It seems there are several steps to the calculation of color contrast:

• Convert from 0-255 values to 0-1 values:
  • R_{sRGB} = R_{8bit} / 255
  • G_{sRGB} = G_{8bit} / 255
  • B_{sRGB} = B_{8bit} / 255
• Adjust that conversion for some non-linearity:
  • If R_{sRGB} \leq 0.03928 then R = R_{sRGB} / 12.92, R=\left(\left(R_{sRGB} +0.055\right)/1.055\right)^{2.4} otherwise.
  • Same adjustment for G and B.
• Calculate the luminosity using this weighting of R, G and B:
  • L_1 = 0.2126\times R + 0.7152 \times G + 0.0722 \times B
• Do the same for the background (or adjacent object with with you want to check the contrast). Calculate the ratio: \displaystyle \frac{L_1 + 0.05}{L_2 + 0.05}. The ratio should be greater than 3 for type 14pt or larger if boldface, or 18pt or larger even if plain, greater than 4.5 otherwise.

As one recognizes from the really low coefficient blue gets (0.0722), blue is relatively “dark”, with just over a tenth the luminance of green. This inspires the question, “can blue be used on a black background? If so, what range of blues can?”

If we take [0,0,0] as black, and note that 0/255=0 and 0/12.92=0, we are interested in B satisfying

• \left(0.0722\times B + 0.05\right)/\left(0.05\right)\geq 3.0, or
• 0.0722\times B \geq 3.0\times 0.05 -0.05 = 2.0\times 0.05 = 0.1
• \Rightarrow B\geq \left(0.1\right)/0.0722 which requires B greater than 1: not possible. Blue needs a little “boosting” with red and green to be bright enough to show well on black. If we use equal amounts of red and green to achieve this, we need
• \left(0.0722\times B + 0.9278\times X +0.05\right)/\left(0.05\right) \geq 3.0
• \left(0.0722\times B + 0.9278\times X \right) \geq 0.15-0.05 \; =\; 0.1 which yields, for the brightest blue (1.0)
• 0.9278 \times X \geq 0.1 - 0.0722 = 0.0278 \Rightarrow X\geq 0.0278/0.9278\approx 0.03
• \left(\left(X_{sRGB}+0.055\right)/1.055\right)^2.4 \geq 0.03
• \Rightarrow \left(X_{sRGB}+0.055\right)/1.055 \geq 0.231=\sqrt[2.4]{0.03}
• X_{sRGB}\geq 0.231\times 1.055 -0.055 = 0.2437 - 0.055 = 0.1887
• 255 \times 0.1887 \approx 48 to we can use [48,48,255].
• Less blue will require more X, to be determined by analagous calculations. Boosting using only green and no red is left as an exercise.

How pale a blue can be used on a white background? White [255,255,255]/255 transforms to [1,1,1], and

• \left(1.055/1.055\right)^{2.4}=1
• (and 0.2126 + 0.7152 + 0.0722= 1.0000) so we are seeking values of B satisfying
• 1.05 \geq 3.0\times\left(0.0722\times B +0.05\right)= 0.2166\times B + 0.15 \Rightarrow 0.90\geq 0.2166\times B \Rightarrow 0.90/0.2166 \geq B
• which holds for all values of B from 0 to 1.0.