Talk:CIE 1931 color space

message: Can someone explain from where the relation Y = 0.2125R + 0.7154G + 0.0721B came from? The matrix equation says Y = (1/0.17697)(0.17697R + 0.81240G + 0.01063B). Am I missing something?-- 19:00, 26 July 2006 MM
--------------------------------------------------
===  15.4 LUMINANCE of sRGB ===
For CIE1931 xyY 2° the LUMINANCE is:
:Y=0.17697r+0.8124g+0.01063b
with:0.17697+.8124+.01063=1
For sRGB the LUMINANCE is:
:Y = 0.2126r + 0.7155g + 0.0721b   
with:0.2126+0.7155+0.0721=1.0002 
:
\mathbf{M}=
\begin{pmatrix}
0,4124564&	0,2126729&	0,0193339\\	
0,3575761&	0,715522&	0,11992	\\
0,1804375&	0,072175&	0,9503041
\end{pmatrix}
Y is given by the second colum of M matrix
Y = 0.2126r + 0.7155g + 0.0721b 
:(x,y,z)=((r,g,b)*M)n1
n1 is the normalization to 1
Divide each term by the sum of the row.
:
\mathbf{(M)n1}=
\begin{pmatrix}
0,412&	0,212&	0,020\\	
0,358&	0,715&	0,120\\
0,180&	0,072&	0,950
\end{pmatrix}
/
\begin{pmatrix}
 0.64 \\
 1.19 \\
 1.20\\
\end{pmatrix}
=
\begin{pmatrix}
0.64 & 0.33 & 0.03 \\
0.30 & 0.60 & 0.10 \\
0.15 & 0.06 & 0.79\\
\end{pmatrix}
Primaries 	                 	
coord.	x   	y   	z
Red   	0.64	0.33	0.03
Green 	0.30	0.60	0.10
Blue  	0.06	0.06	0.79
===15.4 Exemple de Adobe RGB ===
Il s'agit d'un système colorimétrique développé par Adobe en 1998. Il propose un gamut élargi par rapport à l'espace sRGB.
:
Couleurs                           	Rouge R 608nm	Vert G 538nm	Bleu B 468nm	Blanc  D65
Coordonnée x                       	0,640        	0,210       	0,150       	0,3127
Coordonnée y                       	0,330        	0,710       	0,060       	0,3290
Coefficient de luminance L         	0,29734      	0,62736     	0,07529     	total=1
Coef de luminance barycentrique L/y	0,29644      	0,29071     	0,41285     	total=1
Calculs des coefficients de luminance barycentrique
On cherche les luminances barycentriques a,b,c (Y/y) des 3 primaires qui donnent le point W(D65)  (x=0,3127,y=0,3290)(z=0,3590)br/>
a=Lr/yr   b=Lg/yg   c=Lb/yb 
Les formules donnant le barycentre des 3 primaires sont :
x=0,64 a+0,21 b+0,15 c=0,3127 
y=0,33 a+0,71 b+0,06 c=0,3290 
a+b+c=1 
La résolution de ces 3 équations à 3 inconnues donne :
a=0,889332/3 
b=0,872124/3 
c=1,238544/3 
yr*a=0,33*a=k*0,293479490572279/3 
yg*b=0,71*b=k*0,619207839894145/3 
yb*c=0,06*c=k*0,074312669533576/3
S=yr*a+yg*b+yb*c=k*0,987/3
yr*a/S=0,29734497=Lr
yg*b/S=0,62736356=Lg
yb*c/S=0,07529145=Lb 
Ce sont bien les valeurs données par Adobe RGB (arrondies à 5 décimales)  
La luminance visuelle du blanc sur un écran de référence est de 160,00 cd·m (X = 152,07=486,32x, Y = 160,00=486,32y, Z = 174.25=486,32z)
===15.5 Exemple de  sRGB ===
:
Couleurs                           	Rouge \{R~612nm\}	Vert \{G~547nm\}	Bleu \{B~464,5nm\}	Blanc \{ D{65} \}
Coordonnée x                       	0,64             	0,3             	0,15              	0,3127
Coordonnée y                       	0,33             	0,6             	0,06              	0,3290
Coefficient de luminance L         	0,2126           	0,62736         	0,07529           	total=1
Coef de luminance barycentrique L/y	0,2396           	0,3535          	0,4069            	total=1
Vérification des coefficients de luminance barycentrique
Les luminances barycentriques a,b,c (Y/y) des 3 primaires donnent le point W(D65)  (x=0,3127,y=0,3290)(z=0,3590)br/>
a=k*Lr/yr   b=k*Lg/yg   c=k*Lb/yb 
a+b+c=1 après normalisation à l'unité
On vérifie les formules donnant le barycentre des 3 primaires :
x=0,64a+0,3b+0,15c=0,3127 
y=0,33a+0,6b+0,06c=0,3290   — Preceding unsigned comment added by Bercier (talk • contribs) 15:34, 18 June 2020 (UTC)
thread_title: Luminance

message: :
Couleurs                           	Rouge \{R~612nm\}	Vert \{G~547nm\}	Bleu \{B~464,5nm\}	Blanc \{ D{65} \}
Coordonnée x                       	0,64             	0,3             	0,15              	0,3127
Coordonnée y                       	0,33             	0,6             	0,06              	0,3290
Coefficient de luminance L         	0,2126           	0,62736         	0,07529           	total=1
Coef de luminance barycentrique L/y	0,2396           	0,3535          	0,4069            	total=1
Vérification des coefficients de luminance barycentrique
Les luminances barycentriques a,b,c (Y/y) des 3 primaires donnent le point W(D65)  (x=0,3127,y=0,3290)(z=0,3590)br/>
a=k*Lr/yr   b=k*Lg/yg   c=k*Lb/yb 
a+b+c=1 après normalisation à l'unité
On vérifie les formules donnant le barycentre des 3 primaires :
x=0,64a+0,3b+0,15c=0,3127 
y=0,33a+0,6b+0,06c=0,3290   — Preceding unsigned comment added by Bercier (talk • contribs) 15:34, 18 June 2020 (UTC)
thread_title: 15.5 Exemple de  sRGB 

message: If you look at the biography of Grassmann, you'll see it mentions his color work.  The law comes, I believe, from his article "Theory of compound colors", Philosophical Magazine 4 (7), 1854, 254-264.  You can cross-check in a few library catalogues under, e.g. Sources of color science, ed. David L. MacAdam, MIT Press [1970]. --Macrakis 23:35, 11 July 2005 (UTC)
thread_title: Grassmann spelling

message: The outer contour iof the CIE figure lists wavelengths (from 380 to 700 nanometers), but what are the wavelengths of the colors along the straight line from 380 to 700?
How do they fit in, and why is that line straight?
: The wavelengths along the straight line at the bottom are called "non-spectral colors". They are colors which no monochromatic beam can match, they must be made up of a combination of two(only 2 and no more:the min of blue 380nm and the max of red 800nm) monochromatic beams. The outer contour is called the "spectral locus" and contains all of the colors that a monochromatic beam can match. In order to be a physically possible light beam, any beam must be made up of one or more monochromatic beams (or an infinite number for a continuous spectrum.)
: A second fact to note is that if you pick two points inside the range of possible colors (the gamut), every point on a straight line between the two must also lie inside the gamut. That means that every point between the two far ends of the spectral locus must lie inside the gamut. However, if a point were to lie outside the line connecting the two ends, then you would be unable to choose two or more monochromatic beams to specify that point. That means such a point is impossible. PAR 15:31, 6 September 2006 (UTC)
replies: text:  This is probably a silly question, but isn't purple a part of the spectrum? 86.5.107.158 03:08, 20 September 2006 (UTC)
replies: text:  The straight line mentioned above is sometimes called the "line of purples". If we define any color on the straight line as a type of purple, then, no, there is no wavelength that corresponds to purple. The sensation of purple can only be realized by using a combination of 2 monochromatic light sources(lower blue and bigger red). PAR 03:29, 20 September 2006 (UTC)
replies: text: Purple is non spectral, but violet and indigo are spectral, crazy world right?  — Preceding unsigned comment added by Da5nsy (talk • contribs) 22:06, 8 March 2013 (UTC)
thread_title:  Why is the bottom flat? 

