Question about subtractive color mixing

[sophiecentaur]

@pinball My comment about adding a fourth colour was about additive mixing. The results of subtractive mixing can be anyone’s guess. It has to be a matter of suck it and see, rather than mathematical prediction. The precise detail of the secondary filter charactistic has to be known; it’s ‘colour’ / chroma is not enough to predict how it will mix w
ith another secondary.
I would not be “surprised “ about the transmission characteristics of any specialist dye or pigment. That technology has to be a fudge from beginning to end. Far too hard for the simple Physics brain.
I can see that your background is not broadcasting or the reproduction of colour subjects with a choice of subtractive primaries. We have been talking a bit at cross purposes.

 

[JohnnyGui]

Are you confusing the contexts here?

Not really sure which contexts you mean. I couldn't really understand your explanation in of post #65 regarding my question of post #63, sorry.

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 was referring to a particular saturated Cyan. I was thinking that they gave coordinates to colors outside the RGB triangle based on the amount of negative value that one of the RGB primaries would need until the color you get closely resembles the perception of a spectral color (a particular negative value for one of the primaries is then reached). So holding 2 primaries at a fixed amount and making the 3rd increasingly negative would give a certain chroma at increasing saturations (particular direction from the white point) outside the RGB triangle.
I might be completely wrong about this. If so, how do they assign the coordinates of the colors outside the RGB triangle then with these (or imaginary) primaries??

I am impressed with all the reading around you have been doing. Not many people are prepared to do that.

I have this (rather exhausting) ocd to try and understand a subject up to its origins to be fully convinced of its learnings instead just accepting its surface scrapings, if possible. It's time consuming but I find it quite fulfilling at the end.

 

[pinball1970]

My comment about adding a fourth colour was about additive mixing. The results of subtractive mixing can be anyone’s guess. It has to be a matter of suck it and see, rather than mathematical prediction. The precise detail of the secondary filter charactistic has to be known; it’s ‘colour’ / chroma is not enough to predict how it will mix with another secondary.
I would not be “surprised “ about the transmission characteristics of any specialist due or pigment. That technology has to be a fudge from beginning to end. Far too hard for the simple Physics brain.

You can do more than best guess with a decent dye/substrate database, measurement of your target sample and some colour software. It’s more of a technology than a science but it is not a fudge.

 

[sophiecentaur]

You can do more than best guess with a decent dye/substrate database, measurement of your target sample and some colour software. It’s more of a technology than a science but it is not a fudge.

I am sure you can get a very good match for situations where you can choose your dyes to suit. I have yet to see the result of a colour film process that produces accurate matches. Very nice pictures at times but it wouldn't handle the range of scenes and lighting that standard modern TV can. I don't think 'fudge' is too strong a term. Do you remember how telecine inserts in TV programmes looked? And that's not many years ago.

 

[sophiecentaur]

Not really sure which contexts you mean.

If your discussion is in the context of the Maths of the subject then I will agree with you. If you want to bring negative values into the context of a practical colour additive process then I have to challenge you to show how you would actually do it. What do you mean by an "imaginary primary"?

 

[pinball1970]

I am sure you can get a very good match for situations where you can choose your dyes to suit. I have yet to see the result of a colour film process that produces accurate matches.

My area is textiles to a lesser extent paper. Films/photos PC screens and reproducing digital images is not something I know that much about.

We trialled some technology using Camera RGB data for commercial purposes but it was unpredictable. We did this for complex / multi coloured samples where it was impossible to take measurements using a spectrophotometer. The bottom line was that RGB data could throw out an anomaly in different circumstances (presence of optic textured surface metameric sample) but the spectro never did that. If a sample was metameric you could perform a simple calculation using DL Da and Db from the spectral data, I don’t think you can do that with RGB

 

[JohnnyGui]

If your discussion is in the context of the Maths of the subject then I will agree with you. If you want to bring negative values into the context of a practical colour additive process then I have to challenge you to show how you would actually do it. What do you mean by an "imaginary primary"?

That's what I was wondering about as well if my mentioned method is wrong. The way I'd think of to assign coordinates to colors outside the RGB triangle based on certain amounts of negative values is described in my post #72:

I was thinking that they gave coordinates to colors outside the RGB triangle based on the amount of negative value that one of the RGB primaries would need until the color you get closely resembles the perception of a spectral color (a particular negative value for one of the primaries is then reached). So holding 2 primaries at a fixed amount and making the 3rd increasingly negative would give a certain chroma at increasing saturations (particular direction from the white point) outside the RGB triangle.

