I Question about subtractive color mixing

[JohnnyGui]

Hello,

When reading a bit about substractive color mixing, a question came up.

I understood that when mixing Cyan, Magenta and Yellow paint, you would get black because all these 3 colors that get reflected by each corresponding paint are absorbed by the other paint in the mix, leaving no color behind to get out of the mix.

Furthermore, it is stated that when mixing only Cyan and Magenta paint, you would get Blue coming out of the paint mix. This is explained by the fact that Cyan consists of a combination of Green and Blue wavelengths while Magenta paint consists of a combination of Blue and Red wavelengths.
Since Cyan paint absorbs the Red part of the Magenta, and Magenta absorbs the Green part of the Cyan, Blue is left behind and comes out of the mix.

Here’s where my question comes up. What if Cyan is not a mix of Blue and Green wavelengts but a pure Cyan wavelength itself? And the same goes for Magenta, having its own wavelength. Doesn’t that mean that when mixing those two colors, you would get black since Cyan absorbs the Magenta wavelength and Magenta absorbs the Cyan wavelength, leaving no color behind to leave the mix?

 

[QuantumQuest]

I understood that when mixing Cyan, Magenta and Yellow paint, you would get black because all these 3 colors that get reflected by each corresponding paint are absorbed by the other paint in the mix, leaving no color behind to get out of the mix.

Yes, in theory the combination of cyan, magenta, and yellow at 100% creates black, meaning all light being absorbed. However in practice due to imperfections in the inks and other limitations of the process, full and equal absorption of the light is not possible so you can't have true black. So, printers for example use the CMYK process adding black (K).

What if Cyan is not a mix of Blue and Green wavelengts but a pure Cyan wavelength itself? And the same goes for Magenta, having its own wavelength. Doesn’t that mean that when mixing those two colors, you would get black since Cyan absorbs the Magenta wavelength and Magenta absorbs the Cyan wavelength, leaving no color behind to leave the mix?

How can this be? You're essentially changing the color models cancelling / interchanging primary / secondary colors. In CMY model, Blue + Green absorbs Red and creates Cyan, Red + Blue absorbs Green and creates Magenta and Green + Red absorbs Blue and creates Yellow. So, this model has primary colors the secondary colors of RGB model.

 

[Merlin3189]

... What if Cyan is not a mix of Blue and Green wavelengts but a pure Cyan wavelength itself?

You are talking here of subtractive mixing, so you shouldn't talk of the cyan as a mixture of B & G, but as what is left after R (and O & Y) have been removed from white light or from any other mixture which leaves C and a balance of B&G

And the same goes for Magenta, having its own wavelength.

Of course, magenta is not a spectral colour and can be seen only as a mixture of R and B light.

Doesn’t that mean that when mixing those two colors, you would get black since Cyan absorbs the Magenta wavelength and Magenta absorbs the Cyan wavelength, leaving no color behind to leave the mix?

Since there is no magenta wavelength to absorb, the cyan must be absorbing the red component of magenta, leaving some blue.
The magenta pigment absorbs colours other than R&B. So the remaining colour is mainly blue.

 

[Khashishi]

It's all wrong. Colors are not so simple that you can add and subtract them. Rules about color subtraction are at best heuristics and can and will fail.

Color is what your eye perceives due to stimulation of your cone cells and processed by your brain. Most humans have 3 types of cone cells which have different responsivity curves to different wavelengths. Take a look at the responsivity graph in https://en.wikipedia.org/wiki/Cone_cell

In general, the light coming off an object has a combination of many wavelengths of light. This particular spectrum can excite a particular ratio of your cone cells, and your brain processes the ratio into a color. Different spectra of light can produce the same color in your brain. Two pigments could appear the same color but have totally different spectra and mix in totally different ways.

 

[JohnnyGui]

You are talking here of subtractive mixing, so you shouldn't talk of the cyan as a mixture of B & G, but as what is left after R (and O & Y) have been removed from white light or from any other mixture which leaves C and a balance of B&G

Of course, magenta is not a spectral colour and can be seen only as a mixture of R and B light.

Since there is no magenta wavelength to absorb, the cyan must be absorbing the red component of magenta, leaving some blue.
The magenta pigment absorbs colours other than R&B. So the remaining colour is mainly blue.

I'm sorry but I'm more confused now. You're saying in subtractive color mixing, Cyan should not be seen as a mixture of B and G but at the same time Magenta should be seen as a mixture of B and R? And what is O?

Let me ask this stepwise. Apart from subtractive color mixing, if an object is reflecting photons with a wavelength of Cyan, doesn't that mean that it absorbs all other wavelength, including Green and Blue?
Is there such a thing as a pure wavelength of Cyan or is it only possible for an object to reflect Cyan if it reflects both photons of wavelengths Green and Blue together?

 

[Merlin3189]

I'd very much agree with you about subtractive mixing being rough and heuristic - (it's used by artists.)
But I would say additive mixing is very well understood and has been measured in great detail. Whatever the mixture of light in the components the sum can be calculated accurately (for the standard observer*) and generally appear more or less as would be expected. The CIE chromaticity diagram is only roughly triangular, so color triangles are approximations which don't cope accurately with some lights. (* people with colour anomalies or course need special consideration.)

The other problem with all these discussions is the talk of colours as if R,G,B,C,M and Y are defined things. All are a range, even if they are made with a single pure spectral colour (as said B4, M can't be). And as Khashishi says, they are not generally pure spectral colours, but mixtures.

Of course, JohnnyGui could have a point if he were talking about subtractive mixing with narrowband filters rather than pigments.

