# Weathering and elevation (near-surface) corrections

Series Geophysical References Series Problems in Exploration Seismology and their Solutions Lloyd P. Geldart and Robert E. Sheriff 8 253 - 294 http://dx.doi.org/10.1190/1.9781560801733 ISBN 9781560801153 SEG Online Store

## Problem 8.18a

Show that when the source is below the LVL, weathering and elevation corrections for a geophone at the source are given by

{\begin{aligned}\Delta t_{0}=2\left(E_{S}-D_{S}-E_{D}\right)/V_{H}+t_{uh},\end{aligned}} where $E_{S}$ and $E_{D}$ are the elevations of the source point and datum, $D_{S}$ is the depth of the source, and $t_{uh}$ is the uphole time.

### Background

Corrections are necessary to eliminate the effect of changes in the elevation of the surface and in the thickness of and velocity in the LVL. The corrections in effect reduce the traveltimes to those that would be observed if the source and geophones were located on a reference datum, usually a horizontal plane below the base of the low-velocity layer. These corrections are called static corrections because they are the same for all reflections regardless of their arrival times.

### Solution

In Figure 8.18a, the correction to the traveltime for a wave going from the source at $A$ down to the datum is

 {\begin{aligned}\Delta t_{S}=\left(E_{S}-D_{S}-E_{D}\right)/V_{H}.\end{aligned}} (8.18a)

The correction for travel from the datum up to a geophone at the source point is

 {\begin{aligned}\Delta t_{g}=\Delta t_{S}+t_{uh},\end{aligned}} (8.18b)

so the total correction for the traveltime for a geophone at the source point is

 {\begin{aligned}\Delta t_{0}=2\left(E_{S}-D_{S}-E_{D}\right)/V_{H}+t_{uh}.\end{aligned}} (8.18c)

## Problem 8.18b

If the split spread $ACB$ in Figure 8.18a is used to find the dip, what correction must be applied to the dip moveout?

### Solution

The dip moveout is obtained by subtracting traveltimes at sources $A$ and $B$ in Figure 8.18b. If traveltimes have not been corrected for weathering and elevation, the dip moveout must be corrected; this is the differential weathering correction $\Delta t_{d}$ . Using equation (8.18b) we have

 {\begin{aligned}\Delta t_{d}=(\Delta t_{g})_{B}-(\Delta t_{g})_{A}=(\Delta t_{S}+t_{uh})_{B}-(\Delta t_{S}+t_{uh})_{A}.\end{aligned}} (8.18d)

## Problem 8.18c

Derive an expression to correct the traveltime for a geophone at $C$ in Figure 8.18b $A$ and $B$ being source points.

### Solution

We use the first-break traveltimes $t_{AC}$ and $t_{BC}$ that correspond to the paths $A'C''C$ and $B'C'C$ . The velocity contrast at the base of the LVL is usually large enough that the paths $C'C$ and $C''C$ are so close to vertical that the distance $C'C''$ is very small. Hence the sum $\left(t_{AC}+t_{BC}\right)$ is given by

 {\begin{aligned}\left(t_{AC}+t_{BC}\right)\approx \left(A'B'\right)/V_{H}+2t_{w},\\{\mbox{so}}\qquad \qquad t_{w}\approx {\frac {1}{2}}[\left(t_{AC}+t_{BC}\right)-(AB/V_{H})],\end{aligned}} (8.18e)

where $t_{w}$ is the traveltime through the LVL at $C$ , and $D_{W}=V_{W}t_{w}$ . Therefore the correction that effectively places the geophone at $C$ on the datum is

 {\begin{aligned}\Delta t_{C}=t_{w}+\left(E_{C}-D_{W}-E_{D}\right)/V_{H}.\end{aligned}} (8.18f)

We must add to this the correction that locates the source on the datum, namely $\Delta t_{S}$ , given by equation (8.18a). Thus the total correction for a traveltime recorded at $C$ is

 {\begin{aligned}\Delta t_{C}=t_{w}+\left(E_{C}-D_{W}-E_{D}\right)/V_{H}+\Delta t_{S}.\end{aligned}} (8.18g)

## Problem 8.18d

The weathering and elevation corrections given by equations (8.18a) to (8.18g) assume that the source is below the base of the LVL. What changes are required if the source is within the LVL?

### Solution

When the source is within the LVL, the wave traveling down to the datum is in the LVL for the distance $(D_{W}-D_{S}$ , hence equation (8.18a) is changed to

 {\begin{aligned}\Delta t_{S}&=\left(D_{W}-D_{S}\right)/V_{W}+\left(E_{S}-D_{W}-E_{D}\right)/V_{H}\\&=\left(E_{S}-D_{W}-E_{D}\right)/V_{H}+\Gamma _{S},\end{aligned}} (8.18h)

where $\Gamma _{S}=\left(D_{W}-D_{S}\right)/V_{W}$ . Equation (8.18b) is unchanged provided we use the value of $\Delta t_{S}$ in equation (8.18h). Equation (8.18c) becomes

 {\begin{aligned}\Delta t_{0}&=\left(D_{W}-D_{S}\right)/V_{W}+2\left(E_{S}-D_{W}-E_{D}\right)/V_{H}+\left[\left(D_{W}-D_{S}\right)/V_{W}+t_{uh}\right]\\&=2\Gamma _{S}+2\left(E_{S}-D_{W}-E_{D}\right)/V_{H}]+t_{uh}.\end{aligned}} (8.18i)

Equation (8.18d) is unchanged provided we use equation (8.18h) for $\Delta t_{S}$ . Equation (8.18e) becomes

{\begin{aligned}\left(t_{AG}+t_{BG}\right)=AB/V_{H}+2t_{w}+\left(\Gamma _{A}+\Gamma _{B}\right),\end{aligned}} where $\Gamma _{A}=\Gamma _{SA}$ , $\Gamma _{B}=\Gamma _{SB}$ . Thus, equation (8.18e) becomes

 {\begin{aligned}t_{w}={\frac {1}{2}}[\left(t_{AG}+t_{BG}\right)-AB/V_{H}-(\Gamma _{A}+\Gamma _{B})].\end{aligned}} (8.18j)

Equations (8.18f,g) are unchanged except that we must use the values of $t_{w}$ and $\Delta t_{S}$ from equations (8.18j) and (8.18h).