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翻訳と辞書　辞書検索 [ 開発暫定版 ] スポンサード リンク Smoothed octagon ： ウィキペディア英語版 Smoothed octagon The smoothed octagon is a region in the plane conjectured to have the ''lowest'' maximum packing density of the plane of all centrally symmetric convex shapes.〔K. Reinhardt, Über die dichteste gitterförmige Lagerung kongruenter bereiche in der ebene und eine besondere art konvexer Kurven, ''Abh. Math. Sem. Hamburg'' 10, 216-230 (1934).〕 It is constructed by replacing the corners of a regular octagon with a section of a hyperbola that is tangent to the two sides adjacent to the corner and asymptotic to the sides adjacent to these. The smoothed octagon has a maximum packing density given by :$\backslash frac\; \}\; \backslash approx\; 0.902414\; \backslash ,\; .$ This is lower than the maximum packing density of circles, which is :$\backslash frac\}\{5\; +\; 4\; \backslash sqrt\{2\}\}\; \backslash approx\; 0.906163,$ also slightly less than the maximum packing density of circles, but higher than that of the smoothed octagon.〔Steven Atkinson, Yang Jiao and Salvatore Torquato, Maximally dense packings of two-dimensional convex and concave noncircular particles, ''Phys. Rev. E'' 86 (2012), 031302. (Also available online ).〕 The smoothed octagon achieves its maximum packing density, not just for a single packing, but for a 1-parameter family. All of these are lattice packings.〔Yoav Kallus, Least efficient packing shapes, ''Geometry and Topology'', in press. (Also available online ).〕 In three dimensions, Ulam's packing conjecture states that no convex shape has a lower maximum packing density than the ball. ==Construction== By considering the family of maximally dense packings of the smoothed octagon, the requirement that the packing density remain the same as the point of contact between neighbouring octagons changes can be used to determine the shape of the corners. In the figure, three octagons rotate while the area of the triangle formed by their centres remains constant, keeping them packed together as closely as possible. For regular octagons, the red and blue shapes would overlap, so to enable the rotation to proceed the corners are clipped by a point that lies halfway between their centres, generating the required curve, which turns out to be a hyperbola. thumbllowing details apply to a regular octagon of circumradius $\backslash sqrt$ with its centre at the point $(2+\backslash sqrt,0)$ and one vertex at the point $(2,0)$. We define two constants, ''ℓ'' and ''m'': :$\backslash ell\; =\; \backslash sqrt\; -\; 1$ :$m\; =\; \backslash sqrt()\}$ The hyperbola is then given by the equation :$\backslash ell^2x^2-y^2=m^2$ or the equivalent parameterization (for the right-hand branch only): : The portion of the hyperbola that forms the corner is given by : The lines of the octagon tangent to the hyperbola are :$y=\; \backslash pm\; \backslash left(\backslash sqrt\; +\; 1\; \backslash right)\; \backslash left(\; x-2\; \backslash right)$ The lines asymptotic to the hyperbola are simply :$y\; =\; \backslash pm\; \backslash ell\; x.$ 抄文引用元・出典: フリー百科事典『 ウィキペディア（Wikipedia）』 ■ウィキペディアで「Smoothed octagon」の詳細全文を読む スポンサード リンク 翻訳と辞書 : 翻訳のためのインターネットリソース Copyright(C) kotoba.ne.jp 1997-2016. All Rights Reserved.