Question about subtractive color mixing

[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)

 

[Drakkith]

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?

No, it turns out that different wavelengths don't interfere with each other. Or, if you prefer, it turns out that the EM wave, which is composed of a mix of different wavelengths, interferes with itself in such a way as to behave as if each individual wavelength interferes with itself and no others. As far as I know at least. I know my thin-film hydrogen-alpha filter doesn't behave differently if I'm looking at different sources of light.

 

[JohnnyGui]

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.

Thanks. I indeed deduced that there is an extra luminance axis. Trying to understand what you said regarding the subtractive mixing of more saturated colors; does this mean that when you try to get the same perception of a saturated color but through subtractive mixing, you'd get a less luminant version of that saturated color than the original saturated color? Please correct me if I'm wrong.

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)

Yeah, I initially tried searching for more sources to see if saturation can also be explained by physical properties but couldn't find any, so I thought perhaps someone could verify my statement here. After searching a bit more, it seems that you are correct though. Although, I can't really see how saturation can not be considered as the percentage of photons with a certain wavelength w.r.t. other wavelengths in a color. If saturation is truly a perceptive measure and non-physical, shouldn't there exist a 100% saturation of Magenta?
EDIT: Just found a presentation which shows physical explanations regarding hue, saturation and brightness in the context of photons here, starting from slide 21. Not sure how reliable this is though.

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.

Sorry, I was wording my question poorly. I do get that all wavelengths are reflected from both boundaries, but the videos are saying that construction and destruction only occur between the exact same wavelength from both boundaries. A different wavelength can not construct or destruct another wavelength; it gives a different color instead. I just noticed now that @Drakkith somehow understood my question.

No, it turns out that different wavelengths don't interfere with each other. Or, if you prefer, it turns out that the EM wave, which is composed of a mix of different wavelengths, interferes with itself in such a way as to behave as if each individual wavelength interferes with itself and no others. As far as I know at least. I know my thin-film hydrogen-alpha filter doesn't behave differently if I'm looking at different sources of light.

This actually amazes me in a way. It urges me to try and understand why it is like that. One could ask if there's a different color perception (or even photon count) when a wavelength should destruct another wavelength than when it should construct it.

 

[sophiecentaur]

No, it turns out that different wavelengths don't interfere with each other. Or, if you prefer, it turns out that the EM wave, which is composed of a mix of different wavelengths, interferes with itself in such a way as to behave as if each individual wavelength interferes with itself and no others. As far as I know at least. I know my thin-film hydrogen-alpha filter doesn't behave differently if I'm looking at different sources of light.

They add in just the same way (superposition happens) but there is no recognisable pattern.
The result of two waves of different frequencies can, in fact be looked upon as a 'moving' interference pattern. Think of the fringes from the two slits experiment. Instead of two slits, consider two monochromatic (very coherent) sources with on source 1Hz different in frequency from the other. What you would see is a set of fringes that would be moving sideways by one fringe width per half second. The 'null condition' of λ/2 phase difference would be in different positions as time progresses because the phase difference between the two signals would be changing by 360° every second. The received level at any chosen position would be changing at 2HZ as the fringes move past that spot. But, of course, if the frequency difference were much bigger than 1Hz, the pattern would be moving much to fast to observe.
Your thin film H alpha filter uses a large number of internal reflections between several (I think) layers and the result is that it will only pass a very limited range of wavelengths; for all others, the multiple reflections cancel out so all that gets through is the very narrow band of green light. To get an even narrower bandwidth, an etalon is used which uses a wide air gap between two films. Wide gap means narrow wavelength response - which is what we always find with diffraction.

 

[sophiecentaur]

does this mean that when you try to get the same perception of a saturated color but through subtractive mixing, you'd get a less luminant version of that saturated color than the original saturated color?

Yes. Thicker and thicker filters are needed in order to take more and more unwanted wavelengths away (the band pass filter that the pigment contains is not monochrome).
I and others already made the point that subtractive colour mixing is a really bad subject to try to nail down - especially if you haven't completely sorted out additive mixing.
.