 

[JohnnyGui]

I'd very much agree with you about subtractive mixing being rough and heuristic - (it's used by artists.)
But I would say additive mixing is very well understood and has been measured in great detail. Whatever the mixture of light in the components the sum can be calculated accurately (for the standard observer*) and generally appear more or less as would be expected. The CIE chromaticity diagram is only roughly triangular, so color triangles are approximations which don't cope accurately with some lights. (* people with colour anomalies or course need special consideration.)

The other problem with all these discussions is the talk of colours as if R,G,B,C,M and Y are defined things. All are a range, even if they are made with a single pure spectral colour (as said B4, M can't be). And as Khashishi says, they are not generally pure spectral colours, but mixtures.

Of course, JohnnyGui could have a point if he were talking about subtractive mixing with narrowband filters rather than pigments.

Not sure if I'm indeed taking about narrowband filters or pgiments. I'm talking about an object giving off a very specific wavelength. (a very narrow range as possible)

Even if e.g. Green is a range of wavelengths, doesn't that mean that an object can give off Green in two ways:
1. It reflects a specific wavelength range of 495–570 nm and absorbs all other wavelenghts (including Yellow and Cyan)
2. It reflects two wavelengths (each having a certain range) of Yellow and Cyan together.

Is this statement correct?

 

[sophiecentaur]

You're saying in subtractive color mixing, Cyan should not be seen as a mixture of B and G

B and G are not subtractive primaries. The primaries which relate to blue and green are -B and -G, which are Yellow and Magenta.
A Khashishi pointed out,

It's all wrong. Colors are not so simple that you can add and subtract them. Rules about color subtraction are at best heuristics and can and will fail.

Even additive colour mixing is full of misconceptions and that lends itself quite well to numerical treatment. Colour TV is pretty damned good for most peoples' colour perception. Colour film - even the very best quality, with the greatest care taken was pretty dire for fidelity. They got away with it mainly because people watched movies in a darkened cinema with nothing to compare purple / yellow / red faces with. The used to pick up on far less glaring errors on their old Colour TV sets. If you want to match some particular colours as well as possible with pigments, you have to mix it specially with particular raw colours. An artist will carry many different tubes of paint to cope with this.
The first thing to remember about colorimetry is that colour is not wavelength and no one would ever think of analysing a scene with three single wavelength sensors - they would waste most of the available light energy and also they would fail to analyse most colours properly.

And the same goes for Magenta, having its own wavelength.

Magenta is not a single wavelength. There is no single wavelength that 'looks magenta'.

 

[Merlin3189]

I'm sorry but I'm more confused now. You're saying in subtractive color mixing, Cyan should not be seen as a mixture of B and G but at the same time Magenta should be seen as a mixture of B and R? And what is O? Orange.

There is no pure spectral colour corresponding to Magenta. Magenta is a colour mixture of reds and blues. R and B are at opposite ends of the spectrum and all the spectral colours between them are not magenta but hues of reds, oranges, yellows, greens and blues.
Cyans can be a pure spectral colour, or a mixture of blues and greens (and cyans) provided that the average is around the cyans.

Let me ask this stepwise. Apart from subtractive color mixing, if an object is reflecting photons with a wavelength of Cyan, doesn't that mean that it absorbs all other wavelength, including Green and Blue?
If indeed an object did reflect only a narrow band of light around cyans then of course it would not reflect blues and greens outside that band.
Is there such a thing as a pure wavelength of Cyan or is it only possible for an object to reflect Cyan if it reflects both photons of wavelengths Green and Blue together?
There certainly is such a thing as a pure spectral cyan. It would be purer than any mixture of blue and green.

Most objects don't reflect a pure spectral colour, but a broad mixture of colours. Pure spectral colours often look supernatural, because we don't generally experience them.

You might like to take a look at a CIE chromaticity diagram, which sheds some light on this.
The spectrum of pure colours lie along the curved perimeter from red via O, Y, G, C to blue. Magentas lie along the straight edge from red to blue. This edge is straight because all hues of magenta are a mixture of red and blue. There are no pure spectral magentas,.
All other colours are mixtures and the geometry of the diagram shows how the result of combining (adding) lights (pure or mixtures) is obtained. Roughly, the 'centre of gravity' of a set of lights is the result of adding them. The 'CoG' of all colours is roughly at the centre and that is where the whites are found. Any mixture spread evenly around the centre will also be a white, even if it does not contain "all colours" - eg. a mixture of a pure orange and a pure magenta.
You can thus see, that any colour, other than a pure spectral colour, can be made from many different mixtures.
===================

Not sure if I'm indeed taking about narrowband filters or pgiments. I'm talking about an object giving off a very specific wavelength. (a very narrow range as possible)

Even if e.g. Green is a range of wavelengths, doesn't that mean that an object can give off Green in two ways:
1. It reflects a specific wavelength range of 495–570 nm and absorbs all other wavelenghts (including Yellow and Cyan)
2. It reflects two wavelengths (each having a certain range) of Yellow and Cyan together.

Is this statement correct?

Yes.
I don't know if you can get narrow band pigments, but you can create objects which have narrow band reflectance.

Both 1 & 2 are possible. 3. would be, it reflects any number of wavelengths whose average was a cyan.

 

[JohnnyGui]

Explanation

Thanks a lot for the detailed explanation and for answering my questions!

