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MathGroup Archive 2001

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Re: RotateShape[Cuboid...

  • To: mathgroup at smc.vnet.net
  • Subject: [mg27282] Re: RotateShape[Cuboid...
  • From: Jens-Peer Kuska <kuska at informatik.uni-leipzig.de>
  • Date: Fri, 16 Feb 2001 03:58:38 -0500 (EST)
  • Organization: Universitaet Leipzig
  • References: <96876j$3qr@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Hi,

have alook into the Graphics`Shapes`

source code


RotateShape[ shape_, phi_, theta_, psi_ ] :=
    Block[{rotmat = RotationMatrix3D[N[phi], N[theta], N[psi]]},
	shape /. { poly:Polygon[_] :> Map[(rotmat . #)&, poly, {2}],
		   line:Line[_]    :> Map[(rotmat . #)&, line, {2}],
		   point:Point[_]  :> Map[(rotmat . #)&, point,{1}] }
    ]

and you see ...
no rule for Cuboid[] and so it is not rotated at all.

Regards
   Jens


roelb wrote:
> 
> Hi,
> On Mathematica3 for mac, the following input produces 2 rods crossing
> with a 90° angle, as expected. That's OK.
> 
> Show[Graphics3D[Cylinder[.3,1,10] ],RotateShape[
>          Graphics3D[Cylinder[.3,1,10 ] ],
>                   0,Pi/2,0]]
> 
> But a quite similar RotateSphape input applied to Cuboids doesn't behave
> in the same manner. Why? (I expected the top Cudoid being oriented
> differently than the botton one).
> 
> Show[Graphics3D[Cuboid[{-1,-1,0},{1,1,1}]] ,
>   RotateShape[Graphics3D[Cuboid[{-1,-1,1},{1,1,2}]],
>                  0,1,0]]
> 
> The mathematica book state that cuboids faces are parallel to the axes.
> Should it means that RotateShape doesn't apply?


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