[math-fun] Calculus of variations problem: the one-wire cage.
Calculus of variations problem: the one-wire cage. Allan Wechsler: Which three-dimensional continuous differentiable curve of unit arc-length encloses the greatest volume in its convex hull? --WDS: This problem actually can be posed in any number of dimensions. In 2D as Salamin said the answer (Dido problem) is a semicircle-arc. The problem also can be posed for closed curves. In 2D for closed curve answer is circle (isoperimetric problem). The problem can also be posed for arbitrary connected "networks" of curve segments (such as a tree) having total length=1. Tilli (cited below) claims answer in 2D again is a semicircle arc. TRY #1: On a spherical earth, go up the greenwich meridian from S pole to N pole (semicircle). Then go down another meridian at longitude=120 back to S pole. Then up another meridian at longitude=240 back to N pole. Smooth the corners by epsilon since Wechler wants differentiable curve. The curve I have described if earth-radius=1/(3*pI), has length=1. Its convex hull is somewhat less than the whole volume of the earth, which would have been Vearth=4/(81*pi^2) = 0.0050035. In fact, I think the volume ratio is the same as the area ratio of an equilateral triangle to its circumcircle, namely sqrt(27) / (4*pi) = 0.413497. If so, then the convex hull volume is Vconvhull = sqrt(3) / (27*pi^3) = 0.0020689. But one can do better by chopping off some length on each of the two ends of this curve (then rescale back to unit total length), it is not clear what optimizes that, and also it seems doubtful the result is optimal. TRY #2a and 2b: Go round a circle in XY plane centered at origin, then line segment up to a point (0,0,Z). Optionally(b) also could put a line segment on the other end to (0,0,-Z). The convex hull is a cone (or intersection of two opposite-facing cones). If radius of circle is R, then curve length is 2*Pi*R + sqrt(R*R+Z*Z) = 1 or with the option(b) taken 2*Pi*R + 2*sqrt(R*R+Z*Z) = 1. The volume is Pi*R*R*Z/3 or if option(b) taken then multiply that by 2. Optimize to find: 2a: R=0.1015659929, Z=0.3472952850, Vcone=0.003751666 2b: R=0.09290346185, Z=0.1862503544, Vdoubcone=0.003366817 [there are actually exact formulas for all those which I omit] 2a is however not optimal because bowing the line segment outward a bit will be superior. TRY #3: segment of helix. Melzak quoting Egervary 1949 says this is best, and the answer is 1 turn of the Helix at pitch 1/sqrt(2). That is (up to scaling) x(t)=cos(t), y(t)=sin(t), z(t)=t/sqrt(2), 0<t<2*pi. Alleged resulting volume is Vhelix = 1/(18*Pi*sqrt(3)) = 0.010209794. TRY #4: Two semicircular arcs hinged at the equator. But in this case it would be better to smooth the corners. TRY #5: For a closed curve in 4D, use a circumf=1/2 circle in XY plane and another in ZT plane. Schoenberg says this is best (and the analogous construction is best for closed curve in every even dimension for 6D you'd use 3 circles of circumf=1/3 each in 3 orthogonal planes; this is a nice generalization of isoperimetric theorem) TRY #6: for closed curve in 3D, Melzak claims the optimum under certain restrictive assumptions is the "Baggins constant" which he evaluates as Vbaggins=0.0031816877. He thinks this has no simple closed form expression. There is 2010 paper by Paolo Tilli on the network problem version: Isoperimetric inequalities for convex hulls and related questions, Trans. Amer. Math. Soc. 362 (2010) 4497-4509 http://www.jointmathematicsmeetings.org/journals/tran/2010-362-09/S0002-9947... Tilli says it was posed in a 1991 book "unsolved problems in geometry." Other papers: I.J. Schoenberg: An isoperimetric inequality for closed curves convex in even-dimensional Euclidean spaces, Acta Math. 91 (1954) 143-164. A.A. Nudel’man: Isoperimetric problems for the convex hulls of polygonal lines and curves in multidimensional spaces, (Russian) Mat. Sbornik (N.S.) 96,138 (1975) 294-313. Michele Gori: On a maximization problem for the convex hull of connected systems of segments, J. of Convex Analysis 14,1 (2007) 49-68. Z.A. Melzak: The isoperimetric problem of the convex hull of a closed space curve, Proc. Amer. Math. Soc. 11 (1960) 265-274. Z.A. Melzak:Numerical evaluation of an isoperimetric constant, Maths of Comput. 22 (1968) 188-190 http://www.ams.org/journals/mcom/1968-22-101/S0025-5718-1968-0223976-4/ Z.A. Melzak: The isoperimetric problem of the convex hull of a closed space curve, Proc. Amer. Math. Soc. 11 (1960) 265-274 http://www.ams.org/journals/proc/1960-011-02/S0002-9939-1960-0116263-0/ E. Egervary: On the smallest convex cover of a simple arc of space-curve, Publ. Math. Debrecen 1 (1949) 65-70.
TRY #4: Two semicircular arcs hinged at the equator. But in this case it would be better to smooth the corners --sorry, I forgot to include the numbers in this try: is the hinge angle is 90 then the volume one gets is Vtwoarcs=1/(12*pi^3) = 0.0026876. This is as expected, not quite as good as "Baggins constant" Vbaggins=0.0031816877. ---- 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. I do not know if that is optimal.
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.
With a good convex-hull algorithm, this would make a nice exercise in genetic programming: create a bunch of random polyline critters and let the ones with the biggest convex hulls reproduce and mutate. On Fri, Aug 30, 2013 at 4:04 PM, Dan Asimov <dasimov@earthlink.net> wrote:
I'd certainly think that would have the maximum for a 4-segment closed polygon: 4 consecutive edges of a regular tetrahedron that returns to where it started.
--Dan
On 2013-08-30, at 12:45 PM, Cris Moore wrote:
the path on the edges of a tetrahedron I mentioned?
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Warren D Smith