Although, I can't really see how saturation can not be considered as the percentage of photons with a certain wavelength

By that argument, you could only have saturation from a monochromatic source. There are NO photons corresponding to a nearly saturated yellow which has been formed by monochromatic red and monochromatic green primaries. You must stop confusing wavelength with colour or chroma. It only lands you up a blind ally like that one! The only time that photons are of any consequence in colour imaging is in the detailed workings of colour sensor itself (Retina or photo cell). Once the sensor has produced its output, it's just a number (or nerve equivalent). I have a feeling that you are just no willing to be 'wrong about this particular photon thing and that you are trying to bend the argument to fit your mental picture.

 

[JohnnyGui]

I have a feeling that you are just no willing to be 'wrong about this particular photon thing and that you are trying to bend the argument to fit your mental picture.

Not at all. Keep in mind that, as someone who is learning something new, especially in a (not so easy) subject in which terms are quite often interchangeably used by some sources and treated as separate in others, as well interpretations differing among them, confusion can arise easily to that new learner. It's not that he is not willing to be wrong, it's that he finds sources that intepret things differently from, for example, this forum (such as the association with photons in the presentation and the meaning of saturation itself that I linked you to). He will try to eliminate that confusion by asking questions and trying to find a way in which those different interpretations are compatible with each other because, after all, a new learner can not distinguish immediately the correct interpretation from the incorrect one from the start.

Here's an example. Notice that the Wiki, along with several other sources (https://www.colorado.edu/physics/phys1230/phys1230_fa01/topic45.html), is describing saturation as "how much it is distributed across the spectrum of different wavelengths" and that the purest (most saturated) color is achieved by using just one wavelength at a high intensity, such as in a laser light..
Beneath that explanation comes a term that I think you are actually describing as "saturation" but Wiki is interpreting it differently, which is the excitation purity. The formula for the excitation purity also makes sense to me because it shows a ratio of the distance of a color from the white point relative to the distance up to its corresponding spectral locus at the edge (which btw, looks different from the formula for saturation on the Wiki page). Excitation purity apparently does not depend on a color being a single wavelength or made out of primaries while saturation apparently does according to sources. This could all be wrong and excitation purity could be very well the same as saturation regarding definition and calculation but this would redirect me to my first paragraph of this post again.

 

[sophiecentaur]

, confusion can arise easily to that new learner.

Yes. I totally agree.

"how much it is distributed across the spectrum of different wavelengths"

That's just too wishy washy to mean anything and is clearly aimed at spectral colours. Our life (pigment based, mainly) is full of non-spectral colours.

the purest (most saturated) color is achieved by using just one wavelength at a high intensity, such as in a laser light..

I have already made the point that this description would not allow for any colours along the bottom straight line portion of the CIE chart (Magentas) to be 'saturated' because they are only perceived when there are two wavelengths involved and are not spectral. A definition that doesn't include that significant range of colours can't really be worth much.
I cannot imagine how the term 'Excitation Purity' can be taken seriously to mean Saturation, the way it seems to be defined in many documents because it fails to include those non-spectral colours in the chart. But I guess it's all in the small print. As this link puts it, "The excitation purity of any color possessing a dominant wavelength is an exactly defined ratio of the distances in the chromaticity diagram". For non-spectral colours, a modified definition is used. Ah well.
When RGB colour signals are (/were) coded for Analogue TV transmission (NTSC and PAL), they were split into a high resolution luminance signal and a low resolution chroma signal , modulated onto a 'subcarrier' with amplitude and phase relative to a reference zero. The Phase corresponded to the radial direction from the white point to the transmitted colour and the saturation corresponded to the length of the radial line. There was a long history behind this but the basic requirement was for a signal that would be compatible with existing black and white sets, which would just pick up the Luminance signal. The subcarrier appeared as a fine crawling patterns which would not be seen at the right viewing distance. I mention this because the 'vector' nature of the chroma signal may be of interest to you.

