On Tuesday, July 8, 2003, at 09:42 PM, Tim Atherton wrote:
> well I've got to tell you, I couldn't make head or tail of the first
> post,
> never mind the follow up - now I'm going to have to wait for my copy
> of View
> Camera to arrive and see what the article actually says...
>
>
> "I thought I understood the relation of bit depth to dynamic range, but
> now I'm not sure. In the latest issue of View Camera, George DeWolfe
> says that the maximum possible dmax for an 8 bit gray scale is 2.4
> while
> the maximum possible dmax of a 16 bit gray scale is 4.8. I
> understand
> mathematically where these figures come from.
>
> log_10(2^8) = 2.41 and log_10(2^16) = 4.82.
>
> But I don't see what this has to do with dmax. The light detecting
> devices in scanners are physical devices capable of responding to some
> range of light intensities. Below a certain level, say I_min, nothing
Tim,
The formulas above also are based on physical values i.e. the formula
is log10(intensity) = density. So saying that 8-bits implies a max of
255
and therefore log10(255) = 2.4 implies a dMax of 2.4 ONLY works if
the 8bit values are directly proportional to the intensity of the light
passing
through the film. For the RAW output of the CCD and A/D this is pretty
close to true. But once you set endpoints and gamma correction etc,
the numbers in the 8bit grayscale no longer have a direct relation with
the physical values, so there is no way to apply the log equation.
An 8-bit grayscale has no inherent density range or dynamic range i.e.
it's meaningless to try to imply it.
> will be recorded. Above a certain level, say I_max, additional light
> won't produce any more output. So it seems to me the total range of
> densities the device can handle should be log_10(I_max/I_min). It
> seems
> to me that the bit depth just determines how finely that range is
> subdivided. For 8 bit, it will be subdivided into 256 distinct levels,
> while for 16 bit, it will be subdivided into 65536 distinct levels. Of
> course, if there is some minimal ratio of intensities which is
> detectable and we assume the scanner is keys to seperating values
> reflected by that minimal ratio, then the two calculations above would
> be relevant. But why can't a scanner with 8 bit depth just use a
> larger step size.
You can use a larger step size. However, it has no effect on the
density
range. For example, if you double the size, dMax would rise by one
stop, but dMin would also rise by one stop, so the actual range would
remain the same. This is all a result of the way the CCD and A/D map
intensities into digital values. If you could have a different mapping
i.e. logarithmic one it would be possible to map any density and
dynamic range into 8bits -- this is essentially what is done in software
when you ask for an 8bit scan from a 16bit scanner.
> After all, the theory behind all this is that if you
> take the human visual system as a standard, then when viewing a single
> gray scale from deepest black to whitest white, a discrete set of 256
> values separating that range will appear continuous.
>
> I would appreciate any comments from experts about just what is going
> on.
>
> Let me also ask a related question. Just what does my Epson 3200
> scanner and its software do when I use it to scan a b/w negative whose
> maximum density is about 1.4 at 16 bit depth? Does it try to divide
> up the range from 0 to 1.4 into 65536 values, or does it just use
> 1.4/3.4 of that range for about 27000 distinct values. What I get
> using
The mapping of values with the input density is very different than what
most people would expect. The number of values is very heavily
weighted toward the less dense parts of the film -- half of all the 2^N
values come within the first stop of density i.e. between 0.00 and 0.30
This page has a table of where the values are for various N bit
scanners.
The whole thing is good reading but look down toward the end of the page
for the table
.
> Vuescan is values ranging from 0 to 255, and although those may be
> further split up, when I read them into my photoeditor, they also stay
> in that range."
>
Roy
-
Roy Harrington
roy@harrington.com
Black & White Photo Gallery
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