Environmental and Exploration Geophysics II

Tom Wilson, Department of Geology and Geography

Environmental and Exploration Geophysics II

[email protected]

Department of Geology and GeographyWest Virginia University

Morgantown, WV

Layer-to-Thin EffectLayer-to-Thin EffectDipping Layer RefractionsDipping Layer Refractions


C

Snell’s law for multiple layers

1 2

1 2 3

sin sin 1

V V V


The velocity triangle


Expressing trig functions in terms of velocities


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 …


• 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


As x gets larger and larger the reflection from the base of layer 2 and the refraction across the top converge



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


How would you determine the thickness of layer 2 (h2)?

• From reflection arrivals

• From refraction events?


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


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


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.






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.


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




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?



V3=15000f/s


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.



V3=15000f/s


Dipping layer refraction event


Down-dip critical refraction events take longer to arrive at a given offset. This leads to an apparent velocity less than the actual velocity.


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


Some board work for reference


Note that the slant path lengths have the same form as for the horizontal layer


The main difference is that there is a distinction between the up-dip and down-dip

layer depths and slant path lengths


The interface path length also has a familiar look to it accept for the xsintanc


Travel time = distance/velocity = SPL/V1+IPL/V2


Brush the dust off some trig identities

To get another straight line


The time distance (t-x) plot


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


Repeat derivation for the up-dip direction


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.”



Note that the subscript d or u consistently refers to the location of the source as downdip or updip, respectively.


From an “intuitive” perspective - in which direction will the refraction arrivals come in earlier?



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


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.


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


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


Spend time looking over the foregoing problem sets.

We will discuss these assignments further this coming Wednesday.

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Environmental and Exploration Geophysics II

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