RE: filmscanners: Nikon Scan & VS Negative dynamic range

["Austin Franklin" <darkroom@ix.netcom.com>]

> Tony writes ...
>
> > On Wed, 5 Sep 2001 23:29:34 -0400  Austin Franklin
> > (darkroom@ix.netcom.com) wrote:
> >
> > > Mapping the input data into 8 bit values has nothing to do with the
> > > dynamic range of the A/D data
> >
> > Arrgh... not this again! :-)
> >
> > Actually it does, ...
> > ...
>
>   Correct!!! ...

This is a very common misunderstanding of how the scanner system works.
Read what I said carefully.  What the 8 bit data represents is entirely
different (in a dynamic range sense) than the data out of the A/D.  If you
don't understand why, I am happy to attempt to explain it.

First realize that the data from the A/D is 12 bits...which REPRESENTS
integer density ratio values of 1-4096 (a ratio has to start at 1:1).  8
bits represents a NUMBER from 0-255.  Tell me how you believe the 12 bit
integer density ratio value from the A/D gets "mapped" into an 8 bit number?

> "linearity" is the keyword, that is, the scanner's driver
> cannot map the CCD to RGB data non-linearily without losing information.

The scanner's driver "typically" has nothing to do with mapping the data out
of the CCD.  That is typically handled in the scanner hardware and
firmware...which is entirely independent of the driver.  Only in a raw scan
do you get the direct A/D data, which the scanner driver doesn't "map"
either.

> The only method, without losing info, is to map the linear CCD to greater
> RGB depth.  And if you take a the math a bit further you realize
> 12bit depth
> is all that is required to cover the most extreme optical densities

That is only by happenstance.  The 12 bits ONLY represent INTEGER density
ratio values.  If the ratio was 1024.5:1, then 12 bits could not represent
it.  It was an arbitrary decision (not a bad one) and it is simply
representation...but is not "cast in stone" and the only way to do it.