Your answers makes me able to ask the following question;
If there is such a thing as a pigment reflecting a pure spectral wavelength such as Cyan or Green, this means that a pure spectral Cyan paint would absorb all wavelengths except for Cyan and a pure spectral Green paint would absorb all wavelengths except for Green. Doesn't this mean that mixing Cyan and Green together, each color would absorb the other one, which leads to the mix being black? How is this not the case?

 

[Vanadium 50]

If there is such a thing as a pigment reflecting a pure spectral wavelength such as Cyan or Green

Magenta is not a single wavelength.

Same for Cyan and Green.

 

[sophiecentaur]

Same for Cyan and Green.

There are spectral (monochromatic) colours that will be interpreted as Cyan or Yellow (the other two subtractive primaries). Magenta is the one for which there is no monochromatic equivalent.
But there are no 'objects that reflect monochromatic light because condensed materials have a band structure. A gas, glowing as a result of incident light can produce monochromatic light but that's a special case.
@JohnnyGui : You have to be careful in interpreting the CIE diagram. It has no physical reality at all because it just shows the result of how a group of people were able to match 'coloured' objects against combinations of different wavelengths of light (mixing primaries). It's a very shorthand display of our perception and luckily it works! The tristimulus colour vision which humans work with, uses three very wide band analyses, which all pretty much accept all visual wavelengths at different levels and it manages to condense all visible spectra into three 'numbers' for the brain to work with. Evolution has given us a system which is 'just good enough' to let us recognise things like plant types, facial expression of emotions etc.. We never have needed a spectrometer so we never evolved with one.

 

[JohnnyGui]

Same for Cyan and Green.

This would clear up my confusion, but according to @Merlin3189 these are pure colors.

@JohnnyGui : You have to be careful in interpreting the CIE diagram. It has no physical reality at all because it just shows the result of how a group of people were able to match 'coloured' objects against combinations of different wavelengths of light (mixing primaries). It's a very shorthand display of our perception and luckily it works! The tristimulus colour vision which humans work with, uses three very wide band analyses, which all pretty much accept all visual wavelengths at different levels and it manages to condense all visible spectra into three 'numbers' for the brain to work with. Evolution has given us a system which is 'just good enough' to let us recognise things like plant types, facial expression of emotions etc.. We never have needed a spectrometer so we never evolved with one.

Got it. However, I was wondering about the following thing you said:

But there are no 'objects that reflect monochromatic light because condensed materials have a band structure. A gas, glowing as a result of incident light can produce monochromatic light but that's a special case.

Do you meant with this that there is no object, e.g paint, at all that only reflects a single (small range of) wavelength(s)? Could you please enlighten me more on the reason (band structure)?

Suppose there exists paint that only reflects a single wavelength of Cyan and a paint that only reflects a single wavelength of Yellow, would my reasoning about the mix of those 2 being black then be correct? Just like when 2 gases, each emitting a different single wavelength, would look black if they get mixed together?

 

[Khashishi]

Suppose there exists paint that only reflects a single wavelength of Cyan and a paint that only reflects a single wavelength of Yellow, would my reasoning about the mix of those 2 being black then be correct?

So, what you mean is that the paint absorbs all wavelengths except a single wavelength. In that case, yeah, the mix should be black.

Just like when 2 gases, each emitting a different single wavelength, would look black if they get mixed together?

In this case, you should be using additive color, not subtractive.

 

[JohnnyGui]

So, what you mean is that the paint absorbs all wavelengths except a single wavelength. In that case, yeah, the mix should be black.

Thanks for verifying.

In this case, you should be using additive color, not subtractive.

Is this because the emitted photons at those 2 specific wavelengths (C and Y) do not easily "collide" with the gas molecules and thus, both wavelengths of C and Y come out of the gas mix?

 

[Khashishi]

Is this because the emitted photons at those 2 specific wavelengths (C and Y) do not easily "collide" with the gas molecules and thus, both wavelengths of C and Y come out of the gas mix?

It's additive because the gas is emitting the light.

 

[JohnnyGui]

It's additive because the gas is emitting the light.

But shouldn't the emitted wavelength of one gas be absorbed by the other gas, since the other gas emits a different wavelength? And vice versa?

 

[Khashishi]

Typically, gases will emit and absorb at the same wavelengths, and be transparent to other wavelengths. So a gas that emits cyan light would also absorb cyan light.

The paint pigments are subtractive because they are not the source of the light. Rather, they filter the light that came from some other source. So they "subtract" light from the source.

 

[JohnnyGui]

Typically, gases will emit and absorb at the same wavelengths, and be transparent to other wavelengths. So a gas that emits cyan light would also absorb cyan light.

The paint pigments are subtractive because they are not the source of the light. Rather, they filter the light that came from some other source. So they "subtract" light from the source.

Ah, that explains it. But isn't your description about gases called fluorescence and/or phosphorescence though? Can't gases merely have a particular color, just like pigments, instead of being fluorescent/phosphorescent?

 

[Khashishi]

Yes

 

[JohnnyGui]

Yes

In that case, when gases have pigments sending out specific wavelengths, shouldn't they behave just like paint pigments then and become black when mixed together?

 

[Khashishi]

"sending out" wavelengths? Perhaps you need to rephrase that in terms of emission, reflection, transmission, and scattering.

 

[JohnnyGui]

"sending out" wavelengths? Perhaps you need to rephrase that in terms of emission, reflection, transmission, and scattering.

I meant as having a color with specific wavelength as pigments just like paint pigments do. Since paint mixes become black, and you confirmed that gases can have pigments as well instead of being fluorescent, doesn't that mean that mixed gases should become black as well when they have pigments?

