Before that let us understand why are we looking for a correction in our seismic data. Let us consider a situation in which we are obtaining a seismic profile for "horizontal beds", and we are having the array of source and geophones. (as shown in the figure)
Now as soon as there will be blast, the waves would be emitted from it and a part of wave would reflect back and reach to the geophone G0. Let the time recorded by this geophone be T0, and similarly let the time recorded by G1, G2 and G3 be T1, T2 and T3 respectively. One thing is clear that the T0 will be less than T1, because G1 is placed @ a greater distance than G0 w.r.t. the source. So, we can say that T0<T1<T2<T3 and so on (keep one thing in mind that this is only valid for those reflecting surfaces which are horizontal so at this point there should be no debate over this issue). What actually the debatable issue is...how much is the change in time that we are getting ? T1 - To will give some ▲t T1 - To will give some ▲t T2 - To will give some other ▲t' (▲t and ▲t' wont be in linear relationship) AND THIS CHANGE IN TIME WHICH WE ARE GETTING IS CALLED MOVE OUT Now all we need to do is find out these ▲t (general move out term). So, for this purpose we first need to understand the relationship of Time(tx) and Distance(X) from source (offset), and it will be hyperbola.The equation for the hyperbolic relation betn tx and X will be : Let, X = Distance from source '"or" offset V = Velocity of wave tx = Travel time for a wave towards X distance to = Travel time for a wave at X=0 "or" at zero offset
(tx)² = (to)² + (X/V)²
(tx)² = (to)² [ 1 + (X/V*to)² ]( taking to common )
(tx) = (to) [ 1 + (X/V*to)² ]^½
Expanding the term we get :
(tx) = (to) [1 + ½(X/V*to)² + (higher power terms)......... ]
Ignoring higher power terms we get :
(tx) = (to) + (X²/V²*to)( to is cancelled in second term )
(tx) - (to) = (X²/V²*to) where, (tx) - (to) is nothing but T1 - To, T2 - To and so on...
Example:
So our NMO correction is :
▲t = (X²/V²*to) where, ▲t is the required NMO correction.
Thus, we can say that ever point on the hyperbola is incremented by ▲t, with respect to it's preceding point. i.e. to = 0.5 then t1 = 0.5 + ▲t (where ▲t = (X²/V²*to) = (1²/V²*0.5) = (2/V²)
Now as soon as there will be blast, the waves would be emitted from it and a part of wave would reflect back and reach to the geophone G0. Let the time recorded by this geophone be T0, and similarly let the time recorded by G1, G2 and G3 be T1, T2 and T3 respectively. One thing is clear that the T0 will be less than T1, because G1 is placed @ a greater distance than G0 w.r.t. the source. So, we can say that T0<T1<T2<T3 and so on (keep one thing in mind that this is only valid for those reflecting surfaces which are horizontal so at this point there should be no debate over this issue). What actually the debatable issue is...how much is the change in time that we are getting ? T1 - To will give some ▲t T1 - To will give some ▲t T2 - To will give some other ▲t' (▲t and ▲t' wont be in linear relationship) AND THIS CHANGE IN TIME WHICH WE ARE GETTING IS CALLED MOVE OUT Now all we need to do is find out these ▲t (general move out term). So, for this purpose we first need to understand the relationship of Time(tx) and Distance(X) from source (offset), and it will be hyperbola.The equation for the hyperbolic relation betn tx and X will be : Let, X = Distance from source '"or" offset V = Velocity of wave tx = Travel time for a wave towards X distance to = Travel time for a wave at X=0 "or" at zero offset
(tx)² = (to)² + (X/V)²
(tx)² = (to)² [ 1 + (X/V*to)² ]( taking to common )
(tx) = (to) [ 1 + (X/V*to)² ]^½
Expanding the term we get :
(tx) = (to) [1 + ½(X/V*to)² + (higher power terms)......... ]
Ignoring higher power terms we get :
(tx) = (to) + (X²/V²*to)( to is cancelled in second term )
(tx) - (to) = (X²/V²*to) where, (tx) - (to) is nothing but T1 - To, T2 - To and so on...
Example:
So our NMO correction is :
▲t = (X²/V²*to) where, ▲t is the required NMO correction.
Thus, we can say that ever point on the hyperbola is incremented by ▲t, with respect to it's preceding point. i.e. to = 0.5 then t1 = 0.5 + ▲t (where ▲t = (X²/V²*to) = (1²/V²*0.5) = (2/V²)
No comments:
Post a Comment