I On Mixing Colors of Light

[Charles Link]

TL;DR Summary

Doing a discussion with a Gedanken experiment of mixing green (550 nm) and red light (650 nm) and comparing it to that of yellow light at 600 nm

I want to illustrate what the color mixing that is done with a tv screen is all about, and how we can generate the appearance of color using primary colors, even though we don't actually generate any light of the color that we perceive. Consider beginning with a tungsten filament with a current running through it that creates a continuous spectrum that can be approximated by a T= 2500 K blackbody for a typical case where it generates white light.
Let's have that light be collimated by a lens and run it through a prism=basically a prism spectrometer that will split the visible light into the colors of the rainbow, because the index of refraction of the glass of the prism decreases with increasing wavelength.

Now let's sample the output at the locations where the spectrum is green (550 nm) and red (650 nm) and combine these with the necessary mirrors and or lenses onto a sheet of white paper. If we get the proportions right, we should observe some yellow light.

We can also sample the output of the prism at the angle about midway between where the green and red emerge, and there we will find yellow light at 600 nm. If we focus this onto a white sheet of paper, it is likely we can not tell the difference with our eyes between this and the green and red combination.

The next part is to send some of the yellow light that comes off the white paper into a second prism spectrometer. In the first case, we will find green and red light to emerge in the same proportion that we started with at the angles corresponding to 550 nm and 650 nm, but nothing at the 600 nm location. In the second case, everything will emerge at the 600 nm location. The light that is made of green and red light appears yellow, but the light doesn't change its composition when the green is combined with the red. It is still a mixture of green and red, even though it appears yellow.

Thought you might find this of interest. I welcome your feedback. I just came up with this the other day. Generally I don't think they treat this topic in very many of the textbooks in the manner in which I presented above.

 

[Andy Resnick]

TL;DR Summary: Doing a discussion with a Gedanken experiment of mixing green (550 nm) and red light (650 nm) and comparing it to that of yellow light at 600 nm

There's a lot easier way to show this by mixing light from different narrowband sources (e.g. lasers). Color perception occurs at the retina/brain and is due to the broad and partially overlapping absorption bands of different rhodopsin isomers.

 

[Charles Link]

There's a lot easier way to show this by mixing light from different narrowband sources (e.g. lasers). Color perception occurs at the retina/brain and is due to the broad and partially overlapping absorption bands of different rhodopsin isomers.

That certainly would work (using lasers of two different wavelengths), but the concept here is to show that the (secondary) color that is found at a particular wavelength is spectrally different than the composite of two different (primary) colors, even though it looks the same.

I find it interesting that it is possible to generate fairly accurate color screens using just 3 color leds (blue, green, and red). I thought my experiment might provide some insight to the actual physics, since in principle, the entire visible spectrum still consists of the same type of electromagnetic waves. The more advanced student might look at my proposed experiment and say of course that will be the result, (that the composite will split e.g. into red and green when sent through a spectrometer, etc. even though it looks to be light of 600 nm).

The motivation behind all this for me is that I have a younger brother who asked me why did white light (as seen by the eye) appear green when he took a photo of the northern lights with his cell phone camera. It got me thinking that we could in principle make a camera that has a prism spectrometer doing a spectral measurement,(at each pixel), and then have many sources of various wavelengths used to regenerate the measured spectrum. Instead, our systems rely on just 3 color filters for the measurement and 3 led's, and they do fairly well most of the time. :)

 

[DaveC426913]

The motivation behind all this for me is that I have a younger brother who asked me why did white light (as seen by the eye) appear green when he took a photo of the northern lights with his cell phone camera.

This is actually a different phenomenon completely.

Your brother's night vision is comprised mostly of activated rod photoreceptors - which do not detect colour. The cone photoreceptors - which do detect colour - are not activated in dim light.
He is seeing in black, white and shades of grey.

His camera does not suffer from this affliction and sees colour at low illuminations no problem.

 

[hutchphd]

metamer

 

[Andy Resnick]

This is actually a different phenomenon completely.

Your brother's night vision is comprised mostly of activated cone photoreceptors - which do not detect colour. The rod photoreceptors - which do detect colour - are not activated in dim light.
He is seeing in black, white and shades of grey.

His camera does not suffer from this affliction and sees colour at low illuminations no problem.

Good answer, but you reversed rods and cones.

 

[DaveC426913]

Good answer, but you reversed rods and cones.

DANG it! I checked that specifically cuz I always get it wrong! (like bosons and fermions)
(Fixed. Thanks.)

