'Multiplicative' mixing and the 4 primary colors

[thorkil]

Originally Posted by Marc Sabatella

If this paint truly reflected only blue wavelengths, and the yellow paint mixed only yellow wavelengths, the resulting mix would be essentially black.

Not only that, but if the theoretical 'pure' blue & yellow pigments were reflecting only a tiny, tiny slice of the spectrum (what are "pure" colors anyhow?) they would appear very dark, perhaps nearly black, in any normal light. (There is only so much light of a particular wavelength available in a given illumination to hit the pigment, right?) So, the more wavelength-specific a theoretical pigment is, the less reflectance. I think.

In other words, I think our human eyes would see a 'pure' yellow pigment as black. Kinda weird, and not too useful for artists.

(Exception to the above would be a theoretical situation where the pigment is isolated in a non-reflective environment, and the illumination is brought up to a sufficiently high level that the one wavelength reflected to our eyes becomes visible. Again, not too applicable to our everyday reality, right?)

Chuck

[Richard Saylor]

Originally Posted by Michael24

The graphs of the spectra should indicate why any blue and yellow will make a green. Mind you, they may not be a high chroma green or a warm or cool green you desire, but it will make a green. Interesting to note - the blue pigment is cobalt blue which as you can see has a high reflectance in the red region. This mutes any greens made with in and yellow since the high degree of red is subtracting out the chroma of the green that is produced. - thus a muted green, but green none the less.

I took your spectral graphs and superimposed them. (The magenta line is the curve for green.) This shows clearly that the reflectance of the mixture is not the geometric mean. In fact, it follows very closely the minimum of the yellow and blue curves. I wonder whether this is generally true of mixtures.

[Patrick1]

Thanks Richard for superimposing the graphs...much easier to compare.

Originally Posted by cmyguy

This shows clearly that the reflectance of the mixture is not the geometric mean. In fact, it follows very closely the minimum of the yellow and blue curves. I wonder whether this is generally true of mixtures.

I wonder too...at least to get a rough estimate. It's interesting that around the middle, the reflectance of the mix is less than either the blue or yellow.

It looks like Handprint greatly overstates the value of geometric mean or ''the visual average shifted somewhat toward the darker reflectance profile at each wavelength'' to get an estimate.

[Richard Saylor]

Hi, Domer. I'm just trying to get a handle on some sort of rough approximation. I have the feeling that the reflectance of mixtures may generally lie somewhere between the geometric mean and the minimum curve. The Kubelka-Munk formulas are complicated but don't involve anything more mathematically sophisticated than the exponential function. Anyone experienced with a scientific calculator could work them. However, the formulas are practically useless because they use information other than simply the reflectance spectra, such as the absorption spectrum, scattering spectrum, sample thickness, and the reflectance spectrum of the substrate.

[Richard Saylor]

Originally Posted by Domer

It's interesting that around the middle, the reflectance of the mix is less than either the blue or yellow.

Yes, at around 550nm, blue and yellow are at 50% and green is below that. That could be accounted for if the spectrophotometer just takes a finite number of readings at selected wavelengths and a continuous reflectance curve is constructed by interpolation.

[Michael24]

Originally Posted by cmyguy

Yes, at around 550nm, blue and yellow are at 50% and green is below that. That could be accounted for if the spectrophotometer just takes a finite number of readings at selected wavelengths and a continuous reflectance curve is constructed by interpolation.

cmyguy: You are just too clever!!! I never thought about superimposing all three curves. Interesting to see the result. Yes, the Hunter spectrophotometer takes a reading at every 5 nm and interpolation makes the continuous curve via Microsoft Excel.

As you can see its not just the geometric mean. Every single and mixed color has it own spectral response that has a bit of predictability and some unpredictability. I don't know if scatter, surface distortion or luminosity play a role in the spectral, but they do have some surprises.

We note the *red tail* (an abrupt increase in reflectance on the right end of the spectral curve) on many pigments does play an important role in how a color will appear when mixed. It have nearly everything to do with the failure of inorganic pigments (cad yel. and ultramarine or cobalt blue) to make clean high chroma greens. The do grey out but not because of chemical or theoretical issues. They grey because the red reflectance in the cobalt blue neutralizes the green of the cad yel and cobalt mixture. It is as if you are adding a bit of red to the cad yellow cobalt blue mix. That's how much red is being reflected in cobalt blue.

