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- Newsgroups: sci.math
- Subject: Re: Properties of Illumination Functions?
- Message-ID: <a_rubin.724617077@dn66>
- From: a_rubin@dsg4.dse.beckman.com (Arthur Rubin)
- Date: 17 Dec 92 18:31:17 GMT
- References: <1992Dec16.225253.25173@jarvis.csri.toronto.edu>
- Organization: Beckman Instruments, Inc.
- Nntp-Posting-Host: dn66.dse.beckman.com
- Lines: 38
-
- In <1992Dec16.225253.25173@jarvis.csri.toronto.edu> dret@dgp.toronto.edu (George Drettakis) writes:
-
- >I sent this to sci.math.research, but I have a suspicion the big
- >black net hole ate it. I apologize if you see this twice.
- >-----------------------
-
- >I have the following problem in computer graphics. Assume a rectangular
- >area in the plane z=d, with -xa1 < x1 < xa1 and -ya1 < y1 < ya1.
- >This area is assumed to be a diffuse rectangular light source,
- >with the surface normal 0, 0, -1.
- >Assume also the receiving plane z=0, and the function I(x,y) defined on
- >this plane. The function I(x,y) expresses the illumination at this point:
-
- > ya1 xa1
- > / /
- > | | 1
- > I(x, y) := | | ----------------------------- dx1 dy1
- > | | 2 2 2 2
- > / / ((x - x1) + (y - y1) + d )
- > - ya1 - xa1
-
-
- ...
-
- >What I want to show is that the function I(x,y) has only one maximum, and is
- >(radially around 0, 0, 0 ) decreasing everywhere else.
- >Ideally I want to show this for more general sources, but just the rectangle
- >is a good start.
-
- It is easy to show that I(x,y) has only one relative maximum; just
- differentiated under the integral and estimate. I don't think it's as easy
- to show it's radially decreasing elsewhere, but that would be a good start.
-
- --
- Arthur L. Rubin: a_rubin@dsg4.dse.beckman.com (work) Beckman Instruments/Brea
- 216-5888@mcimail.com 70707.453@compuserve.com arthur@pnet01.cts.com (personal)
- My opinions are my own, and do not represent those of my employer.
- My interaction with our news system is unstable; please mail anything important.
-