Questions about multiply

[coelr23]

A is the color created by multiply
I am using the method of changing the color of c to match the color of b as much as possible with my eyes.
But it is not accurate.
Please tell me how to know what color I should use instead of c to make the result color of multiplies b

 

[Rich54]

This is what I think you are asking:

• You have three colors labeled A, B, C.
• Color C has a blend mode of Multiply. When you overlap A and C with the blend mode of Multiply, the result is the small area labeled A x C.
• Your end goal is that you want to change color C so that the small area (A x C) will become the same color as B.

If my interpretation of your question is correct, then I believe this is impossible.

• The Multiply blend mode takes the Luminosity of the first color and multiplies it by the luminosity of the second color. The result of Multiply is always darker than either of the two original colors.
• Color A has Luminosity of 57 and color B has luminosity of 114. Color B is already brighter than color A. Therefore, there is no possibility that the Multiply blend mode could ever create a color that is brighter than color A.
• To get close to what you want, you would need a blend mode of Screen instead of Multiply.

This might be more mathematics than you want to know, but here are some facts:

• To calculate the Luminosity of any color, take the RGB value of that color and apply this formula: Luminosity = (R x 0.3) + (G x 0.59) + (B x 0.11)
• For the Multiply blend mode of two colors A and B, the resulting luminosity of the overlap area comes from this calculation:
  Luminosity_A x Luminosity_B) / 255

 

[IamSam]

Hello.

To add to Rich's post, I'm certain that you can't accomplish what you are try to do using the method you describe.

You must remember basic color theory that shows that all colors are made up of a combination of colors. In this case, your target color using RGB is R:197 G:90 B:14. Any color profile requires, at a minimum, 3 colors combined to make a certain color.

These are the RGB colors that make up your target color on separate layers.
R:197 G:90 B:14


Set the top two layers, red and green, to the Blending mode "Lighten" and you get your desired target color in the center overlap.


Or you can Isolate channels in blending options (Advanced Blending) to get the desired target color in the center overlap.


Above you are trying to match your target color using only 1 known color swatch with an unknown color swatch. You would need at least one more swatch. You would also have to change the known color swatch to make it able to accomplish the desired target color.

You would also need a better understanding of blending modes to know what "multiply" actually does.

Perhaps if you explain what or why you are trying this method, we could offer a solution.

 

[thebestcpu]

Hi @coelr23
I agree with @Rich54 and @IamSam that getting the exact color just with multiply is not possible.
So one question. Must you have the blend mode of multiply to get your result. You can reproduce the desire color with other types of Blends yet not sure that would meet your need.

In the meantime, here are the closest colors which you can use to achieve the same Hue.

Here is the table of desired final color B, starting color C, and Blend Color (two options):


And here is the image (started with yours). Note that the final blend is much lower in Brightness as already explained in the previous posts. Yet this is the best match for Hue and Brightness.




Hope this helps some
John Wheeler

 

[thebestcpu]

HI @coelr23
To follow up with my question about using a different blend:
If you use Hard Mix blending, for any final target color, and any starting color, you can create a blend color that when blended with the starting color you can create the final target color.

I am not sure that is what you need so won't go into the details unless you have the option for alternate blend modes for your problem.

Here is an example blend done with Hard Mix Blend using your final target color and your initial starting color:




Thought is worth mentioning at least
John Wheeler