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Author Topic: Color Gamut RGB Cube  (Read 35683 times)
JeremyLangford
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« Reply #60 on: December 05, 2009, 08:02:41 PM »
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So the sun sends us white light which is a mixture of all the monochromatic colors that our eyes are sensitive to and when we mix both sides of the monochromatic spectrum, we get the extra spectral purples. We can then mix all of the monochromatic colors/extra spectral purples with with white to create less saturated colors. And Black is simply what we see when there is no light present at all. Does this explain how every single color is made? What about mixing the monochromatic colors/extra spectral purples with different grays or even black?
« Last Edit: December 07, 2009, 05:02:36 PM by JeremyLangford » Logged
JeremyLangford
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« Reply #61 on: December 08, 2009, 01:17:31 PM »
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If this question doesn't make sense then could you tell me how and what is wrong with it?
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waynebretl
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« Reply #62 on: February 13, 2010, 03:28:34 PM »
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Quote from: JeremyLangford
Isn't brightness usually measured by the amount if white there mixed into the hue?


You have to be careful about what sort of experiment each diagram is describing.  If you take a certain amount of optical power (watts)and put it all into the wavelength of maximum luminance (with a laser operating at that wavelength), that will be a bright yellow green and will be brighter than a white color with the same amount of power (watts). Therefore, white does not correspond to the maximum luminance function.  Another way to see this is that white must be composed of at least two different wavelengths of light, and therefore at least part of the optical power in white has less luminance than the maximum possible, because it is not that particular wavelength that has maximum lumens per watt.

When you look at the gamut of colors for a display, however, you see that the maximum luminance that the display can produce is for white, because it is the case where all three primary colors are turned on to the maximum available.  The display cannot take all that optical power and put it into the single wavelength with the maximum luminance, it can only produce varying amounts of its three primary colors.
« Last Edit: February 13, 2010, 03:30:19 PM by waynebretl » Logged
waynebretl
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« Reply #63 on: February 13, 2010, 04:02:23 PM »
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Quote from: JeremyLangford
So the sun sends us white light which is a mixture of all the monochromatic colors that our eyes are sensitive to and when we mix both sides of the monochromatic spectrum, we get the extra spectral purples.
Yes


Quote from: JeremyLangford
We can then mix all of the monochromatic colors/extra spectral purples with with white to create less saturated colors.
True, but you can also create a less saturated color by mixing a spectral color with its complementary color.  And if you mix the right amounts of a spectral color and its complement, you will make gray or white.

Quote from: JeremyLangford
And Black is simply what we see when there is no light present at all.
Yes.

Quote from: JeremyLangford
Does this explain how every single color is made?
Sort of, but you have to remember that any non-spectral color can be created by many different spectra, provided that they all give the same stimulation to the three types of cone cells in the eye.  This is the fundamental basis of color "reproduction" (meaning, color matching) by use of three primary colors.  The wonderful thing is that the spectrum of the reproduction does not have to match the original spectrum wavelength by wavelength; only the net stimulation of each of the three types of cones has to match.

Quote from: JeremyLangford
What about mixing the monochromatic colors/extra spectral purples with different grays or even black?

Remember we're talking additive color here (mixing light), so mixing with gray is just mixing with less white, and black is the special case of mixing with nothing, getting the starting color as a result.
« Last Edit: February 13, 2010, 04:03:30 PM by waynebretl » Logged
joofa
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« Reply #64 on: February 13, 2010, 06:21:51 PM »
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Quote from: waynebretl
If you take a certain amount of optical power (watts)and put it all into the wavelength of maximum luminance (with a laser operating at that wavelength), that will be a bright yellow green and will be brighter than a white color with the same amount of power (watts). Therefore, white does not correspond to the maximum luminance function.

