[filmscanners] Re: Dynamic range question

["" <lists@lazygranch.com>]

Sorry, it's Gary. I just use "lists" for, well, lists. ;-)

This part of your reply I can answer without much effort. As far as I
have been able to tell, when endpoints are set on a scan, the only
difference is in the post processing. That is, nothing different has
happened with the scanner. Programs like vuescan demonstrate this quite
well since you can see the raw image (what the CCD "sees") is not
changing as you alter the limits  Thus when you scan negative film and
zoom in on that narrow range, the post processing software (i.e. vuescan
or your twain)  is mapping that range of intensity into a range that
spans the full bit width of the word. Effectively, this is increasing
the contrast of the image. Also, this is truncating bits. That is, if
you only use 1/4 of the range of a 10 bit scan , you are getting 8
"real" bits worth of data (information). Thus negative film needs more
bits in the dataconverter than slide film.

When it is said that negative film has a wide latitude, think about how
that looks on the histogram. What you get is that same narrow region
where most of the samples are located, but the region gets shifted
around depending on the over or under exposure. With slide film, over or
under exposure slams the histogram to either side since there is very
little latitude in the film.

There are a few things that do change how a scanner operates. For
instance,vuescan can do a long pass, which basically uses a longer
illumination time. This is to help with very dense images.


Austin Franklin wrote:

>Hi Unnamed Person Who Made Below Comments,
>
>
>
>
>>If the negative film only uses
>>1/4 of the range of the scanner, then you would expand the range by a
>>factor of 4 before truncating the image to 8 bits (which is all the eye
>>sees).
>>
>>
>
>At this stage, you don't expand anything.  You set your setpoint so that you
>only USE the valid image data within the overall range.  Therefore, say,
>your scanner is 10 bits, and therefore gives you 0-1023...and your image
>data occupies the range of data from 233-876, you set your setpoints at 233
>and 876, and take THAT data and "remap" it to 8 bits.  In this case, it is a
>decimation, and it is rarely an interpolation, since the valid data region
>is almost always more than 8 bits.
>
>
>
>>Thus you really need a 10 bit scan to get 8 bits out.
>>
>>
>
>I'm not sure what you mean by that...  Would you elaborate?
>
>
>
>

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