|
|
Re: Deformable platonic "solids"
Posted:
Feb 28, 2013 1:44 PM
|
|
On Thursday, February 28, 2013 4:48:04 AM UTC-8, Richard Tobin wrote:
> In article <511e7643-79fb-4418-9108-16b317c87dff@googlegroups.com>,
> david petry <david_lawrence_petry@yahoo.com> wrote:
> >Any line segment joining two points reduces the
> >total number of degrees of freedom of the system of points and line
> >segments by one.
> Not in general. Consider a deformable solid with a square face.
> Joining two opposite corners will make that square rigid. Joining
> the other two will have no further effect, while adding a line
> somewhere else in the solid may.
Yes, of course, if the degrees of freedom are already minimal, they can't be reduced further. But the answer I gave does show us how to answer the original question.
"THEOREM" If a platonic solid has V vertices and E edges, then it will be rigid in the sense of Frederick Williams if and only if 3V - E = 6.
Examples
Tetrahedron: V = 4, E = 6, 3V-E = 6 (rigid)
Octahedron: V = 6, E = 12, 3V-E = 6 (rigid)
Cube: V = 8, E = 12, 3V-E = 12 (not rigid)
Dodecahedron: V = 20, E = 30 3V-E = 30 (not rigid)
Icosahedron: V = 12, E = 30, 3V-E = 6 (rigid)
|
|
