Question: will any of this get images on a gallery wall?
-----Original Message-----
From: filmscanners_owner@halftone.co.uk
[mailto:filmscanners_owner@halftone.co.uk] On Behalf Of Roy Harrington
Sent: Tuesday, August 27, 2002 12:11 AM
To: bkubiak@attbi.com
Subject: [filmscanners] Re: Dynamic range
on 8/26/02 3:29 PM, Austin Franklin at darkroom@ix.netcom.com wrote:
> Hi Roy,
>
>>> When you set the setpoints, these 2000 values are then mapped
>> out to occupy
>>> the entire range. You then apply your tonal curves to the high bit,
>>> setpointed data.
>>
>> Ok, Austin, let's go with your numbers.
>> Is the number 2000 now a measure, indication or whatever
>> of the dynamic range of the image at this stage?
>
> The number 2000 is the dynamic range of the captured image by the
scanner,
> yes.
>
>> Read it into PS. Convert 16-bit to 8-bit -- now 256 levels.
>
> That would be a mistake, and severely degrade the image. You would
first
> apply your setpoints and then your tonal curves. The convert to 8
bits. If
Well, of course, we all know that you do that endpoints and basic curves
first and then convert to 8-bit. In fact as I understand your workflow
you do that in the scanner software. In any case just to spell it out
all
tonal corrections and whatever else are done in the high bit mode before
converting to 8-bit.
> you simply do as you suggest, then you would be causing adjacent image
data
> values to be wiped out, combined, so you would be decreasing the
dynamic
> range of the image by doing so, and could get posterization in your
output
> image.
>
>> Is the number 256 now a measure, indication or whatever
>> of the dynamic range of the image at this stage?
>
> Yes, the dynamic range of the resultant image is now 256, but the
density
> range is the same as the density range of the original image.
>
>> If you answered "yes" to both Dyr questions, does the 8-bit image
>> have 1/8 the dynamic range of the 2000 level image?
>
> The scanned image data file had a dynamic range of 33dB, and the
processed
> file is now 24dB. The processed image file has less dynamic range, of
> course.
This is where we differ. Visually we have two files that "look" alike.
There's a 16-bit file (with only 2000 distinct levels) and an 8-bit file
with 256 levels. My contention is that they have identical dynamic
range
as well as identical density range. Your claim is that the 16-bit file
has 9db more dynamic range or in ratio terms 8 times as much dynamic
range.
Now if we actually print these two files (assuming you have a driver
that
can handle 16-bit data) we'll have two prints that are extremely close
if not indistinguishable. Do these two prints also differ in dynamic
range by a factor of 8?
If you answer "yes" they differ by a factor of 8, it seems dynamic range
is
very poor characterization of what we see. I.e. a factor of 8 change in
dynamic range makes a negligible difference in what we perceive. On the
other hand if you answer "no", the prints have the same or at least
similar
dynamic range, then the dynamic range of the files doesn't come close to
translating into a dynamic range of the final print. Either way you get
a
strange paradox of what happened to the dynamic range.
I know my assessment seems to conflict with what you know from "20 years
experience" but I'd like to explain why its perfectly consistent.
With scanners and other digital imagining systems we have CCD's which
respond to intensity of light producing a voltage output. That voltage
output is fed into a A/D converter that digitizes the voltage and
outputs a number representing the voltage. For sake of example, say
its a 10-bit A/D and it measures millivolts from 1mv to 1024mv in 1mv
steps. It's linear in voltage with possible outputs of 1,2,3,4...1022,
1023,
1024. So it's got 1024:1 or 30db dynamic range and 1024 potential
levels.
In this case number of levels and dynamic range go hand in hand,
1024 levels <==> 1024:1 dynamic range. This translates exactly to
the RAW output of a scanner -- the RAW levels and the dynamic range
will track together. The KEY, KEY thing here is that this equivalency
is directly due to the linearity of the A/D tracking with the linearity
of the voltage output of the CCD. The fact that evenly spaced numbers
represent evenly spaced voltages is critical to this equivalency.
The crucial difference is when we applying curves and gammas. For
example let's assume the endpoints are the full range 1mv to 1024mv.
But in applying curves what we want is for evenly spaced numbers to
now represent evenly spaced perceived brightnesses. Think of the
step wedges we print out to calibrate monitors and print output.
Human perception of brightness is very different from the raw voltage
values. Human perception is exponential instead of linear -- so
a possible set of raw voltages that would represent evenly spaced
human perceptions could be 1,2,4,8,16,32,..., 256, 512, 1024.
There are 10 evenly spaces intervals each being one stop apart.
We could represent these 11 values with just 4 bits, the dynamic range
that
they represent is still the same as the original 10-bits i.e. its still
1024:1 or 30db. The important numbers are still 1024mv and 1mv so the
ratio
is still the same. This is analogous to reducing the 2000 levels picture
to
the 256 level picture. The dynamic range is not changed because the top
level and the bottom level still represent the same real-world values so
the
real-world ratios are still the same.
>
> Austin
>
So, in conclusion, Austin, I'm really not trying to give you an
excessively
hard time. What you say about equivalency of dynamic range and number
of levels is true in many cases especially in low level hardware. But
it's a special case situation, not a definition for the general case.
I hope you can spend some considering what I have written, I've spent
a fair amount of time figuring out how to explain my case.
Regards,
Roy
Roy Harrington
roy@harrington.com
Black & White Photography Gallery
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