Digital BW, The Print

----- Original Message -----
From: "Austin Franklin" <darkroom@...>
To: <DigitalBlackandWhiteThePrint@yahoogroups.com>
Sent: Saturday, October 12, 2002 1:35 PM
Subject: RE: [Digital BW] Re: 'combed' histograms in 16 bit ?


> > Can you define or explain low and high bit justification? Maybe an
> > example of
> > each.
>
> Absolutely, Martin.
>
> Say your 12 bits is 1001_0110_1110.  If you want to convert it to 16 bits,
> high justified, you simply add four bits to the "bottom" (Least
Significant
> Bit side, right side) and the result is 1001_0110_1110_0000.  That's high
> bit justification.  Also, since you now added four bits, the 16 bit number
> is 16x larger than the original 12 bit number.
>
> Low bit justification is simply adding the four bits on the "top" (Most
> Significant Bit side, left side) and it would be 0000_1001_0110_1110 and
the
> number would be exactly the same number as you started with, just in a 16
> bit realm.
>
Austin,

Nothing like an example. So right or high bit justification places the
12-bit number in the upper end of the 16-bit space and left or low bit
justification puts it in the bottom end of the range.

It would seem that for raw scanning you would always want to do right bit
justification since the values would not be adjacent in 16-bit space. Each
adjacent pair of 12-bit values winds up 4-bits apart. 1001_0110_1110 is
placed at 1001_0110_1110_0000 and 1001_0110_1111 is placed at
1001_0110_1111_0000.

With Left bit justification all the 12-bit values remain adjacent in 16-bit
space. 1001_0110_1110 is placed at 0000_1001_0110_1110 and 1001_0110_1111 is
placed at  0000_1001_0110_1111.

As an alternative couldn't the programmer who wrote the scanning software do
a combination of both? Place two bits on each end or one on one end and
three on the other? That actually would seem to me to be the logical choice
since the mapped data would then fall more in the middle of the 16-bit
space.

Maybe this is what is being done in Silverfast so that the values are not
adjacent but still close enough together to cause problems if you do a curve
type adjustment without spreading out the end points first. This would also
explain why the raw data occupies a wider portion of the histogram display
that you would expect if it was exactly adjacent.

Martin