「太陽の方角」で，太陽方角ベクトル \( {\bf d} \) を求めた：
\[
\begin{align}
{\bf d} &= ( \ \tau_s ( y^2 + z^2 ) + \tau_c ( x y ), \\
&\ - \tau_s ( x y ) - \tau_c ( z^2 + x^2 ), \\
&\ - \tau_s ( z x ) + \tau_c ( y z ) )
\\ \ \\
| {\bf d} | &= \sqrt { 1 - ( \tau_s \ x - \tau_c \ y )^2 }
\end{align}
\\
\]
「自転軸系経度緯度と公転軸系直交座標の変換式」から，
\[
x = a_c b_c \\
y = n_c a_c b_s - n_s a_s \\
z = n_s a_c b_s + n_c a_s
\\
\]
以上を合わせて：
\[
\begin{align}
& \tau_s ( y^2 + z^2 ) + \tau_c ( x y )
\\
& = \tau_s (1 - x^2 )+ \tau_c ( x y )
\\
& = \tau_s - \tau_s ( a_c b_c )^2
+ \tau_c ( a_c b_c )( n_c a_c b_s - n_s a_s )
\\
& = \tau_s - \tau_s (a_c)^2 (b_c)^2
+ \tau_c a_c b_c n_c a_c b_s
- \tau_c a_c b_c n_s a_s
\\ \ \\
& = \tau_s - \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 ( x y ) - \tau_c ( z^2 + x^2 )
\\
&= - \tau_s ( x y ) - \tau_c ( 1 - y^2 )
\\
&= - \tau_s (a_c b_c) (n_c a_c b_s - n_s a_s)
- \tau_c + \tau_c (n_c a_c b_s - n_s a_s)^2
\\
&= - \tau_s a_c b_c n_c a_c b_s
+ \tau_s a_c b_c n_s a_s \\
& - \tau_c
+ \tau_c ( (n_c)^2 (a_c)^2 (b_s)^2 - 2 n_c a_c b_s n_s a_s + (n_s)^2 (a_s)^2 )
\\
&= - \tau_s n_c (a_c)^2 b_s b_c
+ \tau_s n_s a_s a_c b_c \\
& - \tau_c
+ \tau_c (n_c)^2 (a_c)^2 (b_s)^2
- 2 \tau_c n_s n_c a_s a_c b_s
+ \tau_c (n_s)^2 (a_s)^2
\\ \ \\
&= \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
\\ \ \\
& - \tau_s ( z x ) + \tau_c ( y z )
\\ \ \\
& =
- \tau_s (n_s a_c b_s + n_c a_s) (a_c b_c) \\
& \quad
+ \tau_c (n_c a_c b_s - n_s a_s) (n_s a_c b_s + n_c a_s)
\\ \ \\
& =
- \tau_s (n_s a_c b_s a_c b_c + n_c a_s a_c b_c) \\
& \quad
+ \tau_c ( n_c a_c b_s n_s a_c b_s + n_c a_c b_s n_c a_s - n_s a_s n_s a_c b_s - n_s a_s n_c a_s)
\\ \ \\
& =
- \tau_s (n_s a_c^2 b_s b_c + n_c a_s a_c b_c) \\
& \quad
+ \tau_c (n_s n_c a_c^2 b_s^2 + n_c^2 a_s a_c b_s - n_s^2 a_s a_c b_s - n_s n_c a_s^2 )
\\ \ \\
& =
- \tau_s n_s a_c^2 b_s b_c \\
& \quad - \tau_s n_c a_s a_c b_c \\
& \quad + \tau_c n_s n_c a_c^2 b_s^2 \\
& \quad + \tau_c n_c^2 a_s a_c b_s \\
& \quad - \tau_c n_s^2 a_s a_c b_s \\
& \quad - \tau_c n_s n_c a_s^2
\\ \ \\
& =
\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
\end{align}
\\ \ \\
\]
\[
\begin{align}
| {\bf d} | &= \sqrt { 1 - ( \tau_s \ x - \tau_c \ y )^2 } \\
&= \sqrt { 1 - ( \tau_s \ ( a_c b_c ) - \tau_c \ ( n_c a_c b_s - n_s a_s ) )^2 }
\\ \ \\
&1 - ( \tau_s \ ( a_c b_c ) - \tau_c \ ( n_c a_c b_s - n_s a_s ) )^2
\\ \ \\
&= 1 - (
(\tau_s )^2 (a_c )^2 (b_c)^2 \\
& - 2 \tau_s a_c b_c \tau_c ( n_c a_c b_s - n_s a_s) \\
& + (\tau_c)^2 (\ n_c a_c b_s - n_s a_s )^2 )
\\ \ \\
&= - (a_c)^2 (\tau_s )^2 (b_c)^2 \\
& + 2 n_c (a_c)^2 \tau_s \tau_c b_s b_c \\
& - 2 n_s a_s a_c \tau_s \tau_c b_c \\
& - (\tau_c)^2