 

[hutchphd]

DANG it! I checked that specifically cuz I always get it wrong

I agree that is a good answer. Also the sensitivity of the rods actually peks in the (near) blue at 500 nm. (Perhaps the color of starlight ? Here is a good mnemonic: think of RGB sprinkles on an ice cream cone

 

[Charles Link]

See https://www.physicsforums.com/threads/color-theory-what-does-blue-yellow-make.1008903/

I see with the "Similar Threads" (where I found this "link") that the Physics Forums did indeed cover this topic in depth a couple years ago. I originally thought that I might have had something a little bit new, but I see others have previously come up with basically the same concepts in discussing the mixing of the interference patterns in a two slit experiment where the source contained two colors in the thread I just linked.

 

[Drakkith]

DANG it! I checked that specifically cuz I always get it wrong! (like bosons and fermions)

Bosons are bosom buddies and like to snuggle together when cold. Fermions say, "F that."

 

[sophiecentaur]

the same concepts in discussing the mixing of the interference patterns in a two slit experiment where the source contained two colors

You have to reealise that wavelength and colour are two totally distinct concepts. When you wake up every morning, you should say to yourself "The eye is not a spectrometer". Any apparent paradox to do with what we see can be resolved just by saying that mantra.

 

[Charles Link]

You have to reealise that wavelength and colour are two totally distinct concepts. When you wake up every morning, you should say to yourself "The eye is not a spectrometer". Any apparent paradox to do with what we see can be resolved just by saying that mantra.

It appears that at least several of the PF people, as can be seen by reading through the discussion that I linked in post 9 above, already have a very good handle on the concepts involved in what the eye sees as color. For me I did know about the red,blue, and green LED's of a TV screen, (I've seen them with a magnifying glass), but as you can probably see by reading my OP above, I hadn't previously thought too much about any additional details, and hadn't previously seen any in-depth discussions on the subject such as the one from PF that I linked in post 9. Cheers. :)

Edit: I think it is likely though that there will be at least of few people out there who might think that when you shine red (650 nm) and green light (550 nm) onto a white piece of paper that somehow the wavelengths would combine to change to 600 nm, which is what it does indeed look like, but not what it is. I don't think I had even thought about the answer to that until this last week. :)

 

[DaveC426913]

Edit: I think it is likely though that there will be at least of few people out there who might think that when you shine red (650 nm) and green light (550 nm) onto a white piece of paper that somehow the wavelengths would combine to change to 600 nm, which is what it does indeed look like, but not what it is. I don't think I had even thought about the answer to that until this last week. :)

 

[sophiecentaur]

It appears that at least several of the PF people, as can be seen by reading through the discussion that I linked in post 9 above, already have a very good handle on the concepts

Oh yes. There are some very well informed people on PF but so many of the posts you can read on this subject, over the years and a lot of what you can read in the scientific press make the mistake of hopping between colour and wavelength. Colour mixing (of different kinds) and matching is seriously problematic.

You get similar problems with speed and velocity and between weight and mass but when people get those two pairs confused, there are plenty of people to step in and put them right. But colour vision produces endless confusion with very little help to resolve it. Evolution, with its usual economy of engineering 'design' is to blame; our colour vision is 'just good enough' to help us through life.

 

[Charles Link]

Thank you @DaveC426913 Very good diagram in post 13 above. :)

 

[sophiecentaur]

View attachment 352464

This diagram is good - as far as it goes. The result of adding monochromatic light (i.e. spectral) will never be a yellow of 600nm because your eye will not identify it as exact spectral yellow. The perceived resulting yellow will not be on the spectral locus on the CIEE diagram. It will lie on a straight line between spectral red and green. Your brain analysis will detect a hint of blue to de-saturate the yellow.
We can blame the very first exposure we all get about colour mixing which is over-simplified and confuses colour and wavelength in a very authoritative (the teacher told me) way.
The yellow ray in the diagram only 'looks' yellow. It's the most convincing of such demonstrations, of course and it only works because our kivinv experience very seldom exposes us to actual spectral colours. In the diagram I found, the result of mixing 450nm with 545nm will lie on a line nowhere near the spectral locus and be a very washed out 'cyan'. But, as I pointed out earlier, we just don't need good resolution of all those colours in the top left of the diagram so that's why it turns out the way it does. In a million years, there may be a good reason for us to need colour resolution in that space so we could develop a fourth colour analysis (and a modified TV system) to take care of that. The almost straight locus from 540nm to 620nm gives us a powerful ability to assess 'skin tones' so we can detect facial changes in mood and also the browns and greens that we need to assess the state of vegetation when we're hunting and gathering. Browns are essentially low luminance orange tones.

