Lumondo Photography wrote:
But, even going back to the 3-polarizer example, you can phrase the argument in a way that is purely subtractive, on a photon-by-photon basis.

1. after first polarizer has state |0> = a|-45> + b|45>
2. second polarizer removes |-45> state (subtractive!) and leaves only |45> state
3. now change basis and rewrite |45> as = c|0> and d|90>
4. third polarizer removes |0> state and leaves only |90>

It is a process of repeated waveform collapse - each polarizer is modeled as an operator which "selects" an eigenstate, which then propagates to the next polarizer.

Subtraction requires an amount to be subtracted. A filter cannot be subtractive because it would have an huge impact on dim light, a negligible impact on bright light, and an effect with no ability to be described in terms of f-stops.

If the effect of a filter is proportional to the amount of light it receives then it is fundamentally multiplicative.

Photography by  Roy wrote:
I am not an expert on color theory or the physics and whatever has been discussed in this thread.
From my limited knowledge and experience as a graphic designer I can say that RGB is called "Subtractive Color" because it's that spectrum of light that is reflected back to the viewer. CMYK on the other hand is called "Additive Color" because you add color pigments to get a particular color or hue.

Actually you have the traditional explanations reversed.

Photography by  Roy wrote:
Reversed?

Yes, if you shine a red flashlight in the dark, then overlap that red spot with one from a green flashlight, you are 'adding' the two and will get yellow. Add blue and you will add all the way up to white light - the feeling of all wavelengths present.

If you start with colored paint and 'add' the colors together you are actually 'subtracting' (a term I have some issues with) the amount of light being reflected. With every mix you get a darker and darker color until finally you 'subtract' all the way down to black (in theory, or muddy crap, in practice) - black light not being thought of as the PRESENCE of all paints, but rather as the ABSENCE of reflected photons.

digital Artform wrote:

It's not 'somewhere,' it is in the superstar spotlight location. It is THE operator.

Yes, the operator is there. But, just because the math uses multiplication, doesn't mean that describing the system, as a whole, multiplicative is better.

The manifestation of the system to us is not probabilistic, it's deterministic. That's why the under-the-hood probability calculations are not meaningful (or useful) to anyone without background in the field.

I don't think that anybody is arguing with you that light absorption of mixed pigments is a geometric composition of their absorptions. For anyone still reading this thread, this is a nice writeup

http://www.handprint.com/HP/WCL/color5.html

But the argument is whether, regardless of the underlying method of calculation, the word subtractive or multiplicative is more fitting.

The subtractive moniker is more intuitive and explanatory, because wavelength component amplitude is decreased and less light comes out. It's a good working definition and one that can be quickly grasped. Multiplicative, on the other hand, is an unintuitive word - most do not associate it with decreasing quantities.

Both additive and subtractive color theories (despite the huge fundamental differences between them) were initially rooted in experience. And the experience was best described as additive (light mixtures add) and subtractive (mixed pigments subtract).

The fact that we now have a physical theory that describes this phenomenon adds texture and a predictive framework to the original observations, but because all of modern color theory is based on perceptual properties, terms based on experience will trump terms based on the underlying calculation.

Lumondo Photography wrote:
The subtractive moniker is more intuitive and explanatory, because wavelength component amplitude is decreased and less light comes out. It's a good working definition and one that can be quickly grasped. Multiplicative, on the other hand, is an unintuitive word - most do not associate it with decreasing quantities.

Which is why 'subtractive' is such a great way to introduce the concept to young schoolchildren.

However, you get a little older, and you start to get serious, and . . .

digital Artform wrote:

Subtraction requires an amount to be subtracted. A filter cannot be subtractive because it would have an huge impact on dim light, a negligible impact on bright light, and an effect with no ability to be described in terms of f-stops.

If the effect of a filter is proportional to the amount of light it receives then it is fundamentally multiplicative.

The polarizer example I gave, which you quoted and mis-interpreted, demonstrates that a polarizer is not a multiplicative entity.

The polarizer, in the quantum sense, is an operator which selects one of superposed component states of the photon. The incoming photon is a linear superposition of |-45> and |45> and the polarizer collapses the state down to |45>.

That's how this behaviour is currently described. If you wish to see it as a multiplicative phenomenon, so be it.

Subtraction requires an amount to be subtracted.

