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：
\[
\begin{align}
d_x &=
- \tau_s (a_c)^2 (b_c)^2 \\
& + \tau_c n_c (a_c)^2 b_s b_c \\
& - \tau_c n_s a_s a_c b_c
+ \tau_s
\\ \ \\
d_y &=
\tau_c (n_c)^2 (a_c)^2 (b_s)^2 \\
&- \tau_s n_c (a_c)^2 b_s b_c \\
& - 2 \tau_c n_s n_c a_s a_c b_s \\
&+ \tau_s n_s a_s a_c b_c \\
&+ \tau_c (n_s)^2 (a_s)^2 - \tau_c
\\ \ \\
d_z &=
\tau_c n_s n_c a_c^2 b_s^2 \\
& \quad - \tau_s n_s a_c^2 b_s b_c \\
& \quad + \tau_c ( n_c^2 - n_s^2 ) a_s a_c b_s \\
& \quad - \tau_s n_c a_s a_c b_c \\
& \quad - \tau_c n_s n_c a_s^2
\end{align]
\\ \ \\
\]
そして，つぎのベクトル \( ( d0_x, d0_y, d0_y ) \) が太陽方角ベクトル：
\[
\begin{align}
d0_x &=
- \frac { n_s a_s a_c \tau_s \tau_c } {\sqrt{ 1 - (n_s)^2 ( \tau_c )^2 } } + a_s^2 \tau_s
\\ \ \\
d0_y &=
- ( (n_s)^2 (a_c)^2 + (n_c)^2 (a_s)^2 ) \tau_c \\
& \quad + \frac {( (n_c)^2 (\tau_c)^2 + ( 1 - (n_s)^2 (\tau_c)^2 ) ) n_s a_s a_c }
{\sqrt{ 1 - (n_s)^2 ( \tau_c )^2 } }
\\ \ \\
d0_z &=
n_s n_c ( (a_c)^2 - (a_s)^2 ) \tau_c \\
& \quad + \frac {
( (n_s)^2 (\tau_c)^2 - (1 - (n_s)^2 (\tau_c)^2 ) ) n_c a_s a_c }
{ \sqrt{1 - (n_s)^2 ( \tau_c )^2} }
\end{align}
\]
この二つのなす角度θは，つぎの式に表される：
\[
cos(\theta\) = \frac{ ( d_x, d_y, d_z ) \cdot ( d0_x, d0_y, d0_z )}{ |( d_x, d_y, d_z )| \ | ( d0_x, d0_y, d0_z ) | }
\]
以下，これを計算する。
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