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日の出・日の入りの座標計算

作成: 2020-09-03 更新: 2020-09-20

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問題 公転角がτのときの，緯度ａの日の出・日の入りの地点の座標は？ はじめに，つぎを条件として措くことが必要になる (以下の推論の中でこのことがわかってくる)： \[ a > n \\ a + n < \frac{\pi}{2} \\ \] 簡単のため \[ n_s = sin(n), \ \ n_c = cos(n) \\ a_s = sin(a), \ \ a_c = cos(a) \\ \tau_s = sin(\tau), \ \ \tau_c = cos(\tau) \] とおく。 さらにこの簡略記法 \[ x_s = sin(x) \\ x_c = cos(x) \] を，一般のｘに対しても用いるとする。 (1) τ＝ 1/2π, 3/2π のとき (1.1) τ＝ 1/2π (春分) のとき 日出は \[ x = 0 \\ y = - (a - n)_c \\ z = (a - n)_s \] 日入は， \[ x = 0 \\ y = (a + n)_c\\ z = (a + n)_s \\ \\ \] (1.2) τ＝ 3/2π (秋分) のとき 日出は \[ x = 0 \\ y = (a + n)_c \\ z = (a + n)_s \] 日入は， \[ x = 0 \\ y = - (a - n)_c\\ z = (a - n)_s \\ \\ \ \\ \] {2) τ≠ 0, π のとき 日出・日入点の位置： 日出点Ｐの位置ベクトルを ｘ とする。 ベクトルt をつぎのようにとる： このとき， \[ {\bf t} = ( \tau_s, -\tau_c, 0) \\ {\bf x} \cdot {\bf t} = 0 \] そしてこれより \[ \begin{align} \Longrightarrow & (x, y, z) \cdot ( \tau_s, -\tau_c, 0) = 0 \\ \Longrightarrow & \tau_s \ x - \tau_c \ y = 0 \\ \ \\ \Longrightarrow & y = \frac{\tau_s}{\tau_c} x \\ \end{align} \\ \] 緯度ａの点の座標 (x, y, z) では，つぎが必要条件になる ( 自転軸系経度緯度と公転軸系直交座標の変換式)： \[ a_s = -n_s \ y + n_c \ z \\ \] よって， \[ \begin{align} z &= \frac{n_s \ y + a_s}{n_c} \\ &= \frac{n_s \tau_s \ x + a_s \tau_c}{n_c \tau_c} \end{align} \\ \ \\ \] さらに \[ \begin{align} & x^2 + y^2 + z^2 = 1 \\ \Longrightarrow & x^2 + (\frac{\tau_s}{\tau_c} x)^2 + (\frac{ n_s \tau_s \ x + a_s \tau_c }{ n_c \tau_c} )^2 = 1 \\ \Longrightarrow & x^2 + \frac{(\tau_s)^2}{(\tau_c)^2} \ x^2 \\ & \qquad + \frac{ (n_s \ \tau_s)^2 \ x^2 + 2 n_s \ \tau_s \ \tau_c \ a_s \ x + (a_s)^2 (\tau_c)^2 } { (n_c)^2 (\tau_c)^2 } = 1 \\ \ \\ \Longrightarrow & ( ( n_c \tau_c )^2 + ( n_c \tau_s)^2 + (n_s \ \tau_s)^2 ) \ x^2 \\ & \qquad + 2 ( n_s \ a_s \ \tau_s \ \tau_c ) \ x \\ & \qquad + (a_s \tau_c)^2 = ( n_c \tau_c)^2 \\ \ \\ \Longrightarrow & ( ( n_c \tau_c )^2 + (\tau_s)^2 ) \ x^2 \\ & \qquad + 2 ( n_s \ a_s \ \tau_s \ \tau_c ) \ x \\ & \qquad + (a_s \tau_c)^2 - ( n_c \tau_c)^2 = 0\\ \ \\ \Longrightarrow & ( ( n_c \tau_c )^2 + 1 - (\tau_c)^2 ) \ x^2 \\ & \qquad + 2 ( n_s \ a_s \ \tau_s \ \tau_c ) \ x \\ & \qquad + ( (a_s )^2 - ( n_c)^2 ) (\tau_c)^2 = 0\\ \ \\ \Longrightarrow & ( 1 - (n_s)^2 (\tau_c)^2 ) \ x^2 \\ & \qquad + 2 ( n_s \ a_s \ \tau_s \ \tau_c ) \ x \\ & \qquad + ( (a_s )^2 - ( n_c)^2 ) (\tau_c)^2 = 0\\ \ \\ \Longrightarrow &x = \frac{ - n_s \ a_s \ \tau_s \ \tau_c \pm \sqrt{D}}{1 - (n_s)^2 (\tau_c)^2} \\ \ \\ &D = ( n_s a_s \tau_s \tau_c )^2 - (1 - (n_s)^2 (\tau_c)^2) ( (a_s )^2 - ( n_c)^2 ) (\tau_c)^2 \\ \ \\ &\quad = ( n_s)^2 (a_s)^2 (\tau_s )^2 (\tau_c)^2 - \\ &\quad \quad ( (1 - (n_s)^2 (\tau_c)^2 ) (a_s)^2 - ( 1 - (n_s)^2 (\tau_c)^2 ) (n_c)^2) (\tau_c)^2 \\ \ \\ &\quad = ( n_s)^2 (a_s )^2 (\tau_s )^2 (\tau_c)^2 - \\ &\quad \quad ( (a_s)^2 - (n_s)^2 (a_s)^2 (\tau_c)^2 - ( n_c )^2 + (n_s)^2 (\tau_c)^2 (n_c)^2 ) (\tau_c)^2\\ \ \\ &\quad = ( ( n_s)^2 (a_s )^2 (\tau_s)^2 - (a_s)^2 + (n_s)^2 (a_s)^2 (\tau_c)^2 + ( n_c )^2 - (n_s)^2 (\tau_c)^2 (n_c)^2 ) (\tau_c)^2\\ \ \\ &\quad = ( ( n_c )^2 + ( n_s)^2 (a_s )^2 (\tau_s )^2 + (n_s)^2 (a_s)^2 (\tau_c)^2 - (a_s)^2 - (n_s)^2 (\tau_c)^2 (n_c)^2 ) (\tau_c)^2\\ \ \\ &\quad = ( ( n_c )^2+ ( n_s)^2 (a_s )^2 - (a_s)^2 - (n_s)^2 (\tau_c)^2 (n_c)^2 ) (\tau_c)^2\\ \ \\ &\quad = ( ( n_c )^2 - ( 1 - (n_s)^2) (a_s )^2 - (n_s)^2 (\tau_c)^2 (n_c)^2 ) (\tau_c)^2\\ \ \\ &\quad = ( ( n_c )^2 - (n_c)^2 (a_s)^2 - (n_s)^2 (\tau_c)^2 (n_c)^2 ) (\tau_c)^2\\ \ \\ &\quad = ( (1 - (a_s)^2) - (n_s)^2 (\tau_c)^2 ) (n_c)^2 (\tau_c)^2\\ \ \\ &\quad = ( (a_c)^2 - (n_s)^2 (\tau_c)^2 ) (n_c)^2 (\tau_c)^2\\ \end{align} \\ \] はじめに a + n < π/2 を条件として措いたが，このとき \[ \begin{align} & a < \frac{\pi}{2} - n \\ \Longrightarrow \ \ & a_c > cos(\frac{\pi}{2} - n) = n_s \\ \Longrightarrow \ \ & D > 0 \\ \end{align} \] ｘが \[ x = \frac{ - n_s \ a_s \ \tau_s \ \tau_c \pm n_c \tau_c \sqrt{(a_c)^2 - (n_s)^2 (\tau_c)^2}}{1 - (n_s)^2 (\tau_c)^2} \] と確定したところで， \[ \begin{align} y & = \frac{\tau_s}{\tau_c} x = \frac{ - n_s \ a_s \ (\tau_s)^2 \pm n_c \tau_s \sqrt{(a_c)^2 - (n_s)^2 (\tau_c)^2}}{1 - (n_s)^2 (\tau_c)^2} \\ \ \\ z & = \frac{n_s \ y + a_s}{n_c} \\ \ \\ & = \frac{ - (n_s)^2 \ a_s \ (\tau_s)^2 \pm n_s n_c \tau_s \sqrt{(a_c)^2 - (n_s)^2 (\tau_s)^2} + a_s (1 - (n_s)^2 (\tau_c)^2) } {n_c \ (1 - (n_s)^2 (\tau_c)^2)} \\ \ \\ & = \frac{ - (n_s)^2 \ a_s \ (\tau_s)^2 \pm n_s n_c \tau_s \sqrt{(a_c)^2 - (n_s)^2 (\tau_s)^2} + a_s ( 1 - (n_s)^2 (1 - (\tau_s)^2 ) } {n_c \ (1 - (n_s)^2 (\tau_c)^2)} \\ \ \\ & = \frac{ - (n_s)^2 \ a_s \ (\tau_s)^2 \pm n_s n_c \tau_s \sqrt{(a_c)^2 - (n_s)^2 (\tau_s)^2} + a_s ( 1 - (n_s)^2 + (n_s)^2 (\tau_s)^2 ) } {n_c \ (1 - (n_s)^2 (\tau_c)^2)} \\ \ \\ & = \frac{ - (n_s)^2 \ a_s \ (\tau_s)^2 \pm n_s n_c \tau_s \sqrt{(a_c)^2 - (n_s)^2 (\tau_s)^2} + a_s (n_c)^2 + a_s (n_s)^2 (\tau_s)^2 } {n_c \ (1 - (n_s)^2 (\tau_c)^2)} \\ \ \\ & = \frac{ n_c a_s \pm n_s \tau_s \sqrt{(a_c)^2 - (n_s)^2 (\tau_s)^2}} {1 - (n_s)^2 (\tau_c)^2} \end{align} \] まとめ a ＜ π/2 - n であるａに対し， 日出 τ＝ 1/2π (春分) のとき \[ x = 0 \\ y = - (a - n)_c \\ z = (a + n)_c \] τ＝ 3/2π (秋分) のとき \[ x = 0 \\ y = (a + n)_c \\ z = - (a - n)_c \] τ≠ 0, π のとき \[ x = \frac{ - n_s \ a_s \ \tau_s \ \tau_c - n_c \tau_c \sqrt{(a_c)^2 - (n_s)^2 (\tau_c)^2}}{1 - (n_s)^2 (\tau_c)^2} \ \\ y = \frac{ - n_s \ a_s \ (\tau_s)^2 - n_c \tau_s \sqrt{(a_c)^2 - (n_s)^2 (\tau_c)^2}}{1 - (n_s)^2 (\tau_c)^2} \\ \ \\ z = \frac{ n_c a_s - n_s \tau_s \sqrt{(a_c)^2 - (n_s)^2 (\tau_s)^2}} {1 - (n_s)^2 (\tau_c)^2} \\ \] 日入 τ＝ 1/2π (春分) のとき \[ x = 0 \\ y = (a + n)_c\\ z = (a + n)_s \\ \\ \] τ＝ 3/2π (秋分)のとき \[ x = 0 \\ y = - (a - n)_c\\ z = (a - n)_s \\ \\ \] τ≠ 0, π のとき \[ x = \frac{ - n_s \ a_s \ \tau_s \ \tau_c + n_c \tau_c \sqrt{(a_c)^2 - (n_s)^2 (\tau_c)^2}}{1 - (n_s)^2 (\tau_c)^2} \\ \ \\ y = \frac{ - n_s \ a_s \ (\tau_s)^2 + n_c \tau_s \sqrt{(a_c)^2 - (n_s)^2 (\tau_c)^2}}{1 - (n_s)^2 (\tau_c)^2} \\ \ \\ z = \frac{ n_c a_s + n_s \tau_s \sqrt{(a_c)^2 - (n_s)^2 (\tau_s)^2}} {1 - (n_s)^2 (\tau_c)^2} \\ \]