Re: Normalization XYZ -> xyz

Gernot Hoffmann wrote:

Thomas,

the transform(ation) x=X/(X+Y+Z) etc. is a planar perspective
transform.
The view plane is the plane through (1,0,0), (0,1,0), (0,0,1).
New coordinates are X-->x, Y-->y, Z-->z in the plane x+y+z=1.
The origin X=Y=Z=0 is center of projection.
According to Grassmann's laws we can expect: k*X, k*Y, k*Z delivers
the same chromaticities x,y,z for any k>0. That's indeed the case.

Yes, I agree. The transformation chosen by the CIE has a couple of properties one would require from such a transformation. My second "suggestion" would be a projection onto a sphere in R^3, again having the same property (namely, the constant-chromacity ray \lambda(x,y,z) is mapped to a single point on the sphere), the same holds for the first suggestion. (All of them are "projective", but map onto a different surface).

The transform x'=X/Y etc. is a perspective transform as well. The view
plane
is the plane through (0,1,0) parallel to X,Z (a plane normal is given
by the
coefficients in the denominator in either case).
Again, k*X, k*Y, k*Z delivers the same chromaticities x',y',z' for any
k>0.
Surprising !
The only possible chromaticity diagram would be z'(x'). Very nice,
because
Y is left out.

Yup, I agree.

But alas, this somewhat reasonable transform fails for X-->oo or Z--
oo.
In either case we'll get infinity for x' or z', whereas the classical
transform
results in one instead of infinity. This is IMO a good reason, not to
use
the suggested transform.

This is of course because the object mapped to is an infinite plane instead of one surface of a simplex (the CIE suggestion). I would believe (but please correct me if I'm wrong) that Z->infty or X->infty without also Y->infty is physically impossible (i.e. there is no part of the spectrum where the spectral density of Y is zero, which would be another way of saying that Y is the luminance. If a color is visible, it has to have a luminance in first place). IOW, the gammut will of course look different in this coordinate system, but it would nevertheless cover a finite area.

Back to the classic transform: so far we have mapped all values X,Y,Z
onto the plane x+y+z=1 (as explained above). The spectral locus on
this plane is already a kind of horseshoe shape. Which one is the best
(simplest) view onto this plane, from 3D to 2D ?
x,y (view in negative z-direction)
y,z (view in negative x-direction)
z,x (view in negative y-direction) or x,z (view in positive y-
direction)

The CIE consortium decided to use the x,y-coordinates, but the
reason
isn't obvious (for me).

*Which* of the coordinate transform you pick to express coordinates on the simplex surface is, indeed, arbitrary; I don't see any obvious reason here, except that x and y are the first two available coordinates gained by the projection, and sufficient to define the locus.

About CIE (1931):
http://www.fho-emden.de/~hoffmann/ciexyz29082000.pdf

About planar projections (especially chapter 11):
http://www.fho-emden.de/~hoffmann/project18032004.pdf

Thanks for your questions. Years ago I've been struggling with
this stuff because the formula apparatus appeared in all (available)
sources as deus ex machina.

You're welcome, thanks for the answer. (-: I think we can conclude that this is simply a convention, and one out of several possible ways to "normalize" a "light impression" to "a color impression".

Greetings,
Thomas
.

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