[filmscanners] RE: Dynamic range question

["Austin Franklin" <austin@darkroom.com>]

Hi Unnamed Person Who Made Below Comments,

> When you scan negative film, the histogram is narrow. So I would say
> negative film has a low dynamic range.[Yeah, I know slide and negative
> film is really the same.]

Negative film has a lower DENSITY range, no doubt...but that does not mean
it necessarily has a lower DYNAMIC range.  In fact, negative film typically
has a higher dynamic range than slide film.  Dynamic range is related to the
amount of discernable information contained within stated bounds, where
density is simply a measurable difference between two samples.  You can have
a very high density range, but no dynamic range, and vice versa.

A, IMO, good example of this is a two pixel scanner.  It is thresholded such
that the binary value 00 represents a density value of less than or equal to
.1D (D being absolute density value).  Binary values 01 and 10 represent the
same thing, the range of density values greater than .1D and less than 4.8D,
and binary value 11 represents density values of 4.8D or more.  These two
bits give you a very high density range, but no dynamic range.

Conversely, you could have a scanner that represents a density range of .1D
to .2D in .001D increments, which would give you a much larger dynamic range
than the two pixel scanner.

You can relate that to slide vs negative film, slide having a larger density
range, but lower dynamic range.

> The dynamic range of
> a dataconverter is related to the number of bits, since the smallest
> signal the dataconverter can resolve is 1 lsb out of 2^N bits.

Correct.

> The
> dynamic range of the CCD spans from the  point near where it saturates
> at maximum brightness to where the photons are lost in the noise.

Though true that that is the limit of the CCD, that doesn't (necessarily)
tell the whole story, that only sets the low and high bounds, not
necessarily the discernability between...and there are techniques for
quantifying noise in this region to increase the dynamic range.  IOW, the
low noise threshold does not necessarily equal the noise in it's active
range.

> When you scan negative film, the dynamic range of the film is narrow.

They DENSITY range of the film is narrower.

> However,  you expand this dynamic range during the inversion process to
> restore the contrast in the final image.

Again, substitute density range for dynamic range and your statement is,
IMO, then correct.

> 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?

> Using more
> bits during the scan of slide film doesn't buy you nearly as much as you
> probably aren't going to increase the overall contrast in the
> final product.

Actually, you DO need more bits (if your density representation is linear,
as typically scanners are designed to be) to scan slide film, as the density
range is higher, and the way a typical scanner "represents" higher density
is with larger numbers (unlike my 2 bit scanner example above).

Regards,

Austin


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