[filmscanners] RE: scanner dmax discussion

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

Hi Karl,

I believe Roy and I are in agreement on this...

> 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.

Correct, but that is not how scanners are designed.  Specifically, the A/D
input voltage is matched to the CCD output voltage...and no matter how many
bits the A/D is, a one bit increment (therefore the lowest value) is
basically the "smallest discernable signal", or noise.  The top number that
can come out of the A/D is equal to the largest voltage the A/D can
"receive" (and therefore the largest voltage the CCD can output)...  so...

If you have an 8 bit A/D, you'll get values 0-255.  Say your input to the
A/D is -3V to +3V.  1 will represent 1/256th of 6V, and 255 will represent
+3V.

Now, you have a 16 bit A/D....SAME CCD.  You get values 0-65535.  Same A/D
voltage input, let's say.  1 will represent 1/65536th of 6V, and 65535 will
represent +3V...but...note, nothing has changed in the density range, the
CCD is still detecting the same overall density range.

Regards,

Austin

----------------------------------------------------------------------------------------
Unsubscribe by mail to listserver@halftone.co.uk, with 'unsubscribe 
filmscanners'
or 'unsubscribe filmscanners_digest' (as appropriate) in the message title or 
body