# 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 ${\displaystyle V}$, the traveltime ${\displaystyle t}$ for the reflection path SCR in Figure 4.1a is

 {\displaystyle {\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 ${\displaystyle S}$ to the receiver ${\displaystyle R}$ after being reflected at ${\displaystyle C}$, the angle of incidence ${\displaystyle \alpha }$ being equal to the angle of reflection. The image point I (or virtual source) is the point on the perpendicular from ${\displaystyle S}$ to the reflector as far below the reflector as ${\displaystyle S}$ is above. The line IR is equivalent to the actual path SCR.

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

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

 {\displaystyle {\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 ${\displaystyle \left|\phi \right|<1}$ and is infinite except when ${\displaystyle n}$ 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), ${\displaystyle V}$ being replaced by a suitable velocity such as the average velocity ${\displaystyle {\bar {V}}}$ (see problem 4.13), the root-mean-square velocity ${\displaystyle V_{\rm {rms}}}$ (problem 4.13), or the stacking velocity ${\displaystyle V_{s}}$ (problem 5.12).

Figure 4.1a.  Geometry of NMO.

### Solution

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

{\displaystyle {\begin{aligned}(Vt)^{2}=x^{2}+(2h)^{2},\end{aligned}}}

i.e., the ${\displaystyle (t,x)}$ curve is an hyperbola. Taking the square root, we get equation (4.1a).

## Problem 4.1b

Show that when ${\displaystyle 2h>x}$ the normal moveout is approximately

 {\displaystyle {\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

{\displaystyle {\begin{aligned}t=\left(2h/V\right)[1+(x/2h)^{2}]^{\frac {1}{2}}.\end{aligned}}}

If ${\displaystyle 2h>x}$, we use equation (4.1b) to expand this expression to get the series

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

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

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

## Problem 4.1c

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

### Solution

From equation (4.1c) we have for ${\displaystyle x=600}$ m:

{\displaystyle {\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 ${\displaystyle x^{2}}$, the value for ${\displaystyle x=1200}$ m will be 4 times that for ${\displaystyle x=600}$ m, that is, 36 ms, and for ${\displaystyle x=3600}$ m ${\displaystyle (x\approx h)}$, ${\displaystyle \Delta t_{\rm {NMO}}=328}$ ms. We have ${\displaystyle h=(2.358/2)2.90=3420}$ m.

## Problem 4.1d

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

### Solution

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

## Problem 4.1e

4.1e Show that a more accurate normal moveout can be written

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

### Solution

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

{\displaystyle {\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}}}

{\displaystyle {\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 ${\displaystyle \Delta t_{\rm {NMO}}/2t_{0}=(x/4h)^{2}}$. To affect the result significantly, this increase in accuracy should be at least 1%, i.e., ${\displaystyle (x/4h)^{2}\geq 0.01}$ or ${\displaystyle x\geq 0.4}$ h. Accordingly, equation (4.1f) is useful when the offset ${\displaystyle x}$ is greater than about one-half the reflector depth. When ${\displaystyle x=1200}$ m, the difference in ${\displaystyle \Delta t_{\rm {NMO}}^{*}}$ is less than 1 ms but for ${\displaystyle x=3600}$ m, the difference is 22 ms. However, the effect of velocity errors in real situations is usually more important than the approximation error.