Re: 8 verses 16 bit

[David Chien <chiendh@xxxxxxx>]

> A print, reflecting the image in the presence of illumination.
> To the best of my understanding, neither a color print, (whether chemical
> or inkjet) nor a monitor, whether CRT or LCD (or plasma display) have a
> gamut that exceeds the 8-bit image. If this is so, then the best/only(?)
> reason for going to 16 bits per primary color is to have spare pixels to
> maintain the quality of the image through the editing/optimization process,
> before dropping down to 8 bits for display. Am I wrong in this? If so, I'm
> sure I'm not the only one who'd like to know wherein.
> Hersch

  The problem here is not just how wide the color gamut is/can be.  In
fact, that's totally seperate from the issue of bits used!!

  Here's the example.  If you have a 2-bit system, you can easily cover
the entire Black and White gamut of any print/output system in the world
by simply specifying 0 (off) to represent the darkest black in
existance, and 1 (on) to represent the whitest white in existance.

  There, you've got then entire color gamut encompassed within a 2-bit
system.

  ----

  What the # of bits in a color system really does is let you represent
that many # of shades of a particular color correctly at the same time.

  Thus, a 8-bit system can only represent and show 256 shades of red,
let's say, on screen at any one time.  It can't show or display another
257th shade of red on screen with the rest, and any images captured and
processed into 8-bit mode can't hold/contain more than 256 shades of
red, even if the original image of let's say a beautiful rose & pink
flower show had more than 256 shades of red.

  --

  at the same time, when editing images, more bits allow you to make
subtle and gross changes to the entire image more smoothly.

  Let's pretend we're working in a 2-bit mode where 0 binary represents
the number 0, and 1 binary represents the number 10.

  Okay, so let's say we decrease brigthness by 50%, thus, 50% of 0 = 0,
and 50% of 10 = 5.

  But how do you represent 5 in a bit-system that doesn't have enough
in-between values due to limited bits?

  You round off and turn that 5 into a 0 or keep it as 10.  

  for example, Going back to our image, if we have a very bright red,
decrease brightness by 50% and don't have enough bits, it either goes to
a super-dark red, or stays as a very bright red.

  --

  Now, if you have a few more bits, but not enough to be visually
transparent, what occurs during editing is that some shades can't be
represented properly, and are pushed to the next nearest color shade --
and what results is the often seen, ugly color banding effect as many
shades get assigned to the only available shade.
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