Digital BW, The Print

Austin,

I'm going to take a different approach here. The point/counter-point has got
my head spinning. We've spanned too many posts for me to put it all
together. Clearly you possess an understanding of this I don't, but often,
probably due to the sprawling nature of these discussions, you appear to me
to be contradicting yourself. Perhaps you don't but there are too many
little pieces of info, spanning too many posts, for me to tie it together.

BTW, when I accused you of taking sides of an argument I only meant that we
were each arguing A CASE, in the dispassionate sense. I never felt either of
us were being quarrelsome.

I'm starting from scratch. Below is a draft of a question I will post
elsewhere . I hope it will help you see what I'm left with from our
discussion. I'm thinking we might both benefit from seeing how the topic is
addressed from other's perspective. If you'd like to discuss any of it's
points that'd be fine, or perhaps you could just help me craft it better as
a question.

********

What determines where the endpoints of a RAW highbit scans (no tonal
manipulation by the driver or image editor, and no profile assigned) get
written to Photoshop's histogram in 16-bit mode? In other words, why does it
appear to be of such low dynamic range, and bunched into the dark end of the
histogram? If you scan a chrome which has a dynamic range of 3.7 on a
scanner which has a true dynamic range of 3.7, why wouldn't the data span
the histogram? It would still be bunched up.

This is a cerebral pursuit for me, I just want to be sure I'm
conceptualizing this properly. Often people say "it's because you haven't
set your end points, or corrected it's gamma". Yes, I understand about
toning the data, but my question is, what determines where that RAW data
will sit in the histogram prior to toning?

I've heard a couple of different explanations, but they are contradictory or
incomplete. 

One premise suggests that as the bit depth of the capture device increases
so will it's scans occupy a larger portion of the histogram. IOW, a 10-bit
capture would contain less tones than a 14-bit capture, and thus would
occupy a smaller portion of the histogram. This suggests that dynamic range
is directly a function of bit depth, which I don't believe to be the case
(though someone always throws in a snippet about A/D converters, which only
confuses me).

Another theory proposes exactly the opposite of that, though it still tries
to tie dynamic range to bit depth. It suggests that the reason the tones are
compressed is that it is a representation of how the dynamic range of the
capture device is greater than that of the film. In this scenario, as the
bit depth of the capture device increases, so will it's dynamic range, and
as the dynamic range of the scanner exceeds that of the film, this will be
represented by "empty space" in the histogram -- IOW, empty space reveals
unutilized dynamic range of the scanner. Again, I don't buy this because
it's still trying to tie DR to bit depth.

Yet another suggests it is because scanners write out raw data in linear
gamma and the histogram is representing a working space of a higher gamma.
But gamma is a function of the distribution of the tones inside of the
endpoints, not where the endpoints lie, no? Furthermore, if I convert the
file to a 1.0 gamma working space, the histogram remains essentially
unchanged.

So what is it? Just why does the raw data appear so compressed, and how is
it's shape within the histogram affected by the bit depth of the capture
device?

Todd Flashner