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When you make a colour partly transparent, it lightens (assuming it's on a white background), naturally. I would like to counter this behaviour by "darkening" a colour, so that when I make it transparent, when on a white background, the end result is the same as the original colour.

Basically, I would like to compensate the effect of transparency. And I have no idea of how I would do that. I've tried playing with some numbers in the colour's parameters, but I haven't been able to find what I should do.

This might be hard to understand in words alone, so I've made a visual explanation of what I would like to achieve.

enter image description here

What I want to achieve is the highest square.

Oh, before a clever one tells me to simply do what I did on that image, of course the example was faked by cutting a piece of transparent orange to put over the area where the "stick" overlaps with a solid-coloured square. Of course, that solution can't be applied to more irregular or changing shapes, and won't handle the white areas of a picture.

In case it's important, what I'm trying to make with this is a web design in Adobe Illustrator, which would of course eventually end up being made out of HTML and CSS.

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You can try the various blending modes if one of those gets close enough, but barring that, you'll need separate objects. You can't really alter blending on only a portion of a single object.

RGB CMYK

Of course in HTML/CSS there are no such things as blending modes so you'll have to either use transparency or not and probably resort to multiple objects there as well.

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  • css does have blend modes but browser support is limited. See html.adobe.com/webplatform/graphics/blendmodes/browser-support – Tims Oct 21 '13 at 9:28
  • Uhm, I'm not talking about altering blend on part of an object. That's just how I faked the visual example. What I want to do is to essentially counteract how transparency lightens my colour on a white background. Expressed a bit like algebra, it would be BaseColor-10%opacity+x+WhiteBackground = BaseColor, and the goal is to find x, and the goal is to make x not be opacity. For example, I did something KIND OF like what I wanted by substracting 10% from each R, G and B value, but that wasn't quite it. Saturation and lightness didn't do it either. – Ariane Oct 21 '13 at 12:17
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    Did you try shift-clicking with the Eyedropper Tool in the area which contains the color you want? Shift-clicking with the Eyedropper picks up the color on screen as opposed to the color of the object itself. – Scott Oct 21 '13 at 16:41
  • @ariene: Christopher John's answer hints at this a little, but doesn't make it explicit: he is using HSB color model and compensating the saturation by the amount of opacity (similar to your RGB algorithm) – horatio Oct 22 '13 at 17:23
  • @Scott Sorry for the long delay. I completely forgot about this question. Sorry, it looks like the eyedropper tool doesn't make the same color. So far, the only working method I see is to put a square of the full-opacity color behind the 90 % opacity one. – Ariane Oct 2 '14 at 22:14
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If you have an object that you make 90% opacity, you are introducing 10% of the color below that object. In theory, if you were to take your original object at 100% opacity (no transparency) and put the same background color in a layer in front of it at 10% opacity, you'd end up with the same 90%-10% ratio, resulting in the same color as the semi transparent original color.

If you take the result of the non-transparent original with the 10% above it, that is your "darkened" or "lightened" version of the original color.

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It can work with some blending modes sometimes...colorKeeperExample: Hard Light
...as you can see on the "Rainbow-Strip" it's not really flexible.
You do need them though. I doubt this works with just opacity.

edit: It does kind of work with opacity... colorKeeperExample: Opacity
...only I did it the other way around. Here apparently it works by upping the saturation of the 90% transparent picture by 10. Meaning if the original color already has 100% saturation it won't work, (also when using a different color BG, as John mentioned.)
That's what I found out anyways..

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  • Sorry for the delay. I completely forgot about this question. – Ariane Oct 2 '14 at 22:10

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