PROOF FOR 15-DEGREE PARTIAL NORMAL MOVEOUTS

Let us write

$${\cal T}_{i+1}=\sqrt{{\cal T}_i-{x^2 \over Nv^2}}$$ (15)

for the nontruncated case and  

$$\tilde{{\cal T}}_{i+1}=\tilde{{\cal T}}_i-{x^2 \over 2Nv^2\tilde{{\cal T}}_i}$$ (16)

for the truncated case. We will estimate the difference $(T^2_N-\tilde{T}^2_N)$and show that this difference tends to zero as $N \to \infty$.We can write

$$\begin{eqnarray} {\cal T}^2_{i+1}-\tilde{{\cal T}}^2_{i+1}&=&({\cal T}^2_i-{x^2 ... ...e{{\cal T}}^2_i)-{x^4 \over 4N^2v^4}{1 \over \tilde{{\cal T}}^2_i}\end{eqnarray}$$ (17) (18) (19)

As we did in the previous section, we can rewrite this equation into a system of equations:

$$\begin{eqnarray} {\cal T}^2_1-\tilde{{\cal T}}^2_1&=&({\cal T}^2_0-\tilde{{\cal ... ...}}^2_{N-1})-{x^4 \over 4N^2v^4}{1 \over \tilde{{\cal T}}^2_{N-1}} \end{eqnarray}$$ (20) (21) (22) (23)

By summing up all these equations we get  

$$T^2_N-\tilde{T}^2_N=-{x^4 \over 4N^2v^4}\sum_{i=0}^{N-1}{1 \over \tilde{{\cal T}}^2_i}$$ (24)

Now we need to prove that there exists a constant K independent of N such that

$${1 \over \tilde{{\cal T}}^2_i} < K$$ (25)

for all $i=0,1,\cdots,N-1$.From equation (16) we can see that

$$\tilde{{\cal T}}_{i} \gt \tilde{{\cal T}}_{i+1}$$ (26)

Similarly from equation (24) we can see that

$$\tilde{T}_N \gt T_N$$ (27)

Putting both equations together we have

$$\tilde{{\cal T}}_1\gt\tilde{{\cal T}}_2\gt\cdots \gt\tilde{T}_N\gt T_N$$ (28)

or

$${1 \over \tilde{{\cal T}}^2_i} < {1 \over T^2_N}= {1 \over {t^2-x^2/v^2}}$$ (29)

where the last equation follows from equation (14). If we define

$$K={1 \over {t^2-x^2/v^2}}$$ (30)

then from equation (24) we have  

$$\vert T^2_N -\tilde{T}^2_N \vert < {x^4 \over 4N^2v^4} NK={x^4 K \over 4Nv^4}$$ (31)

which tends to zero as $N \to \infty$.This proves that the N^th partial normal moveout applied N times is equal to the normal moveout for large N.

The convergence of cascaded NMO to a semicircle can be clearly seen on Figure a. The proof of convergence holds for $x/v \neq t$. The behavior for x/v = t can be seen from sequences at the sides of Figure b. The figure convinces us that there is convergence even for x/v=t, but that it is extremely slow.