Digital BW, The Print

> You said "if you are getting a full range of 16
> bit data, you can NOT do any level shifts to it in PS without
> losing data."
>
> This suggests you'd prefer a more contracted set of tones, inside
> of a lower
> bit depth, rather than a full set of tones in a higher bit depth.

No.  It means that you need to have headroom within the space to make
adjustments.

> It doesn't make sense to be worrying about loosing data when you
> already had
> a nicely distributed histogram via a "full range of 16
> bit data" (assuming we are both NOT assuming clipped endpoints of course).

I don't know what you mean.  Be careful when you say "16 bit data".  There
are different types of 16 bit data.  One where the data was expanded into a
16 bit range, such as you do with 12/14 bit data, by seting the setpoints,
then the data gets expanded into the entire 16 bit range...and once
expanded, has holes between every data value.  Second, REAL 16 bits out of
the A/D, with no headroom, with no holes between any of the image data. And
third, the actual raw data from the scanner, typically with no holes in the
actual image data values

> I'd be far more fearful of an image getting degraded, or a histogram
> combing,

Taking 12/14 bit data and putting it into a 16 bit data will comb a 16 bit
histogram...many values between data points have no pixels that have that
value.  That is fine, if your goal is ultimately 8 bit data...as only the
top 8 bits are used, and the 8 bit data won't be "combed".

> due to expansion of too few tones, than contraction of too many,
> for precisely the reason you demonstrated above.
>
> See how you are presenting two sides of the argument at once?

No, sorry, I don't.  I'm merely explaining how things work...I don't see
that as an argument.

> Not trying to trap you in a corner (well, maybe a little <g>),
> just showing
> why I find the conversation confusing.

I don't see anything as being two sides of an argument.  It all depends on
what your "goal" is.  There's nothing to "trap", and I don't understand your
point.  As I said, I am only stating how things work...so I don't understand
how you could argue with how they work.  Perhaps not understand them...or
perhaps my explanations are bad...but I'm really only relaying my direct
knowledge of how this works.

> >> It only means you need less
> >> levels adjustment from the get go; less need to loose data through
> >> adjustments.
> >
> > Hum.  I don't know that I believe that.  I think the level
> adjustments are
> > the same no matter what the bit depth
>
> Yes, that is my point! Why then wouldn't you prefer to have that 16-bit
> scan? You've got the same levels adjustment to make, why not do
> it on a file
> that has more tones to throw away without significant loss?

It depends on what you mean by 16 bit data, and what your goal is.  As I
keep saying, you need headroom, and 16 bit data in a 16 bit space gives you
NO headroom.

> >...what you gain by higher bit depth
> > (in the file, not from the scanner...that is a very important
> point BTW) is
> > more discernable tones.
>
> Right, more discernable tones between the end points (still talking raw
> data), but I don't believe bit depth is what determines where
> your endpoints
> fall.

No, I didn't say it does.  It CAN determine one or both endpoints mind you,
but not always.

> You've been arguing
> (or so it seems as evidenced below) that bit depth
> determines where your endpoints fall. And you suggest that as bit depth
> increases, so does the distribution of tones across the histogram. Higher
> bit depth = a larger portion of the histogram filled.

No, that is not what I said.  As I said above, bit depth CAN determine one
or both of the endpoints.  Also, as bit depth increases the distribution of
the SAME IMAGE does NOT give more tones in the raw scan.  That is what is
misunderstood.  IF and only IF the image is up against one side of the bit
depth, will higher bit depth give you more tones.

These numbers aren't hard, but they are just an example.  Say you have a B&W
image that gives you a range of 1:1 to 1:100 tones (dynamic range of 2.0).
In the 16 bit scanner, the data will only occupy a range of 100 values out
of 65k, and in the 8 bit scanner it will occupy a range of 100 values out of
256.

The point is, bit depth does NOT mean you get more tones out of the same
image.  It ONLY means you CAN get more tones on ONE end of the scale (dark
side) IF and ONLY IF the image you are scanning has detail in that "area".

> We are talking
> about how it gets mapped to the histogram.

I don't understand the question/issue.  There is no mapping.  Mapping means
values get changed.  No values are changed in a histogram.  The histogram is
nothing more than a graphical representation of a count of the values of the
upper 8 bits of the data in the image buffer.

> >> Your scenario makes a case that Dynamic Range is directly a
> >> function of bit
> >> depth, which it is not. As you've stated before you only need one
> >> (or was it
> >> two) bit data to have a large dynamic range.
> >
> > It's a matter of representation.  It JUST SO HAPPENS that by
> design, yes,
> > most every film scanner made DOES represent density ratio
> values as simple
> > integer ratio values...so yes, number of bits does limit the dynamic
> > range...again, this is by design.  But, a scanner does not have to be
> > designed that way.
>
> Could you elaborate?

I've given this explanation before, you can check the archives.  The way the
CCD works is quite simple.  It's all relative.  It gives values of 1 to
N...and a value of 1 means a density ratio of 1:1.  When the CCD gives you a
value of 2, that means it is twice as dark as 1.  These just so happen to be
the exact same thing as integer density ratio values.

> And elaborate how what you are saying fits in with my
> point that DR is simply a ratio of Dmin to Dmax, and thus just one bit of
> data can represent a high DR if it's two points are far apart (black/Dmin
> and white/Dmax). If one bit can describe it, how does more bits make it
> wider?

I have no idea what your asking here, sorry.

> > The specific reason the data is "bunched up" in raw scans is because the
> > dynamic range of the film is less than the dynamic range of the scanner.
>
> Okay, but this is confusing again. This seems to fly in the face
> of what you
> stated above.

Not at all.  It fits exactly with everything I've been saying.  You're
missing something, and I just don't know what it is you're missing.

> Remember your premise was that 10-bit data will map to the
> histogram, by default, due to lower bit depth, and hence lower
> Dynamic Range
> (again your premise that DR is directly a function of bit depth), more
> narrowly than 14-bit data.

I really don't understand your term "map to the histogram".  You're missing
something here...  No matter WHAT the bit depth is, only the top 8 bits are
used for the histogram.  Nothing is mapped.  Also, dynamic range is LIMITED
by bit depth.

If you take a 10 and 14 bit scan of the SAME image, using the same
setpoints, and expand the data into the 16 bit range...you will get the
exact same histograms...for two reasons.  One, the image data will be the
same whether the scan is 10 or 14 bits, providing you have not exceeded the
dynamic range of number of bits, and two, because only the top 8 bits of the
data are used for the histogram.

> But, the logic presented immediately above would suggest the
> opposite;

Not trying to be snyde, but misunderstanding would suggest the opposite.
You're not understanding what density ratio values are, and that they are
not dependant on bit depth, unless you are clipping.

> So where you started the conversation with the premise that a 16-bit raw
> capture would by default fill more of the histogram

ONLY if the image being captured HAD the dynamic range to do so.

> you seem to
> be implying
> at the end that as bit depth goes up, so does DR

As bit depth goes up, you can represent more dynamic range...but unless the
image is clipping, more bits will not give you more DR.  You will only get
more DR by adding more bits if you are clipping.

> and the more the DR
> potential of the capture device exceeds the DR of the film, the more
> narrowly it should map to the histogram.

No.  As I said above, if you have an image that fits in 10 bits, no matter
how many bits your scanner is, you're going to get the exact same data
values from the scanner...and the histogram will be exactly the same.  It's
that "integer density ratio value" thing again ;-)  You've GOT to get that
understanding down for this all to make sense.

I need to ask you a favor.  Can you please try to cut this down to fewer
questions/points?  I just don't have time (near an hour!) to respond ...