Dipping Layer Refraction Problem Moveout and Coincident Source-Receiver Format

Transcription

1 Environmental and Exploration Geophysics II Dipping Layer Refraction Problem Moveout and Coincident Source-Receiver Format tom.h.wilson Department of Geology and Geography West Virginia University Morgantown, WV The time distance (t-x) plot

2 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

3 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.

4 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?

5 We can now determine θ c and δ How do you determine V 2? V sinθ c = V 1 2 V 2 = V1 sinθ c θc = sin V1 1 V 1 ( V u ) + sin V sinθ c = V 1 2 ( V d )

6 Questions about problems 1-4? Some problems to consider: Problem set 4 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 downdip? Explain! (Robinson and Coruh, 1988)

7 Here is some shot data collected in Marshall Co. WV Enhanced display We d like to turn this into geology. Why do the amplitudes drop off below 200ms? How do we get here? The short story The effective source receiver geometry for the records shown at right across the east margin of the Rome Trough is corrected so that the source and receivers share the same surface location. But - this is not the way the data was collected.

8 Note that the reflection point coverage spans half the distance between the source and receiver The split spread provides symmetrical coverage about the source

9 Moveout and the moveout correction Δt Redefine the reflection time equal to the 0-offset arrival time (t 0 ) plus the Δt (drop from t 0 or moveout ).

10 Assume Δt 2 is small relative to other terms and can be ignored to approximate the moveout Δt is the normal moveout correction Look at the reflection time distance relationship in terms of t 2 versus x 2 Square both sides of this equation

11 The hyperbola becomes a straight line In the t 2 -x 2 form, the slope is 1/V 2

12 The moveout velocity V is derived from the slope of the reflection event as portrayed in the t 2 -x 2 plot. The derived velocity is referred to as the Normal Moveout Velocity, NMO velocity, or, just V NMO. The V NMO is used as a correction velocity 2 2 τ = t x 2 V 2 NMO If the velocity is accurately determined the corrected time τ equals t 0

13 Fun with hyperbolas and ellipses If the correction velocity (V NMO ) is too high then the correction is too small and we still have a hyperbola

14 If V < V then NMO < V V NMO And we have

15 NMO correction of the reflection events appearing in the shot records across relatively horizontal strata yields a more accurate image of subsurface geology.

16 These data sets are referred to as being single fold data. Single fold implies only one trace per mid-point. Single fold data are also sometimes referred to as 100% data Continue your work on problem sets 3 and 4 Hand in Exercises I-III and the Attenuation problem today Chapter 4, pages 165 to 199 and 206 to 229.