That would bring out an interesting point. The radiant power amounts required for matching equi-energy white using a usual set of CIE primaries (R=700, G=546.1, B=435.Cool would be in the ratio R:G:B = 72.1:1.4:1, which in photometric units of luminance would be in the ratio of R:G:B = 1:4.6:0.06. Hence, to match 74.5 watts of equi-energy white one would require 72.1 watts of red, 1.4 watts of green, and 1 watt of blue. Most of the contribution in wattage is coming from Red, i.e., 72.1 watts. Therefore, perhaps, one can make a statement that at least in this case the wattage of equi-energy white is close to a certain red as opposed to some green.
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waynebretl
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« Reply #65 on: February 14, 2010, 11:55:22 PM »
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Quote from: joofa
That would bring out an interesting point. The radiant power amounts required for matching equi-energy white using a usual set of CIE primaries (R=700, G=546.1, B=435.Cool would be in the ratio R:G:B = 72.1:1.4:1, which in photometric units of luminance would be in the ratio of R:G:B = 1:4.6:0.06. Hence, to match 74.5 watts of equi-energy white one would require 72.1 watts of red, 1.4 watts of green, and 1 watt of blue. Most of the contribution in wattage is coming from Red, i.e., 72.1 watts. Therefore, perhaps, one can make a statement that at least in this case the wattage of equi-energy white is close to a certain red as opposed to some green.

Hmmm - could you rephrase that last statement?  I'm not following what it means, "the wattage is close to a certain red."  Red is a sensation, so I don't follow what you mean by watts being close to a certain red.  

This particular choice of red primary is far down the sensitivity curve of even the long wavelength cones, and therefore lots of watts are needed to get a long-wavelength cone stimulus equal to that of the equi-energy white source. Using lots of power for this deep red primary does not imply that the color being reproduced is subjectively "close" to that primary.  Another color that is "closer," for example a shade of pink, will require even more watts of this red primary to match it. In other words, with this choice of primaries, the 72.1 is just a scaling factor required because you are so far down the slope of the cone response.  In the published chromaticity diagrams, these power factors of 72.1, 1.4, and 1 are divided out so that the amounts of the primaries can be stated as R=1, G=1, and B=1 for the match to equi-energy white.  If the red wavelength is moved to something shorter and more reasonably up on the long cone sensitivity curve, these ratios are different and much less extreme.
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joofa
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« Reply #66 on: February 15, 2010, 11:25:57 PM »
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Quote from: waynebretl
Hmmm - could you rephrase that last statement?  I'm not following what it means, "the wattage is close to a certain red."  Red is a sensation, so I don't follow what you mean by watts being close to a certain red.

Just saying that for that particular choice of primaries (which is not a random set and one of those actually used in colorimetry publications) most of the actual physical power in matching an equi-energy white is coming from the red color. Of course one can derive the equivalent photometirc luminance units and they are found to be in ratio of R:G:B = 1:4.6:0.06. Now one can see the green having a sensation of higher luminance. Further more, one can do a rescaling of the size of units for red, green, and blue so that equal-energy white is matched by unit quantities of R,G and B, and derive tristimulus values.
« Last Edit: February 16, 2010, 12:03:38 AM by joofa » Logged

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JeremyLangford
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« Reply #67 on: February 16, 2010, 02:45:15 AM »
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I learned a lot from this thread, the main thing being how much I don't know. I only started this thread because I was planning on buying a new monitor to use with my new film scanner and because I'm such a perfectionist, I tried to learn everything I could about color before letting my self decide on a monitor. I now see how ridiculously complicated this stuff really is. I am really curious about why and how some of you guys know as much as you do about color? Do most people here have careers that require it?
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waynebretl
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« Reply #68 on: February 16, 2010, 09:10:16 AM »
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Quote from: joofa
Just saying that for that particular choice of primaries (which is not a random set and one of those actually used in colorimetry publications) most of the actual physical power in matching an equi-energy white is coming from the red color.
Oh, OK, totally agree.  Just didn't understand your original phrasing.
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waynebretl
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« Reply #69 on: February 16, 2010, 09:22:53 AM »
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I learned a lot from this thread, the main thing being how much I don't know. I only started this thread because I was planning on buying a new monitor to use with my new film scanner and because I'm such a perfectionist, I tried to learn everything I could about color before letting my self decide on a monitor. I now see how ridiculously complicated this stuff really is. I am really curious about why and how some of you guys know as much as you do about color? Do most people here have careers that require it?

