0-255 in stops?

Every stop is double so
256/2 = 128
128/2 = 64
etc..

8 stops
 
It's not actually that simple as it depends on the gamma curve as well - 64 might be half of 128 but in terms of luma probably won't be half as bright.

To confuse things further 16bit won't give you 16 stops either, just better tonality.
 
No - start with 256 and halve it 10 times, do you get below 0?
 
The problem is the question you asked in the first place, you're comparing apples with oranges and so the answer doesn't really make much sense.

In simple terms 8bits represent 8 stops - doubled each time but what's important is how the image data actually represents the brightness in a non-linear way.
 
8 'halving downs' only gets you to 2 - don't forget that past 1 there are quite a few stops as halved fractions are quite small.
 
ah its getting clearer in my head, so 0-255 is just a representation of what can be displayed on media and entirely different from actual dynamic range in any given scene?

Yes.

A 3D app I use actually uses 64bit floats per channel to represent colours, that's 192bit and 1.0 equates roughly to 255 in 8bit but it means there's no upper limit to brightness and very smooth tonality too (y)
 
About as many kettles as there are in a fish bun.

The numbers 0-255 are references for "darkest point" and "lightest point" respectively for an 8 bit system (such as jpgs). Most cameras have a range that's much greater but can't be displayed accurately so some range compression has to take place, but the image can still be under or over exposed.

Remember that f stops aren't a real measure of the amount of light coming through the lens, so you're trying to convert an analogue concept into a very small digital number.
 
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