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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 epicycloid ： ウィキペディア英語版 epicycloid In geometry, an epicycloid is a plane curve produced by tracing the path of a chosen point of a circle — called an ''epicycle'' — which rolls without slipping around a fixed circle. It is a particular kind of roulette. If the smaller circle has radius ''r'', and the larger circle has radius ''R'' = ''kr'', then the parametric equations for the curve can be given by either: :$x\; (\backslash theta)\; =\; (R\; +\; r)\; \backslash cos\; \backslash theta\; -\; r\; \backslash cos\; \backslash left(\; \backslash frac\; \backslash theta\; \backslash right)$ :$y\; (\backslash theta)\; =\; (R\; +\; r)\; \backslash sin\; \backslash theta\; -\; r\; \backslash sin\; \backslash left(\; \backslash frac\; \backslash theta\; \backslash right),$ or: :$x\; (\backslash theta)\; =\; r\; (k\; +\; 1)\; \backslash cos\; \backslash theta\; -\; r\; \backslash cos\; \backslash left(\; (k\; +\; 1)\; \backslash theta\; \backslash right)\; \backslash ,$ :$y\; (\backslash theta)\; =\; r\; (k\; +\; 1)\; \backslash sin\; \backslash theta\; -\; r\; \backslash sin\; \backslash left(\; (k\; +\; 1)\; \backslash theta\; \backslash right).\; \backslash ,$ If ''k'' is an integer, then the curve is closed, and has ''k'' cusps (i.e., sharp corners, where the curve is not differentiable). If ''k'' is a rational number, say ''k=p/q'' expressed in simplest terms, then the curve has ''p'' cusps. If ''k'' is an irrational number, then the curve never closes, and forms a dense subset of the space between the larger circle and a circle of radius ''R'' + 2''r''. Image:Epicycloid-1.svg| ''k'' = 1 Image:Epicycloid-2.svg| ''k'' = 2 Image:Epicycloid-3.svg| ''k'' = 3 Image:Epicycloid-4.svg| ''k'' = 4 Image:Epicycloid-2-1.svg| ''k'' = 2.1 = 21/10 Image:Epicycloid-3-8.svg| ''k'' = 3.8 = 19/5 Image:Epicycloid-5-5.svg| ''k'' = 5.5 = 11/2 Image:Epicycloid-7-2.svg| ''k'' = 7.2 = 36/5 The epicycloid is a special kind of epitrochoid. An epicycle with one cusp is a cardioid. An epicycloid and its evolute are similar.〔(Epicycloid Evolute - from Wolfram MathWorld )〕 ==Proof== We assume that the position of $p$ is what we want to solve, $\backslash alpha$ is the radian from the tangential point to the moving point $p$, and $\backslash theta$ is the radian from the starting point to the tangential point. Since there is no sliding between the two cycles, then we have that :$\backslash ell\_R=\backslash ell\_r$ By the definition of radian (which is the rate arc over radius), then we have that :$\backslash ell\_R=\; \backslash theta\; R,\; \backslash ell\_r=\backslash alpha\; r$ From these two conditions, we get the identity :$\backslash theta\; R=\backslash alpha\; r$ By calculating, we get the relation between $\backslash alpha$ and $\backslash theta$, which is :$\backslash alpha\; =\backslash frac\; \backslash theta$ From the figure, we see the position of the point $p$ clearly. :$x=\backslash left(\; R+r\; \backslash right)\backslash cos\; \backslash theta\; -r\backslash cos\backslash left(\; \backslash theta+\backslash alpha\; \backslash right)\; =\backslash left(\; R+r\; \backslash right)\backslash cos\; \backslash theta\; -r\backslash cos\backslash left(\; \backslash frac\backslash theta\; \backslash right)$ :$y=\backslash left(\; R+r\; \backslash right)\backslash sin\; \backslash theta\; -r\backslash sin\backslash left(\; \backslash theta+\backslash alpha\; \backslash right)\; =\backslash left(\; R+r\; \backslash right)\backslash sin\; \backslash theta\; -r\backslash sin\backslash left(\; \backslash frac\backslash theta\; \backslash right)$ 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「epicycloid」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.