695 | 0.1386 | 0.2128 | 0.2867 | 0.3606 | 0.4388 | |
1496 | 1658 | 1775 | 1914 | 2060 | 2190 | |
1500 | 1637 | 1768 | 1894 | 2017 | 2137 | |
0.413 | 0.349 | 0.305 | 0.273 | 0.248 | 0.231 | |
0.413 | 0.762 | 1.067 | 1.340 | 1.589 | 1.820 | |
155 | 189 | 224 | 262 | 302 | 346 | |
155 | 344 | 568 | 830 | 1132 | 1478 | |
0.103 | 0.208 | 0.316 | 0.430 | 0.549 | 0.673 | |
1508 | 1652 | 1795 | 1928 | 2060 | 2196 | |
1500 | 1637 | 1770 | 1898 | 2024 | 2147 |
*Not enough significant figures to calculate with sufficient accuracy. |
layer 1 | layer 2 | layer 3 | layer 4 | layer 5 | layer 6 | |
---|---|---|---|---|---|---|
1500 | 1635 | 1770 | 1890 | 2010 | 2130 | |
1500 | 1643 | 1780 | 1920 | 2060 | 2190 | |
Stacking velocity calculations: | ||||||
* (0) | 1370 (4) | 1707 (6) | 1924 (8) | 2029 (9) | 2170 (14) | |
1496 (6) | 1658 (13) | 1775 (22) | 1914 (32) | 2060 (43) | 2190 (56) | |
1508 (13) | 1652 (29) | 1795 (48) | 1928 (71) | 2060 (98) | 2196 (129) | |
Average velocity along raypaths: | ||||||
1500 | 1636 | 1767 | 1892 | 2013 | 2131 | |
1500 | 1637 | 1768 | 1894 | 2017 | 2137 | |
1500 | 1637 | 1770 | 1898 | 2024 | 2147 |
*Not enough significant figures to calculate with sufficient accuracy. |
Values in parentheses are traveltime differences. |
layer 1 | layer 2 | layer 3 | layer 4 | layer 5 | layer 6 | |
---|---|---|---|---|---|---|
0.400 | 0.333 | 0.286 | 0.250 | 0.222 | 0.200 | |
106 | 128 | 151 | 173 | 197 | 221 | |
0.426 | 0.366 | 0.326 | 0.303 | 0.288 | 0.274 | |
219 | 270 | 328 | 393 | 469 | 561 | |
0.462 | 0.417 | 0.401 | 0.424 | 0.511 | * | |
346 | 452 | 591 | 620 | 1245 | * |
*A head wave is generated at the base of layer 5. |
layer 1 | layer 2 | layer 3 | layer 4 | layer 5 | layer 6 | |
---|---|---|---|---|---|---|
1500 | 1630 | 1750 | 1860 | 1960 | 2050 | |
1500 | 1640 | 1750 | 1880 | 1990 | 2110 | |
1500 | 1650 | 1790 | 1920 | 2100 | * |
* Head wave generated. |
We note that the stacking velocity increases with the offset . The calculated velocities are summarized in Table 5.12b.
5.12b Repeat part (a) for the case where rays make angles of incidence at the free surface of , , and .
Solution
The case where is the same as that for so that we need to calculate only for the results are given in Table 5.12c.
We now calculate a stacking velocity for reflections for each layer for each of the angles (Table 5.12d).
As before, we note that the stacking velocity increases with the offset .
5.12c Assume the 300-m-thick layers dip as shown in Figure 5.12b and determine arrival times for a zero-offset ray and one that leaves the free surface at an angle of and is reflected at .
Solution
The raypath for a zero-offset trace makes a angle in the updip direction at the surface and angles at all of the interfaces so that after reflection the raypath will return to the sourcepoint. The traveltimes are the same as calculated in part (a).
A ray that leaves the free surface at in the updip direction is incident on the interface at and thus makes the same angles with other interfaces as calculated for the case in part (b). The time spent in each of the layers will also be the same as in part (b) but the distances are now measured along the bedding planes. Thus, to determine the locations of the source and the emergent location, these have to be corrected by The geometry is shown in Figure 5.12c. We have from part (b), e = 435 m, g = 488 m, one-way time from top of layer to , time from to the base of layer 0.672 s.