message: The CIE XYZ color space was deliberately designed so that the Y  parameter was a measure of the brightness of a color.
Should not be "Z parameter"?
:No, its the Y parameter. PAR 15:03, 6 September 2006 (UTC)
replies: text: because we see the most bright color in the medium of wavelenght, and Y is the present of this wave lenght, so it's the best choise.  —Preceding unsigned comment added by 81.12.9.2 (talk) 20:31, 4 November 2007 (UTC)
replies: text: No; the CIE XYZ color space was most definitely not designed so that the Y coordinate (it's a coordinate, not a parameter) is the measure of brightness. The coordinate Y was assigned to the luminance of a color, which is a physical, not a perceptual (like brightness) attribute. It was believed that luminance was a linear combination of cone fundamentals, and the objection to existing color matching functions, such as König's, was that V(\lambda) could only be approximated by a linear combination of color matching functions. See Wright's anniversary paper, for example. In any case, brightness and luminance are two different, but related things, just as sound pressure and loudness are. Hope this helps! Lovibond (talk) 01:34, 17 January 2012 (UTC)
thread_title:  mistake 

message: Is the xy chromaticity diamgram on page 2 plotted at a fixed brightness?  Could one display any number of such diagrams with different brightnesses values or is this diagram always plotted at some special brightness?
:If you click on the diagram, it will take you to the diagram page. There is an explanation there of how the diagram was generated. PAR 01:54, 2 June 2006 (UTC)
thread_title:  chromaticity diagram 

message: Should
:\overline{y}(\lambda)=V(\lambda)
be instead something like
:\int \overline{y}(\lambda)\,d\lambda=\int V(\lambda)\,d\lambda
? As far as I understand, \overline{y} and V are separately and empirically determined up to a scalar constant, so it doesn't make much sense to set one equal to the other. Jcreed 04:32, 29 September 2006 (UTC)
replies: text: That certainly looks right. I have changed it, and will check that it is in fact true. PAR 00:25, 20 October 2006 (UTC)
thread_title:  normalization 

message: Bercier (talk) 13:50, 4 March 2013 (UTC)
thread_title: http://www.cis.rit.edu/mcsl/research/1931.php

message: The statement bothers me somewhat. Its true, if you can't find 3 points in the gamut that contain the gamut, then its not a triangle, but its also not some other things too. In other words, the converse is not true - If you can find 3 points that contain the gamut, that doesn't mean it IS a triangle. Only when you introduce the other fact, that the gamut must be convex, does the converse become true as well. PAR 15:59, 16 April 2007 (UTC)
replies: text: True; I considered mentioning convexity after I realized my statement was incomplete.  Feel free to remove or fix it. Dicklyon 17:06, 16 April 2007 (UTC)
thread_title:  Human gamut not a triangle 

message: I'm trying to understand how the CIE 1931 chromacity diagram is created and mapped to the horseshoe gamut. I experimented in MATLAB and wrote the following code with the output image at right. The output looks somewhat different from :Image:CIExy1931.png (for example, the luminosity bands in :Image:CIExy1931.png from the 460nm point on the border to 550nm and from 470nm to 605nm are not present in my image).
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% CIE 1931 color space
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
A = [256:256:3];
for i = 1:256,
	for j = 1:i,
		A( i, j, 1 ) = ( ( j - 1 ) / 256 ) * 255;
	end;
end;
for i = 1:256,
	for j = 1:i,
		A( i, j, 2 ) = ( ( 256 - i ) / 256 ) * 255;
	end;
end;
for i = 1:256,
	for j = 1:i,
		A( i, j, 3 ) = 255 * ( 1 - ( ( 256 - i + j ) / 256 ) );
	end;
end;
for i = 1:256,
	for j = 1:i,
		rgb_max = max( A( i, j, : ) );
		A( i, j, 1 ) = ( A( i, j, 1 ) / rgb_max ) * 255;
		A( i, j, 2 ) = ( A( i, j, 2 ) / rgb_max ) * 255;
		A( i, j, 3 ) = ( A( i, j, 3 ) / rgb_max ) * 255;
	end;
end;
A = uint8( A );
figure, imshow( A );
Mainly I'm curious about how an image such as mine is mapped to the horseshoe gamut. Perhaps this information should be in the article?
Are the regions outside the horseshoe simply clipped, or is there a geometric transformation which occurs? There is far more pure red, green and blue in my image than in the official CIE 1931 chromacity diagram. Thanks! ~MDD4696 23:04, 20 April 2007 (UTC)
replies: text: Interesting, my image looks very similar to the first "rogue" image at http://www.techmind.org/colour/rogues.html. ~MDD4696 23:08, 20 April 2007 (UTC)

text: So, if I interpret you correctly, I'd need a table of wavelengths which specifies values for X, Y, and Z (red, green, and blue) for each wavelength to create the horseshoe. I'm confused about what you mean by choosing a good X,Y,Z from the normalized x,y values though. Thanks for responding. ~MDD4696 23:31, 20 April 2007 (UTC)
replies: text: The simplest thing is to choose X=x, Y=y, and Z=1-x-y.  But you might want to scale them differently; or convert to RGB and then scale the largest to 255.  Find CIE XYZ data through the links at the top of this page. Dicklyon 00:03, 21 April 2007 (UTC)
replies: text: Click on the chromaticity diagram to get information about the image. It has a description of how it was generated. The r,g,b numbers that you use are assumed to define a particular color or chromaticity in the sRGB color space. There is then a transformation (described in that article) on how chromaticities in sRGB can be converted to CIE XYZ coordinates. The sRGB gamut is smaller than the CIE XYZ gamut, and when the color in the CIE diagram steps outside the sRGB gamut, the color is set to a presumably "close" value. (I think if you draw a line from the D65 white point, the color  on that line outside the gamut will be a constant, equal to the color where it first steps outside the gamut). Also - I *DO* see  a white  star in your diagram! I don't know why you dont. PAR 06:30, 21 April 2007 (UTC)
replies: text: There is a white star in mine, but there are also a few "bands" not present in mine in addition to the star. Thanks for trying to explain this all to me, I think I'm slowly figuring it out. ~MDD4696 02:35, 22 April 2007 (UTC)
thread_title:  Creating the horseshoe gamut 

message: The introduction says "X, Y, and Z [are] roughly red, green and blue, respectively". But section "The CIE xy chromaticity diagram" says: "the Y parameter was a measure of the brightness or luminance of a color". Which one is it? The two are certainly not the same as can be seen by the fact that yellow (586 nm) appears brighter than green (546 nm). — Sebastian 00:32, 10 June 2007 (UTC)
replies: text: It's both.  It's exactly luminance, by definition, and it's highly correlated with, or "roughly", green. Dicklyon 00:58, 10 June 2007 (UTC)
replies: text:  Thank you. How are the other two, X and Z, defined, then? If one can see them as weighing functions, do they have negative values at blue resp. red? (Maybe the answer is somewhere in our articles or maybe I could ask this in a better way; I'd have to study this some more.) — Sebastian 02:30, 10 June 2007 (UTC)
replies: text: They were picked to be a non-negative basis for the 3D subspace of spectral space spanned by the human cone sensitivities.  So, no, they are not negative anywhere, though most other sets of color-matching functions are.  The Z is picked to be the narrowest possible non-negative function on the blue end, and then X is left as the smallest possible non-negative final dimension, which ends up with a bump on the blue end as well. Dicklyon 05:34, 10 June 2007 (UTC)

