Packing and isoperimetrics - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T06:33:26Z http://mathoverflow.net/feeds/question/101460 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/101460/packing-and-isoperimetrics Packing and isoperimetrics David Feldman 2012-07-06T04:52:19Z 2013-04-09T21:43:12Z <p>Suppose a manufacturer bottles small units of liquid and ships them via very large trucks.</p> <p>If the transportation cost nothing, spherical bottles would minimize the packaging cost (isoperimetric inequality); if packaging cost nothing, cubic bottles (say) would minimize the transportation cost, because they would pack on the trucks with no wasted space.</p> <p>How would the ideal container vary with the relative cost of packaging (measured as the surface area of one bottle) and transportation (measured by the packing density on an infinitely capacious truck)?</p> <p>[My "applied" formulation notwithstanding, I mean this as a pure mathematics question, so please idealize and ignore any distracting side issues.]</p> <p>A reasonable conjecture might seem that one gets the correct family of shapes by starting with a tight lattice packing of spheres and then over-inflating and rescaling the spheres until asymptotically they assume the form of Voronoi cells of a lattice packing. So one would get constant mean curvature surfaces up to the circular disks at the interfaces between adjacent spheres. Optimal or not, this would give lower bounds. But how to compute them?</p> http://mathoverflow.net/questions/101460/packing-and-isoperimetrics/101475#101475 Answer by Joseph O'Rourke for Packing and isoperimetrics Joseph O'Rourke 2012-07-06T10:11:20Z 2012-07-06T11:53:26Z <p>As an aside, there is a Swedish food packaging company, <em><a href="http://en.wikipedia.org/wiki/Tetra_Pak" rel="nofollow">Tetra Pak</a></em>, that was built on the idea of packaging milk in tetrahedral containers: <br />&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<img src="http://upload.wikimedia.org/wikipedia/commons/thumb/b/b1/Tetra_pak_2008.jpg/305px-Tetra_pak_2008.jpg" alt="TetraPak"> <br /> Practically speaking, there is another criterion that is important: the surface should unfold flat to a polygon that tiles the plane, so that they can be cut out easily. Of course the regular tetrahedron unfolds to an equilateral triangle, but also to a $2 \times 1$ parallelogram; I wouldn't be surprised if Tetra Pak used the latter.</p> http://mathoverflow.net/questions/101460/packing-and-isoperimetrics/101481#101481 Answer by jc for Packing and isoperimetrics jc 2012-07-06T11:16:04Z 2012-07-06T11:16:04Z <p>You probably already know this, but the problem of dividing 3-dimensional space into equal volumes with minimal interfacial area is called Kelvin's problem. It may be one limit of the problem you're considering.</p> <p>You can read about the currently best-known solution in 3D on Wikipedia <a href="http://en.wikipedia.org/wiki/Weaire%E2%80%93Phelan_structure" rel="nofollow">here</a> and there are links to pages with 3D models and data. The Weaire-Phelan structure uses two types of cells though, which I guess would be rather inconvenient from your applied perspective. Kelvin's original conjectured solution uses only one type of cell.</p> <p>In 2D the minimal perimeter solution is the honeycomb, as bees know and <a href="http://arxiv.org/abs/math/9906042" rel="nofollow">Thomas C. Hales proved</a>.</p> <p>While googling I found some <a href="http://www.icms.org.uk/downloads/soapbubble/Hales.pdf" rel="nofollow">talk slides by Hales on some recent work on the 3D problem</a> which I found interesting as well.</p> http://mathoverflow.net/questions/101460/packing-and-isoperimetrics/127005#127005 Answer by Yoav Kallus for Packing and isoperimetrics Yoav Kallus 2013-04-09T17:55:34Z 2013-04-09T21:43:12Z <p>László Fejes Tóth considered the planar version in his book <em>Regular Figures</em>. On page 175, he gives the following theorem, which verifies your guess (discs here does not necessarily mean circular discs):</p> <p>If P denotes the total perimeter of n convex discs, each of area a, lying in a convex hexagon of area H without mutual overlapping, then $$\frac{P}{H}\ge \sqrt{\frac{n}{H}} p(na/H)\text.$$</p> <p>The function $p(a')$ is defined as follows: it is equal to the perimeter of a circle of area $a'$ for $a'\le\pi/\sqrt{12}$ and equal to the perimeter of a rounded hexagon of area $a'$. A rounded hexagon being a regular hexagon of unit area, whose corners have been rounded to arcs of circles.</p> <p>If there was no restriction to convex discs, this would easily imply the honeycomb conjecture. Fejes Tóth writes "It may be assumed that this proposition remains valid without the restriction to convex faces. In the case of isoperimetric faces this conjecture turns out to be true, but for faces of equal area its proof seems to involve considerable difficulties". These difficulties were eventually overcome by Thomas Hales, as pointed out by jc.</p> <p>Regarding your last question "how to compute them", you can use <a href="http://www.susqu.edu/facstaff/b/brakke/evolver/evolver.html" rel="nofollow">Surface Evolver</a> to simulate a bubble confined to the interior of a rhombic dodecahedron. Here, for example, is the result of a simulation of a bubble confined to a cube:</p> <p><img src="http://yoavkallus.com/data/bubble2.png" alt="bubble in a cube"></p> <p>If you're interested, I can provide the input file I used to do the simulation.</p>