Digital BW, The Print

> Okay, this whole ratio thing: when you talk about each next value being
> assigned is the next value that is read as being twice as bright
> as the one
> before it,

Not quite right.  The value of 2 is twice as bright as the value of 1, and
the value of 3 is three times a bright as 1 etc.  100 is twice as bright as
50.  But notice that it takes more values in-between to get 2x the further
up the scale you go.

> I think of f-stops. How can there be thousands of f-stops of
> brightness in the film.

There are intermediate f-stops in film.  One f-stop is only 2x as
bright/darker than the next successive one.  It's the area of a circle (the
aperture).  Twice the area means twice the light, 1/2 the area, means 1/2
the light.

f1 for a 50mm lense has an aperture width of 50mm, which is an area of
pi*r**2 (pi r squared), or 3.14 times 25**2 or 1,963 square mm.  f1.4 for a
50mm lense has a aperture width of 50 divided by 1.4 or 35mm.  The area of a
circle with a radius of 17.5 is 962.1....or, half of the area of the f1
aperture, and therefore lets in 1/2 as much light.  And...f2 has an area of
490 sq mm, and f2.8 has an area of 250 sq mm etc.

Again, it's a ratio thing...

> If DR is the height of the staircase, and bit depth is the number of steps
> to it,

That works for me...

> I would think in the above scenario that the bottom few steps
> (shadow) would contain a lot of noise, but once you got past them
> you'd have
> more steps (discrete tones) throughout the rest of the spectrum.

The noise is less significant the higher up you go.  It's quite simple.  As
I said, two is twice 1, obviously...and that is only one value
away...but...10,476 is one more than 10,475...but the difference is far less
significant.

> For
> instance, if an 8-bit scanner gives you 256 tones of which say the first 6
> may contain noise, you have 250 discreet clean tones further up the scale.
> If you have a 10-bit scanner which gives you 1024 tones, and the
> first 6 (or
> would it be the first 24?) contain noise, won't that give you
> 1018 (or 1000)
> discreet clean tones further up the scale?

Well, not quite...I'm actually trying to figure out a way of detailing this
exact scenario right now.  I'd rather hold off answering this until I work
it all the way through.

> Isn't this what you
> were talking
> about when you suggested that higher bit scans should span more of the
> histogram than lowbit scans of the same material?

That's only true if the data isn't high bit justified.  If the data is high
bit justified, then it will occupy the same span.

> Getting back to dynamic range, I see how bit depth can limit DR in that it
> is tied in with noise, but what else determines dynamic range?

NOTHING.  The pure definition of dynamic range is largest signal divided by
noise, period.  There are some other issues depending on what you are
talking about, but that's just a multiplication by some number...but it's a
constant and really doesn't effect the true definition.

> As a for
> instance, if Margulis is correct, drum scanners will typically have better
> DR than CCDs, even though they might both be of the same bit depth.

I don't know what you are referring to that Margulis said, but drum scanners
give cleaner data, which in turn does mean they have a better dynamic range.
They give cleaner data for a very simple reason.  They sample one pixel at a
time...and there is no cross talk between sensors.

> Furthermore, even on CCD units, what the unit's DR tests out to be may be
> very different than what it's spec sheet might suggest, even when they do
> list it's S/N ratio; which, granted, is probably an optimistic
> estimation of
> it's true S/N. But my point is, I would imagine two scanners from
> different
> manufacturers could use the same CCD and the same A/D, and yield different
> DR. Is it all related to noise suppression?

Pretty much, and design.

> Another thing I need clarification on is low-bit vs high-bit
> justification.

Phew...an easy one ;-)

I assume you know something about the binary system?  You know 8 bits is a
byte, and two bytes is (typically considered) a word (which is 16 bits)?  A
bit can either be 0 or 1, and each bit is a power of two...

1 = 1
10 = 2
100 = 4
1000 = 8

Therefore, 1010 is ten etc.  Anyway, say we have a 12 bit value from the A/D
of 1010_1010_1010 (the "_" are just nibble separators...a nibble is 4 bits).
When we put that value into a 16 bit number (16 bits is 4 nibbles
0000_0000_0000_0000) if we:

left justified, the number 1010_1010_1010 becomes 1010_1010_1010_0000 and
the decimal equivalent is 43,680.  If we right justify it, it is
0000_1010_1010_1010 and the decimal equivalent is 2,730.

As you can see, the left (high bit) justification gives a much larger number
than the low bit justification.  If you high bit justify, you basically
spread the numbers out, and they will occupy a wider range in the histogram.
If you just leave them right justified, then the will occupy a very small
range, but be contiguous values (if they were in the first place that is).

Is that kind of what you were looking for?