 

[Khashishi]

Yes, but only for ideal pigments.

 

[sophiecentaur]

This would clear up my confusion, but according to @Merlin3189 these are pure colors.

There is no such thing as a "pure colour". Any colour on the CIE chart can be synthesised in a whole range of combinations of wavelengths. It is only the spectral colours that are monochromatic. In real life, we hardly ever see 'spectral colours' except from (only) some fluorescent tubes and from lasers. The principle of additive colour synthesis is that any colour on the CIE chart can be produced with weighted combinations of three primaries as long as the target colour lies within the triangle with the primaries at the corners. So a particular coloured patch on two tv screens could be produced with different primaries and they would be indistinguishable. As display technology has advanced, primaries have been developed which are 1. Brighter and 2. encompass a bigger triangle. The bigger the triangle, the more colours that can be reproduced. Note that the primaries will not be spectral colours because, up till now we have ever produced as much light out of a spectral source and the screen would not be bright enough,
There is a similar process with adding proportions of 'negative' colour pigments to produce colours but the range and brightness of colours that can be achieved is much less.

Suppose there exists paint that only reflects a single wavelength of Cyan

Such a paint would be absorbing more or less all the light falling on it. It would look very dark cyan. To be any use as a pigment, you need a broad range of reflected wavelengths so that the paint or filter actually looks bright.
Sometimes a filter is needed that passes only a very narrow bandwidth and that cannot be made with pigments. An Interference Filter is needed for that job and it consists of very thin layers of dielectric material which work like oil films on water. Birds' feathers and insect wings do not use pigments either but use interference for producing their brilliant colours.

 

[JohnnyGui]

Yes, but only for ideal pigments.

Thanks a lot for your answers.

There is no such thing as a "pure colour". Any colour on the CIE chart can be synthesised in a whole range of combinations of wavelengths. It is only the spectral colours that are monochromatic. In real life, we hardly ever see 'spectral colours' except from (only) some fluorescent tubes and from lasers. The principle of additive colour synthesis is that any colour on the CIE chart can be produced with weighted combinations of three primaries as long as the target colour lies within the triangle with the primaries at the corners. So a particular coloured patch on two tv screens could be produced with different primaries and they would be indistinguishable. As display technology has advanced, primaries have been developed which are 1. Brighter and 2. encompass a bigger triangle. The bigger the triangle, the more colours that can be reproduced. Note that the primaries will not be spectral colours because, up till now we have ever produced as much light out of a spectral source and the screen would not be bright enough,
There is a similar process with adding proportions of 'negative' colour pigments to produce colours but the range and brightness of colours that can be achieved is much less.

Apologies for the confusion. With "pure color" I meant the monochromatic spectral colors. Your explanation about the diagram makes sense to me. Thanks for that.

I was asking all these questions about mixing spectral monochromatic colors to see first if my undestanding regarding spectral color mixing is correct or not. From that I can now see where my confusion was regarding the substractive color mixing of paint. The whole reason that substractive color mixing gives other colors rather than black as I erroneously expected in my OP, is because in substractive color mixing, the colors that are being mixed are considered as a combination of other colors, rather than spectral monochromatic wavelenghts.

Is this correct?

 

[sophiecentaur]

From that I can now see where my confusion was regarding the substractive color mixing of paint. The whole reason that substractive color mixing gives other colors rather than black as I erroneously expected in my OP, is because in substractive color mixing, the colors that are being mixed are considered as a combination of other colors, rather than spectral monochromatic wavelenghts.

The 'colour' of a pigment is a confusing term because it has no colour until it is actually illuminated. That sounds obvious but I think you originally did not actually take it on board. Think of a pigment as a Filter, which passes some wavelengths and stops others. If the filter pass bands of two pigments happen to share a common range of wavelengths then those wavelengths will pass through and be seen as another colour. Many filter profiles can produce the same visual colour and a wider passband will be brighter but can still produce the same chrominance value. I would expect the 'best' pigments for mixing will have wide passbands but this could affect the 'purity' or accuracy. I try not to worry too much about those things, though. Artists who actually mix paints live in a different world from simple Colour TV Systems designers.

 

[JohnnyGui]

The 'colour' of a pigment is a confusing term because it has no colour until it is actually illuminated. That sounds obvious but I think you originally did not actually take it on board. Think of a pigment as a Filter, which passes some wavelengths and stops others. If the filter pass bands of two pigments happen to share a common range of wavelengths then those wavelengths will pass through and be seen as another colour. Many filter profiles can produce the same visual colour and a wider passband will be brighter but can still produce the same chrominance value. I would expect the 'best' pigments for mixing will have wide passbands but this could affect the 'purity' or accuracy. I try not to worry too much about those things, though. Artists who actually mix paints live in a different world from simple Colour TV Systems designers.

Yes, I think I understood this. The filter is a good analogy. If 2 filters pass another color than they do seperately, that means each filter must pass two different wavelengths (one wavelength being common between the 2 filters) in the first place, right?

From your explanation, it seems that in order to be able to create more colors (in other words, enlarging the RBG triangle in the diagram towards the edges), it is at a cost of accuracy regarding the saturation of the colors. Is this correct?

 

[sophiecentaur]

each filter must pass two different wavelengths

Not two different wavelengths but two different ranges of wavelengths. We are not dealing with single wavelengths here.

From your explanation, it seems that in order to be able to create more colors (in other words, enlarging the RBG triangle in the diagram towards the edges), it is at a cost of accuracy regarding the saturation of the colors. Is this correct?

