Date: Mar 1, 2013 7:58 AM
Subject: Re: Deformable platonic "solids"
On Feb 28, 8:44 pm, david petry <david_lawrence_pe...@yahoo.com>
> On Thursday, February 28, 2013 4:48:04 AM UTC-8, Richard Tobin wrote:
> > In article <email@example.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.
> 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.