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Author Topic: Color Gamut RGB Cube  (Read 35497 times)
JeremyLangford
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« Reply #20 on: September 23, 2009, 04:23:53 PM »
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I think I pretty much understand everything except for this:

"Now divide each of the X,Y,Z by their sum (X+Y+Z), and you get another set of 3 numbers. Plot them again and that gives you a point on spectral locus on the third diagram."

What exactly are the second set of 3 numbers? Do you ignore the Z number in order to plot this set into the 2d spectral locus?
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joofa
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« Reply #21 on: September 23, 2009, 04:40:31 PM »
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Quote from: JeremyLangford
What exactly are the second set of 3 numbers? Do you ignore the Z number in order to plot this set into the 2d spectral locus?

Yep, sorry, you are right, to get the 3rd diagram you ignore the Z number and get the 2D spectral locus. If you don't want to ignore the Z number, then that is fine and you would get a 3D plot of spectral locus instead of 2D. The 3D spectral locus shape will be similar to 2D and won't look like the wiggly plot of the 2nd diagram.
« Last Edit: September 23, 2009, 04:40:56 PM by joofa » Logged

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JeremyLangford
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« Reply #22 on: September 23, 2009, 08:49:47 PM »
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Trying again. Pick any single color (monochromatic) from your first diagram and note its radiant power (used later). Then measure the amount of X, Y, and Z primaries, which are calibrated using a certain process, that you need to match that color, and this gives you 3 numbers. Plot the 3 numbers and you get a point on the wiggly curve in your second diagram. Now divide each of the X,Y,Z by their sum (X+Y+Z), and you get another set of 3 numbers. Plot them again and that gives you a point on spectral locus on the third diagram. Now go back to the first diagram and pick another color at the same radiant power as the first color picked and repeat the process. Keep on doing until you have done for all/enough colors in the first diagram. Now you have a whole bunch of points on the second and third diagrams and you just join them together to get the curves shown.

Do the X, Y and Z primaries used to form the 3d spectral locus correspond to the 3 cones in the human eye?
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joofa
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« Reply #23 on: September 23, 2009, 10:05:18 PM »
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Do the X, Y and Z primaries used to form the 3d spectral locus correspond to the 3 cones in the human eye?

No, they are different, but people have tried to find a transformation relating the two responses to varying degree of success.
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JeremyLangford
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« Reply #24 on: September 23, 2009, 10:31:38 PM »
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No, they are different, but people have tried to find a transformation relating the two responses to varying degree of success.

Then what are the 3 primaries for the spectral locus? (Not the CIE XYZ space)
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joofa
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« Reply #25 on: September 23, 2009, 10:44:40 PM »
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Then what are the 3 primaries for the spectral locus? (Not the CIE XYZ space)

In theory any 3 primaries can be used as long as they are not in the same direction and all three don't lie in a plane. Recall, we want to define any color with 3 numbers, i.e. any color would be represented by a point in 3D, therefore we just need 3 vectors in a 3D space to represent any point. The primaries are these vectors. We don't want to point them in the same direction or lie in the same plane otherwise we would have effectively a 2D space (a plane), or a line (1D) embedded in a 3D space as we have suppressed one or two degrees of freedom, respectively. So if you want to use a different set of primaries it is just a reorientation of the original set of primaries, i.e., why we multiply primaries by a matrix to transform one set of primaries to another.

In practise it is advisable to have them spaced farther apart so that they cover as many colors that can be physically displayable (i.e., those with positive values). Primaries in the regions of Red, green and blue fulfill that criteria.

And, as I mentioned before, if we want to measure the values for the spectral locus, then for any set of "real" primaries (i.e., those we can display later) it would mean at least one primary has negative value. But that is fine, we are not going to display it at this stage, we are just measuring it. This is how CIE measured these numbers, including negative numbers, using RGB. But CIE did not want to work in negative numbers thinking that people would make mistakes with negative numbers. Therefore a transformation was found to convert RGB in such a way that resulted in all positive numbers, i.e., RGB->XYZ.
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JeremyLangford
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« Reply #26 on: September 24, 2009, 03:50:02 PM »
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Do the X, Y and Z primaries used to form the 3d spectral locus correspond to the 3 cones in the human eye?

I though that because of this picture.



But now this picture is just confusing me.
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joofa
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« Reply #27 on: September 24, 2009, 06:25:50 PM »
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I though that because of this picture.



But now this picture is just confusing me.

Please don't let the graph confuse you. I have to ascertain how close are LMS primaries to XYZ, however, typically, the shape of the spectral locus in that "beauty pass" plot (that horse-shoe like spectral locus) is going to look similar for various different primaries that are close by. It would seem to get stretched, elongated, etc., but overall similar shape.

BTW, which software did you use to generate that "wiggly" spectral locus shape?
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JeremyLangford
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« Reply #28 on: September 24, 2009, 06:38:13 PM »
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Quote from: joofa
Please don't let the graph confuse you. I have to ascertain how close are LMS primaries to XYZ, however, typically, the shape of the spectral locus in that "beauty pass" plot (that horse-shoe like spectral locus) is going to look similar for various different primaries that are close by. It would seem to get stretched, elongated, etc., but overall similar shape.