The source is farther from the zero-offset location than the emergent point, so that the data are not suitable for stacking velocity calculations unless a DMO correction (Sheriff and Geldart, 1995, section 9.10.2) has been applied. Calculating arrival times for dipping reflections for split-dip situations is often done by trial and error.
5.13 “Quick-look” velocity analysis; Effects of errors
5.13a Velocity analysis usually results in a plot of stacking velocity against traveltime. Bauer (private communication) devised a “quick look” method of determining the interval velocity, assuming horizontal layering and that the stacking velocity equals the average velocity. The method is shown in Figure 5.13a. A box is formed by the two picks between which the interval velocity is to be picked; the diagonal that does not contain the two picks when extended to the velocity axis gives the interval velocity. Prove that the method is valid and discuss its limitations.
Solution
We extend the diagonal of the box as shown in Figure 5.13a, thus giving . The interval velocity is given by
( )
In Figure 5.13a the triangle with apices at the points , , and is similar to the triangle with sides and , so we have
Substituting in equation (5.13a), we get
Thus the method gives the interval velocity provided the stacking velocity equals the average velocity. For horizontal velocity layering the stacking velocity is often about 2% higher than the average velocity, but the two may differ considerably if the reflectors are dipping.
5.13b This method can be used to see the influence of measurement error. Discuss the sensitivity of interval-velocity calculations to
- errors in picking velocity values from this graph,
- errors in picking times,
- picking events very close together, and
- picking events late.
Solution
- Errors in or change the slope of the diagonal and hence change and ; the error is proportional to where is either or .
- Changes in or has an effect similar to that in (i).
- When both and are small, the slope of the diagonal and are very sensitive to errors.
- Picking each event late by the same amount will increase by an amount proportional to
We now derive mathematical expressions for the changes in (i) to (iv).
i) Equation (5.13a) is
Hence the error is directly proportional to either or and inversely proportional to ; it increases rapidly as approaches .
ii)
Likewise,
Since , errors in are more serious than errors in ; also, the errors increase rapidly as approaches .
iii) Let , . Then using equation (5.13a),
In general , so the error in depends mainly on the factor .
iv) We assume that both events are late by the same amount . Then equation (5.13a) becomes
where is usually fairly constant over a moderate range of depths; in this case the error in is proportional to the error
5.14 Well-velocity survey
5.14a Figure 5.14a shows data from a well-velocity survey tabulated on a standard calculation form. Calculate the average velocity and interval velocity, and plot graphs of time, average velocity, and interval velocity versus depth using a sea-level datum.
Solution
Figure 5.14a shows only the measured data on a standard form whereas Figure 5.14b shows also calculated values. The successive columns in this form list
1 - Record number
2 - Source (shothole) location
3 - , geophone depth with respect to well datum
4 - , depth of source
5 - , uphole time
6 -, arrival time at reference geophone
7 - , arrival time at well geophone plus polarity and quality grades
8 - , geophone depth with respect to source elevation;
9 - , horizontal distance of source from wellhead
10,11 - tangent, cosine of angle between straight raypath and vertical;
12 - , vertical traveltime from source to geophone
13 - , source to datum elevation difference: ; a minus sign means that the shot was above datum
14 - time correction for
15 - , vertical traveltime from datum to geophone
17 - , depth of geophone below datum
18 - , depth difference between successive geophone depths
19 - , time difference between successive geophone arrivals
20 - , interval velocity
21 - , vertical traveltime from source to geophone
Depths below the datum are positive. The velocity used to correct is (also obtainable from ). Note that the column headed is in milliseconds whereas all other times are in seconds. Column #16 headed average is not used.
Figure 5.14c shows average velocity, interval velocity, and time plotted against depth.
5.14b How much error in average velocity and interval velocity values would result from (i) time-measurement errors of 1 ms, and (ii) depth-measurement errors of 1 m?
Solution
- A 1-ms time error produces an error in the average velocity of 0.1% to 1.5% and an error in the interval velocity of 1.5% to 8.3%.
- A depth error of 1 m produces an error in the average velocity of 0.03% to 0.9% and an error in the interval velocity of 0.5% to 1.5%.