 

[Rob Lewis]

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.

Exactly. Which is why two different color samples can appear identical to the eye under, say, incandescent light, but look quite different when taken outside in the daylight. The different spectra of the illumination cause the proportions of reflected wavelengths to be different. What we call "color" is entirely subjective.

 

[Fooality]

Question: If two guitar strings are very close in pitch but not quite the same, they "beat", as the two waves go between reinforcement and cancellation. Do light waves do the same? If so - and I know the frequencies are many orders of magnitude higher - how slow could it get? Could we ever make a substance that is a beating color, that pulses in and out under natural light?

 

[sophiecentaur]

Question: If two guitar strings are very close in pitch but not quite the same, they "beat", as the two waves go between reinforcement and cancellation. Do light waves do the same? If so - and I know the frequencies are many orders of magnitude higher - how slow could it get? Could we ever make a substance that is a beating color, that pulses in and out under natural light?

I described a thought experiment, higher up the thread, in which two monochromatic light sources with a very close frequently (1Hz separation) the 'interference pattern' would move and the brightness would pulse at twice the difference frequency. That would not involve colours as we couldn't perceive such a narrow frequency shift for light. The two ideas would not be appropriate for the same thread.

 

[sophiecentaur]

What we call "color" is entirely subjective.

However, it could be described as lucky that relatively simple Maths can produce very acceptable colour reproduction with additive processes.
'Nuff said about high quality subtractive mixing. That is much more of an Art than a Science.

 

[darth boozer]

Hello,

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

View attachment 219068

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?

In the subtractive color system, or CMYK (subtractive), which can be overlaid to produce all colors in paint and color printing, cyan is one of the primary colors, along with magenta, yellow, and black. In the additive color system, or RGB (additive) color model, used to create all the colors on a computer or television display, cyan is made by mixing equal amounts of green and blue light. Cyan is the complement of red; it can be made by the removal of red from white light. Mixing red light and cyan light at the right intensity will make white light.

So your question, "What if Cyan is not a mix of Blue and Green wavelengts but a pure Cyan wavelength itself?", is meaningless because, by definition, cyan is the complement of red.

 

[sophiecentaur]

So your question, "What if Cyan is not a mix of Blue and Green wavelengts but a pure Cyan wavelength itself?", is meaningless because, by definition, cyan is the complement of red.

Cyan is a colour. The way it can be produced is irrelevant. It is 'defined' in one way when it suits the system.

 

[JohnnyGui]

I cannot imagine how the term 'Excitation Purity' can be taken seriously to mean Saturation, the way it seems to be defined in many documents because it fails to include those non-spectral colours in the chart.

There are actually sources, confusingly enough, that apply the same calculation of excitation purity also to non-spectral colors, making the distinction between excitation purity and your (along with other sources') definition of saturation a bit vague. This link and this link say that the exact same formula for excitation purity can be used for the bottom edge of the CIE diagram, even if there is no dominant wavelength. And then there is this pdf (try not to get distracted by the font ) and https://www.opt.uh.edu/onlinecoursematerials/stevenson-5320/ColorVision2.pdf that say that saturation is a perceptual term while excitation purity is the physical term.

When RGB colour signals are (/were) coded for Analogue TV transmission (NTSC and PAL), they were split into a high resolution luminance signal and a low resolution chroma signal , modulated onto a 'subcarrier' with amplitude and phase relative to a reference zero. The Phase corresponded to the radial direction from the white point to the transmitted colour and the saturation corresponded to the length of the radial line. There was a long history behind this but the basic requirement was for a signal that would be compatible with existing black and white sets, which would just pick up the Luminance signal. The subcarrier appeared as a fine crawling patterns which would not be seen at the right viewing distance. I mention this because the 'vector' nature of the chroma signal may be of interest to you.

It did. Thanks for sharing!

