Given the triplet R, G, B and factor F, how can I calculate the “watermark” of a color?

Question

I have a triplet (R, G, B) where each color is between 0.0 and 1.0. Given the F factor (0.0 means the original color and 1.0 means white), I want to calculate a new triplet, which is the “watermark” version of the color.

I use the following expression (pseudo-code):

for each c in R, G, B:
    new_c ← c + F × (1 - c)

This creates something that looks fine, but I understand that it introduces deviations from the color cast (checking the HSV equivalent before and after the conversion), and I don't know if this should be expected.

Is there a "standard" (with or without quotation marks) algorithm for calculating the "watermark" color version? If so, what is it? If not, what other algorithms can you tell me?

Answer 1

In fact, it looks like it should give the right shade, minus the small variations for arithmetic rounding errors.

This is certainly reasonable, it was just to achieve the effect of watermarks. I do not know other "standard" ones, there are several ways to do this.

Alternatives:
Mix with white, but do it non-linearly at F, for example. new_c = c + sqrt(F)*(1-c), or you can use other non-linear functions - this can help the watermark look more or less "flat."

You could do this more efficiently by doing the following (where F takes a range of 0..INF):

new_c = 1 - (1-c)/pow(2, F)

for real pixel values ​​(0..255) this translates to:

new_c = 255 - (255-c)>>F

, , 32b- .

Answer 2

?

new_c = F*c

, , .

new_c = 1- * (-1)