text: Can the article be updated to reflect this?  I believe it is good practice to define terms before using them.  I do not understande this topic enough to do that myself - that was why I was reading it! 81.178.102.250 15:34, 10 July 2007 (UTC)
replies: text: Good idea.  I'll see if I can find a ref... Dicklyon 15:41, 10 July 2007 (UTC)

text: At constant radiance, yellow is not brighter than green. It looks that way on a computer because "pure yellow" is red+green, which will be brighter than just green. If we normalised the xy diagram to be even vaguely constant-intensity in sRGB, most of it would look pretty dark — the xy diagram is only supposed to approximate chromaticity. ⇌Elektron 02:26, 30 August 2007 (UTC)

text: Y, or luminance, was just an incorrect term. Because the human eye is so much more responsive to greens (look at the charts), the Y often correlates to brightness. Brightness should actually be X+Y+Z, or intensity. Y should not be used, and Intensity should be. In other words, Y is "perceived luminance"  —Preceding unsigned comment added by 70.42.127.107 (talk) 11:07, 7 June 2010 (UTC)
replies: text: There are two concepts here, the luminance of a light beam and the intensity of a light beam. The luminance of a color is a measure of how "bright" it appears to the human eye, the intensity is a measure of the total physical energy being delivered in a light beam, a concept which does not involve the human eye. The Y chromaticity value gives the luminance of any color. Y does not "often correlate" to brightness, it is the scientific definition of brightness, as measured by experiment. It is the integral of the spectral power density I(\lambda) times the \overline{y}(\lambda) function over all wavelengths \lambda. The intensity is simply the integral of I(\lambda) over all wavelengths. X+Y+Z is NOT intensity, because \overline{x}(\lambda)+\overline{y}(\lambda)+\overline{z}(\lambda) does not equal unity at all wavelengths. For example, the X, Y, and Z values for a beam of light in the microwave region are all zero, yet the intensity is not. The \overline{x}(\lambda),\overline{y}(\lambda),\overline{z}(\lambda) functions are not the only possible functions that could be used. Any set of functions that covered the visible spectrum could have been used. The developers of the CIE functions decided to make the \overline{y}(\lambda) function serve double duty, in order to save on computation. It would not only represent a particular band in the greenish area of the visible spectrum, but would also give the luminance of the color being evaluated. PAR (talk) 15:38, 7 June 2010 (UTC)
thread_title:  What is Y? 

message: The sets of RGB-primaries along with white point are often defined with chromaticities solely, i. e. 4 xy-pairs for red, green, blue and white. For example, this is the only information stored in EDID, as well as numerous reference tables of standard RGB spaces throughout the Wikipedia and many (but not all) other sources.
How these chromaticity sets can be converted to XYZ- or Yxy-sets?
Of course, within common mathematics we can't just add 4 single values for luminance — there will always be some 4 degrees of freedom. But with the concept of that R+G+B=W and the assumption that we don't need absolute Y-values, just relative ones (i. e. fixing luminance of white point to Y=1 or Y=100), it sounds possible.
As a result, I developed some equations, starting with Xr+Xg+Xb=Xw, etc.
* The source chromaticity data: (x_{W},y_{W}),\ (x_{R},y_{R}),\ (x_{G},y_{G}),\ (x_{B},y_{B}).
* Additionally, we fix Y_{W} = 1 (or any other positive value, maybe 100).
* Some intermediate variables (as in Yxy->XYZ, but without Y-multiplier):
** X'_{W}=\frac{x_{W}}{y_{W}}, \  \ldots \ , \ X'_{B}=\frac{x_{B}}{y_{B}}
** Z'_{W}=\frac{1-x_{W}-y_{W}}{y_{W}}, \  \ldots \ , \ Z'_{B}=\frac{1-x_{B}-y_{B}}{y_{B}}
* Some auxiliary variables (differ only in first X'; the subscript G is referring to the RGB-primary we start with):
** a_{G}=\frac{X'_{W}-X'_{B}}{X'_{R}-X'_{B}}, \ b_{G}=\frac{X'_{G}-X'_{B}}{X'_{R}-X'_{B}}
* After some reductions, we finally have the answer:
** Y_{G} = \frac{Z'_{W} - a_{G} \times Z'_{R} - a_{G} \times Z'_{B}}{Z'_{G} - b_{G} \times Z'_{R} - b_{G} \times Z'_{B}} \times Y_{W}
** Y_{R} = a_{G} \times Y_{W} - b_{G} \times Y_{G}
** Y_{B} = Y_{W} - Y_{R} - Y_{G}\,
As you can see, all this seems to give unambiguous results. But, despite of looking very credible and passing all the checks (by definitions), the results for well-known RGB spaces, such as sRGB or Adobe RGB, don't match predefined values of Y (refer to Bruce Lindbloom). Could someone tell me what I'm missing? Which degree of freedom was not eliminated, and how could this happen with such precise formulae?
For your convenience, I've prepared a small Excel spreadsheet which does these calculations automatically. It can be downloaded here: avsco.nm.ru/EDID2XYZ_eng.xls (only 17 Kbytes in size, no archiving).
Any hints are appreciated. Thanks in advance. 213.234.235.82 11:26, 17 October 2007 (UTC)
replies: text: I haven't studied it in detail yet, but where you say R+G+B=W and Xr+Xg+Xb=Xw, you probably mean to use R=G=B at whitepoint or something like that.  That is, spaces are specified such that equal amounts of the primaries add to form the color defined as white.  Dicklyon 17:30, 17 October 2007 (UTC)
replies: text:  This topic is about primaries, i. e. red, green, blue and white at maximum intensity. 213.234.235.82 17:42, 17 October 2007 (UTC)
replies: text: Primaries and whitepoints are chromaticities, independent of intensities.  Dicklyon 04:17, 18 October 2007 (UTC)
replies: text:  Color stimulus is unthinkable without intensity. My question is about primary stimuli of a real display device, which has certain luminocity aside from chromaticity. Only this way the sum of red, green and blue can render the specified white point (no matter whether Y is normalized to 1 for white or not). Otherwise, if we just take some arbitrary "reddish", "greenish", "bluish" and "neutral" stimuli, which do not form a single system, the conversion from 4×xy to 4×XYZ would not be possible, of course. 62.118.220.182 08:40, 18 October 2007 (UTC)
replies: text: But if you're going to asking for help, listen to the answers (and preferably not on an article talk page, since this is supposed to be about article content; but I'll assume that's where you're trying to go with it).  A chromaticity is a color with intensity factored out.  An xy specification of the primaries doesn't tell you how intense those primaries are, and in general they are variable, the max limit notwithstanding.  The whitepoint is a specification of the neutral chromaticity that you get by producing "equal" amounts of the three primaries; a color at the whitepoint chromaticity can be white, gray, or even nearly black.  By "equal" is meant that the colorspace specifications of the primary intensities, generally called the R, G, and B values, are numerically equal; these multiply whatever the intensities of the primary sources are; for example, if you scale the R, G, and B range to be 0 to 1, then they multiply the maximum intensity of the primaries.  If you just want to look at brightest white, that's R=G=B=1, but in general any neutral color has R=G=B and generates a color with chromaticity equal to the white point.  There is no sense in which the sum R+G+B is a meaningful quantity in colorimetry as I know it.  Hope this helps. Dicklyon 14:54, 18 October 2007 (UTC)

text: As for help figuring out why your specific formulas are giving you wrong results, you might try emailing Bruce Lindbloom directly.  He has in the past been quite responsive and helpful.  Or perhaps for a broader response, try the Apple ColorSync users mailing list.    --jacobolus (t) 20:14, 18 October 2007 (UTC)
thread_title:  Converting a full set of chromaticities (''xy'' for RGB+W) to a full set of triplets (''Yxy/XYZ'') 