This doesn't make sense. I cannot imagine how you got that idea? I think it's time that you read something about this colorimetry business. It is not straightforward and doesn't lend itself to arm waving. Try this link. It contains a lot of what you need to know but there isn't a lot of actual chat. There are a number of videos available - look at this one and YouTube will lead you to others.

 

[JohnnyGui]

This doesn't make sense. I cannot imagine how you got that idea?

I got that idea from this quote:

I would expect the 'best' pigments for mixing will have wide passbands but this could affect the 'purity' or accuracy

I understood from this that mixing wide passbands would give a larger variety of pigments, but since wide passbands could affect the purity or accuracy, this is the cost. Not sure what else you meant with this.

I'll check the links.

 

[sophiecentaur]

(in other words, enlarging the RBG triangle in the diagram towards the edges),

What "RGB triangle" would be involved with pigments? The vast difference between additive and subtractive mixing has been stressed several times in the thread. You need to take that on board from the start. The source of inaccuracy I referred to was due to the shape of the 'overlap' of two (or three) secondary colour pigments.

I'll check the links.

Good idea. You can't risk extrapolating on what you know already if you want to understand this better. It's not an easy topic and you clearly need to do some independent study if you want to be able to ask 'the right questions'.

 

[Vanadium 50]

There are spectral (monochromatic) colours that will be interpreted as Cyan or Yellow (the other two subtractive primaries). Magenta is the one for which there is no monochromatic equivalent.

That is correct. The point I am trying to convey is that the OP seems to think there is a 1:1 correspondence between perceived color and wavelength. The sooner he drops that idea, the sooner he will understand. It's manifestly true for magenta, but it's even true for yellow. I could present two different spectra that will both be perceived as "yellow".

 

[sophiecentaur]

That is correct. The point I am trying to convey is that the OP seems to think there is a 1:1 correspondence between perceived color and wavelength. The sooner he drops that idea, the sooner he will understand. It's manifestly true for magenta, but it's even true for yellow. I could present two different spectra that will both be perceived as "yellow".

I have to like++ this. The crazy interchangeable use of the words Colour and Wavelength is so deeply ingrained in the language that it's embarrassing. Even people with otherwise good technical knowledge seem guilty of this heinous crime. There are so many web pages which come from the Creative / Painting side things that maintain that confusion. Their descriptions and 'explanations' are sooo sloppy that any newcomer should really steer clear until they have done the quantitative approach associated with Colour TV.

 

[JohnnyGui]

That is correct. The point I am trying to convey is that the OP seems to think there is a 1:1 correspondence between perceived color and wavelength. The sooner he drops that idea, the sooner he will understand. It's manifestly true for magenta, but it's even true for yellow. I could present two different spectra that will both be perceived as "yellow".

Is it therefore possible to say that one spectral wavelength (range) can evoke the exact same tristumulus value as a combination of 2 or more spectral wavelengths (ranges)?

 

[Drakkith]

Is it therefore possible to say that one spectral wavelength (range) can evoke the exact same tristumulus value as a combination of 2 or more spectral wavelengths (ranges)?

Yes. Some newer light bulbs, either of the LED or the CFL type (I can't remember which), emit light in about a dozen different discrete bands, yet they appear white, just like a regular incandescent light bulb (or the Sun) that emits a continuous spectrum.

 

[Vanadium 50]

Is it possible to say? Sure. Is it correct? No. For it to be correct it would have to be always true, and it is only sometimes true. Sometimes you can't get a single wavelength that gives the same response as two.

 

[Drakkith]

For it to be correct it would have to be always true

I can't say I agree with this. Why would it always have to be true to be a correct statement?

 

[Vanadium 50]

Consider the statement: Triangles have three equal sides. It's not always true, so it is not a true statement. Agree?

 

[Drakkith]

Consider the statement: Triangles have three equal sides. It's not always true, so it is not a true statement. Agree?

To make a fair comparison with the OP's previous statement, you'd need to modify it to say: Triangles can have three equal sides. There's a reason that the OP used "can evoke" and not "will evoke" or just "evokes".

 

[JohnnyGui]

For it to be correct it would have to be always true, and it is only sometimes true.

I disagree with this, for it depends on the statement itself. First, my statement (in a question form) says "is it possible", secondly it asks if " a wavelength can evoke the same tristumulus". Yes, it is possible and yes it can, even if not necessarily. The statement doesn't say or imply that it has to. Therefore, it is correct.

 

[sophiecentaur]

Is it therefore possible to say that one spectral wavelength (range) can evoke the exact same tristumulus value as a combination of 2 or more spectral wavelengths (ranges)?

It's the other way round really; you synthesise a single colour by combining two or more other colours. If you have looked at 'the links' they tell you that you can get a match (totally subjective, of course) for any colour on a line between two colours That can be extended to producing any colour within a triangle of three colours. But you can't get a match for colours outside 'the triangle'. The spectral colours lie along a curve on the CIE chart and they can never lie inside a triangle; there will always be an error.
The limited 'gamut' of colours that any TV system can produce (particularly a three colour system) is very well demonstrated if you look at scenes of sporting events. People tend to wear bright coloured clothing. Those colours lie near the outside of the CIE chart.They are 'saturated' colours. An RGB system runs out of steam and places those all those colours along the sides of the triangle (best it can do). You will notice that there appear to be many jackets that appear to have the same colour because the system has brought colours that are on a radial to the same region on one of the lines. In a live scene, your vision would allow you to see all those colours as different. 'Good' TV productions will choose a palette of colours in a scene which don't test the system too hard.

