Digital BW, The Print

Usually shoots digital raw to preserve 16 bits per channel and avoid
deterioration when processing the images. However, it occurs to me
that if one is shooting color with 8 bits/channel, there are already
24 bits of information. If the image is then converted to 16 bit
grayscale mode, all bits contain information and there should be no
empty bins in the histogram due to lack of intensity resolution. The
quality of the 16 bit image should be adequate to tolerate processing. 

Have anyone tested this? Any ideas?

Manuel Toledo

> From: Manuel Toledo Quinones [mailto:mtoledo@...]
>
> Usually shoots digital raw to preserve 16 bits per channel and avoid
> deterioration when processing the images. However, it occurs to me
> that if one is shooting color with 8 bits/channel, there are already
> 24 bits of information. If the image is then converted to 16 bit
> grayscale mode, all bits contain information and there should be no
> empty bins in the histogram due to lack of intensity resolution. The
> quality of the 16 bit image should be adequate to tolerate processing.

Not true. With three channels of 8 bits, that's roughly equal to 9.5 bits,
because the maximum value you can get by adding three 255's together is 765,
which takes about 9.5 bits to represent.

--

Ciao,               Paul D. DeRocco
Paul                mailto:pderocco@...

If one were 'adding' the values together, you'd be on the right track.  However, it is per color channel that these values are applied.  So, 8 bits is 256 values per channel.  However, if you have a value of 220 in a channel and convert it to 16bit, you still don't have any more real data than you did before.  You only have a 16 bit value with the same real value.  
 
Tom Baker

"Paul D. DeRocco" <pderocco@...> wrote:
> From: Manuel Toledo Quinones [mailto:mtoledo@...]
>
> Usually shoots digital raw to preserve 16 bits per channel and avoid
> deterioration when processing the images. However, it occurs to me
> that if one is shooting color with 8 bits/channel, there are already
> 24 bits of information. If the image is then converted to 16 bit
> grayscale mode, all bits contain information and there should be no
> empty bins in the histogram due to lack of intensity resolution. The
> quality of the 16 bit image should be adequate to tolerate processing.

Not true. With three channels of 8 bits, that's roughly equal to 9.5 bits,
because the maximum value you can get by adding three 255's together is 765,
which takes about 9.5 bits to represent.

--

Ciao,               Paul D. DeRocco
Paul                mailto:pderocco@...


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[Non-text portions of this message have been removed]

> From: Tom Baker [mailto:tbaker1328@...]
>
> If one were 'adding' the values together, you'd be on the right
> track.  However, it is per color channel that these values are
> applied.  So, 8 bits is 256 values per channel.  However, if you
> have a value of 220 in a channel and convert it to 16bit, you
> still don't have any more real data than you did before.  You
> only have a 16 bit value with the same real value.

From a strictly numeric point of view, eight bits is eight bits. The
original poster thought he could somehow get 24 bits of "resolution". My
point was that you do indeed get a little bit more resolution, perceptually,
because the amount of quantization noise you perceive in an 8bpc color image
is probably better than a 9-bit B&W image, for the reason I explained. A
more extreme example of this is that an inkjet print, even a B&W one, made
with color inks will have less visible noise than a black ink only print,
even though each ink "channel" is still only one bit.

--

Ciao,               Paul D. DeRocco
Paul                mailto:pderocco@...