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Re: filmscanners: About 12 or 16 bits



English is not my native language and sometimes I have problems with the
exact meaning of the words. Putting the question with an example:

step 0 (12 bits) = step 0 (16 bits)
step 2000 (12 bits) = step 32000 (16 bits)
step 2001 (12 bits) = step 32016 (16 bits)
step 4096 (12 bits) = step 65536 (16 bits)

Is this correct? When I set the white and black points in the scanning
software, the scale between these points remain with 4096 (12 bits) steps or
the number of steps
available (not the number of steps effectively used from a particular image)
is lowered?

Mrio Teixeira
mjteixeira@yahoo.com


----- Original Message -----
From: "Austin Franklin" <darkroom@ix.netcom.com>
To: <filmscanners@halftone.co.uk>
Sent: Saturday, 27 October, 2001 4:07 PM
Subject: RE: filmscanners: About 12 or 16 bits


|
| > > > I
| > > > suppose that the white point is converted from step 2^12 in
| > the 12 bits
| > > > scale to 2^16 in the 16 bit scale and the intermediary steps are
just
| > > > interpolated (without any gain in the image quality). Am I right?
| > >
| > > Sorry, you're not right.  I believe the 12 bit data is raw data with
no
| > > setpoints at all...those have to be manually set in PS.  The 12
| > > data will be
| > > high bit justified in the 16 bit word, and no intermediate
| > values will be
| > > "interpolated" and there will be gaps between valid data points.
| > > These gaps are not relevant.
| >
| >   I think you're both right, but said it different ways.  IOW, 4096
values
| > in 12bits become 4096 values in 16bits, but '4095' is scaled to
| > '65535' and
| > all other values scaled accordingly.  Naturally, this would leave gaps.
|
| Yes, it does leave gaps...which is not what the original post says, it
says,
| and I quote, "the intermediary steps are just interpolated", which they
are
| not.  We did not say the same thing.
|


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