# Question about subtractive color mixing

• I
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?

#### Attachments

• 1200px-SubtractiveColor.svg.png
9.8 KB · Views: 1,210

Gold Member
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.

Homework Helper
Gold Member
... 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.

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.

Zachary Sheldon, Andy Resnick and Merlin3189
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?

Homework Helper
Gold Member
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?

Gold Member
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
Homework Helper
Gold Member
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
JohnnyGui
Explanation

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

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?

Staff Emeritus
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.

Gold Member
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.

Merlin3189
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?

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?

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?

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?

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?

"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?

Last edited:
Yes, but only for ideal pigments.

Gold Member
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.

Merlin3189
JohnnyGui
Yes, but only for ideal pigments.

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?

Gold Member
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?

Gold Member
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.

Last edited:
Gold Member
(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.
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'.

Staff Emeritus
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 and Bystander
Gold Member
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)?

Staff Emeritus