POV-Ray : Newsgroups : povray.binaries.images : Fractal (100k) : Re: Fractal (100k) Server Time
4 Oct 2024 05:19:41 EDT (-0400)
  Re: Fractal (100k)  
From: Peter Pfahl
Date: 17 Apr 1999 12:43:09
Message: <3718ac0d.0@news.povray.org>
Thanks...but since I didn't do vector algebra in school yet (we'll start in
a few weeks :-( ), that doesn't help much...I'd really appreciate it if
you'd send me some kind of term that I can use in delphi directly. do i need
any information other that the height of the points above "0"?

Jerry Anning <nos### [at] despamcom> schrieb in im Newsbeitrag:
3717f4d6.44175699@news.povray.org...
> On Sat, 17 Apr 1999 02:08:20 +0200, Peter Pfahl <ppf### [at] rz-onlinede>
> wrote:
>
> >Yeah, I'll try that...have to think of a way to calculate the normals of
the
> >corners....
>
> Suppose that you want to calculate the normal at point a, which is
> connected to points b, c, d and e thus:
>
> b____________c
> |\                      / |
> |  \                  /   |
> |    \              /     |
> |      \          /       |
> |        \      /         |
> |          \  /           |
> |           a            |
> |          /  \           |
> |        /      \         |
> |      /          \       |
> |    /              \     |
> |  /                  \   |
> |/                      \ |
> d___________e
>
> Where X stands for the (vector) cross product, calculate:
> N = ((b-a) X (c-a)) + ((c-a) X (e-a)) + ((e-a) X (d-a)) + ((d-a) X
> (b-a))
> Then calculate the (scalar) length of N where n.x means the
> x-component of N, etc.:
> L = sqrt(N.x * N.x +N.y * N.y + N.z * N.z)
> Then normalize N (set it's length to 1) thus
> N = <N.x / L, N.y / L, N.Z / L>
> assuming, of course, that L != 0
> The cross product is a simple calculation that I am omitting for
> reasons of laziness.  You can find it in any vector algebra text or
> mail me and I'll send it.  Be sure to keep the direction of
> calculation around points (that is clockwise or counterclockwise)
> consistent.
>
> Jerry Anning
> clem "at" dhol "dot" com


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