# Accuracy of normal-moveout 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 4.1a

Show that, for constant velocity $V$ , the traveltime $t$ for the reflection path SCR in Figure 4.1a is

 {\begin{aligned}t=\left(1/V\right)(x^{2}+4h^{2})^{1/2}.\end{aligned}} (4.1a)

### Background

Where the velocity is constant, raypaths are straight lines. In Figure 4.1a, a wave travels from the source $S$ to the receiver $R$ after being reflected at $C$ , the angle of incidence $\alpha$ being equal to the angle of reflection. The image point I (or virtual source) is the point on the perpendicular from $S$ to the reflector as far below the reflector as $S$ is above. The line IR is equivalent to the actual path SCR.

The difference between the traveltime $t_{0}$ for a receiver at the source $S$ and the traveltime $t$ for a receiver at an offset $x$ ($x$ being the source-to-geophone separation) is called the normal moveout, $\Delta t_{\rm {NMO}}$ . We can get an approximate value of $\Delta t_{\rm {NMO}}$ by expanding equation (4.1a) in an infinite series to get $t$ , then subtracting $t_{0}$ .

An expression $(1+\phi )^{n}$ , $\left|\phi \right|<1$ , can be expanded as a binomial series [see, e.g., Sheriff and Geldart, 1995, equation (15.40)]:

 {\begin{aligned}(1+\phi )^{n}&=1+n\phi +{\frac {n\left(n-1\right)}{2!}}\phi ^{2}+{\frac {n\left(n-1\right)\left(n-2\right)}{3!}}\phi ^{3}+\cdots \\&\quad +{\frac {n\left(n-1\right)\ldots \left(n-r+1\right)}{r!}}\phi ^{r}+\cdots .\end{aligned}} (4.1b)

This series converges for $\left|\phi \right|<1$ and is infinite except when $n$ is a positive integer.

While the velocity is assumed to be constant in this problem, equations such as (4.1 a,c) are also used when the velocity varies (usually increasing with depth), $V$ being replaced by a suitable velocity such as the average velocity ${\bar {V}}$ (see problem 4.13), the root-mean-square velocity $V_{\rm {rms}}$ (problem 4.13), or the stacking velocity $V_{s}$ (problem 5.12).

### Solution

The virtual path $IR$ equals $SCR$ and thus equals $Vt$ . Since $SI$ is normal to the $x$ -axis, we have

{\begin{aligned}(Vt)^{2}=x^{2}+(2h)^{2},\end{aligned}} i.e., the $(t,x)$ curve is an hyperbola. Taking the square root, we get equation (4.1a).

## Problem 4.1b

Show that when $2h>x$ the normal moveout is approximately

 {\begin{aligned}\Delta t_{\rm {NMO}}\approx x^{2}/2V^{2}t_{0}\approx x^{2}/4Vh.\end{aligned}} (4.1c)

### Solution

We write equation (4.1a) as

{\begin{aligned}t=\left(2h/V\right)[1+(x/2h)^{2}]^{\frac {1}{2}}.\end{aligned}} If $2h>x$ , we use equation (4.1b) to expand this expression to get the series

 {\begin{aligned}t={\frac {2h}{V}}\left\{1+{\frac {1}{2}}\left({\frac {x}{2h}}\right)^{2}+\left[{\frac {1}{2}}\left(-{\frac {1}{2}}\right){\frac {1}{2!}}\left({\frac {x}{2h}}\right)^{4}\right]+\cdots \right\}.\end{aligned}} (4.1d)

Neglecting terms higher than $(x/2h)^{2}$ (i.e., taking the first approximation) and noting that $t_{0}=2h/V$ , we get

 {\begin{aligned}t\approx t_{0}\left[1+{\frac {1}{2}}(x/2h)^{2}\right].\end{aligned}} (4.1e)

Then, $\Delta t_{\rm {NMO}}=t-t_{0}\approx t_{0}x^{2}/8h^{2}=x^{2}/2V^{2}t_{0}=x^{2}/4Vh$ .

## Problem 4.1c

Calculate the normal moveout $\Delta t_{\rm {NMO}}$ for geophones 600, 1200, and 3600 m from the source for a reflection at $t_{0}=2.358$ s, given that the velocity ${}=2.90$ km/s. What is the depth $h$ ?

