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?