I thought the density range related to the film in some way, and
scanner density would correspond more to the value of the MSB (most
significant bit), not the number of bits? Perhaps your analogy to
speedometer is saying the same thing. You can have 16 bits but if the
16 bits only covers a narrow density range, you will not see all the
dynamic range in the film. Is the scanner density then the maximum
light to dark range before clipping? Of course, more bits in that range
will give better tonal quality (signal to noise), but not more range.
Correct me if I am wrong. Or is the density range also a signal-
to-noise (signal-to-quantization) number? To me, 12 bits with more
density range (no clipping) is better than 16 bits with clipping.
So much discussion has been given to ppi information but relatively
little to the Z-axis information on film. I have used the LS-2000 with
slide feeder a lot, but I don't use it with paper slide mounts, only
plastic mounts. Paper mounts tend to catch, at least old paper mounts
do.
Some old kodachromes from 1955 from my Dad that I am trying to
preserve, the LS 2000 still does not pull out the shadow information
that is there. On these slide, there is just not any more information
beyond 2700 ppi (lens and technique limits), but there certainly is
more shadow and highlight detail that is being clipped in the scanner.
See for example: where
the eyes shaded by the hat are being clipped. Trying to archive a lot
of old film is a lot of work. You can also see evidence of fungus
removed by the ice on the side of the building. Ice saves a lot of
work, but doesn't always work with kodachrome.
There are a lot of considerations to archiving film digitally, 4000ppi
is more than adequate for me, but density and SNR are more important
and are not as well specified as ppi. I would rather have a 3000ppi
scanner that had 12-16 bits that covered the slide density, and with
enough bits that I can later adjust the levels in photoshop. When the
histogram hits both ends, I know I am missing information.
Wayne
At 09:27 AM 5/9/2002 +0200, you wrote:
>> but I do think that it was a calculation
>> as to how that 4.8 was arrived as a
>> theoretical 16 bit value.
>
>log10(2^16)=4.81
>
>The catch is that this reflects only the theoretical maximum range that can
>be represented in the 16-bit output of the scanner; it doesn't guarantee
>that the scanner actually manages to extract that range from negatives or
>slides. It's rather like saying that if you have a digital speedometer on
>your car with four digits, your car's maximum speed is 9999 miles per hour.
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