日夜的方程 精选

$\cos\theta = \cos\beta \ \cos L \ \cos d + \sin \beta \ sin L$

$\cos\beta \ \cos L \ \cos d + \sin \beta \ sin L=0$

$\cos d = -\tan \beta \tan L$

$\sin\beta^\prime = \cos\phi\ \sin\beta$

$\cos d = -\tan[\sin^{-1}(\cos\phi\sin\beta)]\tan L = -\frac{\sin\beta\cos\phi\tan L}{\sqrt{1-\sin^2\beta\cos^2\phi}}$

$\hat{x} = \hat{x^\prime}\\ \hat{y}=\cos\beta \ \hat{y^\prime} +\sin\beta \ \hat{z^\prime}\\ \hat{z} = -\sin\beta \ \ \hat{y^\prime} + \cos\beta \ \hat{z^\prime}$

$\cos\theta = [\cos d \ \cos L \cos\beta + \sin L \sin\beta] \cos\phi + \sin\phi \sin d \cos L$

$\cos\theta = \left[ \sin L \sin\beta + \cos L \cos\beta \ \cos (\omega \ t) \right]\ \cos (\Omega\ t) + \cos L \sin (\omega\ t) \sin(\Omega\ t)$

$\left[ \sin L \sin\beta + \cos L \cos\beta \ \cos (\omega \ t) \right]\ \cos (\Omega\ t) + \cos L \sin (\omega\ t) \sin(\Omega\ t) =0$

$\frac{ \sin L \sin\beta + \cos L \cos\beta \ \cos (\omega \ t)}{\cos L \sin (\omega\ t)}=-\tan\Omega t$

http://www.zhenzhubay.com/home.php?mod=space&uid=2&do=blog&id=36892

http://blog.sciencenet.cn/blog-684007-1122999.html

全部精选博文导读

GMT+8, 2019-10-23 21:01