I'ts kind of like the following:
If a certain amount of negative value X is needed for one of the RGB primaries to get a color that closely resembles a spectral color outside the RGB triangle, then any other amount of negative value of that same primary that is less than X, which we can call C, would give a color with coordinates that is $\frac{C}{X}$ less far than the coordinates of the spectral color. I'm merely using proportionality here. I think this is a problem though if you reach an amount of C = 0 for that primary, because how wouldy you then use proportionality to assign coordinates to a color with a needed amount of 0? Also, I recall that distances in the chart are not proportional to the amounts of the used primaries.

Therefore, I'm not sure if that's the correct (or alternative) way it is done.
If that method is completely wrong (which it probably is), then I'd wonder how else it is done. The video and article you linked me to say that imaginary primaries are used but I have no idea what they are or how they are derived and how they give the ability to assign coordinates to colors outside of the RGB triangle to the extent of them closely resembling spectral colors..​

 

[sophiecentaur]

@JohnnyGui You can re-arrange a formulae to get a result that you want. But that doesn't imply that you could actually produce a colour that way. I ask you again how would you implement this added negative colour in practice on a TV display? If you are trying to produce this 'extreme' colour on an image of a scene, your method would need to be applied to the whole image, in places with and without that colour. In specialist printing you can do what you like in particular areas with 'spot colours', dodging and burning etc. etc. but you have stepped outside the mechanics of a TV or even a general Film system. That's not wrong but it's not in the terms of the thread, as I perceived them.
@pinball1970 introduced the concept of dye production and it is clearly a well formalised technology that has progressed way beyond 'a bucket of this and a cupful of that, according to taste'. Matching a colour, rendered on two different fabrics or materials is a serious business - particular when that colour is associated with a well known brand.

 

[Mister T]

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

You can filter sunlight and produce a beam of light that one, can be shown to be of a single wavelength, and two, can be identified by most people without a deficiency in their ability to see color as blue. You can do the same with red. But you cannot do that with cyan. But what you can do is take the two previously-mentioned beams, mix them, demonstrate that it consists of two discrete wavelengths, and is recognized by people as being cyan.

This is all, of course, connected to the way retina cells respond to different colors. There are cells that respond to red and only red, Likewise for blue. But not for cyan.

 

[sophiecentaur]

But you cannot do that with cyan.

A Cyan filter can be produced that will let through just a narrow band of wavelengths and give a Cyan coloured patch on a white board. However, that filter would be no use as a pigment for mixing with other monochromatic filter pigments in a subtractive way. You would get black. Subtractive mixing only works because of the Overlap of two or three filters
You are mixing the concepts of additive and subtractive mixing which easily results in invalid conclusions.

 

[Drakkith]

You can filter sunlight and produce a beam of light that one, can be shown to be of a single wavelength, and two, can be identified by most people without a deficiency in their ability to see color as blue. You can do the same with red. But you cannot do that with cyan.

So what color would most people identify light with a wavelength of ~505 nm as? Would it not be cyan?

 

[JohnnyGui]

@JohnnyGui You can re-arrange a formulae to get a result that you want. But that doesn't imply that you could actually produce a colour that way. I ask you again how would you implement this added negative colour in practice on a TV display? If you are trying to produce this 'extreme' colour on an image of a scene, your method would need to be applied to the whole image, in places with and without that colour. In specialist printing you can do what you like in particular areas with 'spot colours', dodging and burning etc. etc. but you have stepped outside the mechanics of a TV or even a general Film system. That's not wrong but it's not in the terms of the thread, as I perceived them.

I understand that it's not possible to use this "negative" value method for TVs or Film systems. My question is about how coordinates are assigned to the colors outside the RGB primaries during the development of the CIE chart in the first place. If you're using other "imaginary" primaries, wouldn't those primaries give different coordinates w.r.t. the RGB primaries? How can you define coordinates for colors outside the RGB triangle by using totally different "imaginary" primaries and yet put these colors in the same coordinate system as the colors that have coordinates based on the RGB primaries?

It's the (mathematical) method on how these colors outside of the RGB triangle are coordinated on the CIE chart that I'm asking about.

 

[Mister T]

A Cyan filter can be produced that will let through just a narrow band of wavelengths and give a Cyan coloured patch on a white board.

I did not know that. Do you know what the value of that wavelength is?

 

[Mister T]

So what color would most people identify light with a wavelength of ~505 nm as? Would it not be cyan?

Well, you answered the question I just asked @sophiecentaur . I was not aware cyan existed as a monochromatic beam.

I suppose I should stick to stuff I know and teach. Having a color deficiency I find it almost impossible to teach the thing, but I thought I at least understood it.

Is it true that to perceive monochromatic cyan both the red and green cones would need to be activated, whereas to perceive monochromatic red or green only one set would be activated? I think that might make the point I was originally trying to make.

