RE: Re[3]: [cgmo-webcgm] Drawing model (take 2)

[Dieter Weidenbrück <dieter@itedo.com>]

I have no problem to leave the examples out, now
that we all know what was meant.

Thanks Benoit, good stuff!

Dieter 

> -----Original Message-----
> From: Benoit Bezaire [mailto:benoit@itedo.com] 
> Sent: Thursday, April 20, 2006 8:33 PM
> To: cgmo-webcgm@lists.oasis-open.org
> Subject: Re[3]: [cgmo-webcgm] Drawing model (take 2)
> 
> Hi Lofton,
> 
> My personal guess... It's because of the Porter-Duff paper:
> 
> Compositing Digital Images, Thomas Porter, Tom Duff, Computer 
> Graphics Project, Lucasfilm Ltd.
> 
> Computer Graphics Volume 18, Number 3 July 1984
> 
> That paper had a lot of influence in the computer graphics industry.
> In the paper, Porter and Duff, introduce the concept of 
> pre-multiplied RGBA quadruple (mainly to save unnecessary 
> mulitiplications).
> 
> Implementations that support the Porter-Duff operators often 
> refer to premultiplied color values.
> 
> Yes,  Esc.45 is not premultiplied.
> 
> Note, if it were up to me, I wouldn't put any examples at all 
> (it's too much).
> 
> -- 
>  Benoit   mailto:benoit@itedo.com
> 
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> 
> Thursday, April 20, 2006, 11:30:53 AM, you wrote:
> 
> > 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.
> 
> 
> 
>