Date: Mar 1, 2013 7:58 AM
Author: dan.ms.chaos@gmail.com
Subject: Re: Deformable platonic "solids"
On Feb 28, 8:44 pm, david petry <david_lawrence_pe...@yahoo.com>

wrote:

> 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_pe...@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)

This is true , but the situation is more subtle than that . If you

have a 'portion of the solid' that is rigid , adding further

connections 'withing that portion' will not reduce the global number

of degrees of freedom .

Consider a cube connected by a bar to an octahedron . If we add an

edge connecting two opposite vertices of the octahedron , the whole

system will have the same number of degrees of freedom . In order to

calculate the correct number of degrees of freedom we need to

eliminate 'redundant edges' . But it helps that the Platonic solids

have a high degree of symmetry , and appear to have no 'redundant

edges' (edges who's presence of absence does not affect the number of

degrees of freedom , or corresponds to less degrees of freedom than

expected ) . Therefore, your deduction is correct.