Am 29.11.2016 um 13:52 schrieb scott:
>> Okay, sorry, I didn't think it was possible to compare different color
>> spaces like sRGB or Adobe RGB without selecting a white point.
> 
> The "basis" of most colour spaces is XYZ. XYZ defines a colour exactly
> (in terms of human perception) with just three numbers. The numbers are
> linear in terms of physical power, so if you have two light sources in
> very close proximity, you can simply add the two XYZ values to get the
> resulting colour. What this means though is that doubling XYZ does not
> give a perceived brightness twice as bright. Note there is no dependency
> on any "white point", the three numbers alone are enough.
> 
> sRGB in turn defines exactly what "colour" full red, green and blue are
> in terms of XYZ (you can look up the values). I say "colour", because
> obviously you can scale all the values to get a brighter or dimmer
> result, which is still technically sRGB. There is no white point assumed
> or needed in the definitions of sRGB either. The same goes for Adobe
> RGB, that defines a different set of RGB physical colours.
> 
> However, given that RGB are defined in sRGB exactly in terms of physical
> colour, this then defines exactly what physical colour "white" in sRGB
> is, ie if you add up equal ratios of the defined R,G and B, you get what
> is "white" in the sRGB colour space. This happens to be very close to
> D65, I assume by design.

This is not quite true.

sRGB does /not/ explicitly specify what full red, green and blue are.

What sRGB does define is the _xy coordinates_ - i.e. the absolute hue
and saturation - of red, green and blue (aka the "primaries"). This can
be visualized as defining the direction (but not the "length") of the
red, green and blue axes in XYZ space.

sRGB also /does/ explicitly define the _xy coordinates_ of the so-called
"illuminant whitepoint" - i.e. which defines what colours are nominally
"neutral", i.e. entirely desaturated - and also defines that such
neutral colours are to be represented by the red, green and blue
channels all set to the same value (which is typical for RGB colour
models). This can be visualized as defining the direction (but again not
the "length") of the RGB "cube"'s diagonal in XYZ space.

sRGB also /does/ explicitly define the _Y coordinate_ - i.e. the
luminance - of the brightest representable colour: 80 cd/m^2; however,
not everyone adheres to this.

sRGB also defines a host of other things that make matters even more
complicated, namely the "image surround", as well three properties for
both the "encoding" and "typical viewing" conditions: "ambient
illuminance level", "ambient white point" (which is D50, not D65), and
"viewing flare". (And no, don't ask me to explain what these parameters
mean - I barely understand those myself.)


>> Still, I would like to plot the horseshoe in 3D, anyway, using a
>> specific white point. (Or maybe multiple images, each with a different
>> white point.)
> 
> Sorry this just doesn't make sense. The "horseshoe" shape is totally
> independent of any white point, "using" a white point to plot it is
> meaningless, the shape is the same and fixed by the XYZ values for each
> wavelength of light only.

It /does/ make some sense to plot the 3D extension of the horseshoe for
pigment colours under a given illuminant.

It should be noted however that such a plot is non-trivial: It does
/not/ suffice to specify the xy coordinates of the illuminant; the
entire colour spectrum needs to be considered instead. For example with
illuminants from the F series (fluorescent lighting) we can probably
expect the 3D shape to show a distinctive "fingerprint" of the
illuminant's spectral emission lines (though I'm not sure how exactly
this fingerprint would look like).

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