# Waveshapes as hydrocarbon accumulation thickens

Series Geophysical References Series Problems in Exploration Seismology and their Solutions Lloyd P. Geldart and Robert E. Sheriff 10 367 - 414 http://dx.doi.org/10.1190/1.9781560801733 ISBN 9781560801153 SEG Online Store

## Problem 10.18a

10.18a Using the minimum-phase wavelet of problem 9.14a [11, 14, 5, –10, –12, –6, 3, 5, 2, 0, –1, –1, 0], determine the waveshapes for a gas-bearing sand encased in shale when the two-way traveltime through the water-sand is 12 ms, the traveltime through the gas-sand portion being successively 0, 2, 4, 6, 8, 10, and 12 ms. Plot the traces side-by-side shifted successively as would be the case for a horizontal gas-water contact. This illustrates a bright-spot–flat-spot situation. Take the reflection coefficients for shale to gas-sand as –0.10, gas-sand to water-sand as +0.15, and water-sand to shale as –0.05. Neglect small time differences because of velocity differences.

### Background

Minimum-phase and zero-phase wavelets are discussed in Sheriff and Geldart, 1995, section 9.4 and section 15.5.6, bright spot and flat spot in problem 10.17.

### Solution

A two-way traveltime of 12 ms (i.e., ${\displaystyle 6\Delta }$ for ${\displaystyle \Delta =2}$ ms where ${\displaystyle \Delta }$ is the sampling interval) corresponds to about 12 m at 2000 m/s velocity. Since we are only interested in the wavelet shape, we scale the reflection coefficients up to –2, +3, and –1 for the shale to gas-sand, gas to water-sand, and water-sand to shale, respectively. Where the water-filled portion of the reservoir thins to zero we have just two contacts, +1 and –1 at the reservoir top and bottom, and the reservoir is six samples thick. Thus,

 11, 14, 5, –10, –12, –6, 3, 5, 2, 0, –1, –1, 0 –11, –14, –5, 10, 12, 6, –3, –5, –2, 0 1, 1, 0 11, 14, 5, –10, –12, –6, –8, –9, –3, 10, 11, 5, –3, –5, –2, 0, 1, 1, 0.

Where we have 2-ms round-trip time in the reservoir, we have

 –22, –28, –10, 20, 24, 12, –6, –10, –4, 0, 2, 2, 0 33, 42, 15, –30, –36, –18, 9, 15, 6, 0, –3, –3, 0 –11, –14, –5, 10, 12, 6, –3, –5, –2, 0, 1, 1, 0 –22, 5, 32, 35, –6, –24, –35, –15, 6, 16, 14, 5, –6, –5, –2, 0, 1, 1, 0.

For 4-ms round-trip time in the reservoir,

 –22, –28, –10, 20, 24, 12, –6, –10, –4, 0, 2, 2, 0 33, 42, 15, –30, –36, –18, 9, 15, 6, 0, –3, –3, 0 –11, –14, –5, 10, 12, 6, –3, –5, –2, 0, 1, 1, 0 –22, –28, 23, 62, 39, –18, –53, –42, 0, 25, 20, 8, –6, –8, –2, 0, 1, 1, 0.

For 6-ms round-trip time in the reservoir,

 –22, –28, –10, 20, 24, 12, –6, –10, –4, 0, 2, 2, 0 33, 42, 15, –30, –36, –18, 9, 15, 6, 0, –3, –3, 0 –11 –14 –5 10 12 6 –3 –5 –2 0 1 1 0 –22 –28 –10 53 66 27 –47 –60 –27 19 29 14 –3 –8 –5 0 1 1 0.

For 8-ms round-trip time in the reservoir,

 –22, –28, –10, 20, 24, 12, –6, –10, –4, 0, 2, 2, 0 33, 42, 15, –30, –36, –18, 9, 15, 6, 0, –3, –3, 0 –11, –14, –5, 10, 12, 6, –3, –5, –2, 0, –1, –1, 0 –22, –28, –10, 20, 57, 54, –2, –54, –45, –8, 23, 25, 3, –5, –2, 0, –1, –1, 0.

For 10-ms round-trip time in the reservoir,

 –22, –28, –10, 20, 24, 12, –6, –10, –4, 0, 2, 2, 0 33, 42, 15, –30, –36, –18, 9, 15, 6, 0, –3, –3, 0 –11, –14, –5, 10, 12, 6, –3, –5, –2, 0, 1, 1, 0 –22, –28, –10, 20, 24, 45, 25, –9, –39, –26, –4, 17, 12, 1, –2, –3, –2, 1, 0.

Where the reservoir is completely gas filled, we have

 –22 –28 –10 20 24 12 –6 –10 –4 0 2 2 0 22 28 10 –20 –24 –12 6 10 4 0 –2 –2 0 –22 –28 –10 20 24 12 16 18 6 –20 –22 –10 6 10 4 0 –2 –2 0.