I usually don't like to tout myself, but "how are you qualified" is a fair question.  My curriculum vitae is here:
http://www.bretl.com/curricvit.htm

Some details that aren't in the CV: I learned the basics as a necessary adjunct of doing TV set chroma IC design.  I became the go-to guy in Zenith's advanced development lab for colorimetry issues, in conjunction with their phosphor and color instrumentation specialists. Later I participated in the SMPTE monitor colorimetry work and also in the camera standardization for the system comparison shoots used by the FCC Advisory Committee on Advanced Television Systems (ACATS).  

I am still learning a lot about the "non-ideal" (or at least, not simple) aspects of color processing that are necessary to make a camera's results really look good.
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joofa
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« Reply #70 on: February 16, 2010, 10:51:13 AM »
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I am really curious about why and how some of you guys know as much as you do about color? Do most people here have careers that require it?

Hi Jeremy, my interest in color science developed when I was a student in computer vision and pattern recognition while working on a certain problem in high-dimensional multimedia databases requiring color-based query mechanism. Later I held jobs in HD video compression and communications, and currently in making advanced digital cameras for low-noise scientific imaging.

I have an interest in color science as derived from traditional colorimetry, color TV, analog and digital communications, and also along the lines of connections with digital signal processing and pattern recognition.

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Joofa
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joofa
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« Reply #71 on: February 16, 2010, 12:43:58 PM »
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My curriculum vitae is here:
http://www.bretl.com/curricvit.htm

Hi, welcome here. You have great qualifications and experience. Did you by chance ever meet Donald Fink?
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ejmartin
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« Reply #72 on: February 16, 2010, 02:17:00 PM »
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Quote from: joofa
That would bring out an interesting point. The radiant power amounts required for matching equi-energy white using a usual set of CIE primaries (R=700, G=546.1, B=435.Cool would be in the ratio R:G:B = 72.1:1.4:1, which in photometric units of luminance would be in the ratio of R:G:B = 1:4.6:0.06. Hence, to match 74.5 watts of equi-energy white one would require 72.1 watts of red, 1.4 watts of green, and 1 watt of blue. Most of the contribution in wattage is coming from Red, i.e., 72.1 watts. Therefore, perhaps, one can make a statement that at least in this case the wattage of equi-energy white is close to a certain red as opposed to some green.

Where did you get these figures 72.1:1.4:1 ?  That seems out of line with the usual plots of the CIE standard observer spectral response functions; unless their normalizations are not as usually plotted.
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emil
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« Reply #73 on: February 16, 2010, 02:24:37 PM »
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Quote from: JeremyLangford
So the sun sends us white light which is a mixture of all the monochromatic colors that our eyes are sensitive to and when we mix both sides of the monochromatic spectrum, we get the extra spectral purples. We can then mix all of the monochromatic colors/extra spectral purples with with white to create less saturated colors. And Black is simply what we see when there is no light present at all. Does this explain how every single color is made? What about mixing the monochromatic colors/extra spectral purples with different grays or even black?

The eye (barring issues like color blindness) has three separate types of photoreceptors.  Each has a distinct response to the visible spectrum:

http://upload.wikimedia.org/wikipedia/comm...nctions.svg.png

If one takes the intensity of a light source as a function of frequency, multiplies it separately by each response function and adds up the result over frequency, one derives three numbers usually called X,Y, and Z.  Then given one more piece of information -- the X0,Y0,Z0 values of white -- there is then a well-defined map between (X/X0,Y/Y0,Z/Z0) and the RGB values in your favorite color space, say sRGB.  For details, see

http://www.brucelindbloom.com/index.html?W...gSpaceInfo.html
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« Reply #74 on: February 16, 2010, 02:34:57 PM »
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Quote from: ejmartin
Where did you get these figures 72.1:1.4:1 ?  That seems out of line with the usual plots of the CIE standard observer spectral response functions; unless their normalizations are not as usually plotted.

