Post on 03-Jan-2016
description
Tom Wilson, Department of Geology and Geography
Environmental and Exploration Geophysics II
tom.h.wilsontom.wilson@mail.wvu.edu
Department of Geology and GeographyWest Virginia University
Morgantown, WV
Layer-to-Thin EffectLayer-to-Thin EffectDipping Layer RefractionsDipping Layer Refractions
Tom Wilson, Department of Geology and Geography
C
Snell’s law for multiple layers
1 2
1 2 3
sin sin 1
V V V
Tom Wilson, Department of Geology and Geography
The velocity triangle
Tom Wilson, Department of Geology and Geography
Expressing trig functions in terms of velocities
Tom Wilson, Department of Geology and Geography
2 2 2 21 23 1 3 2
3 3 1 3 2
2 2time=
h hxV V V V
V VV VV
21 1
3 1 2
2 cos2 costime= chhx
V V V
3
1dx
dt V
The end result, where 2 2
3 11
3
cosV V
V
and …
Tom Wilson, Department of Geology and Geography
• What is the critical distance
• What is the relationship of the reflection from the base of layer 2 to the critical refraction from the top of layer 2
Tom Wilson, Department of Geology and Geography
As x gets larger and larger the reflection from the base of layer 2 and the refraction across the top converge
Tom Wilson, Department of Geology and Geography
Tom Wilson, Department of Geology and Geography
In the three layer problem the number of possible terms that could potentially be measured directly from the shot record includes -
•V1, V2 and V3
•two reflection time intercepts
•two refraction time intercepts
•one crossover distance, and
•two critical distances
Tom Wilson, Department of Geology and Geography
How would you determine the thickness of layer 2 (h2)?
• From reflection arrivals
• From refraction events?
Tom Wilson, Department of Geology and Geography
What variables can be determined from an analysis of the shot record
V1, V2, V3, ti1, & ti2, where the ti’s refer to the reflection time intercepts
2 2 11 11 2 &
2 2i ii
V t tV th h
Tom Wilson, Department of Geology and Geography
What variables can be determined from an analysis of the shot record
V1, V2, V3, ti1, & ti2, where the ti’s refer to the refraction time intercepts
1 21 12 2
2 1
2 23 2 12 2 3 12 2
3 13 2
& 2
2
2
i
i
VVh t
V V
VV hh t V V
VVV V
Tom Wilson, Department of Geology and Geography
1) 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?
2) Another assumption we have made here is that the refraction from the top of the second 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.
Tom Wilson, Department of Geology and Geography
Tom Wilson, Department of Geology and Geography
Tom Wilson, Department of Geology and Geography
Tom Wilson, Department of Geology and Geography
Tom Wilson, Department of Geology and Geography
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 the higher velocity of 15000fps.
Tom Wilson, Department of Geology and Geography
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
Layer-to-Thin
Tom Wilson, Department of Geology and Geography
h1=10 feetV1=4000f/s
h2=20feetV2 =8000f/s
V3=15000f/s
Tom Wilson, Department of Geology and Geography
The record appears to have only one refraction with time-intercept = 0.0069 seconds. What depth would be calculated for that refractor?
h1=10 feetV1=4000f/s
h2=10feetV2 =8000f/s
V3=15000f/s
Tom Wilson, Department of Geology and Geography
itVV
VVh
21
23
311
2
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.
h1=10 feetV1=4000f/s
h2=10feetV2 =8000f/s
V3=15000f/s
Tom Wilson, Department of Geology and Geography
Dipping layer refraction event
Tom Wilson, Department of Geology and Geography
Down-dip critical refraction events take longer to arrive at a given offset. This leads to an apparent velocity less than the actual velocity.
Tom Wilson, Department of Geology and Geography
h, the thickness of the layer is defined as the distance from a point on the surface to the dipping interface along a line drawn normal to the interface.