message: The article manages to imply that there are objective color values that are universally recognized as red, green, and blue. It should be more clear about the fact that the primary colors used in any color model are chosen to make the model work, not derived from objective fact. --Isaac R 17:58, 18 October 2007 (UTC)
thread_title:  Red, Green and Blue Not Objective Concepts, 

message: From what I understand, the XYZ- or xyz-space has been chosen so that the so-called "desirable properties"
would be met. For instance, the Y value had to be equivalent to the experimental luminosity of the different wavelength to the human eye. And the X and Z values were defined in a purely mathematical way so as to fit the gamut as best as possible, and so that the other conditions would be met, such as being positive everywhere, etc. For instance, if I was to take the values on the graph here, normalise them so that the sum is 1 for each wavelength, I would get a parametrisation of the horsehoe shape on the chromacity diagram.
Now, on the Tristimulus page, it is said that tristimulus values are "the amounts of the three primary colors in a three-component color model needed to match that test color. The tristimulus values are most often given in the CIE 1931 color space, in which they are denoted X, Y, and Z."
Something is wrong. From the definition, tristimulus values is the amount of the three primary colors needed to match a test color, (sometimes "negative", unless the gamut is a perfect triangle, which it is not). In this case, there is a color corresponding to each positive set of values, since they correspond to the superposition of (existing) primary colors. However, in the case of X,Y,Z, there are triples not corresponding to any color, such as (1,0,0), (0,1,0) and (0,0,1). X,Y,Z are not supposed to correspond to any primary colors. Or do they correspond to convenient "imaginary" colors that are of no particular interest, except for generating a nice parametrisation? If this is the case, I'll add something to that effect on the Tristimulus page. Ratfox 23:29, 31 October 2007 (UTC)
thread_title:  Isn't the word tristimulus counterintuitive? 

message: What research is done that results the images in the article (the curves)? There must have been some research project behind the images with the curves in, from which the images then is generated. Is it possible to get the result of that research, for further researching? Or at least the data the images is generated from?
Since the images could be analyzed and data extracted from the curves, the data should be publicated somewhere to get. An analyze would be such a work around... :-\
-Kri 23:25, 11 November 2007 (UTC)
thread_title:  Is there any tables to generate the images in the article? 

message: What I mean is, what range of wavelengths are considered to be part of the visible colour spectrum, as opposed to infrared or ultra-violet?  I can see from the article that it's roughly 400-700 nm, but I wondered whether there is a more precise definition used by the CIE, or other standards body. For example, the article visible spectrum uses the range 380-750 nm. Do those values come from the CIE? (a reference would be appreciated).  Thanks, Thunderbird2 (talk) 12:03, 12 January 2008 (UTC)
replies: text: There are a number of different tabulations of the CIE color matching functions, beginning with the 1931 standard version which covered 360-830 nm. The "abridged" values were a more "user friendly" version (back when calculations were done by hand) and they ran from 380-780 nm. There are others, 1964 improved version, etc. etc. I don't know if there is a set of numbers that the recent literature has settled on, but the 360-830 probably covers everything. Check out Wyszecki and Stiles ("Color Science") for tables of about twenty different versions of color matching functions. PAR (talk) 17:10, 12 January 2008 (UTC)
replies: text: But with respect to the original question, parts of that extended range would certainly be considered IR and UV as opposed to visible, even though it has some (very small) visible effect; 400 and 700 are the most common "conventional" boundaries, but I don't know that they have any special status. Dicklyon (talk) 18:12, 12 January 2008 (UTC)
replies: text: Thanks for your replies - both helpful. Thunderbird2 (talk) 18:44, 12 January 2008 (UTC)
thread_title:  What are the extremes of the visible colour spectrum? 

message: Why?
* The so called (→External links: clean up) removed the links I typically use on this page
* But dead links were NOT removed
* This is exactly what a link clean-up NOT should do
http://cvrl.ioo.ucl.ac.uk/  —Preceding unsigned comment added by 80.156.4.26 (talk) 19:24, 27 February 2008 (UTC)
thread_title:  Undo removal of external links 

message: "The numbers below all have the correct number of significant digits per CIE standards." Does this mean that all of the digits defined by the CIE standards are given (and thus all source information is kept) or that the numbers have been rounded (and thus source information is lost) according to error limits given in the standards? Please disambiguate this sentence!--SiriusB (talk) 21:29, 13 April 2009 (UTC)
thread_title:  Significant digits in Matrix 

message: How exactly has this image been created? The description page only says that the data have been retrieved from here, but on that page there is no RGB CMF, just XYZ CMFs. Has the inverse of the RGB-to-XYZ conversion marix been used? If so, please add this info to the description page. On the cvrl site are other RGB CMFs that differ greatly from this one. In particular, the red curve of the Stiles & Burch (1955) (text file) CMF peaks near 3 while blue and green are both around 1.--SiriusB (talk) 21:49, 13 April 2009 (UTC)
thread_title:  RGB color matching function 

message: I have reverted the replacement of the CIE 1931 Chromaticity Diagram for the following reasons:
* This was not a fixed picture, it is a damaged picture. Removing the "sRGB line artifacts" destroys the color accuracy inside the sRGB triangle. In the original, these colors were carefully reproduced from the xyz values of the sRGB colors, and the "line artifacts" are the unavoidable result of this accuracy.
* The description of the original (CIExy1931.png) gives an exact explanation of how the original color diagram was generated. If the image is replaced, the description should be altered to reflect the exact method by which the new image was generated. This .png image was then replace by a somewhat inferior .svg file (CIExy1931.svg) which nevertheless reproduced the colors accurately. I have copied the description of how the colors were generated over to the description of the .svg file.
* sRGB cannot accurately represent colors outside the sRGB triangle, so the statement that the colors outside the sRGB triangle are "more accurately represented" is questionable at best, and wrong at worst. The method by which the colors outside the triangle are generated were stated in the original. The same reasons should be explicitly stated for any replacement and the reasons why they are more accurate than the original should be given.
PAR (talk) 23:42, 11 June 2009 (UTC)
replies: text: BenRG has now remade the new svg now, with good explanations of exactly what he did.  I agree with him that eliminating the various line artifacts is more important than trying to minimize the pointwise error inside the sRGB gamut region, as it makes the image a visually better representation of what it's trying to represent, within the gamut limitations of sRGB.  Dicklyon (talk) 03:13, 25 September 2009 (UTC)
replies: text: The numbers on the left cause the entire work to be rather unbalanced. In fact, I find that all the numbers are visually rather annoying, conveying an aura of materialistic obsession. Also, the sharp red corner on the right is really at odds with rest of the diagram and should be rounded a bit, don't you think?. There are still rather sharp gradations of color, but these are acceptable, indeed desirable, since they are indicative of the artist's existential struggle with his inner self within the context of the multifaceted aspect of cosmic reality. :)  PAR (talk) 09:10, 25 September 2009 (UTC)
thread_title:  Revert to previous image of color space 

message: Is it just me or is
"The CIE XYZ color space was derived from a series of experiments done in the late 1920s by W. David Wright[3] and John Guild.[4] Their experimental results were combined into the specification of the CIE RGB color space, from which the CIE XYZ color space was derived."
really saying CIE XYZ -> CIE RGB -> CIE XYZ ?  —Preceding unsigned comment added by 93.133.208.217 (talk) 17:33, 23 August 2009 (UTC)
replies: text: Looks to me like it says measurements -> CIE RGB -> CIE XYZ, which is about right. Dicklyon (talk) 03:14, 25 September 2009 (UTC)
thread_title:  Irritated by the 2nd paragraph 