The results are plotted in Figure 10.18a the two outside curves are repetitions of the curves for gas-sand thicknesses of 0 and 12 ms. The first, fairly large trough indicates the top of the gas-filled reservoir and it changes to a peak where water fills the reservoir. The strong horizontal alignment indicates the flat spot and the dipping peak, which gets lost in the tail of the flat spot, the base of the reservoir.

## Problem 10.18b

Repeat using the zero-phase wavelet of problem 9.14e [1, 1, ${\displaystyle -1}$, ${\displaystyle -4}$, ${\displaystyle -6}$, ${\displaystyle -4}$, 10, 17, 10, ${\displaystyle -4}$, ${\displaystyle -6}$, ${\displaystyle -4}$, ${\displaystyle -1}$, 1, 1].

### Solution

Where the reservoir is completely filled with brine, the reservoir being six samples thick, we have:

 1, 1, –1, –4, –6, –4, 10, 17, 10, –4, –6, –4, –1, 1, 1 –1, –1, 1, 4, 6, 4, –10, –17, –10, 4, 6, 4, 1, –1, –1 1, 1, –1, –4, –6, –4, 9, 16, 11, 0, 0, 0, –11, –16, –9, 4, 6, 4, 1, –1, –1.

Where there is 2-ms round-trip time in the reservoir at the top of the structure, we have:

 –2, –2, 2, 8, 12, 8, –20, –34, –20, 8, 12, 8, 2, –2, –2 3, 3, –3, –12, –18, –12, 30, 51, 30, –12, –18, –12, –3, 3, 3 –1, –1, 1, 4, 6, 4, –10, –17, –10, 4, 6, 4, 1, –1, –1 –2, 1, 5, 5, 0, –10, –33, –5, 32, 42, 6, –6, –20, –22, –9, 7, 6, 4, 1, –1, –1.

Figure 10.18a.  Waveshapes as a function of hydrocarbon thickness for a minimum-phase wavelet.

For 4-ms round-trip time in the reservoir,

 –2, –2, 2, 8, 12, 8, –20, –34, –20, 8, 12, 8, 2, –2, –2 3, 3, –3, –12, –18, –12, 30, 51, 30, –12, –18, –12, –3, 3, 3 –1, –1, 1, 4, 6, 4, –10, –17, –10, 4, 6, 4, 1, –1, –1 –2, 1, 5, 11, 9, –4, –39, –47, 11, 63, 46, 0, –26, –31, –15, 7, 9, 4, 1, –1 –1.

For 6-ms round-trip time in the reservoir,

 –2 –2 2 8 12 8 –20 –34 –20 8 12 8 2 –2 –2 3 3 –3 –12 –18 –12 30 51 30 –12 –18 –12 –3 3 3 –1 –1 1 4 6 4 –10 –17 –10 4 6 4 1 –1 –1 –2 –2 2 11 15 5 –33 –53 –31 42 69 42 –20 –37 –24 1 9 7 1 –1 –1.

For 8-ms round-trip time in the reservoir,

 –2, –2, 2, 8, 12, 8, –20, –34, –20, 8, 12, 8, 2, –2, –2 3, 3, –3, –12, –18, –12, 30, 51, 30, –12, –18, –12, –3, 3, 3 –1, –1, 1, 4, 6, 4, –10, –17, –10, 4, 6, 4, 1, –1, –1 –2, –2, 2, 8, 15, 11, –24, –47, –37, 0, 48, 63, 22, –31, –30, –8, 3, 7, 4, –1, –1.
Figure 10.18b.  Waveshapes as a function of hydrocarbon thickness for a zero-phase wavelet.

For 10-ms round-trip time in the reservoir,

 –2, –2, 2, 8, 12, 8, –20, –34, –20, 8, 12, 8, 2, –2, –2 3, 3, –3, –12, –18, –12, 30, 51, 30, –12, –18, –12, –3, 3, 3 –1, –1, 1, 4, 6, 4, –10, –17, –10, 4, 6, 4, 1, –1, –1 –2, –2, 2, 8, 12, 11, –18, –38, –31, –6, 6, 42, 43, 11, –24, –16, –6, 1, 4, –2, –1.

If the reservoir is completely gas filled, we have

 –2, –2, 2, 8, 12, 8, –20, –34, –20, 8, 12, 8, 2, –2, –2 2, 2, –2, –8, –12, –8, 20, 34, 20, –8, –12, –8, –2, 2, 2 –2, –2, 2, 8, 12, 8, –18, –32, –22, 0, 0, 0, 22, 32, 18, –8, –12, –8, –2, 2, 2.

The results are plotted in Figure 10.18b. The evidences are much the same as in Figure 10.18a despite the change in waveshape.

Previous section Next section
Hydrocarbon indicators Salt lead time as a function of depth
Previous chapter Next chapter
Data processing Refraction methods