I think you are talking about photometric response curves, in which case I said the Luminances for that particular set of R, G, and B are in ratio of R:G:B = 1:4.6:0.06. The corresponding radiometric response is based upon R:G:B=72.1:1.4:1. You can find these numbers (72.1:1.4:1) in the book,  "Color Science: Concepts and Methods, Quantitative Data and Formulae, 2nd ed," by Gunter Wyszecki and Walter Stanley Stiles, among several other sources.

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ejmartin
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« Reply #75 on: February 16, 2010, 04:54:44 PM »
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I think you are talking about photometric response curves, in which case I said the Luminances for that particular set of R, G, and B are in ratio of R:G:B = 1:4.6:0.06. The corresponding radiometric response is based upon R:G:B=72.1:1.4:1. You can find these numbers (72.1:1.4:1) in the book,  "Color Science: Concepts and Methods, Quantitative Data and Formulae, 2nd ed," by Gunter Wyszecki and Walter Stanley Stiles, among several other sources.

A simple explanation would help more than sending me off to the library.
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emil
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« Reply #76 on: February 16, 2010, 05:16:39 PM »
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A simple explanation would help more than sending me off to the library.

Can't believe you said that Emil. A physics professor at a major university wants to avoid going to a library   I've right now over a hundred technical books checked out from the library  .

Here is the deal: In the color matching experiments the actual physical amounts of R=700 nm, G=546.1 nm, and B=435.8 nm primaries used to match an equi-energy white are in the ratio of R:G:B = 72.1:1.4:1. In such experiments you have 3 different radiometric response curves for matching spectral colors. Now you can transform this set of curves using the Luminosity curve to the photometric set of numbers, which will give you another 3 curves.  You can further renormalize each curve so that equal amount of primaries are needed for matching the white point -- and you have your tristimulus curves, which were perhaps what you had in mind when you mentioned the CIE standard observer. In the photometric set of numbers the luminances are in the ratio of R:G:B=1:4.6:0.06 when matching the equi-energy white.
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JeremyLangford
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« Reply #77 on: February 16, 2010, 06:16:29 PM »
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I will be starting college next semester at the University of Tennessee. I plan to major in Journalism and Electronic Media and either minor or double-major in Studio Art. I love film photography more than anything but I guess I hope that I can get a good job in some type of video.
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waynebretl
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« Reply #78 on: February 16, 2010, 07:21:31 PM »
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Hi, welcome here. You have great qualifications and experience. Did you by chance ever meet Donald Fink?
Nope, never met him.  It might have been possible, but never happened.  I was born in 1944, so he was the preceding generation, really.  He was writing his first TV engineering books before I had ever seen television.  His last notable contribution to TV was a paper arguing that there was no practical way to get to analog high definition in a reasonable bandwidth, which spurred others to develop our digital HDTV systems.
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joofa
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« Reply #79 on: February 16, 2010, 07:49:39 PM »
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Quote from: waynebretl
Nope, never met him.  It might have been possible, but never happened.  I was born in 1944, so he was the preceding generation, really.  He was writing his first TV engineering books before I had ever seen television.  His last notable contribution to TV was a paper arguing that there was no practical way to get to analog high definition in a reasonable bandwidth, which spurred others to develop our digital HDTV systems.

What about the NHK analog HDTV in Japan? Though, I understand that it was discontinued.

At my previous job with HD video communications, during development phase of our HDMI signals we more more successful in displaying them on LG HDTVs compared to more expensive Sharps.

« Last Edit: February 16, 2010, 07:49:46 PM by joofa » Logged

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