 

[JohnnyGui]

It's the other way round really; you synthesise a single colour by combining two or more other colours. If you have looked at 'the links' they tell you that you can get a match (totally subjective, of course) for any colour on a line between two colours

Yes, I have seen the links, the video was very helpful. But my question wasn’t referring to the diagram. I was asking about an aspect regarding the statement about color and wavelengths not being 1:1, to see if I understand the meaning of color from the perspective of wavelengths in that particular aspect.

Your further detailed explanation helped me along with the links, thanks. I've got 2 questions since we're talking about the CIE chromaticity diagram.
If I understand correctly, the midpoint of a drawn straight line between two colors in the diagram would be the color that you'd get (like you said) if the 2 original colors were mixed in equal "amounts". From that, I'd deduce that if the 2 colors that are to be mixed are spectral colors (2 different wavelengths), then one should draw a line from one wavelength to the other wavelength (at the edges of the diagram) and look at the midpoint of that line.

1. If there is a spectral color that can have the same tristumulus value as a combination of 2 other spectral colors, then I'd expect that the drawn line between those 2 mixed spectral colors would be overlapping the edge of the diagram. I can see that this case is mainly possible for wavelengths on the right edge of the diagram since it's straight. Is this correct?

2. Is it possible to see the end result of 3 colors that are mixed using the diagram? For example, the video says at 2:35 that the perception of a spectral color of 580nm (Yellow) is the same for the human eye as mixing spectral colors of wavelengths of 700nm, 546.1nm and 435.8nm. Is it possible to draw that on the diagram? Looking at the shape of the diagram, I suspect each of the 3 wavelengths has to be in different "amounts" in order to land near the 580nm tristumulus value but I'm not sure how to draw this and which intersection of lines to use.

 

[Drakkith]

If I understand correctly, the midpoint of a drawn straight line between two colors in the diagram would be the color that you'd get (like you said) if the 2 original colors were mixed in equal "amounts".

Doesn't look like it. Per wiki: An equal mixture of two equally bright colors will not generally lie on the midpoint of that line segment.

From that, I'd deduce that if the 2 colors that are to be mixed are spectral colors (2 different wavelengths), then one should draw a line from one wavelength to the other wavelength (at the edges of the diagram) and look at the midpoint of that line.

From the same wiki page: If one chooses any two points of color on the chromaticity diagram, then all the colors that lie in a straight line between the two points can be formed by mixing these two colors.

This makes sense, as you can add varying intensities of each wavelength to give you different perceived colors.

1. If there is a spectral color that can have the same tristumulus value as a combination of 2 other spectral colors, then I'd expect that the drawn line between those 2 mixed spectral colors would be overlapping the edge of the diagram. I can see that this case is mainly possible for wavelengths on the right edge of the diagram since it's straight. Is this correct?

It looks like you can get very close to being able to do so on the right side of the diagram. It's still convex, so you can't get an exact spectral color, but you can get so close that it mostly doesn't matter.

2. Is it possible to see the end result of 3 colors that are mixed using the diagram?

Yes. Draw a triangle with each color as a vertex. Every color within that triangle can be formed from the combination of the colors at the vertices.

For example, the video says at 2:35 that the perception of a spectral color of 580nm (Yellow) is the same for the human eye as mixing spectral colors of wavelengths of 700nm, 546.1nm and 435.8nm. Is it possible to draw that on the diagram?

Yes, just draw a triangle using those three wavelengths and you'll see that one side of the triangle runs extremely close to "spectral" yellow. Close enough that most people wouldn't notice a difference. And if they can't notice a difference, then it's the same color to them, isn't it?

Looking at the shape of the diagram, I suspect each of the 3 wavelengths has to be in different "amounts" in order to land near the 580nm tristumulus value but I'm not sure how to draw this and which intersection of lines to use.

You don't draw anything. Drawing lines or triangles just gives your a range of possible colors that can be made from the chosen starting colors. If you've chosen a color, there's nothing to draw.

However, after having chosen a color you want and the corresponding colors you want to mix, you could figure out how much of each to mix by delving into the math behind the chart. I won't even try to dive into that myself.

 

[JohnnyGui]

Thanks a lot for the answers @Drakkith

Doesn't look like it. Per wiki: An equal mixture of two equally bright colors will not generally lie on the midpoint of that line segment.

Ah, the video says it should be the midline and I leaned towards that.

Yes. Draw a triangle with each color as a vertex. Every color within that triangle can be formed from the combination of the colors at the vertices.

Yes, I indeed understood this. I was wondering though if someone gives you 3 colors with specific amounts of each, if you could deduce from the diagram which color you'd end up with. But I see you answered that with the following:

However, after having chosen a color you want and the corresponding colors you want to mix, you could figure out how much of each to mix by delving into the math behind the chart. I won't even try to dive into that myself.

So this was secretly what I actually wanted to know . I'll see if I can understand the maths behind it.

 

[sophiecentaur]

I'll see if I can understand the maths behind it.

