If this has been explained somewhere else, please just point me to a link. I'd like to understand a comment I have been seeing on the main rumors section when people talk about the d800. How its ISO in FX would be equivalent to a higher ISO in DX.
Ehh, define noise. ;)
Not to derail the question off the bat, but a clear definition of noise is crucial to understanding the question I believe you are asking. Think of a 24MP D3X and a 12MP D300. Use both cameras equipped with equivalent field of view lenses, let's say a 35mm lens on the D3x and a 24mm lens on the D300. Take the same picture, print the same size.
Now you have two photos, both 8"x10" let's pretend. One is made of 24 MP, one is made of 12 MP.
It is not hard to visualize how the photo with twice as many pixels could have significantly more noise per pixel than the photo with less pixels, yet still have less noise per picture. Because each pixel is that much smaller, and therefore less significant to the overall picture.
So what is noise? ;)
FX is a larger sensor, so I would think that it would need more ISO to have equivalent noise (or lack thereof) than a smaller sensor. Or am I missing something?
Sensor size does not matter in the pixel noise game. Individual photosite size matters. Historically this has been indistinguishable from sensor size as Nikon has never made a FX sensor with photosites as small as their DX sensors. The D800, if it comes as rumored, however will basically have D7000-sized photosites. Meaning pixel-noise should be pretty similar between the two (assuming similar sensor technology). Picture-noise should be lower on the D800, however as the larger sensor (again think of the two photos I described before) allows more light, more signal, per photo than the DX one would.