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*Subject*: **Re[2]: [cgmo-webcgm] Drawing model (take 2)**

*From*:**Lofton Henderson <lofton@rockynet.com>***To*: Benoit Bezaire <benoit@itedo.com>*Date*: Thu, 20 Apr 2006 09:30:53 -0600

At 10:51 AM 4/20/2006 -0400, Benoit Bezaire wrote: >BTW, I forgot to mention (again), that the examples are using a 'argb' >notation instead of 'rgba'. Right, I'm fine with that. >Any way, it doesn't seem to be the source of your question. As I said >in this version of the wording "all color values use premultiplied >alpha". Okay, I missed the implication of that in the equations. So "all color values" -- does that mean that Cr,Cg,Cb are also premultiplied by Ca, before plugging into the equations? (None of the examples answer that question.) Btw, just out of curiosity... I have come across "premultiplied alpha" before. Why is that a popular way to express the equations? From a mathematical and intuitive perspective, non-premultiplied Pr,Pg,Pb is clearer, at least to me: Cr' = (1 - Pa)*Cr + Pa*Pr It looks like a linear interpolant, of the Cr and the Pr at the an intermediate point defined by the alpha. (You can probably tell that my graphics history does not include writing a rasterizer!). Finally, Esc.45 says, result = alpha*foreground + (1-alpha)*background So the expression in Esc45 is NOT pre-multiplied, correct? -Lofton. >So fully transparent red is not (0,1,0,0) it is (0,0,0,0). That should >work.

**Follow-Ups**:**Re[3]: [cgmo-webcgm] Drawing model (take 2)***From:*Benoit Bezaire <benoit@itedo.com>

**References**:**Re: [cgmo-webcgm] Drawing model (take 2)***From:*Lofton Henderson <lofton@rockynet.com>

**Re[2]: [cgmo-webcgm] Drawing model (take 2)***From:*Benoit Bezaire <benoit@itedo.com>

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