Crop factor = ? x lens mm ?

I have it understood that the DX captor brings a 1.5x factor applied to the framing of the equivalent in mm of lenses rated in terms of 35mm film.
Doesn't sound that clear...
That is a 50mm lens will render an image in a frame equivalent to that of a 75 mm.
I would like to know what is the math behind that could also be applied to calculate the transformation when applying a crop to an existing image. I would like to express that in terms of pixels.
I'm trying to figure out how this would translate if I would crop a 10.2 mp image from the coming D200 to one that has 6.1, like the D100 or D70 have.
From Tom Hogan's site, I read concerning the D2x that it has: " A high-speed crop mode with a reduced angle of view of 2x from full frame (as opposed to the usual 1.5x) and reduced number of pixels (6.9mp from 12.2)."
6.9/12.2 = 0.57. Is it this .57 that is added to the 1.5 x factor to attain the 2x or so ?
If I will extract a 6.1 mp image from a 10.2 delivered in the D200, it would bring an additional factor of 6.1/10.2 = 0.6, thus having the framing factor of 2.1x.
That would mean that a 300mm shot in 10.2 mp but cropped to about 6 (with a resolution equal to that of the D100 or D70) would give a framing close to that of a 600 mm on a standard 35mm camera.

Is this pure nonsense or what ?

 

I agree Dang, that's why I talked about "framing" equivalent, and not optical. For example, obviously, cropping cannot change the look of the background (DOF and perspective). That said, is the math OK ?

 

Interestingly, if you take .75 horizontal times .75 vertical, you get .56 which applied to the 10.2 mp (a square value) brings you to 5.7 mp.

 

Right, x by y is what it takes to get the exact figure.