Luckily, the Maths is not too hard because the result mixing colours can be done (accurately enough) with linear formulae. But the measurement of those eye sensitivity curves was pretty clever and they have been good enough for many decades. But you need to realize that all of this is very much based on measurement and fitting results to suit many observers. Also, the CIE chart is for Chrominance only and the Luminance also affects what we 'see' colours as. Take Sodium Yellow and display it at a low luminance in amongst other, brighter colours and you will call what you see, Brown. The pigment in the skins of people whose races developed near the equator has very similar Chrominance to that of Nordic races - the dark skin is obtained by an almost neutral added pigment and all skins are pretty Pink if the neutral pigment is ignored. It's good to play with simple Photo Processing software, using the 'eye dropper' colour sampling tool and see the ratios of R:G:B values for apparently different colours. (This is just too add a bit more confusion.
I don't know why you still seem to attach so much importance to the results of mixing spectral colours because it is not really very relevant in practice. I guess it reflects what you (and the rest of us) were told about producing Sodium Yellow with a red and a green light. That would require two primaries that really wouldn't be much use except for producing Sodium Yellow. But that's the old 'colour = wavelength' thing which will never just lay down and die.
The RGB based primaries are chosen to get a good working system for as many colours as possible with just three primaries. Nothing you see on your computer screen can be relied on to tell you anything in detail about colours. For a start, most things that you see on a display are not actual light sources and the spectrum of the illumination will affect what things look like (of course) and both the camera and the display should both be adjusted to take this into account. Your camera 'auto' setting for colour balance is only doing its best, for instance.

 

[JohnnyGui]

Also, the CIE chart is for Chrominance only and the Luminance also affects what we 'see' colours as.

Yes, that's what I read indeed. From what I understand, the CIE chart shows the different combinations of saturation and hue (together called chroma) but not the different luminances of each chroma. Luminance is determined by 3 components if I'm right, the sensitivity of the eye, the amount of white or dark that we mix in a chroma ourselves (= lightness) and the amount of power that a fixed chroma contains (source). I'm trying to put the difference of these properties into words but I need verification on this:
1. Does this mean that there is a specific fixed "lightness" chosen for all of the chroma's in the CIE chart?
2. Can the saturation be considered as the density of photons of a particular spectral wavelength in a hue?
3. Is then the total amount of "photons" (per unit time on a surface) of that wavelength the luminance dependence of power?

I don't know why you still seem to attach so much importance to the results of mixing spectral colours because it is not really very relevant in practice.

I wanted to merely understand the concept and workings of mixing of spectral colours, regardless of its relevance in practice. It did help me also to deduce what my confusion was in my OP since I was apparently thinking in terms of subtractive spectral colors mixing instead of subtractive pigment mixing.

The pigment in the skins of people whose races developed near the equator has very similar Chrominance to that of Nordic races - the dark skin is obtained by an almost neutral added pigment and all skins are pretty Pink if the neutral pigment is ignored.

I like these interesting things that you share. You explained before how feathers and wings of insects give color through interference. For a while before, I've been having a hard time understanding the exact mechanism of this. Because I understand interference as the destruction and construction of wavelengths together but I can't see how this produces a different color.
Does this mean that the cones translate the degree of construction of destruction of different wavelengths together into a perception of certain colors, in addition to translating a combination of different wavelengths (ranges) without any interference also into certain colors (pigments)?

 

[Drakkith]

1. Does this mean that there is a specific fixed "lightness" chosen for all of the chroma's in the CIE chart?

I think so. The $Y$ variable is defined as the luminance, so once you choose a $Y$, the other two ($X$ and $Z$) can be varied to create a the plot. Per wiki again: The CIE model capitalises on this fact by defining Y as 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. The XYZ tristimulus values are thus analogous to, but different from, the LMS cone responses of the human eye. Defining Y as luminance has the useful result that for any given Y value, the XZ plane will contain all possible chromaticities at that luminance.

2. Can the saturation be considered as the density of photons of a particular spectral wavelength in a hue?

Not directly, no. Saturation just defines the "intensity" of the color. The less saturation, the more the color approaches white. You'd need to have both the saturation and the luminance. But given a set luminance value, you might be able to work backwards to arrive back at the spectral radiance functions and from there find the photon density.

I like these interesting things that you share. You explained before how feathers and wings of insects give color through interference. For a while before, I've been having a hard time understanding the exact mechanism of this. Because I understand interference as the destruction and construction of wavelengths together but I can't see how this produces a different color. Does this mean that the cones translate the degree of construction of destruction of wavelengths into a perception of certain colors, in addition to translating a combination of wavelengths without any interference also into certain colors (pigments)?

Not really. Destructive interference creates regions on the retina where little-to-no light of that wavelength falls. So the cones don't have anything to do with it, they just aren't receiving any light of that wavelength. The light that would have fallen into those regions is instead sent to regions created by constructive interference. Since the exact position of the regions depends heavily on wavelength, you get regions where all the blue is missing, but that has all the green, making you see bright green.

 

[sophiecentaur]

1. Does this mean that there is a specific fixed "lightness" chosen for all of the chroma's in the CIE chart?

The two dimensional graph of the CIE chart is the result of eliminating the L component from the XYZ tristimulus. You are getting it back to front again

2. Can the saturation be considered as the density of photons of a particular spectral wavelength in a hue?

Not at all. Saturation is best described as how far from the white point that a colour is. High saturated colours will be the result of mixing just two primaries. Adding the third primary will reduce the saturation.

3. Is then the total amount of "photons" (per unit time on a surface) of that wavelength the luminance dependence of power?

Try reading this link. Its message is that CIE colorimetry is only a small part of human colour perception. You are onto a hiding to nothing if you try to sum up this whole thing with what you have gleaned from what you have read so far. (Also, bringing photons into this will not help you one iota. They do not represent any additional knowledge of macroscopic matters.)

Because I understand interference as the destruction and construction of wavelengths together but I can't see how this produces a different color.

