Re: Normal Disappear Problem
- To: mathgroup at smc.vnet.net
- Subject: [mg55336] Re: Normal Disappear Problem
- From: "Steve Luttrell" <steve_usenet at _removemefirst_luttrell.org.uk>
- Date: Sat, 19 Mar 2005 04:46:22 -0500 (EST)
- References: <d1ecv2$evi$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
You need to make sure your normal is scaled so that it has unit length.
{x,y,z}={Cos[u]Sin[v],Sin[u]Sin[v],Cos[v]};
norm=Cross[D[{x,y,z},u],D[{x,y,z},v]]//Simplify
{(-Cos[u])*Sin[v]^2, (-Sin[u])*Sin[v]^2, (-Cos[v])*Sin[v]}
normlength=Sqrt[Apply[Plus,norm^2]//Simplify]//PowerExpand
Sin[v]
norm/normlength
{(-Cos[u])*Sin[v], (-Sin[u])*Sin[v], -Cos[v]}
Steve Luttrell
<gouqizi.lvcha at gmail.com> wrote in message news:d1ecv2$evi$1 at smc.vnet.net...
> Hi, All:
>
> I have the following parametric equation for an unit sphere:
>
> x = cos(u)sin(v)
> y = sin(u)sin(v)
> z = cos(v)
>
> 0<=u<2*Pi ; 0<=v<=Pi
>
> Then I use
>
> normal = (Dx/Du, Dy/Du, Dz/Du) CROSS (Dx/Dv, Dy/Dv, Dz/Dv) to get the
> normal vector.
>
> I get the follwoing after calculation (with normalization):
>
> normal = [sin(v) ^2 cos(u), sin(v)^2 sin(u), cos(u)^2 cos(v) sin(v)
> + sin(u)^2 cos(v) sin(v)]
>
> Now when u=0, v=0 , Normal = (0,0,0)! How can it be? We know the fact
> that a sphere should have normal everywhere.
>
> Rick
>