[filmscanners] Re: Dynamic range

["Todd Flashner" <tflash@earthlink.net>]

Hi Austin

Thanks for your replies.

Well I guess this really breaks down to the simple difference as to whether
DyR is a resolution vs a range. I think your opponents agree that IF it were
a resolution then your approach would be correct, they just believe it is a
range, as its name would indicate. I'm in no position to know who is right,
but I do like to understand both sides rational as best I can. So, I'd like
to pursue this one step further.

I think it was a mistake for me to bring bit depth into the discussion,
because it is by definition a number of steps or chops. But if we were
considering the DyR of an analog component we wouldn't be forced to consider
bits and steps (not by the nature of digital equipment I mean).

Perhaps if we break down the mechanism of determining the DyR of a scanner
into discreet steps I'll get this better. I'll explain how I presume this is
done and you can correct.

First off, no knowledge of the bit depth of the scanner is required.

A film with a density range greater than that of the scanner is scanned. I
don't know if this film will be a step wedge, a grid, or a gradient, but for
accuracy I'd think you'd need some sort of discreetness to the denser tones
to know at what density your scanner is bottoming out.

I assume the entirety of the film is scanned before encoding begins, and the
highest recorded voltage by the CCD would be assigned a value that when read
in PS would read zero. I assume this because my RAW scans, which include
empty film base with a relatively high degree of base+fog density (Konica
infrared, for those who know it, has a dense film base) still reads with a
value of zero, even though there is density to it. If the encoding were done
on a line by line basis, how would the AD know what value to assign zero? If
it picked one of greater density than will be found elsewhere on the film,
all densities below that will be clipped.

So, the encoding begins at zero. If it is a 14 bit scanner, the next highest
voltage (assuming a gradient here) will be assigned a value 1/16383 higher
(since there are 16384 possible values it can assign), and so on down the
line.

Now if this film is a step wedge there will be a significant jump from one
voltage recorded to the next. While the scanner is capable of 16384 discreet
tones, if it's an 21 step wedge the resultant scan will have 21 density
steps (at least if this hypothetical wedge had NO soft transitional edges
between steps). Let's assume that the scanner can accurately distinguish the
second to most dense step of the wedge from noise, but not the most dense
step. Let's assume the most dense step had a density of 4.0 and the one
before 3.2. (not saying those would be the right values for a 21 step wedge,
just guessing) You'd assume the Density Range of that scanner to be
somewhere between 3.2 and 4.0. While others would say it's Dynamic Range
were between 3.2 and 4.0.

If we used a target with much finer increments we could better determine at
which density the scanner actually can not distinguish from noise. Lets say
it was found to be 3.7. I believe this to be within the range of our 14 bit
scanner. So we can assume that noise was the limiting factor, not bit depth.
Lets assume that the data of this RAW scan would span about 3/4 of the
histogram in PS.

If the same film target were scanned on a 15 bit scanner of equal noise, the
max discernable density would still be 3.7 but fewer of the potential
"steps" were used to get there. Each sequential voltage could be designated
values 1/32768 apart. Thus the RAW data might only span 1/4 of the histogram
(again, just rough guesses). You'd still say this scanner has a density
range of 3.7, others would say its DyR is 3.7.

That you guys use different calculations to get there is ironic. But if
clear film base establishes one endpoint, the distance of density out from
that is what determines DnR/DyR. IOW, zero density establishes one endpoint
and max discernable density determines the other. Both scanners will have
been limited by noise, and the fact that the 15-bit scanner chopped it up
finer is immaterial to what density can be accurately distinguished (min
discernable signal).

on 6/13/02 11:12 AM, Austin Franklin wrote:

> The ISO spec states:
>
> "DR = Dmax - Dmin"
>
> Since density values are in log form, this is the correct equation.
>
> The ISO spec defines:
>
> "Dmax = Density where the Signal to noise ratio is 1"
>
> Which is correct, as it is the lowest signal level that will come from the
> scanner electronics when measuring the highest density, and where the signal
> and noise will be equal.  This value obviously takes into account noise.
> They make the assumption that the noise is the same as the "minimum signal
> level", which is fine for the case they are using it for (more below).

Okay so in my example above, the dmax of the scanner coincides with the
densest patch on my target film that the scanner can distinguish over noise.
I said it to be a density of 3.7.

> "Dmin = Minimum density where the output signal of the luminance OECF
> appears to be unclipped"
>
> Again, correct, as that is when you get the lowest density, when you are
> reading the strongest signal from the electronics.

This corresponds to my film base+fog. Let's assume it could theoretically be
zero density (maybe the film doesn't fill the film holder).

So what don't I get?

If the density of the film base+fog were 0, and the readable dmax  occurred
at a film density of 3.7, using THIS spec, how could the DyR be anything
other than 3.7 - 0 = 3.7 for BOTH scanners?

If this approach is right it obviates the need to know the bit depth of the
scanners, or the number of pieces the image is being chopped into, as it is
based upon what densities the scanner can record.

Sorry that this was so long and rambled, just don't quite know how to get my
head around this...

Todd




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