236 | 2200 | 255 | 1970 | 5400 | 65 |
441 | 2400 | 155 | 2038 | 3700 | 70 |
551 | 2600 | 65 | 2118 | 3900 | 90 |
661 | 2500 | 155 | 2228 | 6500 | 130 |
820 | 3100 | 165 | 2360 | 4000 | 135 |
973 | 2700 | 140 | 2506 | 4400 | 155 |
1101 | 3600 | 115 | 2643 | 4300 | 120 |
1203 | 2400 | 90 | 2783 | 4100 | 160 |
1350 | 4800 | 205 | 2928 | 4500 | 130 |
1508 | 4200 | 110 | 3043 | 4800 | 100 |
1603 | 2200 | 80 | 3145 | 5200 | 105 |
1730 | 7600 | 175 | 3243 | 6000 | 90 |
1878 | 5000 | 120 |
5.14c Determine and for a velocity-function fit to the data in (a) assuming the functional form , where is the interval velocity and the depth.
Solution
We can find and by (i) plotting the data and measuring the slope and intercept of the best-fit straight line, or (ii) using the least-squares method (see problem 9.33). The former method is difficult because of the large, irregularly spaced jumps in the curve, and therefore we shall use the latter method. We take as the depth in meters below datum to the center of each interval and give each data pair the weight (see problem 9.33b) . Using the data in Table 5.14a, we get , as shown in Figure 5.14c.
5.15 Interval velocities from analyses
An survey gives the stacking velocity results in Table 5.15a. Calculate the interval velocities.
1 | 1.20 | 1.100 | 2.18 |
2 | 2.50 | 1.786 | 2.80 |
3 | 3.10 | 1.935 | 3.20 |
4 | 4.10 | 2.250 | 3.64 |
Background
In equation (4.13a), the sum is the traveltime down to the base of the layer, that is, . Therefore equation (4.13a) can be written
( )
If we write and for the rms velocity and traveltime down to the base of the layer, and for the same quantities down to the top of the layer, we can get an equation for the interval velocity in the bed by subtracting expressions for and obtained from equation (5.15a). The result is
( )
Solution
We assume horizontal velocity layering so that the values of in Table 5.15a are approximate values of (see Table 5.12b) and therefore we can use equation (5.15b) to find interval velocities. We take 2.18 km/s as the interval velocity from the surface to the first reflector. We now calculate the interval velocities to :
5.16 Finding velocity by the method
Determine the velocity by the method using the data given in Table 5.16a, being for a horizontal reflector and for a reflector dipping toward the source. (The two reflections are observed on different records.)
Background
Using equation (4.3a), we have for a dipping horizon,
where is the offset, is the dip, and the slant depth at the midpoint between source and receiver.
(km) | (s) | (s) | (km) | (s) | (s) | (km) | (s) | (s) |
---|---|---|---|---|---|---|---|---|
0.0 | 0.855 | 0.906 | 1.4 | 1.005 | 0.977 | 2.8 | 1.330 | 1.202 |
0.1 | 0.856 | 0.902 | 1.5 | 1.017 | 0.991 | 2.9 | 1.360 | 1.234 |
0.2 | 0.858 | 0.898 | 1.6 | 1.037 | 1.004 | 3.0 | 1.404 | 1.253 |
0.3 | 0.864 | 0.898 | 1.7 | 1.068 | 1.019 | 3.1 | 1.432 | 1.272 |
0.4 | 0.868 | 0.899 | 1.8 | 1.081 | 1.037 | 3.2 | 1.457 | 1.296 |
0.5 | 0.874 | 0.902 | 1.9 | 1.105 | 1.058 | 3.3 | 1.487 | 1.304 |
0.6 | 0.882 | 0.903 | 2.0 | 1.118 | 1.066 | 3.4 | 1.513 | 1.334 |
0.7 | 0.892 | 0.909 | 2.1 | 1.151 | 1.083 | 3.5 | 1.548 | 1.356 |
0.8 | 0.900 | 0.916 | 2.2 | 1.166 | 1.102 | 3.6 | 1.580 | 1.377 |
