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