 

[DaveC426913]

The yellow ray in the diagram only 'looks' yellow.

Yes, that is the entire message of the diagram.

Things that look yellow are yellow. Until and unless we have a way of examining their constituents individually (such as a prism).


(I was going to add 'or a digital camera' but a digital camera fails the same way our eyes do. Even if we caned the setup so that we shone a true spectral 575nm yellow on paper, the camera would only be able to pick that up by artificially breaking it into red and green components - corrupting the original signal.)

Followup question: Are there any devices other than a prism that can distinguish the true spectral nature of a beam of light? Many spectrometers use (or at least used to use) prisms. I suppose modern, solid state spectrometers don't, do they?

 

[sophiecentaur]

Things that look yellow are yellow.

Totally agree.
Funny thing is that the one and only example 'they' quote of R+R=Y compares what you 'see' with a spectral doublet.

Are there any devices other than a prism that can distinguish the true spectral nature of a beam of light?

Indeed: a diffraction grating works a treat and its dispersion is not dependent on oddities of the glass in the prism- it's all down to the geometry.

 

[sophiecentaur]

a digital camera fails the same way our eyes do

Again - indeed. It tries to behave exactly as our eyes; spectral accuracy is lower on the list of priorities than signal to noise ratio, pixel size, refresh rate blah blah. And what would the electronics do with a load of highly resolved spectral data? But a digital camera sensor is great for measuring the output lines of a spectrometer. That film stuff was a nightmare for doing the same job.

Something that I only recently cottoned on to is that, in the early days, camera film had really bad red sensitivity so spectral measurements had very little information about red (or IR). The universe looked very different in them thar days.

 

[Charles Link]

It might be interesting to see a demonstration sometime of how true yellow (at or near 600 nm) compares visually to the composite of 550 green and 650 red. One additional thing that is contained in the OP that is not in @DaveC426913 's diagram above is that the yellow at 600 nm that comes out of a first prism from white light will emerge from a second prism in the same place, (basically midway between where the green and red would emerge, but with no green and red from the second prism).

Meanwhile we can analyze things with a CIE color chart, but it would still be worthwhile to see a live demo=my guess is even the live demo will not look exactly the same to everyone. Some might be able to tell the composite yellow from red and green apart from the pure yellow a little easier than others if there is a noticeable difference=in my OP I think I inferred they would look identical, but the CIE chart as others have mentioned shows some slight differences, etc.

For other things besides a prism to give the spread of colors, the diffraction grating spectrometer is indeed a very good one, and in general has much higher resolution than a prism spectrometer. Otherwise, we can also get the same spectral spreading when we see a rainbow in the sky. Thin film interference filters that are based on the Fabry-Perot effect can also be used to select or pass various wavelengths of light.

I really think this topic makes a good one for discussion, because as mentioned by @sophiecentaur above, there is a fair amount of incomplete information on the topic that in many cases I think is presented by people who have a very limited physics background. Cheers. :)

Edit: and I still need to learn how to read the CIE chart. The sodium doublet at 589.0 and 589.6 nm should show up as yellow, but the chart seems to suggest it may be red. I find the chart a little confusing...

 

[Charles Link]

and a follow-on: The part I was missing with the CIE graph is that the blue cone response z=1-x-y, certainly not self-explanatory, but I found it in a google. Meanwhile the red cone is x which also has some response from light around 480 nm=something that I pretty much knew from instinct, (i.e. violet and red are similar in appearance), and the y is the green cone.

I think that some of the coloring in the diagrams above may only be approximate=589 nm (sodium doublet) should come in as yellow. (also see post 18 by @sophiecentaur =I think the coloring in the diagram is incorrect=you do not get R+R=Y).

and if my arithmetic and graphing is correct, it looks to me like neither the composite of 550 nm and 650 nm nor the pure 600 nm yellow will have very much blue in it.

 

[DaveC426913]

Something that I only recently cottoned on to ...

Er.. we don't use that expression that anymore.

 

[Charles Link]

One additional comment on the above: If I have it right, on the CIE diagram $x+y+z=1$. This makes it so the diagram, even though in 3-D space, is simply a plane in 3 dimensions. It is not a 3-D structure, so it is often shown as an x-y graph.

Given the x and y coordinate, (the red and green response), you simply do a little arithmetic to calculate the z=blue (normalized) response, where $z= 1-x-y$. The tongue shaped figure, with its colors painted in, actually lies in the plane $x+y+z=1$, rather than in the x-y plane.