message: I have noticed that this has been brought up before, but the current wording in the article is (presumably) as confusing now as it was when this was discussed before. Section CIE_XYZ#The_CIE_xy_chromaticity_diagram_and_the_CIE_xyY_color_space reads "Less saturated colors appear in the interior of the figure with white at the center." I understand the text is meant to talk about white in a "zero-chromatic", "white point" sense – but it doesn't do the job properly. Especially since the word "white" is actually linked to the article white, which obviously talks about white-white, not "white point" white. Given the number of purists around here I was reluctant to be bold, since I was afraid I'd risk ending up with something even worse – but please, cut regular folk some slack and reword that for clarity!
Having said that, is there a given top level in the xyY shape of the human gamut, at least in theory? (In practice there must be a limit, since your retina would surely burn if you pushed the luminance too high, but I haven't found any definitive answer regarding the theory.) I assume that if there is one, the top level should be a point, just like the section at Y=0, since there's no possible chromatic variation with pure white -- is that correct? Speaking of which, is there any representation of that 3D shape in its entirety? I understand it would serve little practical purpose, especially since from the outside you'd see mostly colors outside the gamut of any device, but I think it would be a good helper for putting the concept into perspective, if you'll excuse the pun. Oh, and finally, in such a three-dimensional representation, what would be the shape of the section we're used to see? Would it be a plane parallel to xy? A spherical cap? A paraboloid? None of the above? --Gutza T T+ 02:44, 27 September 2009 (UTC)
replies: text: It looks to me like White describes the concept as described in this article.  White, in terms of chromaticity, just means neutral; there's no intensity implied, and there's not max white in xyY or XYZ space.  There is no bounding 3D shape for these spaces.  A chromaticity diagram is a slice through the space, anywhere that cuts in with a finite cross section, and not including the origin; xy and uv and rg are examples; they all look fairly similar, just stretched differently.  Dicklyon (talk) 03:09, 27 September 2009 (UTC)

text: It depends on what you’re interested in. For colors of surfaces, often a perfectly diffusing reflector is taken to be “white”. For light sources, there’s no upper limit to luminance, but when the eye has adapted to some particular surroundings, there is some brightness beyond which the eye’s photodetectors will no longer notice any differences. Take a look at http://www.scholarpedia.org/article/Color_spaces. –jacobolus (t) 22:30, 8 June 2010 (UTC)

text: It's pretty arbitrary; I think they just picked the first two of the over-complete set of three.  Dicklyon (talk) 17:29, 27 September 2009 (UTC)
replies: text:  At the beginning, using x and z in a chromaticity diagram was also somewhat common. I believe (but am not positive) that the x–y diagram is from Judd. –jacobolus (t) 22:32, 8 June 2010 (UTC)

text: There's no need for us to put our speculations or feelings into the article; find a source or don't mention it. Dicklyon (talk) 17:55, 27 September 2009 (UTC)
thread_title:  "White" 

message: "One might ask: "Why is it possible that Wright and Guild's results can be summarized using different primaries and different intensities from those actually used?" One might also ask: "What about the case when the test colors being matched are not monochromatic?" The answer to both of these questions lies in the (near) linearity of human color perception. This linearity is expressed in Grassmann's law."
The linearity in Grassmann's law is due to the fact that it uses linear arithmetic to add amounts of light energy. By definition, the total light from two equal sources is twice that of one of the sources. In fact, human perception of light intensities is very non-linear, but this is irrelevant in color matching, since the two colors that match produce the same cone excitations. The relative visual response of unequal excitations is never compared in the determination of matches. Matching is the only situation to which Grassman's law applies, and it must apply by definition of the experimental conditions (matching) and the measurement units (intensity) involved.
Because of the linear arithmetic, other things also hold, such as the chromaticity of a mixture of two colors lying on a line connecting the chromaticities of the two components.
Old tv nut (talk) 01:26, 21 April 2011 (UTC)
replies: text: Doesn't the statement "(near) linearity of human color perception" rather than human perception of intensity make the statement correct? PAR (talk) 05:23, 21 April 2011 (UTC)

text: It has to do with the (near) linearity of the independent cone responses with respect to different wavelengths. That is, if two different stimulus spectra produce the same color sensation, Grassman's law says that when you double the intensity of each stimulus, the two resulting stimuli will still produce the same color sensation (that is, the same as each-other; different obviously from the first one). It turns out that this statement is not quite empirically true. However, it is close enough to true to be useful in constructing models; without it, it is impossible to summarize the color sensation induced by some spectrum in terms of coordinates like XYZ or similar. –jacobolus (t) 05:37, 21 April 2011 (UTC)

text: It's not linearity of the visual system as a whole, but it is linearity of the conversion from spectrum to effective stimulus.  That is, the effective stimulus is a weighted intensity, weighted via the spectral sensitivity curves of each rod and cone type.  Grassman's law essentially means that it's sensible to treat this as a dot product.  Also, for it really to be true, the different cone types need to respond with similar nonlinearities or adapted gains at least, and the rods need to be out of the picture; so it doesn't work at very low (mesopic and scotopic) levels.  Dicklyon (talk) 01:38, 23 April 2011 (UTC)
thread_title:  Grassmann's law linearity is not due to visual system linearity 

message: The article does not mention any reference to the modifications of CIE 1931 by Judd (1951) and Vos (1978) . And apparently, these modifications are essentially not used in any standard (including modern ones). Are these modifications officially rejected for some reason?--SiriusB (talk) 12:13, 5 September 2011 (UTC)
replies: text: I think you’d be best served by reading some basic color science or colorimetry book, instead of asking questions piecemeal across 5 different talk pages. These updated CMFs were never “rejected”, but they never gained much use, because there is no way to convert data collected assuming the CIE 2° standard observer to another set of CMFs, and so any official modification would make a break in measurement compatibility. Basically, the 1931 standard observer is “good enough” that maintaining compatibility between measurements is more important than the quite slight changes that would result from corrections. –jacobolus (t) 18:16, 5 September 2011 (UTC)
replies: text:  If Wikipedia, including its discussion pages, requires the user to read all relevant literature first, before being able to understand the articles, rather than providing exactly this basic information (and the question why an 80 years old standard is still not yet replaced by an improved one is exactly such a basic and obvious one), then WP is doing a bad job and is rendering itself irrelevant. You are the second one replying to me in such a manner "read the books (which?) before consulting WP", and therefore the second argument to me to leave WP forever. The third may be the final one. WP has its best days passed. Really. BTW If you know any references to this "the 1931 standard is still good enough", why don't you just add this in the article?--SiriusB (talk) 10:30, 7 September 2011 (UTC)
replies: text:  I’ll agree with you that Wikipedia’s articles on these topics are quite lacking in many ways. The problem is that it’s hard to synthesize any kind of coherent introductory material given that the organizational structure here demands articles strictly limited to one subject, and no one has really put much effort into writing clear and sufficiently complete basic pages that can be linked to from all the others. I’m not suggesting you find a book just to chase you away (indeed the articles here need all kinds of work if you’re willing to do the research and writing that it takes to improve them). I just actually think you’d get more out of time spent on a book than trying to learn only from Wikipedia pages. In re books, I think Hunt’s Reproduction of Colour is pretty good, or if you’re interested in CIE colorimetry more specifically, there’s a recent book called Colorimetry: Understanding the CIE System, edited by Schanda. Really though any book would probably be helpful, even ones that are 30 or 40 years old: any local public library probably has a couple. Why the CIE 1931 color space persists is actually quite a deep and tricky problem, and answering it properly would take a few thousand words. The reason I haven’t added everything I know about anything to Wikipedia is that time is quite a limited resource and the number of such things to be added is large.  –jacobolus (t) 20:20, 7 September 2011 (UTC)
thread_title:  Modifications by Judd (1951) and Vos (1978) 

message: HOW TO FIND 0.49 0.31 0.20  0.17697 0.81240 0.01063 0 0.01 0.99?
sbar=xbar+ybar+zbar   
x=xbar/sbar  y=ybar/sbar z=zbar/sbar  
(x,y,z)=(xbar,ybar,zbar)n1      
n1 means normalization to unity   
divide each term by the sum of the line.
V1924=ybar relative luminous efficiency. 
X=.49r+.31g+.2b           
Y=.17697r+.8124g+.01063b  Y is the relative LUMINANCE. 
Z=.01g+.99b                
L=X+Y+Z=X/x=Y/y=Z/z                    
L=Y/y is the relative BARYCENTRIC LUMINANCE .
xbar=.49rbar+.31gbar+.2bbar  
ybar=.17697rbar+.8124gbar+.01063bbar   
zbar=.01rbar+.99bbar      
we put D=d.r+e.g+f.b
V=V1924
rbar=r*V/D
gbar=g*V/D
bbar=b*V/D
so (rbar,gbar,bbar)=(r,g,b)*V/D who is one identity always right.
so S(A)=sum of 5 nm in 5 nm from 360nm  to 830 nm of the function A
S(V1924)=S(xbar)=S(ybar)=S(zbar)=S(rbar)=S(gbar)=S(bbar)=S
By approximations on d,e and f you find:
for S(rbar)=S(gbar)=S(bbar)=S gives 0.17697 0.8124 and 0.01063
x>0  and z>0 gives by approximations 0.49 0.31 0.20 and 0 0.01 0.99
It's very easy with Excell. 
 
