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Re: Normal Disappear Problem
*To*: mathgroup at smc.vnet.net
*Subject*: [mg55341] Re: [mg55302] Normal Disappear Problem
*From*: DrBob <drbob at bigfoot.com>
*Date*: Sat, 19 Mar 2005 04:46:45 -0500 (EST)
*References*: <200503181035.FAA14770@smc.vnet.net> <opsnuhpjmfiz9bcq@monster.ma.dl.cox.net>
*Reply-to*: drbob at bigfoot.com
*Sender*: owner-wri-mathgroup at wolfram.com
Oops, disregard this part of my previous post:
>> 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}
I did the normalization right farther on.
Bobby
On Fri, 18 Mar 2005 12:01:57 -0600, DrBob <drbob at bigfoot.com> wrote:
> 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
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