Mixing two RGB color vectors to get resultant

Question

I am trying to mix two source RGB vectors to create a third "resultant vector" that is an intuitive mix of the first two.

Ideally, I would be able to emulate "real paint mixing characteristics", but for simplicity, I am trying to find a method where the resultant looks intuitively like what you'd get from combining the two source rgb's.

minimally, these characteristics:
RED + BLACK = DARK RED
RED + WHITE = LIGHT RED

optimally, also with real paint characteristics:
RED + BLUE = PURPLE
RED + YELLOW = ORANGE
(etc)

--

I am currently doing this the "lazy way" by adding the two source RGB vectors/255, then normalizing (and multiplying by 255). So, using this: [Red = <1,0,0> * 255] + [Blue = <0,0,1> * 255] gives Magenta=<1,0,1>/sqrt(2) * 255, though the other colors are less intuitive or even visible... I need a better method! Please help :-)

Answer 1

Is what you suggested the same as a weighted average?

Average R = w1*R1 + w2*R2 + w3*R3 + ... + wn*Rn

Average G = w1*G1 + w2*G2 + w3*G3 + ... + wn*Gn

Average B = w1*B1 + w2*B2 + w3*B3 + ... + wn*Bn

The w's are the weighting fractions and the sum of them is 1.

R1 is the red component of the first color to mix, for example.

So if you wanted to mix two colors evenly, it would be:

Average R = 0.5*R1 + 0.5*R2

Average G = 0.5*G1 + 0.5*G2

Average B = 0.5*B1 + 0.5*B2

As for mapping the resultant color to a named color ("dark red") maybe just do a lookup table and pick the closest one?

Answer 2

The problem is that you're trying to relate two different color theories. RGB is additive color theory (colored light is emitted from the source). In this case the more of the spectrum you add to the mix, the closer you get to white.

For "paint mixing" you need to emulate subtractive color theory which governs the way reflected colors are mixed. (i.e. reflected light such as paints, different surfaces etc.) Here's an overview.

The JavaScript on this page is a clue as to how you might proceed in converting RGB to CMYK.