nice. isn't this the path on the edges of a tetrahedron I mentioned? On Aug 30, 2013, at 1:38 PM, Warren D Smith <warren.wds@gmail.com> wrote:
It occurs to me a more tractable version of the problem(s) would be, if the curve is required to be polygonal with E edges. In that case you should be able to find optimum and prove it. Example, in 3D with 3 line segments, all at mutual right angles, all with length=1/3, we would get V=1/162=0.0061728 and I'm pretty sure this is optimal for a 3-edge polygonal curve.
In 3D closed curve using 4 line segments, all length=1/4, formed by folding a square along diagonal at 90 degrees hinge angle, V=1/(384*sqrt(2))=0.0018414239 but better is to start with a 60-120-60-120 rhombus not a square, then V=1/512=0.001953125. But wait, the best of all such rhombi is one for which sharp angle is theta=arctan(1/sqrt(2))=70.528degrees which amazingly enough is the carbon bond angle, and then Vfoldedcarbonrhomb=sin(theta)*cos(theta/2)/384=0.002004688434.