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三角形を見込む立体角計算ノート

三角形を見込む立体角計算ノート（書きかけ）

直角球面三角形の面積

  • \begin{eqnarray*} \Omega &=& \arcsin\frac{\sin\theta}{\sqrt{1-\cos^2\theta\cos^2\varphi}}-\arcsin\frac{\sin\theta\cos\varphi}{\sqrt{1-\cos^2\theta\cos^2\varphi}}\\ &=& \arcsin\left(\frac{\sin\theta}{\sqrt{1-\cos^2\theta\cos^2\varphi}}\sqrt{1-\frac{\sin^2\theta\cos^2\varphi}{1-\cos^2\theta\cos^2\varphi}}-\frac{\sin\theta\cos\varphi}{\sqrt{1-\cos^2\theta\cos^2\varphi}}\sqrt{1-\frac{\sin^2\theta}{1-\cos^2\theta\cos^2\varphi}}\right)\\ &=& \arcsin\left(\frac{\sin\theta}{1-\cos^2\theta\cos^2\varphi}\left(\sqrt{1-\cos^2\theta\cos^2\varphi-\sin^2\theta\cos^2\varphi}-\cos\varphi\sqrt{1-\cos^2\theta\cos^2\varphi-\sin^2\theta}\right)\right)\\ &=& \arcsin\left(\frac{\sin\theta}{1-\cos^2\theta\cos^2\varphi}\left(\sqrt{1-\cos^2\varphi(\cos^2\theta+\sin^2\theta)}-\cos\varphi\sqrt{\cos^2\theta-\cos^2\theta\cos^2\varphi}\right)\right)\\ &=& \arcsin\left(\frac{\sin\theta}{1-\cos^2\theta\cos^2\varphi}\left(\sin\varphi-\cos\varphi\cos\theta\sin\varphi\right)\right)\\ &=& \arcsin\frac{\sin\theta\sin\varphi}{1+\cos\theta\cos\varphi} \end{eqnarray*}

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  • となる。ただし

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    • \begin{eqnarray*} \arcsin u \pm \arcsin v =\arcsin \left(u^2\sqrt{1-v^2}\pm u^2\sqrt{1-v^2}\right) \end{eqnarray*}

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  • を用いた。さらに

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    • \begin{eqnarray*} \arcsin x =\arctan\frac{x^2}{\sqrt{1-x^2}} \end{eqnarray*}

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  • を用いて変形すると

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    • \begin{eqnarray*} \Omega &=& \arcsin\frac{\sin\theta\sin\varphi}{1+\cos\theta\cos\varphi}\\ &=& \arctan\frac{\frac{\sin\theta\sin\varphi}{1+\cos\theta\cos\varphi}}{\sqrt{1-\left(\frac{\sin\theta\sin\phi}{1+\cos\theta\cos\varphi}\right)^2}}\\ &=& \arctan\frac{\sin\theta\sin\varphi}{\sqrt{\left(1+\cos\theta\cos\varphi\right)^2-\sin^2\theta\sin^2\phi}}\\ &=& \arctan\frac{\sin\theta\sin\varphi}{\sqrt{1+2\cos\theta\cos\varphi+\cos^2\theta\cos^2\varphi-(1-\cos^2\theta)(1-\cos^2\varphi)}}\\ &=& \arctan\frac{\sin\theta\sin\varphi}{\sqrt{2\cos\theta\cos\varphi+\cos^2\theta+\cos^2\varphi}}\\ &=& \arctan\frac{\sin\theta\sin\varphi}{\cos\theta+\cos\varphi} \end{eqnarray*}

  • となる。

球面三角形の面積

  • \begin{eqnarray*} \Omega &=& \arctan\frac{\sin\theta\sin\varphi}{\cos\theta+\cos\varphi}+\arctan\frac{\sin\theta'\sin\varphi}{\cos\theta'+\cos\varphi}\\ &=& \arctan\frac{\frac{\sin\theta\sin\varphi}{\cos\theta+\cos\varphi}+\frac{\sin\theta'\sin\varphi}{\cos\theta'+\cos\varphi}}{1-\frac{\sin\theta\sin\varphi}{\cos\theta+\cos\varphi}\frac{\sin\theta'\sin\varphi}{\cos\theta'+\cos\varphi}}\\ &=& \arctan\frac{\sin\theta\sin\varphi (\cos\theta'+\cos\varphi)+\sin\theta' \sin\varphi(\cos\theta+\cos\varphi)}{(\cos\theta+\cos\varphi)(\cos\theta'+\cos\varphi)-\sin\theta\sin\theta'\sin^2\varphi}\\ &=& \arctan\frac{\sin\varphi [\sin\theta\cos\theta'+\sin\theta' \cos\theta + \cos\varphi(\sin\theta+\sin\theta')]}{\cos\varphi(\cos\theta+\cos\theta')+\cos\theta\cos\theta'+\cos^2\varphi-\sin\theta\sin\theta'(1-\cos^2\varphi)}\\ &=& \arctan\frac{\sin\varphi [\sin(\theta+\theta')+ \cos\varphi(\sin\theta+\sin\theta')]}{\cos\varphi(\cos\theta+\cos\theta')+\cos(\theta+\theta')+(1+\sin\theta\sin\theta')\cos^2\varphi}\\ \end{eqnarray*}

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  • となる。ここで

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    • \begin{eqnarray*} a &=& \theta + \theta'\\ \cos b &=& \cos\theta\cos\varphi\\ \cos c &=& \cos\theta'\cos\varphi\\ \end{eqnarray*}