Again, love what you did with the graphs.

Michael Skalka
Conservation, National Gallery of Art, Wash. DC

[Einion]

Originally Posted by cmyguy

The Kubelka-Munk formulas are... practically useless because they use information other than simply the reflectance spectra, such as the absorption spectrum, scattering spectrum, sample thickness, and the reflectance spectrum of the substrate.

I'd go further, they're of no use to artists whatsoever (regrettably!)

Originally Posted by cmyguy

I'm just trying to get a handle on some sort of rough approximation. I have the feeling that the reflectance of mixtures may generally lie somewhere between the geometric mean and the minimum curve.

I can certainly appreciate why you'd want to do this but I'm afraid anything one comes up with can only approximate certain results and if I'm thinking about this correctly won't account for specific 'oddball' mixtures and other effects.

Originally Posted by cmyguy

I took your spectral graphs and superimposed them. (The magenta line is the curve for green.) This shows clearly that the reflectance of the mixture is not the geometric mean. In fact, it follows very closely the minimum of the yellow and blue curves. I wonder whether this is generally true of mixtures

Nope, it's certainly about right in specific instances but it's not widely true: the relationship of subtractive mixing complements is very irregular, hence no simple idea can account for all mixes; off the top of my head there are bowed mixing lines, plus the unexpected hues in some near-complement mixes aren't accounted for. I think, again, this is accounted for by the undercolour, the information for which isn't inherent to reflectance profiles.

I think I've posited this previously in a discussion with Bill, but even if we had measurements of more of the parameters of pigments than just the masstone reflectance, the results still couldn't be collated and point to anything more than broad relationships (much less be a useful predictor) because of the unusual effects one can get at different concentrations of two or more colours, the influence of the binder's colour and refractive index, substrate/underlying colour, film thickness, surface gloss etc. etc. There are just too many variables to take into account in paints to do anything other than, maybe, predict a mix at a given proportion applied at complete opacity with a specific surface finish.


Michael, great to see you here and have your input.

Einion

[Richard Saylor]

Michael and Einion, thanks for the clarification.

[Twinbee]

Hi all,

Originally Posted by Marc Sabatella

To me, this would be like teaching people "If you throw a baseball, it will stay in the air forever. The only reason it falls in the real world is that anti-gravity devices don't exist as yet".

Hehe.. that's not a bad analogy.
But if someone wants to learn the basics of color theory, then I can imagine it might be somewhat confusing to have two different results for the same mix. Especially if they try to equate what's happening with what a colour wheel would imply. The main problem we have is that there are two definitions of blue; one that's true blue, and the other that's: "let's call it blue, because it's close enough blue"

Originally Posted by Marc Sabatella

The real world has gravity, and while it can indeed be useful to discuss what would happen in the absence of gravity, you would normally treat this as the exception. For most purposes, it makes more sense to discuss how balls behave here on earth.

Actually, it's simpler to discuss the effects without gravity, because the mathematics involved would then be easier to learn. How gravity effects things should be taken into account afterwards. The same goes for the 'true blue' argument.

Originally Posted by Marc Sabatella

If this paint truly reflected only blue wavelengths, and the yellow paint mixed only yellow wavelengths, the resulting mix would be essentially black.

True. But judging by the complexity of the Kubelka-Monk model, might it be a dark shade of grey?

Originally Posted by cmyguy

It's still 'subtractive' color mixing (i.e., absorbances get combined, not reflectances), so you would get green.

Okay. This is what I imagined, so I was hoping you'd say that

Originally Posted by cmyguy

By the way, subtractive mixing really just refers to adding negatives (absorbances).

I found this site which might explain how the term 'subtractive' first came into use. For many decades, photographers have always measured logarithmically. And obviously, subtracting (or adding) logarithmic values is equivalent to normally multiplying. For more info, see this site

Originally Posted by cmyguy

Don't blame R-Y=M on Marc. :-) I don't know what sort of subtraction you're doing. If you are doing binary subtraction, 100-110=-010, the complement of green, which is magenta. If you are subtracting number triples (like rectangular coordinates), then (1,0,0)-(1,1,0)=(0,-1,0)=-(0,1,0), which again is the complement of green.

Rectangular coordinates it is...

Originally Posted by Michael24

The visible light spectrum starts at 370- violet, 450 - blue violet, 480 - blue, 590 - cyan, 540 - green, 580 - yellow, 600 - orange, 650 - red. (I go from short waves to longer. You may see a lot of web sites with long to short wave order.) So blue does not appear at the end in terms of wavelength.

