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

[Benoit Bezaire <benoit@itedo.com>]

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

This e-mail and any attachments are confidential and may be protected
by legal privilege. If you are not the intended recipient, be aware
that any disclosure, copying, distribution or use of this e-mail or
any attachment is prohibited. If you have received this e-mail in
error, please notify us immediately by returning it to the sender and
delete this copy from your system. Thank you for your cooperation. 


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.