[filmscanners] RE: scanner dmax discussion

["Austin Franklin" <austin@darkroom.com>]

Hi David,

> > 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,

I didn't imply anything...my statement was simply how things actually work.
The assumption was, that 8 bits was sufficient for the system defined.  YOU
can pick the number of bits, I don't care, but the point was increasing the
number of bits over what is required does not increase the density range or
the dynamic range, period.

> 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.)

IF the number of bits was insufficient, in the first place, to discern down
to noise, then additional bits would increase the density range (and dynamic
range as well).  On the low end, you would be able to discern a value 1/2
the value you previously were able to discern, whether it's logarithmic or
linear.

> 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.

I believe you missed my point...  The number of bits CAN change the density
range, of course, but only if it isn't noise you are detecting.  The
assumption in my statement, obviously, was that 8 bits was sufficient to
discern down to noise.

> 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.

ONLY if the number of bits was insufficient in the first place.

> 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.

Er, right...I know that.  That WAS the point.  You made your assumption in
the wrong direction, you simply could have qualified what I said, or simply
asked for clarification.

Austin

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