: I have no idea what you’re asking about. –jacobolus (t) 20:50, 11 October 2012 (UTC)
thread_title: COLOR MATCHING FUNCTIONS

message: In this model, Y means luminance, Z is quasi-equal to blue stimulation, or the S cone response, and X is a mix (a linear combination) of cone response curves chosen to be nonnegative. Thus, XYZ may be confused with LMS cone responses. But in the CIE XYZ color space, the tristimulus values are not the L, M, and S responses of the human eye, even if X and Z are roughly red and blue. Rather, they may be thought of as 'derived' parameters from the long-, medium-, and short-wavelength cones.
The CIE has defined a set of three color-matching functions, called \overline{x}(\lambda), \overline{y}(\lambda), and \overline{z}(\lambda), which can be thought of as the spectral sensitivity curves of three linear light detectors that yield the CIE XYZ tristimulus values X, Y, and Z.  The tabulated numerical values of these functions are known collectively as the CIE standard observer.
:(l,m,s)=(xbar,ybar,zbar)*Ut-1
:Ut-1 is a 3x3 matrix
:(L,M,S)=(l,m,s)*D1
:D1 is a 3x3 diagonal matrix diagonal terms are 330 286 55.
For CIE1931 2° xyY
:V1924= \overline{y}(\lambda)
LUMINOUS EFFICIENCY
: I(\lambda)=k
EQUIENERGETIC LIGHT
:X=0.49.r+0.31.g+0.20.b        
:Y=0.17697.r+0.81240.g+0.01063.b
:Z=0.r+0.01.g+0.99.b
:Y is the relative LUMINANCE.
:Y/y is the barycentric RELATIVE  LUMINANCE  :
:Y/y=L=X+Y+Z=X/x=Y/y=Z/z=0.66697.r+1.1324.g+1.20063.b
:x=X/(X+Y+Z)
:y=Y/(X+Y+Z)
:z=Z/(X+Y+Z)
:X<>Xa
:Y<>Ya
:Z<>Za
For having :X=Xa,Y=Ya,Z=Za, you have to integrate for dλ ->0 like for a ray of light(monochromatic light).
THEN:
:Xa=  I(\lambda)\,\overline{x}(\lambda)
:Ya=  I(\lambda)\,\overline{y}(\lambda)
:Za=  I(\lambda)\,\overline{z}(\lambda)
then:
If:
Sa=Xa+Ya+Za
:x=Xa/Sa
:y=Ya/Sa
:z=Za/Sa
x=Xa/(Xa+Ya+Za)=xbar/(xbar+ybar+zbar)
y=Ya/(Xa+Ya+Za)=ybar/(xbar+ybar+zbar)
z=Za/(Xa+Ya+Za)=zbar/(xbar+ybar+zbar)
If:
\overline{s}(\lambda)=\overline{x}(\lambda)+\overline{y}(\lambda))+\overline{z}(\lambda)
:x=  \overline{x}(\lambda)/\overline{s}(\lambda)
:y=  \overline{y}(\lambda)/\overline{s}(\lambda)
:x=  \overline{z}(\lambda)/\overline{s}(\lambda)
we can write more simply:
:x=  \overline{x}/\overline{s}
:y=  \overline{y}/\overline{s}
:x=  \overline{z}/\overline{s}
so:
:(x,y,z)= ( \overline{x},\overline{y},\overline{y})n1
n1 normalization to unity
divise x,y,z by the sum x+y+z
Bercier (talk) 10:45, 10 April 2013 (UTC)
thread_title: Meaning of X,Y and Z

message: I removed something like that from the article once before.  It's not clear what the "X a" notation is supposed to represent, or why the non-relevant term "LUMINANCES" is introduced here, or whether there's an important point being made in the change of the limits of integration, or where the coefficients come from in the rgb matrixing (CIE RGB?  sRGB?  LMS? or what?).  WIth some clarification, including links to a relevant source or two, maybe we can see what problem Bercier is trying to solve.  Dicklyon (talk) 19:50, 12 November 2012 (UTC)
thread_title: Discussion

message: Ya is one indefined integral but really one summation by 5 nm to 5nm between the extremities of the visible spectrum like tabulated by V1924=ybar by the CIE 1931-2° also 360nm to 830nm.
For CIE xyY 1931-2° Ya is for calculation
 
1/if you take a ray of wawelenght λ
and for the equi-energetic spectre S(λ)=S 
Ya=xbar*S
also y=Ya/(Xa+Ya+Za)
we have also y=Y/(X+Y+Z)
but Y( LUMINANCE ) is different of Ya:a for ABSOLUTE LUMINANCE
2/For all the spectrum from 360nm to 830nm
and for the equi-energetic spectre S(λ) 
Ya is the area under the curves of the integral and the CIE 1931xyY-2° 
choose   Xa=Ya=Za
3/ You have also:
Ra=Ga=Ba
r=Ra/(Ra+Ga+Ba)=g=b=1/3
4/To return to WRIGHT's(10) and GUILD's(7) experimentations
we have to inverse the matrix Mt
(rbar,gbar,bbar)=(xbar,ybar,zbar)*Mt-1
rbar=2,3635xbar	    -,5151ybar	+,0052zbar
gbar=-,8958xbar    +1,4264ybar	-,0145zbar
bbar=-,4677xbar	    +,0887ybar	+1,0093zbar
sbar=rbar+gbar+bbar
r=rbar/sbar
g=gbar/sbar
b=bbar/sbar
r,g,b are the calculed WRIGHT+GUILD'results with 2nm differencie with the WRIGHT+GUILD 's real experimentations(10+7) 1930.
5/we can re-start with r,g,b
 
Y est déterminé pour que les surfaces sous les courbes rbar,gbar et bbar soient égales entre elles et égales à la surface sous V1924.
soit pour que Ra=Ga=Ba
soit D=dr+eg+fb qui est une identité
				
En calculant :
				
(rbar,gbar,bbar)=(r,g,b)*V1924/D on trouve par ajustements successifs
equivalents à la méthode des moindres carrés ou à la méthode d'approximations successives de NEWTON ou à un calcul itératif:
donc on trouve
				
d=0.17697  e= 0.8124  f=0.01063	
			
			
On détermine ainsi Y=0.17697r+0.8124g+0.01063b la LUMINANCE à un coefficient 1/0.17697=5.65(08) près qui ne joue pas sur les différents calculs.			
				