0.9 | 0.906 | 0.922 | 2.3 | 1.203 | 1.121 | 3.7 | 1.610 | 1.407 |
1.0 | 0.930 | 0.932 | 2.4 | 1.237 | 1.127 | 3.8 | 1.649 | 1.415 |
1.1 | 0.945 | 0.943 | 2.5 | 1.255 | 1.158 | 3.9 | 1.674 | 1.438 |
1.2 | 0.950 | 0.950 | 2.6 | 1.283 | 1.177 | 4.0 | 1.708 | 1.459 |
1.3 | 0.979 | 0.965 | 2.7 | 1.304 | 1.195 |
( | ( | ||||||
---|---|---|---|---|---|---|---|
0.00 | 0.731 | 0.00 | 0.821 | 4.41 | 1.325 | 4.28 | 1.173 |
0.01 | 0.733 | 0.01 | 0.814 | 4.84 | 1.360 | 4.69 | 1.214 |
0.04 | 0.736 | 0.04 | 0.806 | 5.29 | 1.447 | 5.13 | 1.257 |
0.09 | 0.746 | 0.09 | 0.806 | 5.76 | 1.530 | 5.59 | 1.270 |
0.16 | 0.753 | 0.16 | 0.808 | 6.25 | 1.575 | 6.06 | 1.341 |
0.25 | 0.764 | 0.24 | 0.814 | 6.76 | 1.646 | 6.56 | 1.385 |
0.36 | 0.778 | 0.35 | 0.815 | 7.29 | 1.700 | 7.07 | 1.426 |
0.49 | 0.796 | 0.48 | 0.826 | 7.84 | 1.769 | 7.60 | 1.445 |
0.64 | 0.817 | 0.62 | 0.839 | 8.41 | 1.850 | 8.16 | 1.523 |
0.81 | 0.821 | 0.79 | 0.850 | 9.00 | 1.971 | 8.73 | 1.570 |
1.00 | 0.865 | 0.97 | 0.869 | 9.61 | 2.051 | 9.32 | 1.618 |
1.21 | 0.893 | 1.17 | 0.889 | 10.24 | 2.123 | 9.93 | 1.680 |
1.44 | 0.902 | 1.40 | 0.902 | 10.89 | 2.211 | 10.56 | 1.700 |
1.69 | 0.968 | 1.64 | 0.931 | 11.56 | 2.289 | 11.21 | 1.780 |
1.96 | 1.010 | 1.90 | 0.965 | 12.25 | 2.396 | 11.88 | 1.839 |
2.25 | 1.034 | 2.18 | 0.962 | 12.96 | 2.496 | 12.57 | 1.896 |
2.56 | 1.075 | 2.48 | 1.008 | 13.69 | 2.592 | 13.28 | 1.980 |
2.89 | 1.141 | 2.80 | 1.030 | 14.43 | 2.719 | 14.00 | 2.002 |
3.24 | 1.169 | 3.14 | 1.075 | 15.21 | 2.802 | 14.75 | 2.068 |
3.61 | 1.221 | 3.50 | 1.119 | 16.00 | 2.917 | 15.52 | 2.129 |
3.61 | 1.250 | 3.88 | 1.136 |
Solution
The values of and are tabulated in Table 5.16b and plotted in Figure 5.16a. The best-fit lines in Figure 5.16a determined by eye give
Using the least-squares method (see problem 9.33), we get
5.17 Effect of timing errors on stacking velocity, depth, and dip
5.17a Given that the trace spacing in Figure 5.17a is 50 m, determine the stacking velocity, dip, and depth at approximately 0.5, 1.0, 1.5, 2.0, and 2.4 s.
Solution
Using an enlarged version of Figure 5.17a, we measured arrival times on the center traces and on traces symmetrically located left and right of the center, limiting the offsets to where we felt we could pick the events with confidence. The measurements give us , , and and we must find and ; for this we need the velocity . We can find from equation (5.12a) and, since the dip is small, and we can therefore use equation (4.2b) to get an approximate value of . The equations are
(s) | 0.533 | 0.975 | 1.575 | 2.008 | 2.417 |
offset (m) | 575 m | 1125 | 1425 | 1625 | 2375 |
(s) | 0.633 | 1.150 | 1.658 | 2.058 | 2.533 |
(s) | 0.608 | 1.142 | 1.758 | 2.150 | 2.575 |
(s) | 0.0875 | 0.171 | 0.133 | 0.096 | 0.137 |
(m/s) | 1885 | 1950 | 2200 | 2615 | 2920 |
(m) | 502 | 950 | 1730 | 2625 | 3530 |
(ms/m) | 0.0435 | 0.0071 | 0.0702 | 0.0566 | 0.0177 |
0.0410 | 0.0070 | 0.0772 | 0.0740 | 0.0258 | |
The calculated results for five reflections are shown in Table 5.17a.
5.17b What problems or ambiguities do you have in picking these events? How much uncertainty is there in your ability to pick times and how much uncertainty does this introduce into the velocity, depth, and dip calculations?
