1. Environmental and Exploration Geophysics II tom.h.wilson wilson@geo.wvu.edu Department of Geology and Geography West Virginia University Morgantown, WV Layer-to-Thin Effect Dipping Layer Refractions

6. 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.

7. The refraction travel times plotted below were computed for the model at right h 1 =10 feet V 1 =4000f/s h2=30feet V 2 =8000f/s V 3 =15000f/s Layer-to-Thin

8. h 1 =10 feet V 1 =4000f/s h2=20feet V 2 =8000f/s V 3 =15000f/s

9. h 1 =10 feet V 1 =4000f/s h2=10feet V 2 =8000f/s V 3 =15000f/s The record appears to have only one refraction with time-intercept = 0.0069 seconds. What depth would be calculated for that refractor?

10. h 1 =10 feet V 1 =4000f/s h2=10feet V 2 =8000f/s V 3 =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.

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

28. How do you determine V 2 ? Also recall When  is small sin  ~ 

29. This reduces further to The above can serve as a useful approximation of V 2 obtained directly from measurement s of the apparent velocities. The approximation is not too bad. For example - 20 o is 0.349 radians and the sin (20 o ) =0.342 and 30 o is 0.52 radians with sin 30 o = 0.5. We often have large velocity contrasts from alluvia into bedrock so that critical angles are often 30 o or less.

30. 1. A reversed seismic refraction survey indicates that a layer with velocity V 1 lies above another layer with velocity V 2 and that V 2 >V 1. 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 V 1 and V 2 layers dip? i.e. where is down-dip? Explain! (Robinson and Coruh, 1988) 2. Suppose that a reversed refraction survey indicated velocities V 1 =1500 m/s and V 2 2500 m/s from one end, and V 1 =1500 m/s and V 2 =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) Homework Problems

31. Spend time looking over the foregoing problems. We will discuss the assignment further next week.