I think there are two important issues that tend to get glossed over in
thinking
about this. The first is that density is the LOG of the intensity --
not just
the intensity, and the second is that for the most part we're
interested in
the density RANGE -- not just the max.
So with that in mind here's how I see the example:
Say the intensities are: 1,2,3,4,5,6,...15,16
The density range = log(16) - log(1) = 1.2 - 0 = 1.2
Now do the 3 bit mapping:
a) increment by 2: 2,4,6,8,...14,16
density range = log(16) - log(2) = 1.2 - 0.3 = 0.9
b) go half way: 1,2,3,4,...,7,8
density range = log(8) - log(1) = 0.9 - 0 = 0.9
The range is exactly the same both ways. The logarithmic nature
of density throws off our intuitive feel for how the numbers work.
Roy
On Thursday, July 10, 2003, at 09:26 PM, Karl Schulmeisters wrote:
> I'm not sure the below is accurate. Perhaps I'm misunderstanding what
> was
> written, in which case I'm probably in 'violent agreement' with what
> was
> written.
>
> Consider the values coming out of the analogue detector to be
> incremental
> numbers:
> 1,2,3,4,5,6....dMAX(detector)
>
> This actually works fairly well since the detector is essentially
> generating
> voltage proportional to light intensity detected.
>
> So in the above example, lets say dMAX is 16.
> If I map it with 4 bits, I can generate a bit value/ incremental value
> returned. (4 bits can represent 16 numbers).
>
> If I map it with 8 bits, I can represent 1/2 step increments
>
> If I map it with 3 bits (8 possible values) I can either
> a) aggregate each step to represent 2 incremental values - no
> effect on
> dMAX
> b) retain the same level of tonal resolution, but toss out either
> the
> high or low end of the scale, thereby reducing the effective dynamic
> range
> and dMax covered.
>
> So doubling the bit count may or may not increase the dMAX being
> represented.
>
> ----- Original Message -----
> From: "Roy Harrington" <roy@harrington.com>
> To: <karlsch@earthlink.net>
> Sent: Wednesday, July 09, 2003 4:39 PM
> Subject: [filmscanners] Re: scanner dmax discussion
>
>
>
> On Tuesday, July 8, 2003, at 09:42 PM, Tim Atherton wrote:
>
> o o o
>> 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.
>
>
>
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-
Roy Harrington
roy@harrington.com
Black & White Photo Gallery
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