Note that our h is the text’s j
Tom Wilson, Department of Geology and Geography
Some board work for reference
Tom Wilson, Department of Geology and Geography
Note that the slant path lengths have the same form as for the horizontal layer
Tom Wilson, Department of Geology and Geography
The main difference is that there is a distinction between the up-dip and down-dip
layer depths and slant path lengths
Tom Wilson, Department of Geology and Geography
The interface path length also has a familiar look to it accept for the xsintanc
Tom Wilson, Department of Geology and Geography
Travel time = distance/velocity = SPL/V1+IPL/V2
Tom Wilson, Department of Geology and Geography
Brush the dust off some trig identities
To get another straight line
Tom Wilson, Department of Geology and Geography
The time distance (t-x) plot
Tom Wilson, Department of Geology and Geography
Determination of layer properties in the dipping layer case requires shots in the down-dip and up-dip directions
Up-dip shot
Down-dip shot
Tom Wilson, Department of Geology and Geography
Repeat derivation for the up-dip direction
Tom Wilson, Department of Geology and Geography
The t-x plot gets a little more complicated and includes the combining the responses in the up-dip and down-dip direction. Assuming there is no
knowledge of dip these directions are simply referred to as “forward” and “reverse.”
Tom Wilson, Department of Geology and Geography
Tom Wilson, Department of Geology and Geography
Note that the subscript d or u consistently refers to the location of the source as downdip or updip, respectively.
Tom Wilson, Department of Geology and Geography
From an “intuitive” perspective - in which direction will the refraction arrivals come in earlier?
Tom Wilson, Department of Geology and Geography
Tom Wilson, Department of Geology and Geography
How do you determine V2?
2
1sinV
Vc
c
VV
sin1
2
)(sin)(sin
2
1 1111
duc V
V
V
VFirst get c
When is small sin ~
)(sin)(sin
2
1sinsin 1111
duc V
V
V
V
du V
V
V
V
V
V 11
2
1
2
1
Tom Wilson, Department of Geology and Geography
du V
V
V
V
V
V 11
2
1
2
1
ud
ud
VV
VVV 22
This reduces further to
The above can serve as a useful approximation of V2 obtained directly from measurement s of the apparent velocities. The approximation is not too bad. For example - 20o is 0.349 radians and the sin (20o) =0.342 and 30o is 0.52 radians with sin 30o = 0.5. We often have large velocity contrasts from alluvia into bedrock so that critical angles are often 30o or less.
Tom Wilson, Department of Geology and Geography
1. A reversed seismic refraction survey indicates that a layer with velocity V1 lies above another layer with velocity V2 and that V2>V1. We examine the travel times at a point located midway (at C) between the shotpoints (at A and B). The travel time of the refracted ray from end A to midpoint C is less than the travel time of the refracted wave from end B to midpoint C. Show that the apparent velocity determined from the slope of the travel time curve for refracted waves produced from the source at A is less that the apparent velocity for refracted waves produced from the source at B. Toward which end of the layout does the boundary between the V1 and V2 layers dip? i.e. where is down-dip? Explain! (Robinson and Coruh, 1988)
2. Suppose that a reversed refraction survey indicated velocities V1 =1500 m/s and V2 2500 m/s from one end, and V1 =1500 m/s and V2=3250 m/s from the other. Find the dip of the refractor. What would be the changes in velocities if the refractor had a slope 10 degrees larger than the one you computed? (Robinson and Coruh, 1988)
Some problems to consider: Problem set 4
Tom Wilson, Department of Geology and Geography
Please read through Chapter 3, pages 95 to top of 114.
Chapter 4, pages 149 to 164 (as assigned previously.
• Hand in Exercises I-III this Wednesday• Attenuation problem is also due on Wednesday
• Look over problems in problem sets 3 and 4
Tom Wilson, Department of Geology and Geography
Spend time looking over the foregoing problem sets.
We will discuss these assignments further this coming Wednesday.