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On the division of space with minimum partitional area. (On the division of space with minimum partitional area.) (English) JFM 20.0523.01

Bekanntlich wird der in eine Seifenlösung eingetauchte Plateau’sche Drahtwürfel derart in Zellen geteilt, dass in jeder inneren Kante drei Flächen sich schneiden, dass jede dieser Flächen überall gleiche Krümmung (in dem von Thomson und Tait festgestellten Sinne) besitzt, und dass endlich stabiles Gleichgewicht zwischen allen Flächen vorhanden ist. Um nach demselben Gesetze den Raum mit congruenten Zellen auszufüllen, müssten dieselben die Gestalt von Rhombendodekaedern haben; doch würde in diesem Falle kein stabiles Gleichgewicht stattfinden. Der Verfasser findet nun zunächst durch geometrische und mechanische Ueberlegungen, dass die Zellen einer Raumteilung mit stabilem Gleichgewicht die Gestalt eines Vierzehnflachs haben müssen, wie es (abgesehen von der Krümmung einzelner Flächen) aus dem regelmässigen Oktaeder entsteht, wenn zwei Ecken schwach, die übrigen vier gleichmässig stark abgestumpft werden. dieser Körper ist demnach begrenzt von zwei kleinen und vier grossen ebenen Vierecken und acht nicht ebenen Sechsecken. (Die beiden an das innere Viereckdes Plateau’schen Würfels grenzenden Zellen sind Teile von ihm.) Der Verfasser stellt die beiden Differentialgleichungen auf, welche die Form dieser letzteren Flächen bestimmen, und löst dieselben durch Annäherung unter der empirischen Voraussetzung, dass die Flächen nur wenig von der ebenen Form abweichen. (Vgl. den Bericht in Bd. XIX. 1887. 520, JFM 19.0520.02).

Citations:

JFM 19.0520.02
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[1] By “curvature{” of a surface I mean sum of curvatures in mutually perpendicular normal sections at any point; notGauss’s “curvatura integra{”, which is the product of the curvature in the two “principal normal sections{”, or section of greatest and least curvature. (SeeThomson andTait’s Natural Phylosophy, part i. §§ 130, 136.)}}}
[2] The rhombic dodecahedron has six tetrahedral angles and eight trihedral angles. At each tetrahedral angle the plane faces cut one another successively at 120{\(\deg\)}, while each is perpendicular to the one remote from it; and the angle between successive edges is cos I/3 or 70{\(\deg\)}32’. The obtuse angles (109{\(\deg\)}28’) of the rhombs meet in the trihedral angles of the solid figure. The whole figure may be regarded as composed of six square pyramids, each with its alternate slant faces perpendicular to one another, placed on six squares forming the sides of a cube. The long diagonal of each rhombic face thus made up of two sides of pyramids conterminous in the short diagonal, is times the short diagonal.
[3] I see it inadvertently stated byPlateau that all the twelve films are “légère ment courbées{”.}
[4] To do for every point of meeting of twelve films what is done by blowing in the experiment of § 5.
[5] The corresponding two-dimensional problem is much more easily imagined; and may probably be realized by aid of moderately simple appliances. Between a level surface of soap-solution and a horizontal plate of glass fixed at a centimetre or two above it, imagine vertica film partitions to be placed along the sides of the squares indicated in the drawing (fig. 2): these will rest in stable equilibrium if thick enough wires are fixed vertically through the corners of the squares. Now draw away these wires downwards into the liquid: the equilibrium in the square formation becomes uustable, and the films instantly run into the hexagonal formation shown in the diagram;T/a per unit area on either of the pair of vertical walls which are perpendicular to the short sides of the hexagons; and on either of the other pair of walls 2 cos 30{\(\deg\)}XT/a; whereT denotes the pull of the film per unit breadth, anda the side of a square in the original formation. Hence the ratio of the pulls per unit of area in the two principal directions is as I to 1732.
[6] This figure (but with probably indefinite extents of the truncation) is given in books on mineralogy as representing a natural erystal of red oxide of copper.
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