Solution
There are clearly different families of events interfering with each other on this record, which we have not attempted to sort out. The axes of symmetry of some of the data shift to the left with depth, indicating dip to the right. The event at 2.417 s may be a multiple. Clearly many more events could be picked.
We timed the centers of the black peaks, and this involves 5-10 ms uncertainty in this case. At a work station where a best-fit curve can be used to smooth-out noise, uncertainty can be reduced appreciably, and measurements can be accurate to 1 ms. Measurements of and are based on time differences and their errors are probably about 5%. If the offsets had been longer, measured differences would have been larger, giving better accuracy, but then uncertainties in event continuity and interference with other events might have increased the errors. In calculating depths, the onset of reflections should be measured so time measurements are probably 20 ms (1 to 4%) late. This may introduce 1% error in , but other errors involved in are probably more important, including the assumption that it is the correct velocity to use. Dips are almost certainly underestimated by the use of , which does not allow for the fact that the velocity at the reflector is usually larger than because of the usual increase of velocity with depth.
5.18 Estimating lithology from stacking velocity
5.18a A velocity analysis at SP 100 of Figure 5.18a yields the plot shown in Figure 5.18b. Pick stacking velocity versus time pairs and calculate interval velocities.
Solution
Velocity-time pairs are listed in Table 5.18a. Depth and interval velocities . In Table 5.18a, values are for the intervals above the reflection picks.
(s) | (m/s) | (m) | (m/s) |
---|---|---|---|
0.22 | 2450 | 270 | 2450 |
0.38 | 2650 | 495 | 2800 |
0.58 | 2800 | 800 | 3050 |
0.72 | 2750 | 990 | 2715 |
1.12 | 3150 | 1765 | 3875 |
1.35 | 3600 | 2450 | 4890 |
1.58 | 4100 | 3280 | 7215 |
2.22 | 4550 | 5050 | 5530 |
The quality of Figure 5.18b does not permit accurate picking, but the NMO correction is fairly tolerant of errors. We did not tabulate time-velocity pairs for the events at (0.97, 2600), (1.48, 3000), (1.79, 3050), and (1.90, 3000), because we thought these events were multiples. Interpreters usually ignore stacking velocity values lower than those at shallower depths. Structure, faulting, and other features can distort velocity analyses.
5.18b What can you tell about the lithology from this?
Solution
The interval velocity values are plotted on Figure 5.18c with a smooth curve for Tertiary clastics plotted as a reference. The interval-velocity curve is everywhere above the reference, and above 1 km it is roughly parallel to but higher than the reference. The section above 1.8 km is probably mainly clastics whose velocity is increased, perhaps by age, cementation, uplift, or the presence of carbonates. The higher velocities below 1.8 km suggest carbonates. The velocity from 2.5 to 3.3 km is unreasonably high. While overlying reflections are mostly strong and continuous, data quality deteriorates here.
5.18c If the section that is present in the syncline but is missing over the anticline consists of poorly consolidated rocks, what values would you expect for a velocity analysis at SP 45?
Solution
Poorly consolidated rocks will have lower velocities than at SP 100 and this will lower the measured stacking velocities. Interval velocities between the same reflections, which will be deeper at SP 45, will probably be slightly higher than for the same intervals at SP 100.
5.18d Note the downdip thinning of the section to the left of SP 100 between 0.75 to 1.25 s. Suggest an explanation.
Solution
If the interval thickness remains constant, traveltime through it will decrease with increasing depth because the rocks have higher velocities as they are buried deeper and thus are under greater pressure.
5.19 Velocity versus depth from sonobuoy data
Determine velocity versus depth from Figure 5.19a assuming horizontal refractors. The direct wave that travels through the water (assume ) can be used to give source-receiver distances.
Background
A sonobuoy is a free-floating device that radios the outputs of hydrophones to a recording ship. The ship fires its sources as it sails away from the sonobuoy to achieve a refraction profile.
Solution
Figure 5.19b shows the picked events , , , and the waterbreak. This is an old profile where navigation was not as accurate as today. The direct-arrival waterbreak forms a distinct first arrival out to about SP 120 and an alignment with about the same slope can be seen from about SP 140 to SP 190, but it does not quite align with the water break seen at shorter offsets. The disagreement may merely indicate that the recording ship speed varied (note slight slope changes in the waterbreak alignment) and/or the sonobuoy drifted during the recording. We determine 48.5 m/SP or 11.8 km for the maximum offset. The first trace is 400 m from the sonobuoy.