Am I right in saying that a mono-spectral wavelength at the very end of the visible spectrum (deep violet - around 370nm) activates the red cone slightly as well as the blue cone?

Originally Posted by cmyguy

I'm wrong, neither ordinary binary subtraction nor bitwise binary subtraction works. The above computation is a fluke. However, under no circumstances will you get anything like black. In the real world, most red pigments reflect red and (to a lesser extent) yellow. If you remove the yellow reflectance, you just get a redder red.

My reasoning for saying red minus yellow equals black was based on a purely abstract level since yellow itself contains red (in the RGB model anyway). Think of R minus RG, and then count zero as a cutoff point.

Originally Posted by Michael24

Sorry I am either so bad a speed reader or really stupid. I get what you mean about grey and in reality the pure blue and pure yellow (again if one could obtain pure colors) would form blocks, thus a void and reflect nothing - therefore black. But since that can't exist right now in the real word we are stuck with green as a reflectance.

Yep, that's right. Incidentally, I think that a near fully saturated true blue (that is; a blue with equal green and red pollution, say... 100% blue, 20% green, 20% red), is possible with real paints. This would make a dark grey if mixed with yellow. So in other words, we can get a real blue hue with paint, but at the cost of the saturation level

Originally Posted by Michael24

In your first paragraph above, I am not sure what you mean by *And/ that the reason they make green in the world is because true blue paints don't exist as yet?* The graphs of the spectra should indicate why any blue and yellow will make a green.

Yes, they only make green because the 'blue' isn't true blue. It contains a green bias in.

I'll attempt to rephrase the paragraph:
"True blue and yellow will always mix to black/grey, but the closest blue paints have a green bias in them, which is why you get a greenish grey instead."

Just for the record, I realise 'true blue' (100% response of the blue cone, and 0% red and green cone response) isn't possible even with a mono-spectral laser. The red and green cones would need to be disabled completely for anyone to see the truest possible blue.

[Marc Sabatella]

Originally Posted by Twinbee

The main problem we have is that there are two definitions of blue; one that's true blue, and the other that's: "let's call it blue, because it's close enough blue"

True, but given that most people will never see the former in their entire lives, I'm not sure it makes sense to give it that much attention. See below.

Actually, it's simpler to discuss the effects without gravity, because the mathematics involved would then be easier to learn. How gravity effects things should be taken into account afterwards. The same goes for the 'true blue' argument.

Depends on what you're learning. If we're talking about just getting the basics across to a group of schoolchildren, I don't see the point of confusing them with things that don't actually exist in reality. It's not going to helpthem understand the real world one bit. And even to the extent it makes sense to start getting into the math involved with, say, a high school student, I still don't see this as coming down to any more than spending 30 seconds telling them that without gravity, you could throw a ball to the moon, but then spend the next week working out the equations for how gravity factors in. Of course, we're still ignoring the drag of air resistance; this strikes me as perhaps similar to using geometric mean or min or some such approximation to the full calculation of reflectances.

But the main point here is, I don't see any value in giving more than a passing thought on the things that *don't* happen in the real world except if you're making a lifelong study of this. Spending more than 30 seconds talking about situations in blue and yellow don't make green has no value whatsoever to 99.99999% of people on earth, and even 99.9% of artists.

Incidentally, I think that a near fully saturated true blue (that is; a blue with equal green and red pollution, say... 100% blue, 20% green, 20% red), is possible with real paints. This would make a dark grey if mixed with yellow.

I have yet to see any sort of even remotely blue pigment that doesn't create something we'd all recognize as green when mixed with anything we'd recognize as yellow. You'll have to name some actual pigments to convince me otherwise. Even ultramarine mixed with cad yellow deep looks green. Not even a slightly green grey, but a color that a child would look at and instantly call green.

[Richard Saylor]

Originally Posted by Twinbee

My reasoning for saying red minus yellow equals black was based on a purely abstract level since yellow itself contains red (in the RGB model anyway). Think of R minus RG, and then count zero as a cutoff point.

I see exactly what you mean. Rather than subtraction (the inverse of addition), that's what I would think of as the Boolean product (multiplication/set-intersection) of red and not-yellow. I.e., all reflected colors common to red and blue, which would indeed be nuttin' (black).