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  • を用いると

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    • \begin{eqnarray*} \Omega &=& \arctan\frac{\sin\varphi [\sin a+ \cos\varphi(\sin\theta+\sin\theta')]}{\cos a +\cos b+\cos c+(1+\sin\theta\sin\theta')\cos^2\varphi}\\ \end{eqnarray*}

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  • となる。さらに

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    • \begin{eqnarray*} \frac{\cos b}{\cos c} &=& \frac{\cos \theta}{\cos \theta'}\\ \cos \theta &=& \cos(a-\theta')\\ &=& \cos a\cos\theta'-\sin a\sin\theta'\\ \cos \theta' &=& \cos(a-\theta)\\ &=& \cos a\cos\theta-\sin a\sin\theta \end{eqnarray*}

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  • を用いると

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    • \begin{eqnarray*} \tan\theta &=& \frac{1}{\sin a}\left(\frac{\cos c}{\cos b}-\cos a\right)\\ \tan\theta' &=& \frac{1}{\sin a}\left(\frac{\cos b}{\cos c}-\cos a\right) \end{eqnarray*}

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  • となるので

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    • \begin{eqnarray*} \sin\theta &=& \frac{\tan \theta}{\sqrt{1+\tan^2\theta}}\\ &=& \frac{\frac{1}{\sin a}\left(\frac{\cos c}{\cos b}-\cos a\right)}{\sqrt{1+\frac{1}{\sin^2 a}\left(\frac{\cos c}{\cos b}-\cos a\right)^2}}\\ &=& \frac{\cos c-\cos a\cos b}{\sqrt{\cos^2 b+\cos^2 c-2\cos a\cos b\cos c}}\\ \sin\theta' &=& \frac{\cos b-\cos a\cos c}{\sqrt{\cos^2 b+\cos^2 c-2\cos a\cos b\cos c}}\\ \sin\theta + \sin\theta' &=& \frac{(1-\cos a)(\cos b+\cos c)}{\sqrt{\cos^2 b+\cos^2 c-2\cos a\cos b\cos c}}\\ \sin\theta \sin\theta' &=& \frac{\cos b\cos c(1+\cos^2 a)-\cos a(\cos^2b+\cos^2 c)}{\cos^2 b+\cos^2 c-2\cos a\cos b\cos c} \end{eqnarray*}

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  • と書ける。さらに

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    • \begin{eqnarray*} \cos \varphi &=& \frac{\cos b}{\cos\theta}\\ &=& \cos b\sqrt{1+\tan^2 \theta}\\ &=& \cos b\sqrt{1+\frac{1}{\sin^2 a}\left(\frac{\cos c}{\cos b}-\cos a\right)^2}\\ &=& \frac{1}{\sin a}\sqrt{\cos^2 b+\cos^2 c-2\cos a\cos b\cos c}\\ \sin \varphi &=& \sqrt{1-\cos^2\varphi}\\ &=& \frac{1}{\sin a}\sqrt{1-\cos^2 a -\cos^2 b-\cos^2 c+2\cos a\cos b\cos c}\\ \end{eqnarray*}

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  • と書ける。これらを用いると

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    • \begin{eqnarray*} \Omega &=& \arctan\frac{\frac{1}{\sin a}\sqrt{1-\cos^2 a -\cos^2 b-\cos^2 c+2\cos a\cos b\cos c} \left(\sin a+ \frac{1}{\sin a}\sqrt{\cos^2 b+\cos^2 c-2\cos a\cos b\cos c}\frac{(1-\cos a)(\cos b+\cos c)}{\sqrt{\cos^2 b+\cos^2 c-2\cos a\cos b\cos c}}\right)}{\cos a +\cos b+\cos c+\left(1+\frac{\cos b\cos c(1+\cos^2 a)-\cos a(\cos^2b+\cos^2 c)}{\cos^2 b+\cos^2 c-2\cos a\cos b\cos c}\right)\frac{1}{\sin^2 a}(\cos^2 b+\cos^2 c-2\cos a\cos b\cos c)}\\ &=& \arctan\frac{\sqrt{1-\cos^2 a -\cos^2 b-\cos^2 c+2\cos a\cos b\cos c} \left(\sin^2 a+ (1-\cos a)(\cos b+\cos c)\right)}{\sin^2 a(\cos a +\cos b+\cos c)+\cos^2 b+\cos^2 c-2\cos a\cos b\cos c + \cos b\cos c(1+\cos^2 a)-\cos a(\cos^2b+\cos^2 c)}\\ &=& \arctan\frac{\sqrt{1-\cos^2 a -\cos^2 b-\cos^2 c+2\cos a\cos b\cos c} (1-\cos a)(1+\cos a+\cos b+\cos c)}{(1-\cos^2a)(\cos a +\cos b+\cos c)+(1-\cos a)(\cos^2 b+\cos^2 c)+\cos b\cos c (1-\cos a)^2}\\ &=& \arctan\frac{\sqrt{1-\cos^2 a -\cos^2 b-\cos^2 c+2\cos a\cos b\cos c} (1+\cos a+\cos b+\cos c)}{(1+\cos a)(\cos a +\cos b+\cos c)+\cos^2 b+\cos^2 c+\cos b\cos c (1-\cos a)}\\ &=& \arctan\frac{\sqrt{1-\cos^2 a -\cos^2 b-\cos^2 c+2\cos a\cos b\cos c} (1+\cos a+\cos b+\cos c)}{\cos a +\cos b+\cos c +\cos^2 a +\cos^2 b+\cos^2 c +\cos a\cos b+\cos a\cos c +\cos b\cos c- \cos a\cos b\cos c}\\ \end{eqnarray*}

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  • となる。さらにこの式を変形すると

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    • \begin{eqnarray*} \Omega &=& \end{eqnarray*}

last edited 2014-02-26 08:18:28 by NobuyukiKobayashi