for later usage in a converter I'd like to exactly control the transparency of colors.

  • Let's say I have a simple layer (B), filled with #FF0000 and 50% layertransperency.
  • Now I duplicate that layer (C) (again with 50%).
  • In addition to that I have a layer for the background (A) which is filled with #FFFFFF.

I would have expected that the resulting pixel is a perfect #FF0000, but it is a #ff3f3f (still red, but much lighter).

I also tried to play with layer-styles and hoped that Linear Dodge (add) would help me out, but it doesn't.

Is there any simple trick, so that I am able to exactly control the transparency of every pixel (layer)?

  • Thank you Stan (: I just found out about this other SE-site and also flagged this question, so it can be moved by a moderator. I'm sorry for being OT.
    – Gundon
    Jan 12, 2013 at 22:02

1 Answer 1


Think of any non-opaque layer as if it were a filter through which you view the layers below. With only layers A and B visible, you're looking through Layer B at layer A, which has an effect similar to looking at Layer A through a filter that only passes half of the non-red light. (Provided, of course, that Layer A is white. The maths actually used only let this analogy go so far.) Turning on the visibility of Layer C means that you filter out half of the remaining non-red light. It's kind of like Zeno's Paradox—you can only ever get halfway there (although, since the numbers are integers, you'll eventually hit a practical 100% if you stack enough layers).

What you need, then, is a way to modulate the opacity of the filtering layer, and that's done with a layer mask (or alpha channel if you're working with an imported image). A mask fill at 50% grey with a layer opacity of 100% will give you the same effect as a layer with no mask and an opacity of 50%. The difference is that you can add lighter or darker areas to the mask (black blocks, white weeveals) to modulate the strength of the effect from 0 to 100%. Exactly how you'd go about that depends on what you're actually doing (which is not clear from the question as asked).

  • Thank you very much (: I would upvote if I had enough rep ;)
    – Gundon
    Jan 12, 2013 at 22:19

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