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 <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. > > 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.