BTW, which software did you use to generate that "wiggly" spectral locus shape?

I found it on wikipedia. I am still having trouble understanding how the "wiggly" 3d spectral locus shape is made. Heres another one.



"Spectral locus in XYZ and xy (demonstration of chromaticity derivation)"
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crames
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« Reply #29 on: September 24, 2009, 07:32:34 PM »
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Have a look at www.handprint.com, especially this section. The site goes into quite a lot of detail about chromaticity diagrams, etc.

Cliff
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Cliff
JeremyLangford
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« Reply #30 on: September 24, 2009, 07:38:58 PM »
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Quote from: crames
Have a look at www.handprint.com, especially this section. The site goes into quite a lot of detail about chromaticity diagrams, etc.

Cliff

I actually found that site yesterday. It's great.
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JeremyLangford
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« Reply #31 on: September 25, 2009, 12:49:17 PM »
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Do the X, Y and Z primaries used to form the 3d spectral locus correspond to the 3 cones in the human eye?

Here is another diagram that seems to show that the 3d spectral locus dimensions correspond to the 3 human eye cones.

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JeremyLangford
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« Reply #32 on: September 28, 2009, 03:36:39 PM »
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Nevermind, that was a dumb question.
« Last Edit: September 28, 2009, 04:41:26 PM by JeremyLangford » Logged
JeremyLangford
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« Reply #33 on: October 22, 2009, 08:38:14 PM »
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I'm confused about the green and purple lines in this spectrum locus diagram. I'm reading that the purple one is the V luminosity function which is L plus M and that the green line is the contrast between L and M which is L minus M. I don't really get what this means or what the lines are representing.


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JeremyLangford
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« Reply #34 on: October 23, 2009, 09:59:46 PM »
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I'm confused about the green and purple lines in this spectrum locus diagram. I'm reading that the purple one is the V luminosity function which is L plus M and that the green line is the contrast between L and M which is L minus M. I don't really get what this means or what the lines are representing.


Does the green line represent the hues that appear brightest to the human eye in photopic vision?
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crames
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« Reply #35 on: October 23, 2009, 11:23:47 PM »
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Does the green line represent the hues that appear brightest to the human eye in photopic vision?
   
    The colors that extend the farthest in the direction of the green line appear more luminous.
   
    The blue line to 525nm is along the crease where the spectral locus seems to bend, marking a division of color perception. Quoting Handprint:
   
   
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But if we set aside luminance perception defined by the L+M diagonal, then color perception is divided into two parts:       at wavelengths above 525 nm, changes in the relative excitation of the L and M cones define the color response; the S cones are silent.    

      at wavelengths below 525 nm, the relative L,M  excitations are approximately the same as they are at 525 nm (the  dotted line and purple line are equivalent) so it is the relative  excitation of the independent S cone that defines the color response.
   
    Cliff
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Cliff
JeremyLangford
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« Reply #36 on: October 24, 2009, 12:25:50 AM »
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Quote from: crames
The colors that extend the farthest in the direction of the green line appear more luminous.
   
    The blue line to 525nm is along the crease where the spectral locus seems to bend, marking a division of color perception. Quoting Handprint:
   
   
   
    Cliff

Thanks. I think I understand now.
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JeremyLangford
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« Reply #37 on: October 24, 2009, 10:27:34 AM »
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Quote from: crames
The colors that extend the farthest in the direction of the green line appear more luminous.
   
    The blue line to 525nm is along the crease where the spectral locus seems to bend, marking a division of color perception. Quoting Handprint:
   
   
   
    Cliff

Do the colors that extend the farthest in the direction of the green line appear more luminous than the white point?
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crames
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« Reply #38 on: October 24, 2009, 01:10:17 PM »
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Do the colors that extend the farthest in the direction of the green line appear more luminous than the white point?

Yes, quoting further from Handprint:

Quote
A final observation is that the white point is not located on the luminosity function. This simply demonstrates that white is not the same as bright. The perception of white is a form of color sensation, whereas the perception of bright is a unique intensity sensation. The cone excitation space implies that a "bright" stimulus produces more than two times  the cone excitation of a "white" surface, and therefore visual "white"  always has a lower luminosity than visual "bright" under the same  viewing conditions.

I'm not sure, but I think this might be related to the Helmholtz-Kohlrausch effect, where in general, more chromatic colors will appear brighter than a white of the same luminance.

Cliff
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Cliff
JeremyLangford
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« Reply #39 on: October 25, 2009, 03:09:34 AM »
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Quote from: crames
Yes, quoting further from Handprint:



I'm not sure, but I think this might be related to the Helmholtz-Kohlrausch effect, where in general, more chromatic colors will appear brighter than a white of the same luminance.

Cliff

Isn't brightness usually measured by the amount if white there mixed into the hue?
« Last Edit: October 25, 2009, 03:12:25 AM by JeremyLangford » Logged
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