# Blondeau weathering 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

The Blondeau method of making weathering corrections (Musgrave and Bratton, 1967, 231−246) is useful in areas where, because of appreciable compaction within the low-velocity layer, the velocity is given approximately by the equation

 {\begin{aligned}V=az^{1/n};\end{aligned}} (8.21a)

$a$ and $n$ being constants, $n>+1.$ The Blondeau method starts with a curve of first breaks versus offset [Figure 8.21a(ii)] plotted on log-log graph paper; the curve is approximately a straight line with slope $B=\left(1-1/n\right)$ (see Musgrave and Bratton, 1967, 244).

To remove the effect of a surface layer of thickness $z_{m}$ , we find $F$ , a tabulated function of $B$ . Then $x=Fz_{m}$ . Next we use the $x-t$ plot to find the corresponding $t$ . Finally, $t_{v}$ , the vertical traveltime to the depth $z_{m}$ , is given by $t_{v}=t/F$ .

Verify the Blondeau procedure by deriving these relations:

1. $z=z_{m}\sin ^{n}i$ , where $i$ is the angle of incidence measured with respect to the vertical at depth $z$ ;
2. $x=Fz_{m}$ , where $F=2n\mathop {\int } \nolimits _{0}^{\pi /2}\sin ^{n}i\ {\hbox{d}}i=$ function of $n$ , hence also of $B$ ;
3. $t=\left(G/a\right)(x/F)^{B}$ , where $G=2n\mathop {\int } \limits _{0}^{\pi /2}{\hbox{sin}}^{n-2}i\ {\hbox{d}}i$ ;

[Note that $G$ can be obtained from the table for $F$ by writing $m=n-2$ , finding for $F$ , and multiplying the value by $\left(m+2\right)/m=n/(n-2$ 4. ${\hbox{d}}x/{\hbox{d}}t$ , the horizontal component of the apparent velocity at any point $\left(x_{j},\;z_{j}\right)$ of the trajectory, is $V_{m}=x/Bt$ ;
5. $t_{v}=\mathop {\int } \nolimits _{0}^{t_{m}}{\hbox{d}}z/V=t/F$ .

### Solution

i) Note that the quantities $x$ and $t$ are the offset and traveltime for the point of emergence. For intermediate points we write $\left(x_{j},\;t_{j},\;z_{j},\;V_{j},\;i_{j}\right)$ except for the deepest point, where we have $x_{m}=x/2,$ $t_{m}=t/2,$ $z_{j}=z_{m}$ , $V_{j}=V_{m}$ , $i_{m}=\pi /2$ .

Solving equation (8.21a) for $z_{j}$ gives

 {\begin{aligned}z_{j}=(V_{j}/a)^{n}.\end{aligned}} (8.21b)

From Snell’s law we have

 {\begin{aligned}\left(\sin t_{j}/V_{j}\right)=1/V_{m}.\end{aligned}} (8.21c)

Thus, using equations (8.21b,c) we get

 {\begin{aligned}\left(z_{j}/z_{m}\right)=(V_{j}/V_{m})^{n}={\sin }^{n}i_{j}.\end{aligned}} (8.21d)

ii) From equations (4.17d), we get

{\begin{aligned}x_{j}-\int _{0}^{z_{j}}\tan i\ \mathrm {d} z=\int _{0}^{z_{i}}(\tan i)[nz_{m}\sin ^{(n-1)}i\cos i]\mathrm {d} i=nz_{m}\int _{0}^{i_{j}}\sin ^{n}i\ \mathrm {d} i,\end{aligned}} where we have differentiated equation (8.21d) to replace $z$ with $i$ . If we integrate from 0 to $\pi /2$ and multiply by 2, the result is

