Up 秋分の太陽の方角 作成: 2020-09-25
更新: 2020-10-02


    公転角度τ,緯度a,経度bでは,つぎのベクトル \( {\bf d} = ( d_x, d_y, d_y ) \) が太陽方角ベクトルになる ( 太陽の方角): \[ \begin{align} d_x &= - \tau_s (a_c)^2 (b_c)^2 \\ & + \tau_c n_c (a_c)^2 b_s b_c \\ & - \tau_c n_s a_s a_c b_c + \tau_s \\ \ \\ d_y &= \tau_c (n_c)^2 (a_c)^2 (b_s)^2 \\ &- \tau_s n_c (a_c)^2 b_s b_c \\ & - 2 \tau_c n_s n_c a_s a_c b_s \\ &+ \tau_s n_s a_s a_c b_c \\ &+ \tau_c (n_s)^2 (a_s)^2 - \tau_c \\ \ \\ d_z &= \tau_c n_s n_c a_c^2 b_s^2 \\ & \quad - \tau_s n_s a_c^2 b_s b_c \\ & \quad + \tau_c ( n_c^2 - n_s^2 ) a_s a_c b_s \\ & \quad - \tau_s n_c a_s a_c b_c \\ & \quad - \tau_c n_s n_c a_s^2 \\ \ \\ | {\bf d} | &= \sqrt { 1 - ( a_c ( n_c \tau_c b_s - \tau_s b_c ) - n_s a_s \tau_c )^2 } \end{align} \\ \ \\ \] 秋分は,\( \tau = \pi / 2 \) ── \( \tau_s =1, \ \tau_c =0 \) ──なので, \[ d_x = - a_c^2 b_c^2 + 1 \\ d_y = -n_c a_c^2 b_s b_c + n_s a_s a_c b_c \\ d_z = n_s a_c^2 b_s b_c - n_c a_s a_c b_c \\ | {\bf d} | = \sqrt { 1 - ( a_c ( - b_c ) )^2 } \\ \quad = \sqrt { 1 - ( a_c)^2 (b_c)^2 } \\ \]