In the case where the compression struts are incompressible, we get a generalized version of the Law of Sines when the springs & struts are in equilibrium: Label our triangle with vertices A,B,C in the usual way, so that a,b,c are opposite A,B,C, respectively. The edges a,b,c carry the tension with springs of zero rest length and spring constants Ka, Kb, Kc, respectively. Then Ka*sin(A)/a = Kb*sin(B)/b = Kc*sin(C)/c. The 'force diagram' is dual to the spring diagram, so we have sin(A)/FA = sin(B)/FB = sin(C)/FC, where FA, FB, FC are forces on the struts OA, OB, OC respectively. We conclude that the 'effective length' of OA, OB, OC is a/Ka, b/Kb, c/Kc, i.e., we divide the length of a by its opposite spring constant, etc. Now to solve the general problem with all 6 struts having spring constants. At 01:01 PM 7/15/2013, Henry Baker wrote:
Ok, so I guess the overall form of the analytic solution will be something like:
PEmin = min(PE_A, PE_B, PE_C, PE_D),
where PE_i = potential energy when vertex i (i=A,B,C,D) is pushed into the middle (where we now have a _planar_ problem, which should be much easier to solve).
In each planar problem, the 'outer' 3 edges are under tension, while the 'inner' 3 edges are under compression.
Make the central vertex the origin.
I guess that the whole thing is linear (at least in terms of displacements), so we should be able to solve the projections onto the X and Y coordinates separately, and then put the results back together again.
At 12:25 PM 7/15/2013, meekerdb wrote:
On 7/15/2013 10:26 AM, Henry Baker wrote:
I've been playing with a mathematical model of physical tetrahedrons:
The four vertices are joined by edges which are Hookean springs with non-zero rest lengths.
The vertices are 'pins'; i.e., the edges can move freely into different angles -- i.e., the vertices exert no forces other than axial forces along the edges.
Each of the springs can have a different rest length and a different spring constant (non-regular tetrahedra).
Define 'failure' by a tetrahedron turning itself inside out.
Interestingly, Buckminster Fuller played with this problem of inverting tetrahedrons; he called the process of turning inside-out 'dimpling'.
Also, modern computer graphics uses such models for 'deformable' surfaces -- e.g., human skin. However, computer graphics goes to great lengths to make sure that the tetrahedra never turn themselves inside out (never 'dimple').
So the major question is: how much energy is required to invert/'fail' the tetrahedron. Notice that you have 4 choices over which vertex to 'push through'.
I want to find the minimum 'failure energy' in closed form.
Does it even matter which vertex is pushed through?
The maximum potential energy state will be when the tetrahedron is flattened. So it must make a difference which vertex is pushed through; if the tetrahedron is already flat or nearly flat it will take very little energy to push through the vertex that is near the plane of the other three.