The next part, which I think I also have figured out now, is if you have one color of amount $a$ at $(x_1, y_1, z_1)$ and amount $b$ at $(x_2,y_2, z_2)$, then the resultant color (coordinate) is the normalized $(x_3,y_3,z_3)$ that you get by doing a vector addition $\vec{r_3}= a \vec{r_1} + b \vec{r_2}$ and normalizing $\vec{r_3}$.

Edit: One slight addition/correction to this is that this normalized vector can be used to define a line $\vec{r}=\vec{r_{3n}} t$. The color coordinate is then found by where this line intersects the plane $x+y+z= 1$. I think I have this last part correct now, but I am on a learning curve with the CIE map, so I welcome any feedback.

 

[Charles Link]

With my latest entry in post 23, I think I have it correct now. This seems to make sense since if two colors with different coordinates are added together in varying amounts, the result in the 3-D space will be that the two coordinates will make for two vectors of varying lengths (added together) that define a plane. Where this plane intersects the plane $x+y+z=1$ will be a line on the CIE map that yields all the possible results, (i.e. colors), from the two sources.

I welcome any feedback. This topic (the CIE map) is written up in a couple places in the literature, but I think I have been able to pick out a couple important features that aren't emphasized very much in the write-ups.

Edit: One other item is that they could have defined the x-y-z color coordinates by the direction of the normalized unit vector in the 3-D space, but instead they used a slightly different scheme where the color coordinates are found by the location where the line in the direction of the unit vector intersects the plane $x+y+z=1$, (instead of using the point where the line intersects the unit sphere $x^2+y^2+z^2=1$).

One other detail is if we say we have amount $a$ at one color coordinate and amount $b$ at another, how do we know what to reference the $a$ and $b$ to ? It appears from what I read in the literature that they use/specify the amount of the y components to get the multiplication factors $a$ and $b$.

 

[sophiecentaur]

Even if we caned the setup so that we shone a true spectral 575nm yellow on paper, the camera would only be able to pick that up by artificially breaking it into red and green components - corrupting the original signal.

A point of information: RGB are essentially used to describe the Synthesis of colours in a display. The spectra of the 'phosphors ' and the points on the CIE chart are not the same things. The RGB 'points' do not actually represent monochromatic sources because the requirement is for bright sources and different display technologies may well use different spectra to achieve the same RGB points. The eye / camera analysis curves are not the same as the spectra of phosphors; they are essentially very broad band. They have to be, in order to catch all wavelengths and they all overlap. Note the SML labels and not RGB. There is some complicated processing to get from those three signals to appropriate signals to drive the phosphors with.

Er.. we don't use that expression that anymore.

OMG is it verboten now? I wonder why. I imagined it was a fishing term?

(also see post 18 by @sophiecentaur =I think the coloring in the diagram is incorrect=you do not get R+R=Y).

If you look half way along the R-G line, the colour is a fairly convincing yellow to me. If you are looking for an error (not a bad idea, aamof) then I guess there may be discrepancy from source to source - although all the versions of CIE charts on a Goodle search seem to agree largely. If you look in a variety of printed versions, I'd bet there would be a significant spread.
You make a very valid comment that Yellow on the chart is not actually a spectral sodium yellow. I imagine that example of colour matching would have been chosen because it was easy to give a convincing demo, with a sodium source and a couple of very loosely specified primary filters in front of filament lamps.
During the development of colour TV, there have been several specs for RG and B phosphors but the human colour vision system takes care of that, as long as the synthesis formulae are adjusted for each system. The only colours that can be displayed lie within the gamut of the RGB triangle. The result is that very saturated colours cluster on the sides of the triangle. I have noticed that sports crowd scenes often seem to show the same colours of bright outdoor wear. I think that must be due to this ('crushing') effect. Systems have developed to show a bigger range than in the past. Modern TV displays are seriously stunning, compared with early CRT displays with inadequate phosphors.

 

[Charles Link]

A point of information: RGB are essentially used to describe the Synthesis of colours in a display. The spectra of the 'phosphors ' and the points on the CIE chart are not the same things. The RGB 'points' do not actually represent monochromatic sources because the requirement is for bright sources and different display technologies may well use different spectra to achieve the same RGB points. The eye / camera analysis curves are not the same as the spectra of phosphors; they are essentially very broad band. They have to be, in order to catch all wavelengths and they all overlap. Note the SML labels and not RGB. There is some complicated processing to get from those three signals to appropriate signals to drive the phosphors with.
View attachment 352503

OMG is it verboten now? I wonder why. I imagined it was a fishing term?

If you look half way along the R-G line, the colour is a fairly convincing yellow to me.