Ensuite en écrivant que xbar et zbar(donc x et z) ne comportent aucune valeur négative et que le diagramme y(x) est tangent aux axes on trouve par ajustements successifs 				
				
en calculant xbar=arbar+bgbar+cbbar   et zbar=grbar+hgbar+ibbar:
				
a=0.49    b=0.31      c=0.2 
 				
g=0       h=0.01      i=0.99
				
ce qui permet de calculer 
X=0.49r+0.31g+0.2b et
Z=0.01g+0.99b
				
				
et donc la luminance relative barycentrique L=X+Y+Z qui permet avec le gamut de déterminer le mélange de n lumières comme barycentre des points x,y,z affectés du poids de leur luminance relative Ln.
all these calculations are very easy with Libre Office tableur
reference:BROADBENT
Bercier (talk) 12:33, 6 February 2013 (UTC)
replies: text: Bercier, I don't follow.  Maybe you can start by pointing out a source that distinguishes Ya from Y as you do.  And we use the term "luminance" only for Y.  It's not clear what term in English would be better for the "relative luminance Y/y" that you refer to; maybe tristimulus component Y?  Dicklyon (talk) 15:34, 13 November 2012 (UTC)

text: Sorry, you link is not working for me and I don't follow your point.  I don't see anything like the Xs notation.  Are you just worrying about the imperfect precision of the sums of the CMFs?  Dicklyon (talk) 22:54, 13 November 2012 (UTC)

text: You are nevertheless unintelligible.  A couple of working links could help.  Dicklyon (talk) 19:19, 14 November 2012 (UTC)

text: Bercier:  But I’ve been letting Dicklyon talk to you here, because I honestly can’t figure out what you’re trying to say, and therefore I have nothing to add. –jacobolus (t) 01:03, 15 November 2012 (UTC)
thread_title:  Ya integral is ABSOLUTE LUMINANCE

message: http://www.cis.rit.edu/mcsl/research/1931.php  
1. ↑ A.RE-DETERMINATION OF THE TRICHROMATIC COEFFICIENTS OF THE SPECTRAL COLOURS BY W.D.WRIGHT MS,received,7th February,1929 Read and discussed,14th March ,1929. pages 141-164 
dans Transactions of the Optical Society, vol. 30, no 4, 1er mars 1929
2.↑ A.RE-DETERMINATION OF THE MIXTURE CURVES OF THE SPECTRUM MS,received,3 1stJanuary,1930. Read and discussed,10th April,1930. pages 201-218 
 
3.↑ 
http://www-cvrl.ucsd.edu/cmfs.htm
CVRL Color & Vision database CIE (1931) 2 et 10 degré color matching functions λ xbar ybar zbar
4.↑
http://www.cis.rit.edu/mcsl/research/broadbent/CIE1931_RGB.pdf
Calculation from the original experimental data of the CIE 1931 RGB standard observer spectral chromaticity Co-ordinates and color matching functions
By A.D. Broadbent,Département de génie chimique, Université de Sherbrooke,Québec,Canada J1K 2R1
Description of the data obtened by Wright
5.↑ 
http://sellig.sed.myriapyle.net/
Restitution formelle de la théorie sous-jacente à la norme colorimétrique XYZ CIE 1931
Calcul de la matrice de passage de RGB à XYZ par la résolution d'un systeme de 6 équations à 6 inconnues.
6.↑ 
http://www.photo-lovers.org/color.shtml.fr
Bercier (talk) 14:59, 23 March 2013 (UTC)
thread_title:  Références 

message: My French and online translator are not good enough for me to quite understand the point.  Apparently Bercier thinks the article is hash by not distinguishing Y, Ya, and Ys, whatever those are.  If someone can help interpret, that might be useful.  Dicklyon (talk) 19:01, 13 April 2013 (UTC)
thread_title: More discussion

message: There is an important information missing: Does the 2-degree figure refer to the angular radius (i.e. up to 2 degrees off the central visual axis) or the diameter (i.e. 2 degrees rim to rim)? Even the freely available sources cited here do not clarify this. Or is there some convention how fields of view are specified?--SiriusB (talk) 19:56, 29 July 2013 (UTC)
replies: text: Thanks. This should be written explicitely in the article since it is far from being self-explanatory (normally, the radius is used where an approximate radial or axial symmetry is present). I have added it now. The Wright paper is copyright-censored and not available to the library account of my institute, so I cannot look up (unless you refer to another paper than the one the 1931 standard is based on).--SiriusB (talk) 09:50, 30 July 2013 (UTC)

text: Addition: I've found a direct link to the paper. Maybe the journal website was broken when I tried, but not it seems to work again. It would have been a surprise to have a 80 years old paper behind a paywall... So, diameter instead of radius seems to be more widely used in the early 20th century, and since CIE 1931 is a standard (and a starting point for new ones) this has been continued up to now.--SiriusB (talk) 12:59, 30 July 2013 (UTC)
replies: text: Diameter is a convention for optical engineering,see field of view or angle of view.Physicsxuxiao (talk) 15:55, 1 August 2013 (UTC)
thread_title:  2 degrees angular radius or diameter? 

message: So where Greeen, #00FF00 is, there is no color after that at the top in the grey area? Then how come in a different image with some other bluish green there it is visible, should that image be removed?  — Preceding unsigned comment added by 88.104.101.221 (talk) 16:43, 18 October 2013 (UTC)
thread_title:  Are all colours outside the triangle impossible on an RGB screen then? 

message: This seems to be a subset of what is already on the other page. It should be over there. But I'm not an expert.    🍺  Antiqueight confer 12:44, 21 October 2013 (UTC)
replies: text: Oppose – Why do you think a model of audio stimuli should be moved into an article about color? I don't get it. –jacobolus (t) 06:02, 22 October 2013 (UTC)
replies: text: I thought I pulled the transfer - I thought I had rolled that back once I read it again. Obviously I missed it here.--   🍺  Antiqueight confer 10:35, 22 October 2013 (UTC)
thread_title:  
Proposed merge with [[Tristimulus timbre model]]
 

message: Rec. 709 has a bit depth of 8 bits per RGB channel, for a total of 24 bits of color. while Rec. 2020 has a bit depth of 12 bits per channel for a total of 36 bits of RGB color. My question is, how many bits per channel would it take to encode the entire CIE 1931/XYZ color space? Bumblebritches57 (talk) 01:48, 6 December 2013 (UTC)
thread_title:  Bit depth? 

message: the page doesn't define "quasi" blue of Z tristimulus value, I think, it says quasi-blue but doesn't say where or how Z comes about, is the RGB just LMS that is close to an RGB or the RGB of LMS?
the same question for X tristimulus value, is it really just a combination of LMS responses close to RGB, or the RGB of LMS?
the page is not clear of xbar,ybar,zbar derivation, I think
are xbar,ybar,zbar defined to obtain XYZ tristimulus values? the page does not say one way or another as far as I comprehend
is there a device definition of xbar,ybar,zbar like the page indicates for rbar,gbar,bbar (mercury vapor?)?
Dale r kelly (talk) 03:30, 11 January 2015 (UTC)
replies: text: Dale, the X, Y, and Z are linear combinations of L, M, S.  That is they are different bases for the same 3D subspace of spectral space.  I think some of the refs explain the derivation.  Basically, iirc, the Y is first defined to correlate well with luminance.  Then the Z is defined as a compact non-negative blue peak, and then X gets what's left, transform to be as compact as possible while staying nonnegative.  Or something like that.  So Z is not too far from S.  Dicklyon (talk) 03:42, 11 January 2015 (UTC)
thread_title: XYZ relationship to LMS?

message: In the first sentence of the text, under "Tristimulus values", I find "short (S, 420–440 nm)". Just to the right of this text is a diagram that clearly shows the short peak at about 450 nm. These cannot both be correct. The diagram appears to be good; if so, the peak should be stated to be 440-460 nm rather than 420-440. I am reluctant to edit the page, but this is clearly a real issue. Jonsinger (talk) 21:58, 13 November 2015 (UTC)
thread_title:  Apparent error in the text of the page: 

message: The xy chromaticity diagrams in this article do not tell you the value for Y which they used. This should really be mentioned in the image captions. SharkD  Talk  16:38, 26 November 2016 (UTC)
replies: text: Having done many of these kinds of renderings I can tell you that the brightness they used is not uniform in the image. It is instead the brightest possible Y value for each pixel. Near the white point that will be Y = 100 and elsewhere it is lower. 
I have considered replacing this with another style image. Maybe someday I'll get around to it. TDcolor (talk) 03:22, 5 January 2021 (UTC)
thread_title:  Lightness of chromaticity diagrams? 

