Stacking velocity versus rms and average velocities

ADVERTISEMENT
From SEG Wiki
Jump to: navigation, search

Problem 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,


(5.12a)

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, .

Figure 5.12a.  Model of 300-m-thick layers.

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.

Table 5.12a. Calculation of , , and
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.
Table 5.12b. Calculated values of , and
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.


Table 5.12c. Calculation of raypaths for dipping layers.
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.


Table 5.12d. Calculation of stacking velocity (m/s).
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.

Problem 5.12b

Repeat part (a) for the case where rays make angles of incidence at the free surface of , , and .

Solution

Figure 5.12b.  Dipping model.

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 .

Problem 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.

Figure 5.12c.  Geometry of problem 5.12c.

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.

Continue reading

Previous section Next section
Horizontal component of head waves Quick-look velocity analysis and effects of errors
Previous chapter Next chapter
Geometry of seismic waves Characteristics of seismic events

Table of Contents (book)

Also in this chapter

External links

find literature about
Stacking velocity versus rms and average velocities
SEG button search.png Datapages button.png GeoScienceWorld button.png OnePetro button.png Schlumberger button.png Google button.png AGI button.png