There is something I don't quite understand about the construction of the CIE chart, despite watching and reading several links. I understand that for the colors outside the RGB triangle, there are negative values for one or more of the RGB values, which means you have to add that negative value to the color itself that you're trying to match. How then can one achieve the colors outside that RGB triangle by using "imaginary" primaries? Is there a relatively simple way to explain this? What is the exact ability that "imaginary" primaries give you to be able to get these colors that you can't get with RGB initially?

 

[sophiecentaur]

And then there is this pdf (try not to get distracted by the font ) and https://www.opt.uh.edu/onlinecoursematerials/stevenson-5320/ColorVision2.pdf that say that saturation is a perceptual term while excitation purity is the physical term.

I had to fight myself to keep looking at that 'fun' font and to take it seriously. But the link was quite good, really. (enjoyable, even) I had a problem with
"The dominant wavelength may not be the largest component", though. It looked like an attempt to make something work even under inappropriate conditions. It still nags me that wavelength is still considered to be synonymous with colour and we have already established that it's not the whole story. I was thinking of an analogous thing in sound; Colour is more like a chord or a musical phrase, whilst a monochromatic source corresponds to a single, sinusoidal tone. No one (but an Audiographer) would try to describe our hearing experience simply in terms of our ears' frequency response. But, having had all my colorimetric knowledge from work on Colour TV, I have to approve of any quantitative approach the works.
I also have a problem with your contrasting a "perceptual term" with a "physical term" when both are based entirely on a graph that's obtained on a perceptual basis. I do appreciate that the 'Purity' figure is handy as it has a numerical value. I think it's the insistence with using the term 'spectral' in its verbal definition when it could just as easily have come clean and referred to 'boundary of the CIE gamut. Also, of course, it requires that the White Point has been specified (so you can draw a line from it!)
@Khashishi got it about right in his early post (#4?).
There is a history to all this. If you Google "Colorimetry", the majority of hits do not take you to information about TV Imaging and reproduction - they seem to be mainly concerned with chemical analysis and measurement. I imagine that there has been a change in direction of the study of colour and objective measurements done by Chemists etc. will still use the CIE, subjective based approach, because it's already been established. In most cases, I would say that a more complete special analysis of a chemical substance would give 'better' results but it may still be useful for an experimenter to be able to assess results visually.

 

[sophiecentaur]

I understand that for the colors outside the RGB triangle, there are negative values for one or more of the RGB values, which means you have to add that negative value to the color itself

This is just a consequence of the Maths and it doesn't imply that you could actually 'do' it. I guess the nearest thing would be in the context of Subtractive Mixing where you can take more and more of a band of wavelengths away from the white illuminant - with a filter- and end up with a very dim and very saturated colour. But I can't think how that has any relationship with any RGB Phosphor primaries.
Relating secondary colours, produced additively with combinations of just two primaries - i.e. Cyan as White - Red etc and the equivalent subtractive primaries, produced with inks and dyes makes my brain ache. On top of that, they mix those secondary / primary colours and expect to be able to predict what colour you would actually see.

 

[pinball1970]

@Johnny original OP You can learn a lot about mixing paints/dyes rather than comparing RGB data to spectral data subtractive/additive colour and the theory.In terms of dyes it is difficult to manufacture compounds that give a narrow spectral band, you have to think about what is happening once the light source has interacted with the sample.Photons come in, interaction, some loss of energy, excitation and photons emitted just at the wave length you require. The molecules involved are usually quite complex so the chances of all those excitations yielding one specific photon energy when you have single double bonds Nitrogen Hydrogen Oxygen Carbon as well as metals would have to be extremely specific.The maths are just abstract constructs of 3 or 4 dimensional spaces depending on which system you use, these are useful for telling the difference between colours. An objective mathematical assessment usually given as DE (using a spectrophotometer) the close to zero the closer the colours are. DH (tone) DL (depth) DC (brightness) make up the total colour difference.

