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Re: simplifying RotationMatrix
*To*: mathgroup at smc.vnet.net
*Subject*: [mg96366] Re: [mg96321] simplifying RotationMatrix
*From*: DrMajorBob <btreat1 at austin.rr.com>
*Date*: Thu, 12 Feb 2009 06:38:08 -0500 (EST)
*References*: <23667197.1234176058052.JavaMail.root@m02>
*Reply-to*: drmajorbob at longhorns.com
I decided a specific "mean" would do, and simplifications gave me the
following:
Needs["VectorAnalysis`"]
SetCoordinates[Spherical];
center = {1, 0, 0};
pt = {1, t, p};
xy = CoordinatesToCartesian /@ {center, pt};
rot = RotationMatrix@xy;
LeafCount@rot
1059
rot //. {Conjugate -> Identity, Abs[z_]^2 :> z^2} // Simplify
% // LeafCount
{{Cos[p]^2 (Cos[t] + Tan[p]^2), -Sin[2 p] Sin[t/2]^2,
Cos[p] Sin[t]}, {-Sin[2 p] Sin[t/2]^2, Cos[p]^2 + Cos[t] Sin[p]^2,
Sin[p] Sin[t]}, {-Cos[p] Sin[t], -Sin[p] Sin[t], Cos[t]}}
80
Bobby
On Wed, 11 Feb 2009 04:23:42 -0600, DrMajorBob <btreat1 at austin.rr.com>
wrote:
> If I calculate the rotation matrix from one vector to another in
> spherical
> coordinates, the result is HUGELY complicated:
>
> Needs["VectorAnalysis`"]
>
> SetCoordinates[Spherical];
> mean = {1, t0, p0};
> pt = {1, t, p};
> xy = CoordinatesToCartesian /@ {mean, pt};
> rot = RotationMatrix@xy;
> LeafCount@rot
>
> 32798
>
> I don't know what to tell Simplify about this, but it seem there are MANY
> unnecessary Conjugate mentions:
>
> Cases[rot, Conjugate[z_], Infinity] // Length
>
> 1575
>
> and MANY unnecessary cases like this, too:
>
> Cases[rot, Power[Abs[z_], 2], Infinity] // Length
>
> 1980
>
> Each of these could be simply z^2, I think.
>
> Trying to eliminate these with rules doesn't seem to help much, if any.
>
> The first thing I thought of was PowerExpand, but
>
> Map[PowerExpand, rot, {2}] // LeafCount
>
> 32798
>
> What am I missing? How can it be that complicated in the first place?
>
> Bobby
>
--
DrMajorBob at longhorns.com
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