I don't believe this generalization is useful. You can have an operator which subtracts the input quantity from itself, thereby sending all input values to zero. An example, the process of "eating everything on the table" is a subtractive one. I take food away until there is nothing left. If there is no food to be had, the process simply does nothing.

And like I said, nobody is arguing the underlying model used to describe the phenomena of subtractive color.

Lumondo Photography wrote:
The polarizer example I gave, which you quoted and mis-interpreted, demonstrates that a polarizer is not a multiplicative entity.

The polarizer, in the quantum sense, is an operator which selects one of superposed component states of the photon. The incoming photon is a linear superposition of |-45> and |45> and the polarizer collapses the state down to |45>.

That's how this behaviour is currently described. If you wish to see it as a multiplicative phenomenon, so be it.

blah blah blah, black box, who cares how the probabilities are arrived at for filter 1?

blah blah blah, black box, who cares how the probabilities are arrived at for filter 2?

Multiply.

Get answer.

digital Artform wrote:

Which is why 'subtractive' is such a great way to introduce the concept to young schoolchildren.

However, you get a little older, and you start to get serious, and . . .

Yes, it is. You start with basic terms which someone can understand. You then expand the realm of the model you presented, show additional details, and refinements, and use new knowledge and technical ability of your audience to make their understanding more sophisticated.

This is why one does not start with number theory in grade 1, but simple addition.

This is why studies the Bohr model of the atom before Schroedinger's.

Neither addition or Bohr are taught as the supreme principle, simply principles appropriate for a combination of age and background.

Maybe one day, we will start teaching early stages of school differently. I'll be the last to say that the current system is ideal. But until then, using a term which simplifies things for the purpose of illustration and explaining in depth to those who are interested, is a pretty good way to go.

Lumondo Photography wrote:
I don't believe this generalization is useful. You can have an operator which subtracts the input quantity from itself, thereby sending all input values to zero. An example, the process of "eating everything on the table" is a subtractive one. I take food away until there is nothing left. If there is no food to be had, the process simply does nothing.

And like I said, nobody is arguing the underlying model used to describe the phenomena of subtractive color.

No, you are eating 100% - which requires no special knowledge about how much is coming, nor does it require an off switch.

Filers with probabilities don't need to know anything but the odds.

Filters with amounts don't exist. And filters based on subtraction that are also proportional are doing a multiply after all, or are magically attuning themselves to the incoming amount.

Lumondo Photography wrote:

Yes, it is. You start with basic terms which someone can understand. You then expand the realm of the model you presented, show additional details, and refinements, and use new knowledge and technical ability of your audience to make their understanding more sophisticated.

This is why one does not start with number theory in grade 1, but simple addition.

This is why studies the Bohr model of the atom before Schroedinger's.

Neither addition or Bohr are taught as the supreme principle, simply principles appropriate for a combination of age and background.

Maybe one day, we will start teaching early stages of school differently. I'll be the last to say that the current system is ideal. But until then, using a term which simplifies things for the purpose of illustration and explaining in depth to those who are interested, is a pretty good way to go.

Exactly. Then when you get older, you move beyond the simplified introduction.

digital Artform wrote:

blah blah blah, black box, who cares how the probabilities are arrived at for filter 1?

blah blah blah, black box, who cares how the probabilities are arrived at for filter 2?

Multiply.

Get answer.

Look, if you want to argue terminology and dig up back-of-envelope math to try to prove your point, that's fine and keeps the forums going. It's fun to chat about terms and such, and precision and applicability of names of theories.

But don't piss on people who then turn around and show you the actual principle on which is at the heart of the whole system.

Lumondo Photography wrote:
Look, if you want to argue terminology and dig up back-of-envelope math to try to prove your point, that's fine and keeps the forums going. It's fun to chat about terms and such, and precision and applicability of names of theories.

But don't piss on people who then turn around and show you the actual principle on which is at the heart of the whole system.

1) It's irrelevant. We are talking about filter mixing, not filter creation. You keep harping on filter creation.

2) I'm entertaining this excursion into polarizing filters, to begin with. That's hardly pissing.

digital Artform wrote:
The last thing real physical light does is subtract. Try to shine or produce some negative light and see.