 

[Drakkith]

Is it true that to perceive monochromatic cyan both the red and green cones would need to be activated, whereas to perceive monochromatic red or green only one set would be activated?

The red and the green cone cells have response curves that overlap a great deal, so any particular wavelength will likely activate both to some extent, including monochromatic cyan. In addition, the response curve for both types of cells extends about 200-250 nm, so green and red cone cells will respond to both red and green light.

 

[sophiecentaur]

If you're using other "imaginary" primaries, wouldn't those primaries give different coordinates w.r.t. the RGB primaries?

You would first have to define the R primary and then define the white point. The (or 'a') secondary Cyan would be on the line from R through the white point. But I don't understand why you are finding this worth following up. It's just geometry and algebra, starting from the formula that you have already found and will give you 'an answer'. But what significance does that answer have?

 

[sophiecentaur]

Is it true that to perceive monochromatic cyan both the red and green cones would need to be activated, whereas to perceive monochromatic red or green only one set would be activated? I think that might make the point I was originally trying to make.

ALL cones are stimulated by almost ALL frequencies. Eyes are not spectrometers and they use all sensors to classify all the spectral colours. The three signal values are what our brains work with.

 

[Drakkith]

ALL cones are stimulated by almost ALL frequencies. Eyes are not spectrometers and they use all sensors to classify all the spectral colours. The three signal values are what our brains work with.

Indeed. A word of warning to those reading this thread; many of the cone response graphs you can find online appear to be inaccurate or require care in interpreting. For example, almost all of them have been normalized, so each cone cell's peak response appears to be the same as the others. In other words, the curves all reach the same height, despite blue cone cells having significantly less sensitivity than the others. So don't just look at the first graph you find. Dig a little deeper.

 

[JohnnyGui]

But I don't understand why you are finding this worth following up. It's just geometry and algebra, starting from the formula that you have already found and will give you 'an answer'. But what significance does that answer have?

It's just my curiosity and interest on the way the chart was developed since I don't understand how they assigned coordinates to colors with negative values of one or more of the primaries.

You would first have to define the R primary and then define the white point. The (or 'a') secondary Cyan would be on the line from R through the white point.

Are the R primary and the white point in that case defined by the "imaginary primaries" the links are talking about?

 

[sophiecentaur]

It's just my curiosity and interest on the way the chart was developed since I don't understand how they assigned coordinates to colors with negative values of one or more of the primaries.

I think you have this the wrong way round. The chart is not based on the primaries; it started with the analysis curves of the eye. Those sensitivity curves were arrived at with a vast number of subjective tests using a range of monochromatic wavelength mixes and comparing the perceived brightness and 'colours' of combinations of mixtures. The curves came out of some complicated analysis of all the results (A load of simultaneous equations in effect). It was, of course, not possible to put a voltmeter on the outputs of the three sensors and just measure the response.

I would have expected your reading to have given you information about the way the CIE chart was arrived at but it was chosen to fit the eye's appreciation of colour and does its best to eliminate the Luminance factor and its scale is arranged so that the position of a perceived colour on a line between two other colours is given by a simple linear weighted combination of the relative brightnesses of the two mixed colours. This is analogous to the Centre of Mass of two masses on a light rod. Now move to a triangle.You cannot obtain a match outside a triangle of three chosen colours (which we could call Primaries). A colour outside a primary triangle can be only be obtained by adding another contribution, on the other side of the line.

Mechanical analogy: Imagine a large, massless plate with masses at the vertices of a triangle, drawn on the plate . You want to balance it on a point somewhere inside the triangle. You can do this by choosing the right combination of masses. To get it to balance on a point outside the triangle, you would need to LIFT one of the corners. This would correspond to a negative mass at that corner. Same with mixing primaries. But negative masses do not exist and neither is there a corresponding negative primary. It is just a bit of Maths.

 

[JohnnyGui]

I think you have this the wrong way round. The chart is not based on the primaries; it started with the analysis curves of the eye. Those sensitivity curves were arrived at with a vast number of subjective tests using a range of monochromatic wavelength mixes and comparing the perceived brightness and 'colours' of combinations of mixtures. The curves came out of some complicated analysis of all the results (A load of simultaneous equations in effect). It was, of course, not possible to put a voltmeter on the outputs of the three sensors and just measure the response.

I would have expected your reading to have given you information about the way the CIE chart was arrived at but it was chosen to fit the eye's appreciation of colour and does its best to eliminate the Luminance factor and its scale is arranged so that the position of a perceived colour on a line between two other colours is given by a simple linear weighted combination of the relative brightnesses of the two mixed colours. This is analogous to the Centre of Mass of two masses on a light rod. Now move to a triangle.You cannot obtain a match outside a triangle of three chosen colours (which we could call Primaries). A colour outside a primary triangle can be only be obtained by adding another contribution, on the other side of the line.

