I Question about subtractive color mixing

[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.

 

[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.

 

[Herman Trivilino]

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.

 

[Herman Trivilino]

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?

 

[Herman Trivilino]

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 !