The region at 570 nm is yellow to me, but at 590 nm it is already very orange. It is possible though that my computer screen does not have the same colors that yours does. In any case, by 600 nm (from the chart from your post 16) I see it to be very orange and almost red.

Edit: It is possible that my OP is in some error, and that 600 nm is actually orange, and that I should have used 570 nm for yellow. I did not base my OP on the CIE chart, but rather on knowing that the sodium doublet at 590 nm is yellow. Perhaps I made a slight error=not sure at this point. [and a google shows yellow to occur in another chart from about 580 nm to 590 nm. The transition to orange is very rapid and happens by 600 nm. Meanwhile 570 nm in this other chart still looks rather green]. See https://www.lumitex.com/blog/visible-light-spectrum

See https://www.physicsforums.com/threads/color-theory-what-does-blue-yellow-make.1008903/
The exam question presented in this post also calls 600 nm as yellow=perhaps it is also in error.

Meanwhile your inputs and feedback are very good=thank you. :)

 

[sophiecentaur]

The region at 570 nm is yellow to me, but at 590 nm it is already very orange.

I wouldn't disagree. However, your problem with equal R+G not giving Y will basically depend on where your R and G primaries lie. Remember, the original story about R+G being Y comes from a time when they didn't have any form of colour display. It convinced me when I was shown it at school but a good teacher beats a poor demo any time - the idea got across. Billions of people these days are totally fooled by the good colour displays. So that basic work from 100 years ago wasn't too bad.

PS there is a lot of baloney thrown up from Google searches about colour theory and matching, and yet Land's alternative views about colour vision can't just be dismissed. I never followed it through so my opinions don't count.

 

[Charles Link]

In any case, the main concept that I wanted to show in the OP is presented in the diagram by @DaveC426913 in post 13, even if it might take a little fine tuning of the green and red to get yellow, where the yellow when sent through a second prism comes out as green and red again with no yellow in the output.

The other thing I wanted to show in the OP is that if you take the pure yellow from a prism where a blackbody type (2500 K) white light is used, and then send it through a second prism after shining it on a white sheet of paper, it would come out as being the same pure yellow. The wavelength may need to be 580-590 nm instead of 600 nm, but the main concept is still there.

I also found the CIE coordinates to be very interesting. I knew about them many years ago, but only worked out a couple of the details in the last day or two. (See posts 23 and 24 above). Cheers. :)

 

[Ibix]

OMG is it verboten now? I wonder why. I imagined it was a fishing term?

I looked it up because that made me blink too. I think "cotton on" has a perfectly innocent etymology - I gather that it probably refers to the tendency of unprocessed cotton to cling to you, although there are other possibilities. "Cotton pickin'', as often used by Bugs Bunny, may have a slavery-related origin and its acceptability seems to be a topic of some debate. I wonder if @DaveC426913 may have conflated the two?

 

[sophiecentaur]

I looked it up because that made me blink too. I think "cotton on" has a perfectly innocent etymology - I gather that it probably refers to the tendency of unprocessed cotton to cling to you, although there are other possibilities. "Cotton pickin'', as often used by Bugs Bunny, may have a slavery-related origin and its acceptability seems to be a topic of some debate. I wonder if @DaveC426913 may have conflated the two?

Hmmmm.
It can be a minefield. Hopefully this will pass - although i can understand many people feel they've been wronged in the past.

 

[DaveC426913]

"Cotton pickin'', as often used by Bugs Bunny, may have a slavery-related origin and its acceptability seems to be a topic of some debate. I wonder if @DaveC426913 may have conflated the two?

I think I'm not the first or only one to see the conflation.

 

[Charles Link]

I want to reiterate something that for me was a major find with the CIE chart that I mentioned in posts 23 and 24 above. It basically treats the 3 color bands [Edit=color cone stimulus/response] in 3 dimensions with the X, Y, and Z being the intensity of each band. (X=red cone, Y=green cone, Z=blue cone). The light of a given luminance and color is a vector $\vec{R}=X \hat{i}+Y \hat{j}+ Z \hat{k}$ , and where the line $\vec{R} t$ crosses the plane $x+y+z=1$ is designated as the color coordinate. The CIE chart is a color map in this plane which crosses the axes at $(1,0,0)$, $(0,1,0)$ , and $(0,0,1)$. (They usually show it as an x-y graph, but they are actually showing the view from above of the map in the plane $x+y+z=1$.)

If you have two sources of some intensity with different color coordinates on this chart, making vectors of two different lengths, the resultant vector will be in the plane containing these two vectors, and its color coordinates are found by where the line made from this resultant vector passes through the plane $x+y+z=1$. The CIE chart is really a neat piece of mathematics. See also https://en.wikipedia.org/wiki/CIE_1931_color_space , but they don't seem to highlight these details.

