Re: Normal Disappear Problem

*To*: mathgroup at smc.vnet.net*Subject*: [mg55340] Re: [mg55302] Normal Disappear Problem*From*: DrBob <drbob at bigfoot.com>*Date*: Sat, 19 Mar 2005 04:46:41 -0500 (EST)*References*: <200503181035.FAA14770@smc.vnet.net>*Reply-to*: drbob at bigfoot.com*Sender*: owner-wri-mathgroup at wolfram.com

Of course the sphere has a normal everywhere, but that doesn't mean a given method of computing the normal WORKS everywhere. ClearAll[normVec] normVec[f_][u__] := Block[{vars = Table[Unique["x"], {Length@{u}}], result}, result = Cross @@ Transpose[D[f @@ vars, {vars}]] /. Thread[vars -> {u}]; Remove @@ vars; result] f[u_, v_] = {Cos[u]Sin[v], Sin[u]Sin[v], Cos[v]}; normVec[f][u, v] {(-Cos[u])*Sin[v]^2, (-Sin[u])*Sin[v]^2, (-Cos[u]^2)*Cos[v]*Sin[v] - Cos[v]*Sin[u]^2*Sin[v]} (I haven't normalized, but neither did you.) If you substitute u = v = 0, that's certainly the zero vector: normVec[f][0,0] {0,0,0} Now consider the vectors used in the cross product: showCross[f_][u__] := Block[{vars = Table[Unique["x"], { Length@{u}}], result}, result = cross @@ Transpose[D[f @@ vars, {vars}]] /. Thread[ vars -> {u}]; Remove @@ vars; result] showCross[f][0,0] cross[{0,0,0},{1,0,0}] Clearly, that cross product is zero. But why is the first vector all zeroes? Well, it's simply because: D[f[u, v], {u}] % /. v -> 0 {-Sin[u] Sin[v],Cos[u] Sin[v],0} {0,0,0} And that's for ALL values of u, so: normVec[f][u,0] {0,0,0} That doesn't look good, but think about it. You started with a "parameterization" of the unit sphere. But it's not a one-to-one mapping: f[u,0] {0,0,1} If the parameterization f is no good at a point, the normal can't be computed using f at that point. Normalizing reveals the problem in a different way: #/#.# &@normVec[f][u, v] // TrigExpand % /. v -> 0 {-Cos[u],-Sin[u],-Cot[v]} {-Cos[u],-Sin[u],ComplexInfinity} Everything fails when v = 0. So.... What's the simpler, better method? Normalizing actually helps: #.# &@normVec[f][u, v] // TrigExpand // FullSimplify Sin[v]^2 normVec[f][u, v]/Sin[v] // TrigExpand {-Cos[u] Sin[v], -Sin[u] Sin[v], -Cos[v]} We'd probably reverse the sign, and get f[u,v]==-% True In other words, the (outward-pointing) normal at f[u,v] is just f[u,v]. Bobby On Fri, 18 Mar 2005 05:35:18 -0500 (EST), gouqizi.lvcha at gmail.com <gouqizi.lvcha at gmail.com> wrote: > 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 > > > > -- DrBob at bigfoot.com

**References**:**Normal Disappear Problem***From:*"gouqizi.lvcha@gmail.com" <gouqizi.lvcha@gmail.com>