Re: Photoshop Lab values
Re: Photoshop Lab values
- Subject: Re: Photoshop Lab values
- From: Graeme Gill <email@hidden>
- Date: Tue, 10 Jul 2007 10:48:20 +1000
Marco Ugolini wrote:
The way I see it (and I realize it's not a very scientific way to put it -- my apologies to Graeme),
> the blue primary in ProPhoto RGB is *outside* CIELAB for reasons of pure and simple mathematics.>
Well, no. Technically L*a*b* puts no limits on the a* and b* values, so the ProPhoto blue
is perfectly representable in L*a*b*. But the ICC L*a*b* PCS (Profile Connection
Space) does set limits on a* and b*, as does Photoshop, hence the clipping to
128.
If you convert values on the spectrum locus (ie. the span of visible monochromatic colors)
and the purple line (the mixture of monochromatic minimum and maximum visible wavelengths)
into L*a*b*, then you end up with some wild values for a* and b* (over 1000).
You can see that clipping very clearly in ColorThink, where the blue end of the profile's
> gamut looks like it has hit a wall and got flattened by the impact.
This would be a limitation of ColorThink. Any real system will have numerical
limits in it somewhere. Often when dealing with ICC based color a*b* limits
of +/- 128 are assumed. One of the unfortunate (IMHO) changes in ICC V4
is that software such as CMMs are now meant to clip all L*a*b* PCS values to +/- 128,
even when this is not technically necessary (such as when dealing with matrix/shaper
profiles). If software follows this specification, then there may be even more odd
behaviour when dealing with extreme profiles such as ProPhoto.
That seems to cause the RGB blue value at the extreme of the numerical scale (R 0, G 0, B 255)
> to sort of "fall to the floor", and return a nonsensical L* value near 0 for a color whose
> appearance is actually quite bright and saturated.
A colorspace such as L*a*b* doesn't really represent the appearance of such a color
well. Even CIECAM02 is not so good at this, since the appearance model it uses
doesn't incorporate the Helmholtz-Kohlraush effect, which models the way
very saturated colors appear lighter.
I'm not sure how the math is done and why a color with a value of L* 0 doesn't look
> pitch black, as it should in theory. I still haven't figured that one out.
A color with a zero L* and extreme a* and/or b* values is a paradox.
In principle L*a*b* separates different visual attributes. L* is meant to
correspond to the monochromatic lightness. a*b* (or the equivalent C and h)
are meant to represent the hue and Chroma ("Saturation"). But a color
with zero L* and very large a* and/or b* is somehow black but extremely
saturated. It's probably outside the spectrum locus (hence is an imaginary
color), but the best way to clip such a color to represent it as a real world
color can be debated. Should the fact that it's black dominate, or the
fact that it's highly saturated ? Should it be some balance between these two ?
Reasonable people can come to different conclusions on this.
Graeme Gill.
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