Edit: It should be noted that the vector $\vec{R}$ above thereby has color coordinates of $x=X/(X+Y+Z)$, $y=Y/(X+Y+Z)$ , and $z=Z/(X+Y+Z)$.

 

[Charles Link]

See https://www.wtamu.edu/~cbaird/sq/20...lors-red-orange-yellow-green-blue-and-violet/ for what I think is a more accurate color chart.

I would like to return to post 16 above by @sophiecentaur and make an additional comment or two. It does appear from the CIE chart that we should be able to get what would appear to be nearly pure yellow to the eye if we choose the right combination of green at 550 nm and red at 650 nm. (The 590 nm (yellow) does indeed lie on what is very nearly a straight line on the boundary of the CIE chart connecting the green at 550 nm with the red at 650 nm, and there should be almost no blue, (contrary to post 16). (Edit: Note that the border of the CIE chart from 550 nm to 650 nm lies very nearly on the line where $x+y=1$, so that $z=0$ , i.e. no blue content=i.e. no blue cone stimulus to a very good approximation).

Meanwhile though as what can be seen from the "link" I just provided, yellow does indeed occur at 590 nm, and it seems to be a bit of false coloring in this CIE chart of post 16 that I also found on-line where their yellow is around 570 nm.

 

[DaveC426913]

See https://www.wtamu.edu/~cbaird/sq/20...lors-red-orange-yellow-green-blue-and-violet/

I would like to return to post 16 above by @sophiecentaur and make an additional comment or two. It does appear from the CIE chart that we should be able to get what would appear to be nearly pure yellow to the eye if we choose the right combination of green at 550 nm and red at 650 nm. (The 590 nm (yellow) does indeed lie on what is very nearly a straight line on the boundary of the CIE chart connecting the green at 550 nm with the red at 650 nm, and there should be almost no blue, (contrary to post 16).

Meanwhile though as what can be seen from the "link" I just provided, yellow does indeed occur at 590 nm, and it seems to be a bit of false coloring in this CIE chart of post 16 that I also found on-line where their yellow is around 570 nm.

Let's not forget that "yellow" is highly subjective. It's different for everyone, and even different depending on context.

Anyone encountered this little test?
https://ismy.blue/
My score (today) is 172.

 

[Charles Link]

I'm actually looking for somewhat high precision here, since it seems to be available. It does appear from the CIE chart, and its nearly straight line border that we should be able to generate what appears as an almost very precise yellow at 580-590 nm to the eye from wavelengths of green at 550 nm and red at 650 nm in the right combination. I would agree that everyone won't see precisely the same thing, but in this case it should be pretty close.

 

[DaveC426913]

I'm actually looking for somewhat high precision here, since it seems to be available. It does appear from the CIE chart, and its nearly straight line border that we should be able to generate what appears as an almost very precise yellow at 580-590 nm to the eye from wavelengths of green at 550 nm and red at 650 nm in the right combination. I would agree that everyone won't see precisely the same thing, but in this case it should be pretty close.

Pretty close to what?
There is no such thing as "The One True Yellow".

Heck, there's no such thing as "One True Red" or "One True Green"; how could there be a mixture that's true?

The very nature of colour is that occurs only in the mind of the observer. Each observer.

 

[DaveC426913]

I'm actually looking for somewhat high precision here, since it seems to be available. It does appear from the CIE chart, and its nearly straight line border that we should be able to generate what appears as an almost very precise yellow at 580-590 nm to the eye from wavelengths of green at 550 nm and red at 650 nm in the right combination. I would agree that everyone won't see precisely the same thing, but in this case it should be pretty close.

Here's your 580 and 590:

Photoshop thinks 580 is 98% yellow and 29% magenta, while 590 is 60% magenta.

 

[Charles Link]

Pretty close to what?

From what I have in my OP=to actually make a live demo instead of just a Gedanken experiment. From what the CIE chart has with 580-590 nm being on the almost straight like connecting 550 nm and 650 nm. To follow what I am referring to here, you would need to understand the CIE chart in depth=perhaps you already do, but it is something I only figured out in the last day or two=(see also my posts 23 and 24). The primary colors (edit: actually the amount of stimulation of each of the 3 color cones=I think this is now a more accurate description) are made as vectors in an XYZ space with the length being proportional to the intensity. Meanwhile the entire visible spectrum lies on the outside border of the color coordinate curve, which is on the plane $x+y+z=1$ of this space. Any visible light source will have a vector that passes through the interior of the CIE color coordinate curve. Individual sources are all treated as vectors, and the laws of vector addition apply to determine the result. Meanwhile where the resultant vector passes through the plane $x+y+z=1$ determines its color coordinates. Using this, the pure yellow of 580-590 nm (on the border of the CIE curve) is the vector sum of the right combination of green at 550 nm and red at 650 nm, (both also on the border), since the 580-590 nm lies almost on the straight line connecting the 550 nm to the 650 nm.

