Digital BW, The Print

Hi Martin,

> Nothing like an example. So right or high bit justification places the
> 12-bit number in the upper end of the 16-bit space

I think you've confused your self a bit (no pun intended) with left and
right, and I possibly didn't help.

What you describe above would be LEFT bit or HIGH bit or MSB (Most
Significant Bit) justified...but the extra bits would be added on the RIGHT
side, after the LSB (Least Significant Bit).


1001_0110_1110

^            ^
|            |_ LSB, LOW, RIGHT side
|
|_ MSB, HIGH, LEFT side


The original binary number converted to 16 bits, MSB justified, LEFT
justified, HIGH bit justified ;-)


1001_0110_1110_0000

               ^
               |_ Extra 4 bits added on the RIGHT side


> and left or low bit
> justification puts it in the bottom end of the range.

That would be RIGHT bit justification...but the extra bits would be added on
the LEFT side.

The original 12 bit number converted to 16 bits, LSB justified, RIGHT
justified, LOW bit justified...

0000_1001_0110_1110

^
|_ Extra 4 bits added to the LEFT side


> It would seem that for raw scanning you would always want to do right

You mean LEFT bit justification...

> bit
> justification since the values would not be adjacent in 16-bit space.

It really doesn't matter, except remember, PS only actually uses 15
bits...so if you really had a full 16 bits, you are better off LOW bit
justified.

> 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.

Yes, except, again, it doesn't matter, except for possible PS issues.  You
need to set the setpoints and THEN expand the data over the entire 16 bit
range.  I believe the algorithm may work better if it works on contiguous
data (that's just a guess, I'd have to do some work to figure out why I say
that)...as in you set your setpoints on the contiguous data, then spread
them out over the 16 bit space.

>
> With Left bit justification all the 12-bit values remain adjacent

Again, you mean RIGHT, LSB, LOW bit justification...

> 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.

Again, it really doesn't matter...as long as the relativity of the data is
maintained, and no data is lost.

> 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.

Could be, and there should be gaps in the data.

Regards,

Austin