nmo eq

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Tingpingbing
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nmo eq

Post by Tingpingbing »

Why do the nmo eq go from "nmo eq 1" to "nmo eq 2"? (see figure text)
x = offset and h= half offset.
why is there a 4 and not a 2?


:?:
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GuyM
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Re: nmo eq

Post by GuyM »

You are confusing NMO (equation 1) and DMO (equation 2)

https://wiki.seg.org/wiki/NMO_for_a_flat_reflector
https://wiki.seg.org/wiki/Principles_of ... correction

DMO is a partial prestack time migration with a weak velocity dependence, that corrects the velocity for dip and reducing CMP smearing associated with dip. It's much more common to use pre-stack time migration unless you have very poor signal:noise or spatial sampling issues.

Tingpingbing
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Re: nmo eq

Post by Tingpingbing »

Why does eq 2 have NMO velocity? and i still didn't understand why there is 4h^2..

Hope you can help

GuyM
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Re: nmo eq

Post by GuyM »

Did you read the second link?

That walks you through how DMO and NMO are combined when it comes to a dipping interface.

Remember we are working with events in two-way-time; if you want to consider angles, then you have to be dealing with distances (or depths), not times. Think about the method of dimensions - to have time on both sides of the equation and distance term (2h) MUST be divided by a velocity.

Tingpingbing
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Re: nmo eq

Post by Tingpingbing »

Sorry, english is not my first language so it is hard to understand the English.
Thank you for the explanation:)

GuyM
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Re: nmo eq

Post by GuyM »

I'll keep that in mind, if I can.

It is hard when there is so much technical language at play.

Tingpingbing
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Re: nmo eq

Post by Tingpingbing »

Hey again,

How do you relate the NMO equation to Pythagorean theorem?

how does it go from equation 3 to equation 4?
On equation 4 --> why isn't it x/2?
Why do they consider the whole offset and not half of it only in the theorem?
Is it because we have to use the whole offset since we have to find the traveltime of the wave from source to the reflector and then back to the receiver?

Also why is the vtx divided by 2 for each sides of the wave on the figure?


Do the curvature moveout and dip moveout mean the same thing?


https://seabed.software.slb.com/velocit ... locity.htm
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GuyM
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Re: nmo eq

Post by GuyM »

How do you relate the NMO equation to Pythagorean theorem?
In the picture, you have a right angled triangle formed by :

- half the source receiver distance; ie from the source to the mid point
- the vertical depth of the reflector below that midpoint
- the ray-path from the source to the mind point, which is the hypotenuse of the triangle

Now in this case they are, I think, being a bit sloppy about defining those terms, and expressing whether the times are in one-way travel time or two-way travel time. Seismic records are usually in two-way-time.

The bit from the SEG wiki and Yilmaz's book makes this clearer, I think:
Figure 3.1-1 shows the simple case of a single horizontal layer. At a given midpoint location M, we want to compute the reflection traveltime t along the raypath from shot position S to depth point D then back to receiver position G. Using the Pythagorean theorem, the traveltime equation as a function of offset is


{\displaystyle t^{2}=t_{0}^{2}+{\frac {x^{2}}{v^{2}}},}{\displaystyle t^{2}=t_{0}^{2}+{\frac {x^{2}}{v^{2}}},} (1)
where x is the distance (offset) between the source and receiver positions, v is the velocity of the medium above the reflecting interface, and t0 is twice the traveltime along the vertical path MD

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