I suggest you look up "Optical Interference Filters" and find yourself a link that makes sense to you. The simple effect of a single thin layer of oil on water produces different colours, depending on the angle that it's viewed. It is essentially a 'subtractive' effect because the maximum of attenuation will be when there is a half wavelength difference in path length for the rays reflected at the two layers. A 'notch' in the spectrum is produced and that is what causes those 'unreal' colours that characterise oil on roads and many bird feathers. You just do not get that with pigments because the maximum light levels are much higher than when you use overlapping band stop filters (pigment mixing).

I wanted to merely understand the concept and workings of mixing of spectral colours,

Every colour we see is the result of a mixture of spectral wavelengths - of course. A fair to good match may be obtained by using a limited number of 'spot' wavelengths but that is just because a colour is the result of a very limited human sense.

 

[JohnnyGui]

The two dimensional graph of the CIE chart is the result of eliminating the L component from the XYZ tristimulus. You are getting it back to front again

How does getting it back to front make my statement incorrect though? Since color is actually defined by L, along with hue and saturation, and the CIE chart only shows the chromaticity, then all these chromas on the CIE chart must be at certain L values, like the Wiki states.

Not at all. Saturation is best described as how far from the white point that a colour is. High saturated colours will be the result of mixing just two primaries. Adding the third primary will reduce the saturation.

But in addition to that, the Wiki says that saturation is the combination of "light intensity and how much it is distributed across the spectrum of different wavelengths". Aren't intensity and distribution of different wavelengths physical properties? Which for example can be expressed as the amount and density of photons at certain wavelenghts? If a spectral wavelength is a 100% saturation, and consists of only photons of that specific wavelength (band), then the density of those photons at that wavelength w.r.t. photons at other wavelengths, is 100%. Pull it towards the white and the saturation, and consequently the density of photons at that wavelength w.r.t. other wavelengths, will both decrease.

(Also, bringing photons into this will not help you one iota. They do not represent any additional knowledge of macroscopic matters.)

I didn't care about the macroscopic matters with that question. It interests me to ask if there are links between different fields of physics, just out of curiosity.

Every colour we see is the result of a mixture of spectral wavelengths - of course. A fair to good match may be obtained by using a limited number of 'spot' wavelengths but that is just because a colour is the result of a very limited human sense.

Yes, I understood that through the questions that I asked about mixing spectral colors. Perhaps there is some confusion when I say spectral color, which is (according to Wiki) a color at a single wavelength or at a relatively narrow band of wavelengths. Asking questions about subtractive mixing of those spectral colors, and people answering that this would indeed give me black as I expected, helped me to distinguish that the subject I was asking about in my OP must be about subtractive mixing of colors that each consists of a mixture of spectral wavelengths since they give other colors.

I suggest you look up "Optical Interference Filters" and find yourself a link that makes sense to you. The simple effect of a single thin layer of oil on water produces different colours, depending on the angle that it's viewed. It is essentially a 'subtractive' effect because the maximum of attenuation will be when there is a half wavelength difference in path length for the rays reflected at the two layers. A 'notch' in the spectrum is produced and that is what causes those 'unreal' colours that characterise oil on roads and many bird feathers. You just do not get that with pigments because the maximum light levels are much higher than when you use overlapping band stop filters (pigment mixing).

Just looked it up and it makes sense now to me. However, I noticed from several sources (like here and here) that when they explain interference in a soap bubble, they talk about a single wavelength (band) that reflects from both the outer and inner layers of the bubble, resulting in either destruction or construction of that wavelength. Isn't it possible at all for 2 different wavelengths to destruct or construct each other, or will this only lead to additive mixing?The link you gave was a great read, thanks!

 

[sophiecentaur]

How does getting it back to front make my statement incorrect though?

I said "wrong way round" because condensing three variables into two variables will introduce uncertainty if you try to get back from just the two - so you can only go one way.
You have picked up on the fact that Chrominance and perceived colour are two different things. That CIE diagram would actually have an additional vertical axis (out of the page) if it is to include all identifiable colours (including luminance). We are just lucky that our vision system actually allows the use of the (condensed) two dimensional representation for chrominance. Touching on the subtractive mixing method, the way the vertical (luminance) axis works is that the more saturated the colour, the lower the maximum luminance that can be presented. This is sort of obvious when you realize that a highly saturated colour will have dense filtering and none of one of the secondary pigments there, to take light away. So you have a 'prism' shape for additive mixing and a'humped' shape for subtractive mixing. (Sorry for introducing yet more complication. ) Edit: It is the reason that interference filters produce unexpected colours because they do not rely on the overlap of lossy filters and they produce high luminance, high saturation colours.

they talk about a single wavelength (band) that reflects from both the outer and inner layers of the bubble,

I didn't bother going through both of those movies but you must have misunderstood. All wavelengths are reflected from both boundaries. Depending on the path difference, the rays will with enhance, cancel of something in between, giving different perceived colours as the angle changes.

Wiki says that saturation is the combination of "light intensity and how much it is distributed across the spectrum of different wavelengths". Aren't intensity and distribution of different wavelengths physical properties?

This is very muddled and the does not tally with the definition of Saturation in the business of colour description.
I think you are too ready to reach your own conclusions about this topic before reading around enough. I realize that you want to keep this thread going but there is a list to how big a list of questions you can expect answers for. If you can get identical messages from a number of references then you can probably 'believe' them without needing to check with PF for each item. It will give you a much better flavour - as long as you can identify good sources. Avoid 'arty' sites which you cannot rely on to keep to an objective and quantitative approach. (Nothing wrong with them, per se, but they are aimed at a different audience)