# Weathering corrections and dip/depth calculations

Series Geophysical References Series Problems in Exploration Seismology and their Solutions Lloyd P. Geldart and Robert E. Sheriff 4 79 - 140 http://dx.doi.org/10.1190/1.9781560801733 ISBN 9781560801153 SEG Online Store

## Problem 4.16a

Figure 4.16a shows part of a seismic record where the geophone group spacing is 50 m and the offset to the nearest group is 50 m and that to the far group 600 m. What is the apparent velocity of the first breaks?

### Background

The weathered layer or low-velocity layer (LVL) is a shallow layer that usually extends from the surface downward for 4 to 50 m and is characterized by low velocity of the order of 250 to 1000 m/s. Often the base of the LVL is near to or coincident with the water table, and many of the pore spaces in the LVL are filled with air. The LVL is important because of the high absorption in it and because its low (and frequently quite variable) velocity has considerable effect on the traveltime. The velocity change at the base of the LVL is usually large, making it a good reflector and therefore important in the generation of multiples (see problem 3.8) and in P- to S-wave conversion. The large velocity change at its base also bends raypaths from below the LVL so that they are nearly vertical within the LVL regardless of their direction below the LVL.

The first energy from a source to arrive at the geophone groups is called the first break. When the source is below the base of the LVL, as in Figure 4.16b, first-break raypaths are almost parallel to the base of the LVL until they are refracted upwards at approximately the critical angle $\theta _{c}$ (see problem 4.18).Writing $V_{W}$ and $V_{H}$ for the velocities in the LVL and just below it, $\sin \theta _{c}=V_{W}/V_{H}$ .

### Solution

The slope of the first breaks is about 0.200 s/600 m, or a velocity of 3000 m/s.

## Problem 4.16b

Assuming that the source is just below the base of the LVL and that the LVL velocity is 500 m/s, how thick is the LVL?

### Solution

The uphole time is about 31 ms, so the depth of the base of the LVL is $D_{w}=500\times 0.031=16$ m.

## Problem 4.16c

Arrival times at the sourcepoint for two reflections are given as 0.475 and 0.778 s; what is the average velocity to these reflectors?

### Solution

The normal moveout equation [see equation (4.1c)] is $\Delta t_{\rm {NMO}}=x^{2}/2V^{2}t_{0}$ , so $V=(x^{2}/2t_{0}\Delta t_{\rm {NMO}})^{1/2}$ . The NMO measure gives about 0.040 ms/600m = 0.067 ms/km for the 0.475 s reflection and about 0.037 ms/600m = 0.062 ms/km for the 0.778 s reflection; this gives velocities of 2.4 km/s and 1.9 km/s, respectively.

## Problem 4.16d

For these reflections, the arrival-time differences between the far traces in opposite directions from the source point are given as $+0.005$ s for both reflections. What are the reflector dips?

### Solution

For dip moveouts of 0.005 s/600m = 0.0083 s/km, equation (4.2b) gives for the 0.475 s reflection, $\sin \xi =(2.4/2)\times 0.0083=0.010$ , $\xi =0.6^{\circ }$ . For the deeper reflection we get $\sin \xi =(1.9/2)\times 0.0083$ , $\xi =0.5^{\circ }$ .

## Problem 4.16e

What is the dominant frequency of these reflections (approximately)?

### Solution

Counting the number of cycles in 0.1 s at the arrival times of the two reflections, we get approximately 80 and 60 Hz.