Answer to: Is there an easier way to count fringes?
I synthesized an image to resemble your image with
x = linspace(0, 1, 256);
[x,y] = meshgrid(x,x);
image = sin((7*x + 2*y) * 2*pi) ./ (0.1 + (x - 0.5).^2 + (y - 0.6).^2);
Then I can get the vector the 2D FFT
A = abs(fft2(image));
A = A(1:end/2, 1:end/2)
[~, imax] = max(abs(A(:))) % 899
If the watermark is additive and static and, the content is smooth, and the vide long enough you could compute the watermark that minimizes $\sum ||S(F_i - W)||^2$, for the frames $F_i$ and a fixed watermark $W$. The solution $W$ is expected to converge to the watermark.
But it is always better to use the original video.