 

[sophiecentaur]

In terms of dyes it is difficult to manufacture compounds that give a narrow spectral band

What does "narrow spectral band" mean? If you want a Magenta dye, what narrow spectrum would it have? Wouldn't it need to pass blues and reds and exclude Greens?

excitations yielding one specific photon energy

What substances, other than gases, have single line absorption characteristics? Doesn't condensed matter have band energy structures? Also, we have already discussed the futility of having narrow band pigments because they would exclude most of the light and, two together would always result in Black. This is very confusing.

 

[JohnnyGui]

This is just a consequence of the Maths and it doesn't imply that you could actually 'do' it. I guess the nearest thing would be in the context of Subtractive Mixing where you can take more and more of a band of wavelengths away from the white illuminant - with a filter- and end up with a very dim and very saturated colour. But I can't think how that has any relationship with any RGB Phosphor primaries.
Relating secondary colours, produced additively with combinations of just two primaries - i.e. Cyan as White - Red etc and the equivalent subtractive primaries, produced with inks and dyes makes my brain ache. On top of that, they mix those secondary / primary colours and expect to be able to predict what colour you would actually see.

I might have missed your point. Let me try asking the question in an example form:

For example, the used RGB primaries can not give a particular more saturated version of Cyan. The video you linked to says that the RGB primaries used are at respectively 700nm, 546.1nm and 435.8nm and that for that particular Cyan, a certain negative value is needed for 700nm. Does that mean that they just assigned the negative amount value of 700nm for Cyan to a coordinate outside the RGB triangle on the CIE chart, and if e.g. another color needs twice the negative amount of 700nm than for Cyan (but the same values for the B and G primaries), then the coordinate outside the RGB triangle is twice as far, but in the same direction as for Cyan?

 

[sophiecentaur]

for that particular Cyan, a certain negative value is needed for

Are you confusing the contexts here? Could you be trying to apply the ideas that were used to deal with those negative response values to the visual analysis curves to synthesis.There is no reason why the Cyan primary that's used in subtractive mixing has to lie inside the triangle of three phosphor primaries. You can produce a dye that will lie outside the triangle of additive mixing when illuminated with white light. I imagine someone could invent an additive system that would involve four primaries, with a near-Cyan primary on the line joining Red and the chosen white point and that could match many more colours - way out towards the spectral curve around the cyans. I haven't come across one that's been used in an available TV system. Such a system would be driven using calculations using your 'negative values' that would emerge, no doubt, from the system. But would it be worth doing it? Imaging is a very practical subject and it is probably true that the apparently large area of unattainable colours up above the BG line is not a naturally occurring colour. I referred before to jazzy colours used on clothes and we can see bright cyans on many out door clothes. But they are to unnatural for people to judge the shortcomings of their accurate reproduction. It's perhaps just not worth bothering with.

twice as far, but in the same direction as for Cyan?

Wouldn't that take you way outside the curve of the spectral colours? How can the line be any longer than from white to cyan?
I am impressed with all the reading around you have been doing. Not many people are prepared to do that.

 

[pinball1970]

sophiecentaur said: ↑ What does "narrow spectral band" mean? If you want a Magenta dye, what narrow spectrum would it have? Wouldn't it need to pass blues and reds and exclude Greens?That was a response to an earlier post about pure colours 700nm for red etc not being possible.You would be surprised what spectral curves some dyes have, they may look like perfect blues or reds but they may have peaks in there you do not expect.Perhaps I did not get my point across well enough, If you want to find out about colour then it is a good idea to investigate dyes and combinations of on a substrate, then take readings on the spectro, this includes Blacks (which are mixtures) or white bases usually involving optical brightening agents.sophiecentaur said: ↑ could invent an additive system that would involve four primaries, with a near-Cyan primary on the line joining Red and the chosen white point and that could match many more colours

Dyers usually use three dyes and cover about 85% of required colours (trichromats) Chroma is the tricky component to predict adding blue to a combination does not always make the colour appear bluer, it depends on your starting point.