{\begin{aligned}x=2x_{m}=2nz_{m}\int \limits _{0}^{\pi /2}{\sin }^{n}i\ \mathrm {d} i=Fz_{m},\end{aligned}} {\begin{aligned}{\mbox{where}}\qquad \qquad F=2n\int _{0}^{\pi /2}\sin ^{n}i\ \mathrm {d} i.\end{aligned}} (8.21e)

iii) From Figure 4.17a we have $\Delta t=\Delta z/V\cos i$ , so

{\begin{aligned}t_{j}=\int _{0}^{z_{j}}\mathrm {d} z/V\cos i=\left(nz_{m}\right)\int _{0}^{i_{j}}{\frac {\sin ^{\left(n-1\right)}i}{V}}\mathrm {d} i,\end{aligned}} where equation (8.21d) was used to replace $z$ with $i$ . Using equation (8.21c) to eliminate $V$ , we get

{\begin{aligned}t_{j}=(nz_{m}/V_{m})\int _{0}^{i_{j}}\,\sin ^{(n-2)}i\ \mathrm {d} i.\end{aligned}} Integration from 0 to $\pi /2$ gives $t_{m}$ , so

{\begin{aligned}t=2t_{m}=\left(2nz_{m}/V_{m}\right)\int _{0}^{\pi /2}\sin ^{\left(n-2\right)}i\ \mathrm {d} i=\left(z_{m}/V_{m}\right)G.\end{aligned}} Since $V_{m}=az_{m}^{1/n}$ from equations (8.21a) and $z_{m}=x/F$ from equation (8.21e), this result can be written

 {\begin{aligned}t=(G/a)(x/F)^{B}.\end{aligned}} (8.21f)

iv) Solving equation (8.21f) for $x$ gives

{\begin{aligned}x=F(at/G)^{1/B};\end{aligned}} {\begin{aligned}{\mbox{then}}\qquad \qquad {\frac {\mathrm {d} x}{\mathrm {d} t}}=\left(F/B\right)(a/G)^{1/B}t^{\left(1/B-1\right)}.\end{aligned}} (8.2lg)

Equation (8.21f) can be used to eliminate $\left(a/G\right)$ , so equation (8.21g) becomes

 {\begin{aligned}{\frac {\mathrm {d} x}{\mathrm {d} t}}=\left(F/B\right)t^{-1/B}\left(x/F\right)t^{\left(1/B-1\right)}=x/Bt.\end{aligned}} (8.21h)

The angle between the wavefront and a horizontal line equals the angle between a ray and the vertical, both angles being the angle of incidence (see Figure 4.2c), so

{\begin{aligned}\sin i_{j}=\left(V_{j}\Delta t_{j}/\Delta x_{j}\right)\end{aligned}} [compare with equation (4.2d)]. Therefore,

{\begin{aligned}\left(\Delta t_{j}/\Delta x_{j}\right)=\sin i_{j}/V_{j}=1/V_{m}\end{aligned}} using Snell’s law. Thus the apparent velocity $\left(\Delta x_{j}/\Delta t_{j}\right)=V_{m}$ ; since this is a constant, it holds at every point of the trajectory, including the point of emergence. Therefore, from equation (8.21h),

 {\begin{aligned}V_{m}=\mathrm {d} x/\mathrm {d} t=x/Bt.\end{aligned}} (8.21i)

v) We have $t_{v}=$ time to travel vertically from the surface to the depth $z_{m}$ , Thus,

{\begin{aligned}t_{v}=\int _{0}^{z_{m}}dz/V=\int _{0}^{z_{m}}\mathrm {d} z/az^{1/n}={\frac {z_{m}^{\left(1-1/n\right)}}{a\left(1-1/n\right)}}={\frac {z_{m}}{aBz_{m}^{1/n}}}={\frac {z_{m}}{BV_{m}}},\end{aligned}} where we have used equation (8.21a) in the first and last steps. From equations (8.21e) and (8.21i) we have $z_{m}=x/F$ and $V_{m}=x/Bt$ , so

 {\begin{aligned}t_{v}=t/F.\end{aligned}} (8.21j)