Accuracy of normal-moveout calculations

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Problem 4.1a

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


(4.1a)

Background

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

The difference between the traveltime for a receiver at the source and the traveltime for a receiver at an offset ( being the source-to-geophone separation) is called the normal moveout, . We can get an approximate value of by expanding equation (4.1a) in an infinite series to get , then subtracting .

An expression , , can be expanded as a binomial series [see, e.g., Sheriff and Geldart, 1995, equation (15.40)]:


(4.1b)

This series converges for and is infinite except when 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), being replaced by a suitable velocity such as the average velocity (see problem 4.13), the root-mean-square velocity (problem 4.13), or the stacking velocity (problem 5.12).

Figure 4.1a.  Geometry of NMO.

Solution

The virtual path equals and thus equals . Since is normal to the -axis, we have

i.e., the curve is an hyperbola. Taking the square root, we get equation (4.1a).

Problem 4.1b

Show that when the normal moveout is approximately


(4.1c)

Solution

We write equation (4.1a) as

If , we use equation (4.1b) to expand this expression to get the series


(4.1d)

Neglecting terms higher than (i.e., taking the first approximation) and noting that , we get


(4.1e)

Then, .

Problem 4.1c

Calculate the normal moveout for geophones 600, 1200, and 3600 m from the source for a reflection at s, given that the velocity km/s. What is the depth ?

Solution

From equation (4.1c) we have for m:

Because the NMO varies as , the value for m will be 4 times that for m, that is, 36 ms, and for m , ms. We have m.

Problem 4.1d

Typical uncertainties in measurements of , , and be 0.6 m, 0.2 km/s, and 5 ms. Calculate the corresponding uncertainty in . What do you conclude about the accuracy of calculations?

Solution

The uncertainty in is about 0.1%, that in is about 7%, and that in is about 0.2%. The uncertainties in and are 0.2% and 14%. Since the three factors are multiplied or divided to get , the uncertainties add so that the uncertainty in is about . This uncertainty is due mainly to the error in .

Problem 4.1e

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


(4.1f)

How much difference is there between and for and 3600 m? Taking into account the uncertainties in , , , when is this equation useful?

Solution

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

Equation (4.1f) reduces the uncertainty by . To affect the result significantly, this increase in accuracy should be at least 1%, i.e., or h. Accordingly, equation (4.1f) is useful when the offset is greater than about one-half the reflector depth. When m, the difference in is less than 1 ms but for m, the difference is 22 ms. However, the effect of velocity errors in real situations is usually more important than the approximation error.

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