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?
nmo eq

 Silver Member
 Posts: 34
 Joined: Fri May 15, 2020 2:02 pm
nmo eq
 Attachments

 nmo eq 1
 nmoeq1.PNG (5.18 KiB) Viewed 898 times

 nmo eq 2
 nmoeq.PNG (2.88 KiB) Viewed 898 times
Re: nmo eq
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 prestack time migration unless you have very poor signal:noise or spatial sampling issues.
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 prestack time migration unless you have very poor signal:noise or spatial sampling issues.

 Silver Member
 Posts: 34
 Joined: Fri May 15, 2020 2:02 pm
Re: nmo eq
Why does eq 2 have NMO velocity? and i still didn't understand why there is 4h^2..
Hope you can help
Hope you can help
Re: nmo eq
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 twowaytime; 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.
That walks you through how DMO and NMO are combined when it comes to a dipping interface.
Remember we are working with events in twowaytime; 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.

 Silver Member
 Posts: 34
 Joined: Fri May 15, 2020 2:02 pm
Re: nmo eq
Sorry, english is not my first language so it is hard to understand the English.
Thank you for the explanation:)
Thank you for the explanation:)
Re: nmo eq
I'll keep that in mind, if I can.
It is hard when there is so much technical language at play.
It is hard when there is so much technical language at play.

 Silver Member
 Posts: 34
 Joined: Fri May 15, 2020 2:02 pm
Re: nmo eq
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
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
Re: nmo eq
In the picture, you have a right angled triangle formed by :How do you relate the NMO equation to Pythagorean theorem?
 half the source receiver distance; ie from the source to the mid point
 the vertical depth of the reflector below that midpoint
 the raypath 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 oneway travel time or twoway travel time. Seismic records are usually in twowaytime.
The bit from the SEG wiki and Yilmaz's book makes this clearer, I think:
Figure 3.11 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

 Similar Topics
 Replies
 Views
 Last post