[filmscanners] Re: scanner dmax discussion

["David J. Littleboy" <davidjl@gol.com>]

"Austin Franklin" <austin@darkroom.com>

> Hmm. Since the density range is defined logarithmically, you only get the
> same density range if the noise in the CCD output signal is
> 1/256th of 6V or
> greater. If the noise is less, you get a larger dynamic range
> with the extra
> bits.

Agreed, but why does it matter if it's logarithmic or not?
<<<<<<<<<<<<<<<

Because if it's linear, as you implied, additional bits doen't increase the
density range, but if it's logarithmic, additional bits do increase the
density range (assuming noise less than the additional bits.)

You wrote:
>>>>>>>>
If you have an 8 bit A/D, you'll get values 0-255.  Say your input to the
A/D is -3V to +3V.  1 will represent 1/256th of 6V, and 255 will represent
+3V.

Now, you have a 16 bit A/D....SAME CCD.  You get values 0-65535.  Same A/D
voltage input, let's say.  1 will represent 1/65536th of 6V, and 65535 will
represent +3V...but...note, nothing has changed in the density range, the
CCD is still detecting the same overall density range.
<<<<<<<<<<

But "density range" is defined not as absolute values, but as largest value
over smallest value, so changing the number of bits _does_ change the
density range.

So while the CCD is still detecting the same overall density range (that's
defined by the noise), the scanner is _reporting_ a different density range.
When the number of bits exceed the CCD-detected density range, the low order
bits will be noise, and when the number of bits is insufficient, information
is lost and the reported density range is lower than the CCD measured.

David J. Littleboy
davidjl@gol.com
Tokyo, Japan

----------------------------------------------------------------------------------------
Unsubscribe by mail to listserver@halftone.co.uk, with 'unsubscribe 
filmscanners'
or 'unsubscribe filmscanners_digest' (as appropriate) in the message title or 
body