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? --- I assume that this problem has been investigated by people other than architects & computer graphics folks.