 

[DaveC426913]

From what I have in my OP=to actually make a live demo instead of just a Gedanken experiment. From what the CIE chart has with 580-590 nm being on the almost straight like connecting 550 nm and 650 nm. To follow what I am referring to here, you would need to understand the CIE chart in depth=perhaps you already do, but it is something I only figured out in the last day or two=(see also my posts 23 and 24). The primary colors are made as vectors in an XYZ space with the length being proportional to the intensity. Meanwhile the entire visible spectrum lies on the outside border of the color coordinate curve, which is on the plane $x+y+z=1$ of this space. Any visible light source will have a vector that passes through the interior of the CIE color coordinate curve. Individual sources are all treated as vectors, and the laws of vector addition apply to determine the result. Meanwhile where the resultant vector passes through the plane $x+y+z=1$ determines its color coordinates. Using this, the pure yellow of 580-590 nm (on the border of the CIE curve) is the vector sum of the right combination of green at 550 nm and red at 650 nm, (both also on the border), since the 580-590 nm lies almost on the straight line connecting the 550 nm to the 650 nm.

You're confusing the map with the territory, and doing your math on the map instead.
The CIE chart is just a map; it does not define the territory; it can only attempt to mimic it.

Witness the fact that there are dozens of competing colour maps, all vying to achieve the ideal.

 

[Charles Link]

You're confusing the map with the territory, and doing your math on the map instead.
The CIE chart is just a map; it does not define the territory; it can only attempt to mimic it.

Witness the fact that there are dozens of competing colour maps, all vying to achieve the ideal.

I find it a little disappointing that most others don't seem to be able to follow the mathematics behind the CIE chart that I just figured out in the last couple of days and/or they haven't read my posts very carefully, especially posts 23 and 24. I knew of the CIE chart as much as 40 years ago or more, but only in the last couple of days did I figure out enough of what are perhaps somewhat hidden mathematical details that you normally don't even find except in a very good write-up of the CIE chart, which so far I haven't seen. Some others who already know these details simply may not be giving much or any feedback. Cheers. :)

 

[DaveC426913]

I find it a little disappointing that most others don't seem to be able to follow the mathematics behind the CIE chart

This did not start off there. This started off as discussion about the nature of yellow, and wandered into the weeds of what it is to be true yellow. (Mia culpa.)

But that's a far cry from the quantitative mathematics of the CIE chart. Perhaps you could take some time to compose your thoughts, separating out what is subjective from what is objective, so as to help us follow your thesis, and to engage the members you want to engage?

 

[Charles Link]

Here's your 580 and 590:
View attachment 352561
Photoshop thinks 580 is 98% yellow and 29% magenta, while 590 is 60% magenta.

Please see my post 33 where I "linked" another color chart=I do think this "linked" color chart is more correct =see also post 16 where that chart may also have some rather incorrect color to it.

I don't think very many people have tried to follow along my posts through the complete thread. The discussion then got off on a tangent about cotton. I would have preferred if it stuck more to the topic which I think I have presented reasonably well.

The diagram you presented in post 13 is a very good one as I mentioned in post 15. :)

 

[DaveC426913]

The discussion then got off on a tangent about cotton. I would have preferred if it stuck more to the topic

You could report and ask to have that diversion removed. Not sure if they will, but it's not unreasonable to ask.

 

[Charles Link]

You could report and ask to have that diversion removed. Not sure if they will, but it's not unreasonable to ask.

It really didn't matter much to me one way or the other. What I would like to see though as that you and others, including @sophiecentaur who has had some very good inputs, read through and study my posts in complete detail throughout the whole thread. That may be asking a lot, but otherwise the feedback I get may be rather mediocre, because you/they will have missed perhaps a couple things or more that I presented throughout the thread. (Post 33 is also one that I could enjoy getting some feedback on).

