In CSS, when you apply
filter: invert(1) hue-rotate(180deg)
to an image, the color red turns into peachy-pink.
Why is this, and what can be done to use CSS to invert an image and still have red look like red?
Example:
The same image with filter: invert(1) hue-rotate(180deg) applied:
Update:
It was recommended in the initial answer to apply the above filter to the html element and then also apply it the image. The colors still look off. Here is the result:
To understand the why we need to perform some math.
If we start with the invert(1) (the easiest one) we will use f(x) = (255 - x)ref. So
rgb(255,0,0) will become rgb(0,255,255)
For the hue-rotate(180deg) it's more complex. Considering the specification we will get the following matrix:
-0.574 1.43 0.144
0.426 0.43 0.144
0.426 1.43 -0.856
So we will have
R' = -0.574*R 1.43*G 0.144*B = 1.43*255 + 0.144*255 = 401.37
G' = 0.426*R 0.43*G 0.144*B = 0.43*255 + 0.144*255 = 146.37
B' = 0.426*R 1.43*G -0.856*B = 1.43*255 - 0.856*255 = 146.37
Then a final color rgb(255,146.37,146.37) which is not a red one
html {
background: rgb(255, 146.37, 146.37)
}what can be done to use CSS to invert an image and still have red look like red?
It depends on what result you want to get considering the other colors but you can add a staturate() filter to get back your red color:
img {
filter: invert(1) hue-rotate(180deg) saturate(10);
}<img src="https://i.sstatic.net/jikdT.png">Again some Math to understand what is happening. From the same specification and after some simplication we will have the following matrix:
7.87 -7.15 -0.72
-2.13 2.85 -0.72
-2.13 -7.15 9.28
So
R' = 7.87*R -7.15*G -0.72*B = 7.87*255 - 7.87*146.37 = bigger than 255
G' = -2.13*R 2.85*G -0.72*B = -2.13*255 + 2.13*146.37 = negative
B' = -2.13*R -7.15*G 9.28*B = -2.13*255 + 2.13*146.37 = negative
A final color rgb(255,0,0)