Mechanical analogy: Imagine a large, massless plate with masses at the vertices of a triangle, drawn on the plate . You want to balance it on a point somewhere inside the triangle. You can do this by choosing the right combination of masses. To get it to balance on a point outside the triangle, you would need to LIFT one of the corners. This would correspond to a negative mass at that corner. Same with mixing primaries. But negative masses do not exist and neither is there a corresponding negative primary. It is just a bit of Maths.

Apologies for the late reply. I like your mechanical analogy a lot. It did help me understand this a bit better. The sources that I read don't really pay sufficient attention to how coordinates of the colors outside the RGB primaries are derived/calculated, and it was therefore hard for me to grasp the mathematical method since there are no negative primaries, like you said.

There are also no units as far as I understand regarding the "amount" of each primary of RGB used for each color. If there was, then colors outside the RGB triangle would have a certain "negative" amounts of those units (like "- kg" for example in your mechanical analogy) which can then be used to assign coordinates. I might be wrong about this though.

 

[sophiecentaur]

The sources that I read don't really pay sufficient attention to how coordinates of the colors outside the RGB primaries are derived/calculated,

There is a lot of stuff out there. Try including "Tristimulus" and "analysis" in your search terms.
Glad you liked the mechanical bit.
"how coordinates of the colors outside the RGB primaries are derived/calculated,"
I don't know why you are finding this so interesting because what would be the point in talking about producing colours additively when you would have to include subtraction? The sort of thing you seem to be suggesting would involve projecting two primaries through a subtractive filter, I think. What would be the point?
The fictitious (negative) coefficient of one of the primaries is obtainable using the same formulae as for colours within the triangle but there are many examples in Science where the result of a mathematical operation has no real meaning. Imo, there are enough real situations to solve.

 

[pinball1970]

Yes this is how do it in the lab, we choose say 3 4 or 5 dyes (3 usually enough) and this will give us a gamut of possible shades. Each of those individual dyes has an associated set of library dyeings, that is technically viable to produce from lightest to darkest given by a percentage of the weight of the substrate . The target shade say "Blue" is measured using a spectro and the software gives a recipe matching up the library of those dyes against the data generated from our Blue. If our Blue is very bright, brighter than the brightest blue in the library then the predicted recipe will give a negative Chroma value in the predicted outcome. The Flatter (duller) our library dye the greater the negative Chroma value (given as DC). So the software is telling us the colour is unobtainable. In this case this may be what JohnnyGui had in mind. How are the initial points in colour space set? By measuring each library of each dye, point one would be 0.0001% reactive red, this will have an absolute Chroma value (C) absolute Hue angle (H) and lightness Value (L) depending on what sort of software you use. Point 2 would be 0.0002% and so on.

 

[pinball1970]

Just to add this is the colour space we create using applicable colorants on a given substrate.

 

[sophiecentaur]

How are the initial points in colour space set?

Are you referring to the CMY values of three primary pigments?
But the detailed filter characteristic of a pigment must be known if you want to use it in a mixing process, surely. That's why I question the idea of those 'points' in colour space. Looking at it from an RF Engineering point of view, knowing only the centre frequency of a filter can never tell you the result of inserting it in a channel .
I take my hat off to 'colour engineers' in the world of CMY. I wouldn't know where to start to produce a matching procedure that didn't actually involve knowing the band pass characteristics of all the pigments.

 

[pinball1970]

I wouldn't know where to start to produce a matching procedure that didn't actually involve knowing the band pass characteristics of all the pigments.

Re Band pass - There isn't one. There are commercial tolerances but different equations and different weightings on the parameters. There is a useful one called CMC that dye houses use. Measure your target measure, your batch and if you get a value of 1.0 (called Delta E) or below there is a good a chance it is commercially viable. Where it gets interesting is when you split the numbers up into the respective components and then fiddle around with the recipe to see how each parameter varies.

 

[sophiecentaur]

Re Band pass - There isn't one.

We must be talking at cross purposes. I am referring to the frequency (wavelength) response of the filter. It will have a significant bandwidth and the pass band shape is important.

 

[pinball1970]

I take my hat off to 'colour engineers' in the world of CMY

I prefer technologist, we take our hats off to the guys who made predictions without the use of a spectro or colour software and to the guys that used colour space and library dyeings to come up with matching software. Where would the world be if we did not control colour? Chaos.

 

[sophiecentaur]

Where would the world be if we did not control colour?

In the 1970s !