Article: 27208 of comp.graphics.rendering.renderman
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From: katsu@optgraph.com (Katsuaki Hiramitsu)
Newsgroups: comp.graphics.rendering.renderman
Subject: Re: A Possible addition to Advanced Renderman Errata
Date: 9 Mar 2004 21:09:56 -0800
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heniser@yahoo.com (Ryan Heniser) wrote in message news:<43c3327b.0403081137.7f3df12d@posting.google.com>...
> Shouldn't you normalize Sobel's first order differential
> operator with 8 instead of 6? Shouldn't the second order differential
> operator (the Laplacian) be normalized by 3?
#I'm sorry for my poor English.
When I read pg 472, I thought so, too.
As you say, the normalize value of first order differential operator
is 8, and the second order differential operator is 3
in Mr. Saito's paper.
But, I think that a normalize value must be determined with
a final-output-maximum value, which the author hopes.
(in other words, a normalize value is an arbitrary value.)
So, in the case that Mr. Johnston hopes 1.0
as the final-output-maximum value,
the first operator normalize value is 6.0, and the second is 8.0 .
The important thing is not normalize value but filter matrix, I think.
> Is this and error or am I over looking some something?
Thus, these are not errata, I beleive.
p.s -1
After divided a value with 8 or 3,
Mr.Saito normalized the value with the other value again in his paper.
p.s - 2
If the range of pixel value is [0.0 - 1.0],
it is apparent that the maximum value of '8*x-a-b-c-d-e-f-g-h' is 8.0 .
So, its normalize value is 8.0 .
Also, the maximum value of 'abs(a+2b+c-f-2g-h)+abs(c+2e+h-a-2d-f)' is 6.0;
I define the following,
A = a+2b+c-f-2g-h;
B = c+2e+h-a-2d-f;
X = 2b+2c-2d+2e-2f-2g;
So,
A + B = X;
Also,
|X|<=6.0;
Thus,
|A+B| <= 6.0;
Thus,
|A|+|B| <= 6.0;
So, its normalize value is 6.0 .
--katsu
Katsuaki Hiramitsu
katsu@optgraph.com