"Exposing to the right" is over exposed, what now?

On Thu, 15 Oct 2009 01:28:38 -0500, Holy **** are these fools ever the
biggest idiots to ever appear on earth! wrote:


In editing, if you take 0,0,0,0 as the base RAW for pure black, that then
becomes the base 0,0,0 RGB for pure black, in the 8-bit JPG, NO MATTER THE
BIT DEPTH.

If you take a 12-bit RGGB value of 4096, 4096, 4096, 4096 as pure white,
that gets remapped to 256, 256, 256 in 8-bit JPG. If you take the 14-bit
16386, 16384, 16384, 16384 RGGB value that gets remapped to the 8-bit JPG
value of 256, 256, 256. There is NO difference in if the camera properly
converts those whites to 8-bit values or if you do it in editing. But you,
being the amazingly stupid ****wad that you are, think that if you have to
resort to an editor to accomplish what your camera fails to provide, that
somehow this blatant fault of creating the proper JPG file to begin with
makes for a better camera????



The two ends (black and white) are hardly the most interesting parts
of the photograph, though.

It's all that in the middle that get's reduced that I tend to wonder
more about.

So, while (4096,4096,4096,4096) might = (256,356,256), what would be
the equivalent of a shading somewhat between that and the zeros?

Can all of the colors represented between (3152,3152,3152,3153) -
(3308,3308,3308,3309) - to take some random color mix - be represented
faithfully when reduced to the much smaller jpeg color space?

And even if not, does it matter when the devices I'm going to output
to can't even fully represent the entire jpeg color space?