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[filmscanners] RE: Density vs Dynamic range
> > Plain and simple, do you agree that a dynamic range of 5000:1 REQUIRES
> > 13
> > bits to represent every integer value between 1 and 5000? If so, then
> > where's the problem? If not, then plain and simple, why not?
> >
> > Austin
> >
Hi Peter,
Sorry if I sounded a bit surely in my last response. I'll try to straighten
you out without getting surely again ;)
> I think where we differ is the assumption the a 5000:1 dynamic range
> yields 5000 discrete integer values.
Well, it does, that's in and of what a "5000:1" dynamic range means!
> Light intensity is a linear value so
> that 5000:1 range can be divided into as many steps as you'd like to use.
Absolutely not true. The 1 is the noise floor, and you can not discern
below the noise floor, and you can only discern in increments OF the noise
floor. The noise floor IS the increment.
> It is in any case only a ratio and therefore has no units or integer
> values. If we double the number of bits (possible values) that doesn't
> increase the dynamic range of the scanner, only it's ability to represent
> accurately the value coming from the CCD.
No, increasing the number of bits does not increase the dynamic range of the
scanner, NOR does it increase the ability to represent accurately the value
coming from the CCD, providing you already have enough bits TO represent the
full dynamic range of the scanner. Assuming you already have enough bits TO
represent the full dynamic range of the scanner, adding more bits simply
means more bits are "in the noise".
If you believe what you say in the above paragraph, then I'd have to say you
do not understand what dynamic range is, and what it means. If you don't
increase the dynamic range out of the scanner, more bits will NOT give you
more usable data...as I said, they will simply be "in the noise".
> If not, then your statement is still wrong, or at best, an
> oversimplification.
It's (my initial statement you disagree with) hardly an oversimplification,
in fact, it is about as complete and accurate as you can get. It is a plain
and simple fact that a dynamic range of 5000:1 requires 13 bits to represent
that full dynamic range. Again, 1 is the first discernable signal (as in
the noise floor), and since you can only measure in increments OF the noise
floor...a dynamic range of 5000:1 simply means you can have every integer
value from 1 to 5000...and to represent every integer value from 1 to 5000,
you need 13 bits.
> I'm sure you'll agree that it's the noise floor and the saturation level
> that define the true dynamic range.
Yes, and if you know that than I am miffed you say that my statement is
wrong, because it says the EXACT same thing. You can only differentiate
values that are different by the level of the noise....and it appears you
somehow believe you can differentiate to a finer degree than the noise,
which you can not.
Simply put, dynamic range is the (max value  min value) divided by noise.
That is the maximum number of differentiable values you can get. You can't
measure half the noise, because your unit of measure IS noise.
Regards,
Austin

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