Article: 27208 of comp.graphics.rendering.renderman Path: uni-berlin.de!fu-berlin.de!postnews1.google.com!not-for-mail 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 Organization: http://groups.google.com Lines: 60 Message-ID: References: <43c3327b.0403081137.7f3df12d@posting.google.com> NNTP-Posting-Host: 220.107.220.94 Content-Type: text/plain; charset=ISO-8859-1 Content-Transfer-Encoding: 8bit X-Trace: posting.google.com 1078895397 1303 127.0.0.1 (10 Mar 2004 05:09:57 GMT) X-Complaints-To: groups-abuse@google.com NNTP-Posting-Date: Wed, 10 Mar 2004 05:09:57 +0000 (UTC) Xref: uni-berlin.de comp.graphics.rendering.renderman:27208 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