I wasn't the one who introduced the CIE color map in the thread, (I think @sophiecentaur did), but in hindsight it seems to be very useful for discussing how the colors mix. If you follow my posts throughout the thread, you will see that at first I had a very limited understanding of what the CIE color map is all about. In fact, after figuring out some of its finer mathematical details, (see posts 23, 24, 32, and 38), it seems IMO to be a very good way to proceed when discussing the mixing of any combination of colors of light. :)

 

[sophiecentaur]

I wasn't the one who introduced the CIE color map in the thread, (I think @sophiecentaur did), but in hindsight it seems to be very useful for discussing how the colors mix.

IMO, the point of the CIE chart is as a common discussion point when design colour reproduction systems.

One thing we haven't yet mentioned is the specification of the illuminant of a scene that's being reproduced. The CIE system allows colour matching between different technologies and different illuminants. Every modern camera / colour chain has software that tries to match the original scene to what's displayed for 'all' combinations of camera analysis and display synthesis, and the eye. Our brains manage a good job of colour matching for outdoor conditions, in and out of shadow (direct sunlight and light from the sky and surrounding objects (i.e. for different 'white points') but it can be confused. Camera software finds this more different, despite being very clever. Things are even harder now that there is a whole range of colour phosphors.

Pretty close to what?
There is no such thing as "The One True Yellow".

Exactly. this thread has already exposed the nonsensical generalised equivalent that used to be drawn between spectral yellow and 'what you get' when you mix red and green. Nothing is certain about colour but without well specified primaries and white point, you can't say how a coloured patch on an item in a scene will be displayed on a TV screen.

I also have to mention the importance of subtractive mixing of colours in dyes and pigments and also in some CMY displays like colour film. The advertising industry is obsessed with that.

Then there's the fact that the everyone's eyes are totally individual and that there is a huge spread in colour perception. This has allowed the high levels (as with HiFi) of BS that has gone into comparisons of cameras and displays. I recently bought a new telly with an OLED screen and its pictures can be really stunning (source dependent, of course) but I'd hesitate to make an objective comparison between the best displayed pictures and the original scenes. I remember being told (65 years ago , at least) that colour film colourimetry is simply awful but cinema audiences accept it because they are in an isolated darkened room with no comparison with real life.

 

[Charles Link]

Very good inputs @sophiecentaur Thank you. :)

I'm still looking for some feedback on the mathematics though=to me it isn't readily apparent the things that I mentioned in post 32. Perhaps you have already known this concept (that $x+y+z= 1$ as being the plane on which the CIE map is located) and the vector addition for two or more sources of the $X$ , $Y$, and $Z$, etc. for the 3 color cones of the eye, but for me this is something I finally figured out just this past week, and I wouldn't be surprised if it is a new concept for many others. Hope you find this part of interest. :) See also post 38.

[Edit: Note also that the color coordinates $(x,y,z)$ for any source is the location where the vector formed by the intensities $X, Y$ and $Z$ seen by the 3 color cones crosses the plane $x+y+z=1$. Thereby the CIE chart maps out each and every possible color, (each with its color coordinates $(x,y, z)$), right in this plane.]

Note that they could have normalized the $X,Y,Z$ vector and let its direction cosines determine the $(x,y,z )$ color coordinates of the color, but instead they used the location where the vector crosses the plane $x+y+z=1$, (instead of where the vector crosses the unit sphere $x^2+y^2+z^2=1$). It really makes for easier arithmetic=given the $x$ and $y$ on the color map (=red and green cones contributions/coordinates to the color ), we can very easily compute the blue cone contribution/coordinate as being $z=1-x-y$.

 

[Charles Link]

I'm still looking for some feedback on the mathematics of the CIE color chart as I presented in post 46 above, and in other posts in this thread. I think I came up with a good way of explaining it.

I find it somewhat disappointing these days that so few seem to take much interest in any mathematics. When I meet others in my neighborhood Starbucks, it is difficult to find anyone who can even work something simple like finding the hypotenuse of a right triangle whose sides are 35 and 12. I'm hoping at least a couple people are able to follow the explanation in post 46, but I guess if I don't get a whole lot of feedback, that is ok too. Cheers. :)

 

[DaveC426913]

When I meet others in my neighborhood Starbucks, it is difficult to find anyone who can even work something simple like finding the hypotenuse of a right triangle

What? Like strangers?

 

[DaveC426913]

That was a serious question.

 

[Charles Link]

That was a serious question.

Yes, and thank you for asking. Yes, I have a Starbucks about one block away from my house and I get coffee there almost every day. I meet a fair number of people there=not a huge number because many keep to themselves, but in any case I look for people who might understand some mathematics beyond high school algebra, and most of them that I meet have almost zero math skills. If I can get them to even work a right triangle with the Pythagorean theorem, I consider it a victory. Trying to get them to do something more difficult like the law of cosines is like asking for a miracle.