True, light is additive.  PIGMENT is subtractive.  Pigment absorbs light.  As the pigment "subtracts" color, it absorbs more light until finally going to black.  It is subtracting from what is being reflected.  If it were white, all would be reflected and none was subtracted.  This multiply stuff is all computer speak.  Geez, elementary school science anyone?

Since subtractive color refers to pigments, you cannot use that model to explain the red filters... that's light.

Kristine Kreations wrote:
This multiply stuff is all computer speak.  Geez, elementary school science anyone?

You should read the thread in some detail.

Anything I would say in response to your post I've already said at least twice.

I think pretty much everything worth saying on the matter from all directions has been said.

Most people will get along fine no matter what their views on the matter are.

digital Artform wrote:


As I already said a page ago.

What happens to this explanation when the YELLOW contains neither RED nor GREEN?

What happens to it when the CYAN contains neither BLUE nor GREEN?

What do you mean when it "contains" neither color?  True yellow is yellow- light with a wavelength around 570-580 nm.  It doesn't "contain" anything else.  You don't mix red and green to get yellow.  What we see looks red because the pigment is absorbing in the cyan spectrum.

digital Artform wrote:

You should read the thread in some detail.

Anything I would say in response to your post I've already said at least twice.

I think pretty much everything worth saying on the matter from all directions has been said.

Most people will get along fine no matter what their views on the matter are.

If everything has been said, then why are you still asking questions in this thread and replying to responses?  It sounds to me like this is just a thread you opened so you can argue with the world... nobody is right except for you.  Why did I bother?

Kristine Kreations wrote:
What do you mean when it "contains" neither color?  True yellow is yellow- light with a wavelength around 570-580 nm.  It doesn't "contain" anything else.  You don't mix red and green to get yellow.  What we see looks red because the pigment is absorbing in the cyan spectrum.

Take a 'true yellow' filter and hold it up.

Now take a magenta filter and hold it up such that a beam of full spectrum white light can pass through both.

What color is the light coming out the back end?

Kristine Kreations wrote:
If everything has been said, then why are you still asking questions in this thread and replying to responses?  It sounds to me like this is just a thread you opened so you can argue with the world... nobody is right except for you.  Why did I bother?

Give it one last shot

(see above one post)

digital Artform wrote:

Take a 'true yellow' filter and hold it up.

Now take a magenta filter and hold it up such that a beam of full spectrum white light can pass through both.

What color is the light coming out the back end?

You're looking at the filter with your eyes, correct?  Well, how do your eyes percieve color?  The filter looks yellow because it's absorbing its compliment.  When you shine light through the filter, it will absorb just like it was absorbing the ambient light- making the light on the other side the same color as the filter.

Kristine Kreations wrote:

You're looking at the filter with your eyes, correct?  Well, how do your eyes percieve color?  The filter looks yellow because it's absorbing its compliment.  When you shine light through the filter, it will absorb just like it was absorbing the ambient light- making the light on the other side the same color as the filter.

A 'true yellow' filter only allows true yellow 575 nm (plus or minus a few nm) to pass through.

When that true yellow light hits the magenta filter it is completely blocked, since a magenta filter only lets red and blue in roughly equal amounts pass (there is no 'true magenta wavelength in the spectrum)

Net result = no light at the end.

Kristine Kreations wrote:

You're looking at the filter with your eyes, correct?  Well, how do your eyes percieve color?  The filter looks yellow because it's absorbing its compliment.  When you shine light through the filter, it will absorb just like it was absorbing the ambient light- making the light on the other side the same color as the filter.

A 'true yellow' filter only allows true yellow 575 nm (plus or minus a few nm) to pass through.

When that true yellow light hits the magenta filter it is completely blocked, since a magenta filter only lets red and blue in roughly equal amounts pass (there is no 'true magenta wavelength in the spectrum)

Net result = no light at the end.

Kristine Kreations wrote:
Why would anyone have a "true yellow" filter?  What's the point of only having one wavelength pass through?

I'm done with this thread.  Congratulations, you've managed to piss off yet another person.  I'm not a physics major anyways- I do animal science.

I don't know why, but you can by one from Perkin Elmer online if you want

Don't be mad, it's just the internet, plus it's how I think of color, and I thought someone might find it useful.

My goal with the true yellow filter was just to point out that the simple subtractive chart is really only a mnemonic device, and can be made to fail. Mutiplying the transission curves at least gives a pretty good shot at an answer.