message: Where does the 1 / b21 originate from because it's not from the cited paper?
I have no clue either. Searched for hours where this comes from. Cannot find it anywhere (not a single article, or standard).  — Preceding unsigned comment added by 2A02:A03F:E40F:7C00:6156:16F8:3715:683C (talk) 11:05, 15 December 2020 (UTC)
replies: text: Dropping the normalization value is indeed correct.  It is used when transforming the color matching functions r[hat](lambda). g[hat](lambda) and b[hat](lamda) to x[hat](lambda), y[hat](lamda) z[hat](lambda).  This constant arises from the fact that the areas under the r,g,b tristimulus functions were less than the area under the photopic response curve by the factor of 0.17697.  However, when R, G and B are calculated from the r,g,b color matching function, that factor is implicitly incorporated in order for a perfect diffuse reflector of illuminant E to have R, G and B values of 100, 100, 100.  From what I can gather, the photopic response curve was scaled so that its maximum value is 1.0 and the r,g,b color matching functions are scaled so that they all have the same area and so that the r color matching function at 600 nm was equal in value to the photopic response curve at that wavelength.  No idea why they did this rather than have the areas under all four curves (photopic response and the r,g,b color matching functions) equal.  Of course, when XYZ color space was developed, Y was set equal to the photopic response curve, so it of course has the same area.  X and Z are then scaled to also have that area.  I had noticed that the matrix that had been shown with the (1/0.17697) factor was wrong for a while, but am new to Wikipedia and didn't know how to correct it.  I am glad someone beat me to this! 97.78.195.34 (talk) 01:38, 8 April 2022 (UTC)
replies: text: Oops - meant the values at 700 nm, not 600. 97.78.195.34 (talk) 01:39, 8 April 2022 (UTC)
thread_title:  Construction of the CIE XYZ color space from the Wright–Guild data 

message: Wiki: "The unit of the tristimulus values X, Y, and Z is often arbitrarily chosen so that Y = 1 or Y = 100 is the brightest white that a color display supports . The corresponding whitepoint values for X and Z can then be inferred using the standard illuminants."
So in 1931 when CIE defined all these, a "color displays" was in question?
Please stay serious!  — Preceding unsigned comment added by 62.202.181.251 (talk) 02:57, 16 February 2018 (UTC)
replies: text: We could discuss that here, but if you remove the remarks of others it's not going to go anywhere.  Let me know if you'd like to discuss. Dicklyon (talk) 23:03, 16 February 2018 (UTC)
thread_title:  Meaning of X, Y and Z .. please! 

message: I added sources for the equations in the section. This was no big deal. I wonder why this was not done ages ago. My question: did I miss something or is it time to remove the "unreferenced" tag? AndreAdrian (talk) 10:10, 11 December 2018 (UTC)
replies: text: I think it should go, especially since it says "not....any sources".    But the section is still light on sourcing. North8000 (talk) 11:51, 11 December 2018 (UTC)
thread_title:  Remove unreferenced section for CIE xy chromaticity diagram and the CIE xyY color space 

message: CIE 1931 color space#Color matching functions describes an approximation of the color matching functions of terms of elementary functions, and provides a definition of this approximation by way of a Haskell implementation. I have edited this section so that this definition is now given in a mathematical fashion, and have also added a C implementation. Given that a mathematical definition is now present, should the Haskell and/or C implementations remain in the article? —JivanP (talk) 00:38, 10 April 2020 (UTC)
thread_title:  Remove code in §{{nbsp}}Color matching functions? 

message: Apparently, all arguments of the Gaussians, which describe the x, y, z color matching function, except the first argument should have an additional point/comma. For example, the Gaussians should not be centered around 5998, but 599.8, same is true for the sigma 1 and sigma 2 values. This can be easily seen from the linked paper which is publicly available.
Feel free to check this error yourself and, if you come to the same conclusion, potentially change it.  — Preceding unsigned comment added by 91.19.232.210 (talk) 13:59, 18 June 2020 (UTC)
replies: text: The equations use angstroms as the unit, 10 angstroms = 1nm so these equations seem correct to me. TDcolor (talk) 02:33, 18 October 2021 (UTC)
replies: text: It looks like the unit choice here was to eliminate fractions so the code could be written as integers. For a Wikipedia article, it would probably be clearest to just use mathematical notation instead of a code implementation, and use nm as a unit. –jacobolus (t) 16:01, 18 October 2021 (UTC)
thread_title:  Error in Gaussian Equations 

message: y is defined as y = \frac{Y}{X+Y+Z}. Thus Y cannot represent luminance, cause the y coordinate changes if Y changes. Only if we set X = Z = 0, we get a fixed y coordinate in the xy-chromacity diagram. But as soon as X or Z have a non zero value, the y chromacity coordinate changes if you change Y. This is a contradiction. --95.91.242.234 (talk) 10:07, 23 May 2022 (UTC)
replies: text: @95.91.242.234 Something is completely wrong here. I think Z represents luminosity, but x and y, as represented in the article, don't make sense to me either. 92.252.104.29 (talk) 04:47, 19 October 2023 (UTC)
replies: text: @92.252.104.29 To correct myself. I just don't know what's going on. 92.252.104.29 (talk) 05:01, 19 October 2023 (UTC)

text: Y can represent luminance, while still changing the x, y chromaticity coordinates. The reason is that luminance is mostly determined by the amount of green light. Red contributes only a little bit to luminance and blue contributes almost nothing at all. So to increase the luminance, you must add green. Unless you also add red and blue, the hue will of course change. XYZ is designed so that adding X automatically subtracts from green so that luminance stays the same. MTres19 (talk) 21:41, 24 April 2024 (UTC)
thread_title:  How can Y represent luminance if y chromacity coordinate changes in xyY with a change of the Y coordinate 

message: "In XYZ space, all combinations of non-negative coordinates are meaningful, but many, such as the primary locations [1, 0, 0], [0, 1, 0], and [0, 0, 1], correspond to imaginary colors outside the space of possible LMS coordinates; imaginary colors do not correspond to any spectral distribution of wavelengths and therefore have no physical reality."
Firsly what does "meaningful" constitute in the first sentence?
Secondly, "possible LMS coordinates", I assume means "subspace of valid LMS coordinates that can can be achieved in a human with normal eyesight", but it could also be taken to mean invalid LMS coordinates like negative values for a component. If I understand correctly, it is currently only assumed that the CrCgCb triangle is contained within whatever space valid LMS occupies in the rg diagram. If it's the case that the valid XYZ space contains invalid LMS values, then "meaningful" in the first sentence doesn't seem right at all.
I'm new to this so tell me if I got it all wrong.
p.s. I used "subspace" in the colloquial sense, it very well might not meet the mathematical definition. TheOnlyRealEditor (talk) 08:00, 19 August 2022 (UTC)
thread_title:  Meaningful, imaginary colors in XYZ 

message: When I tried converting the normalized LMS color cone response to CIE 1931 XYZ, there seemed to be a big discrepancy from the S cone response to the Zbar graph. Supposedly, the Zbar should only consist of the S cone response, which tops out at 1. The matrix also does not scale the S cone response by any amount. However, the actual Zbar function's maximum is approximately 1.78 at 445nm . Have I made a mistake somewhere? 66.183.66.205 (talk) 07:14, 19 January 2024 (UTC)
thread_title:  Discrepancy between LMS --> XYZ coordinates from the HPE matrix?