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Re: Deformable platonic "solids"
Posted:
Mar 1, 2013 9:28 AM
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On 03/01/2013 07:58 AM, Dan wrote:
> 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.
>
For the dodecahedron, we have 12 faces and 20 vertices.
If we represent the Cartesian coordinates of the
i'th vertex by (x_i, y_i, z_i), i = 1 ... 20
then the 30 edges give 30 algebraic equations
for 60 real variables, such as:
(x_1 - x_2)^2 + (y_1 - y_2)^2 + (z_1 - z_2)^2 - 1 = 0.
Without loss of generality,
x_1 = y_1 = z_1 = 0,
so 57 real variables remain.
We might as well fix an edge in space, so
x_2 = 1, y_2 = z_2 = 0, and
so 54 real variables remain.
A second edge ending at the origin can be assumed
to remain in the x-y plane, with its z-coordinate
fixed at zero:
(x_1 - x_3)^2 + (y_1 - y_3)^2 + (z_1 - z_3)^2 - 1 = 0 or
x_3^2 + y_3^2 + z_3^2 - 1 = 0
and
z_3 = 0.
So 53 real variables remain:
x_3, y_3 and the x_i, y_i and z_i for 4<= i <= 20
3*17 + 2 = 51 + 2 = 53 real variables and
29 distance equations. (one equation is removed due
to fixing the vertices of an edge in space).
53 real variables, 29 algebraic equations.
It doesn't look easy ...
http://en.wikipedia.org/wiki/Dodecahedron
David Bernier
--
dracut:/# lvm vgcfgrestore
File descriptor 9 (/.console_lock) leaked on lvm invocation. Parent PID
993: sh
Please specify a *single* volume group to restore.
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