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MathGroup Archive 2005

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Re: Normal Disappear Problem

  • To: mathgroup at smc.vnet.net
  • Subject: [mg55325] Re: Normal Disappear Problem
  • From: Roland Franzius <roland.franzius at uos.de>
  • Date: Sat, 19 Mar 2005 04:45:35 -0500 (EST)
  • Organization: Universitaet Hannover
  • References: <d1ecv2$evi$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

gouqizi.lvcha at gmail.com schrieb:
> 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.

Use normalized tangent vectors

X = {Cos[u]Sin[v],Sin[u]Sin[v],Cos[v]};

t1 = D[X, #] & /@ {u, v};

t2 = t1/Sqrt[Dot[#, #] & /@ t1] // TrigExpand // FullSimplify // 
PowerExpand;

t3 = Cross @@ t2

	{(-Cos[u])*Sin[v], (-Sin[u])*Sin[v], (-Cos[u]^2)*Cos[v] - Cos[v]*Sin[u]^2}

Dot[t3, t3] // FullSimplify
    1

The u-direction tangent vector t1[[1]] is 0 at the poles v=0,pi 
corresponding to the fact of no east direction there. Normalizing t 
makes the cross product normal unit vector constant even at the poles.

-- 

Roland Franzius


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