### Solution

From equation (4.1c) we have for $x=600$ m:

{\begin{aligned}\Delta t_{\rm {NMO}}=x^{2}/2V^{2}t_{0}=0.600^{2}/\left(2\times 2.90^{2}\times 2.358\right)=9.10\ {\rm {ms}}=9\ {\rm {ms}}.\end{aligned}} Because the NMO varies as $x^{2}$ , the value for $x=1200$ m will be 4 times that for $x=600$ m, that is, 36 ms, and for $x=3600$ m $(x\approx h)$ , $\Delta t_{\rm {NMO}}=328$ ms. We have $h=(2.358/2)2.90=3420$ m.

## Problem 4.1d

Typical uncertainties in measurements of $x$ , $V$ , and $t_{0}$ be 0.6 m, 0.2 km/s, and 5 ms. Calculate the corresponding uncertainty in $\Delta t_{\rm {NMO}}$ . What do you conclude about the accuracy of $\Delta t_{\rm {NMO}}$ calculations?

### Solution

The uncertainty in $x$ is about 0.1%, that in $V$ is about 7%, and that in $t_{0}$ is about 0.2%. The uncertainties in $x^{2}$ and $V^{2}$ are 0.2% and 14%. Since the three factors are multiplied or divided to get $\Delta t_{\rm {NMO}}$ , the uncertainties add so that the uncertainty in $\Delta t_{\rm {NMO}}$ is about $14.4\%\approx 14\%$ . This uncertainty is due mainly to the error in $V$ .

## Problem 4.1e

Show that a more accurate normal moveout can be written

 {\begin{aligned}\Delta t_{\rm {NMO}}^{*}\approx \Delta t_{\rm {NMO}}\left(1-\Delta t_{\rm {NMO}}/2t_{0}\right).\end{aligned}} (4.1f)

How much difference is there between $\Delta t_{\rm {NMO}}^{*}$ and $\Delta t_{\rm {NMO}}$ for $x=1200$ and 3600 m? Taking into account the uncertainties in $x$ , $V$ , $t_{0}$ , when is this equation useful?

### Solution

If we go to the second approximation in the expansion of equation (4.1d), equation (4.1e) becomes

{\begin{aligned}t=t_{0}\left[1+{\frac {1}{2}}\left({\frac {x}{2h}}\right)^{2}-{\frac {1}{8}}\left({\frac {x}{2h}}\right)^{4}\right].\end{aligned}} {\begin{aligned}{\hbox{Then,}}\qquad \Delta t_{\rm {NMO}}^{*}&=t-t_{0}={\frac {t_{0}}{2}}\left({\frac {x}{2h}}\right)^{2}-{\frac {t_{0}}{8}}\left({\frac {x}{2h}}\right)^{4}\\&={\frac {t_{0}}{2}}\left({\frac {x}{2h}}\right)^{2}\left[1-{\frac {1}{4}}\left({\frac {x}{2h}}\right)^{2}\right]=\Delta t_{\rm {NMO}}\left[1-{\frac {1}{4}}\left({\frac {x}{2h}}\right)^{2}\right]\\&=\Delta t_{\rm {NMO}}\left[1-\left({\frac {x^{2}}{4Vh}}\right)\left({\frac {V}{4h}}\right)\right]=\Delta t_{\rm {NMO}}\left(1-\Delta t_{\rm {NMO}}/2t_{0}\right).\end{aligned}} Equation (4.1f) reduces the uncertainty by $\Delta t_{\rm {NMO}}/2t_{0}=(x/4h)^{2}$ . To affect the result significantly, this increase in accuracy should be at least 1%, i.e., $(x/4h)^{2}\geq 0.01$ or $x\geq 0.4$ h. Accordingly, equation (4.1f) is useful when the offset $x$ is greater than about one-half the reflector depth. When $x=1200$ m, the difference in $\Delta t_{\rm {NMO}}^{*}$ is less than 1 ms but for $x=3600$ m, the difference is 22 ms. However, the effect of velocity errors in real situations is usually more important than the approximation error.