Download - Introduction to Geophysics
Introduction to Geophysics
Ali [email protected].
saDepartment of Earth SciencesKFUPM
Thin Layer EffectThin Layer EffectDipping Layer RefractionsDipping Layer Refractions
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Review: Refraction InversionIn
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Some pitfall should be considered-
Low Velocity layer (V2<V1<V3):
We have assumed that our layers have successively higher and higher velocity.
What happens if we have a velocity inversion - let’s say V2 is less than V1 and V3?
Thin layer (V3>V2>V1 but h2 very small): Another assumption we have made here is that the refraction from the top of the third layer, for example, will actually show itself, and not get buried somewhere beneath the earlier refraction and reflections.
This can happen if the 2nd layer is too thin.
Problem SituationsIn
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Fig. 4.10 of Lillie
leads to an over-estimation of the depth to the underlying
Velocity Inversion-Hidden Layers
•No head waves are generated at a boundary where velocity decreases undetectable “hidden” layer
•Only one critical refraction appears in the time-distance plot
The presence of the velocity inversion delays the refraction from interface 2 and leads us to overestimate its depth. In addition, we have entirely missed the presence of the second layer. We overestimate thickness because we incorrectly assume that the refraction event traveled down to the refractor with a single velocity of 15000fps. In
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Velocity Inversion-Hidden Layers
Thin layers may be undetectable, if rays traveling to deeper levels arrive first due the V2-V3 velocity contrast and the thinness of the V2-layer.
Hidden Zones: Thin Layer Problem
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Hidden Zones: Thin Layer Problem
The refraction travel times plotted below were computed for the model at right
h1=10 feetV1=4000f/s
h2=30feetV2 =8000f/s
V3=15000f/s
Hidden Zones: Thin Layer ProblemIn
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h1=10 feetV1=4000f/s
h2=20feetV2 =8000f/s
V3=15000f/s
Hidden Zones: Thin Layer ProblemIn
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h1=10 feetV1=4000f/s
h2=10feetV2 =8000f/s
V3=15000f/s
The record appears to have only one refraction with time-intercept = 0.0069 seconds. What depth would be calculated for that refractor?
Hidden Zones: Thin Layer ProblemIn
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itVV
VVh
21
23
311
2
h1=10 feetV1=4000f/s
h2=10feetV2 =8000f/s
V3=15000f/s
In this case we estimate the depth to the 15000 f/s refractor to be approximately 14.4 feet. We underestimate the depth because the seismic wave did not spend its time traveling only at the 4000 f/s velocity. It sped through the second layer at 8000 f/s thus reducing the time intercept and thus our estimate of h.
As was the case for velocity inversion, we have again missed an entire layer.
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Fig 4.13 of Lillie
Single Dipping Refractor Surface
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t1d ≠ t1u
TAB = TBA
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t1d = T-axis intercept when shooting downdip (f rom A to B)
= T-axis intercept when shooting updip (f rom B to A)t1u
t1d = T-axis intercept when shooting downdip (f rom A to B)
= T-axis intercept when shooting updip (f rom B to A)t1u
=Travel time f rom shoot at A to receiver at BTAB
=Travel time f rom shoot at B to receiver at ATBA
=Travel time f rom shoot at A to receiver at BTAB
=Travel time f rom shoot at B to receiver at ATBA
dd tVV
VVh 12
122
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2
uu tVV
VVh 12
122
21
2
For next time read over pages 73-99 in Chapter 4
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