[Richard Saylor]

Originally Posted by Marc Sabatella

But the main point here is, I don't see any value in giving more than a passing thought on the things that *don't* happen in the real world except if you're making a lifelong study of this. Spending more than 30 seconds talking about situations in blue and yellow don't make green has no value whatsoever to 99.99999% of people on earth, and even 99.9% of artists.

The point is that colors like blue, yellow, and green are abstractions, like numbers. A pound of potatoes is never exactly one pound. Colors in the real world are bluish, yellowish, and greenish, not blue, yellow, and green.

Let's get away from yellow and blue and onto a very similar issue. Kids learn early on (at least I did) that red and blue make purple. Let's say you're teaching an oil painting class, and one of your students notices that one of the most popular reds (cadmium, which came in the oil paint set which s/he got for Christmas) does not make purple when mixed with blue. Are you going to tell her/him that it is just an anomaly, and that s/he is only 1 out of 1000 people who is weird enough to be concerned about this?

I may be strange, but it seems to me that exactly what we mean by 'red' is quite important, especially to artists. (Of course, I'm also strange enough to wonder about what is exactly meant by the number 'six,' so maybe my opinion is automatically disqualified.)

[Marc Sabatella]

Originally Posted by cmyguy

Let's get away from yellow and blue and onto a very similar issue. Kids learn early on (at least I did) that red and blue make purple. Let's say you're teaching an oil painting class, and one of your students notices that one of the most popular reds (cadmium, which came in the oil paint set which s/he got for Christmas) does not make purple when mixed with blue. Are you going to tell her/him that it is just an anomaly, and that s/he is only 1 out of 1000 people who is weird enough to be concerned about this?

Not at all - that's the level at which I *do* believe it makes to sense to think about color for most people. There are very real practical reasons cadmium red makes a violets, and that is precisely what I advocate concentrating on what happens in the real world. We can teach about this - using reflectance diagrams if necessary, or just the Michael Wilcox simplification using the color wheel - without ever resorting to talking how some mythical color that reflected only a single wavelength would behave.

[Richard Saylor]

Originally Posted by Twinbee

I found this site which might explain how the term 'subtractive' first came into use. For many decades, photographers have always measured logarithmically. And obviously, subtracting (or adding) logarithmic values is equivalent to normally multiplying. For more info, see this site

That's a good read. Thanks. What they are calling the density of a filter is called the absorbance in other contexts.

A = -log(T) = log(1/T)

where the transmittance T is the ratio of the output to the input of the filter. (The following is an explanation for the non-mathematically inclined.) When you are mostly interested in blocking certain wavelengths, A is really handy and intuitive. The basic unit is A = 1, the absorbance corresponding to a transmission of .1 (10%). If A = 5, the effect is the same as stacking 5 filters, each having an absorbance of 1. For perfect 100% transmission, A = 0. If the filter completely blocks the wavelength, A = infinity (like stacking an infinite number of A = 1 filters, ha-ha).

Yeah, photographers do use logarithms a lot, especially base 2 (which are proportional to base 10, so there's no difference in the computations) but less and less as photographic equipment becomes more and more automated (sigh). Being a mathematician and an amateur photographer, I like to use about the most non-automated (some would say antiquated) equipment available, my basic camera being a totally manual Leica rangefinder.

[Einion]

Originally Posted by Twinbee

Incidentally, I think that a near fully saturated true blue (that is; a blue with equal green and red pollution, say... 100% blue, 20% green, 20% red), is possible with real paints.

This isn't a particularly useful way of thinking about colour unless you're working in the digital realm, but if you want to examine this colour specificially why don't you plug in the numbers to a graphics program and have a look at it?

This would make a dark grey if mixed with yellow. So in other words, we can get a real blue hue with paint, but at the cost of the saturation level

If you check 51, 51, 255 in RGB values you'll see immediately that it isn't a saturated blue for a start... which should be reasonably easy to see because of the proportion that amounts to white light (20% in case it's not evident). And in paint it won't mix anything like grey with any yellow, even a yellow earth. If you continue your research you should easily be able to learn that you can mix a colour like this pretty easily and it will indeed mix a definite green just as Marc says.

While it's interesting to examine the underlying principles of colour relationships I think one should always work at it from the basis of looking for something with a practical application, so any imaginings of mono-spectral colours and so forth is only of academic interest (pun intended).

Einion