Three distinct refraction (headwave) events can be seen: event , which gives the first breaks beyond SP 150; , which gives the first breaks between SP 120 and 150, and , which has a projected arrival time at SP 240 of about 6.4 s. When is a first break, its velocity is about 2.5 km/s but when it is a second arrival (problem 6.12), its velocity is about 2.9 km/s (the difference may be due to change of dip); we take its velocity as 2.7 km/s. Thus apparent velocities and intercept times for these events are about 5300, 2700, and 2400 m/s and 2.8, 1.7, and 1.4 s, respectively. Bearing in mind the distance uncertainties and timing errors (since first cycles are not clear enough for timing), we get crude answers only.
We get the depth of water by estimating for the sea-floor reflection; since , the water depth is about km.
Next we calculate depths to the refracting horizons using equation (4.18a) for , equation (4.18d) for and . The shallowest refractor is probably the top of the first consolidated rock, the material above it being unconsolidated sediments. We assume that the velocity in the sediments is close to that of water, so we get the depth to as follows:
We get the distance between and using the intercept time difference:
so the depth to is .
For event , we get
To get the distance from to ,
depth to = depth to .
The first reflection (from the sea floor) occurs at 1.45 s and a multiple of the seafloor reflection arrives at about 2.9 s; events below this multiple are so confused that interpretation cannot be done.
5.20 Influence of direction on velocity analyses
Given orthogonal dip and strike seismic lines, will velocity analyses at line intersections yield the same values?
Background
Reflecting points on a CMP gather in the dip direction are located updip by distances which increase with offset [see equation (4.11e)] so that a CMP stack smears the data. A DMO correction (Sheriff and Geldart, 1995, section 9.10.2) remedies this.
Solution
A line in the strike direction should yield reliable velocity values, but depths will be slant distances perpendicular to the bed, and reflecting points will be located updip. A commonmidpoint gather in the dip direction will not involve common reflecting points, and measurements will yield erroneous values unless DMO corrections have been made. The velocity analyses will probably yield somewhat different values.
5.21 Effect of time picks, NMO stretch, and datum choice on stacking velocity
5.21a Because velocity analysis is not made on the wavelet onset, how will this effect stacking-velocity values?
Background
A causal wavelet has zero amplitude for negative time values, that is, when .
A normal-moveout correction is subtracted from the arrival times of a reflection to compensate for the increase of raypath distance with offset. The normal-moveout equation (4.1c) has the factor in the denominator. While the value of is the same for all traces, it generally cannot be measured and the traveltime is used instead of . Also, the velocity usually increases with traveltime, and hence the correction is smaller than it should be. This effectively lowers the frequency, an effect called normal-moveout stretch.
Because of variable weathering (LVL) and the fact that the source and geophones are at various elevations, seismic traces are usually time shifted to effectively locate them on a horizontal datum surface below which conditions are assumed to be constant; this is called applying static corrections (see problem 8.18).
Solution
Equation (4.1a) gives for the velocity
Because the onset of a wave is not obvious, measured times are slightly larger than those associated with the reflector depths , and the calculated velocities will be slightly smaller.
5.21b What will be the effect of NMO stretch?
Solution
All parts of a wavelet should be corrected for the time and velocity that is appropriate for the wavelet onset, but instead they are corrected for a delayed time and the velocity appropriate to it. The undercorrection will increase with offset, making the measured velocity too low.
5.21c What will be the effect if the datum is appreciably removed from the surface?
Solution
The objective of corrections to a datum is to be able to treat the data as if the sources and geophones were all located at the same elevation and there are no horizontal velocity changes below the datum. It effectively corrects arrival times, but it does not change horizontal locations to account for the changes in datum depth. Because migration is often done assuming the same vertical velocity at all locations, errors in horizontal location will create errors in the migrated location of events. The datum should be sufficiently deep in the section so that horizontal velocity changes below the datum are so small that they do not affect the results. Note that corrections can be made for depths below the datum level so that deeper corrections do not imply a deep datum. In fact, the datum sometimes is above the ground level.
A datum that is appreciably deeper than the surface makes time measurements too small and calculated velocities too large. The horizontal component of raypaths is usually small near the surface but becomes larger as depths increase. Thus geophones actually located on the datum would involve smaller horizontal components of the distances to features that are not vertically below and thus create errors when the data are migrated.