[filmscanners] Re: scanner dmax discussion

["Karl Schulmeisters" <karlsch@earthlink.net>]

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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