(\ (n_c)^2 (a_c)^2 (b_s)^2
- 2 n_c a_c b_s n_s a_s
+ (n_s)^2 (a_s)^2 ) + 1
\\ \ \\
&= - (n_c)^2 (a_c)^2 (\tau_c)^2 (b_s)^2 \\
& - (a_c)^2 (\tau_s )^2 (b_c)^2 \\
& + 2 n_c (a_c)^2 \tau_s \tau_c b_s b_c \\
& + 2 n_c n_s a_s a_c (\tau_c)^2 b_s \\
& - 2 n_s a_s a_c \tau_s \tau_c b_c \\
& - (n_s)^2 (a_s)^2 (\tau_c)^2 +1
\\ \ \\
&= - (a_c)^2 ( (n_c)^2 (\tau_c)^2 (b_s)^2 + (\tau_s )^2 (b_c)^2 - 2 n_c \tau_s \tau_c b_s b_c ) \\
& + 2 n_c n_s a_s a_c (\tau_c)^2 b_s \\
& - 2 n_s a_s a_c \tau_s \tau_c b_c \\
& - (n_s)^2 (a_s)^2 (\tau_c)^2 +1
\\ \ \\
&= - (a_c)^2 ( n_c \tau_c b_s - \tau_s b_c )^2 \\
& + 2 n_s a_s a_c \tau_c ( n_c \tau_c b_s - \tau_s b_c ) \\
& - (n_s)^2 (a_s)^2 (\tau_c)^2 +1
\\ \ \\
&= - ( a_c ( n_c \tau_c b_s - \tau_s b_c ) - n_s a_s \tau_c )^2 + (n_s)^2 (a_s)^2 (\tau_c )^2\\
& - (n_s)^2 (a_s)^2 (\tau_c)^2 +1
\\ \ \\
&=1 - ( a_c ( n_c \tau_c b_s - \tau_s b_c ) - n_s a_s \tau_c )^2
\\ \ \\
\end{align}
\]
まとめ
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：
\[
\begin{align}
d_x &=
- \tau_s (a_c)^2 (b_c)^2 \\
& + \tau_c n_c (a_c)^2 b_s b_c \\
& - \tau_c n_s a_s a_c b_c
+ \tau_s
\\ \ \\
d_y &=
\tau_c (n_c)^2 (a_c)^2 (b_s)^2 \\
&- \tau_s n_c (a_c)^2 b_s b_c \\
& - 2 \tau_c n_s n_c a_s a_c b_s \\
&+ \tau_s n_s a_s a_c b_c \\
&+ \tau_c (n_s)^2 (a_s)^2 - \tau_c
\\ \ \\
d_z &=
\tau_c n_s n_c a_c^2 b_s^2 \\
& \quad - \tau_s n_s a_c^2 b_s b_c \\
& \quad + \tau_c ( n_c^2 - n_s^2 ) a_s a_c b_s \\
& \quad - \tau_s n_c a_s a_c b_c \\
& \quad - \tau_c n_s n_c a_s^2
\\ \ \\
| {\bf d} | &= \sqrt { 1 - ( a_c ( n_c \tau_c b_s - \tau_s b_c ) - n_s a_s \tau_c )^2 }
\end{align}
\\ \ \\
\]
例：秋分 ( τ＝π/2 ) の場合
\( \tau_s =1, \ \tau_c =0 \) なので，
\[
d_x = - a_c^2 b_c^2 + 1 \\
d_y = -n_c a_c^2 b_s b_c + n_s a_s a_c b_c \\
d_z = n_s a_c^2 b_s b_c - n_c a_s a_c b_c \\
| {\bf d} | = \sqrt { 1 - ( a_c ( - b_c ) )^2 } \\
\quad = \sqrt { 1 - ( a_c)^2 (b_c)^2 }
\\
\]
南中は \( b_c = 1, b_s = 0 \) のときで，このとき
\[
d_x = - a_c^2 + 1 = a_s^2 \\
d_y = n_s a_s a_c \\
d_z = - n_c a_s a_c \\
| {\bf d} | = \sqrt { 1 - (a_c)^2 } = a_s
\\
\]
検算
\[
|{\bf d}| = \sqrt{ (a_s^2)^2 + (n_s a_s a_c)^2 + (- n_c a_s a_c)^2 } \\
\quad = a_s \sqrt{ a_s^2 + n_s^2 a_c^2 + n_c^2 a_c^2 } \\
\quad = a_s \sqrt{ a_s^2 + a_c^2 } \\
\quad = a_s
\\
\]
\( {\bf d} \) を単位ベクトル化すると：
\[
{\bf d}_u = ( a_s, - n_s a_c, n_c a_c )
\]
この方角での太陽の仰角をβとすると，
\[
\begin{align}
& \beta_c = {\bf d}_u \cdot {\bf s} \\
& \quad = ( a_s, - n_s a_c, n_c a_c ) \cdot ( \tau_s, -\tau_c , 0 ) \\
& \quad = ( a_s, - n_s a_c, n_c a_c ) \cdot ( 1, 0, 0 ) \\
& \quad = a_s
\\ \ \\
\Longrightarrow & \ \ \beta = \frac{\pi}{2} - a
\end{align}
\]
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