主に3DCG関係で自分がよく使用する数式の自分用メモです。
表記と定義
q = ( q_0, q_1, q_2, q_3 )\\
q \cdot p = \sum q_i p_i\\
\begin{vmatrix} q \end{vmatrix} = \sqrt{q \cdot q} = \sqrt{ q_0^2 + q_1^2 + q_2^2 + q_3^2 }\\
\begin{align}
q \times p =
( &q_0p_0 - q_1p_1 - q_2p_2 - q_3p_3,\\
&q_0p_1 + q_1p_0 + q_2p_3 - q_3p_2,\\
&q_0p_2 - q_1p_3 + q_2p_0 + q_3p_1,\\
&q_0p_3 + q_1p_2 - q_2p_1 + q_3p_0 )
\end{align}
逆数
i = (0,1,0,0)\\
j = (0,0,1,0)\\
k = (0,0,0,1)\\
q^{-1} = - \frac{0.5}{ \begin{vmatrix} q \end{vmatrix} } (q + i \times q \times i + j \times q \times j + k \times q \times k)
線形補完
q, p の補完, t=[0,1]
\begin{align}
s &= \left\{
\begin{array}{ll}
1-t & (q \cdot p \geq 0) \\
t-1 & (q \cdot p \lt 0)
\end{array}
\right.\\
r &= s \cdot q + t \cdot p\\
Lerp(q,p,t) &= \frac{r}{\begin{vmatrix} r \end{vmatrix}}
\end{align}
球面線形補間
q, p の補完, t=[0,1]
\begin{align}
a &= \cos^{-1} \begin{vmatrix} q \cdot p \end{vmatrix} \\
s &= \left\{
\begin{array}{ll}
1 & (q \cdot p \geq 0) \\
-1 & (q \cdot p \lt 0)
\end{array}
\right.\\
r &= \frac{1}{\sin(a)} \big( s \cdot \sin(a (1 - t)) \cdot q + \sin(at) \cdot p \big)\\
Slerp(q,p,t) &= \frac{r}{\begin{vmatrix} r \end{vmatrix}}
\end{align}
任意のベクトルを回転軸とする四元数
回転軸ベクトル：a、回転角：θ
\acute{a} = \frac{a}{\begin{vmatrix} a \end{vmatrix}}\\
q = ( \cos(0.5 \theta), \; \acute{a}_0 \sin ( 0.5 \theta ), \; \acute{a}_1 \sin ( 0.5 \theta ), \; \acute{a}_2 \sin ( 0.5 \theta ) )
回転行列へ変換
qと等価な回転行列M
M = \begin{pmatrix}
q_0^2 + q_1^2 - q_2^2 - q_3^2 &
2 (q_1q_2 - q_0q_3) &
2 (q_1q_3 +q_0q_2)\\
2 (q_1q_2 + q_0q_3) &
q_0^2 - q_1^2 + q_2^2 - q_3^2 &
2 (q_2q_3 - q_0q_1)\\
2 (q_1q_3 - q_0q_2) &
2 (q_2q_3 + q_0q_1) &
q_0^2 - q_1^2 - q_2^2 + q_3^2
\end{pmatrix}
回転行列からの変換
回転行列Mと等価なq
\begin{align}
q_0 &= 0.25 \sqrt{chop(M_{00} + M_{11} + M_{22} + 1)}\\
q_1 &= 0.25 \cdot a \sqrt{chop(M_{00} - M_{11} - M_{22} + 1)}\\
q_2 &= 0.25 \cdot b \sqrt{chop(-M_{00} + M_{11} - M_{22} + 1)}\\
q_3 &= 0.25 \cdot c \sqrt{chop(-M_{00} - M_{11} + M_{22} + 1)}\\
chop(x) &= \left\{
\begin{array}{ll}
x & (x \geq 0) \\
0 & (x \lt 0)
\end{array}
\right.\\
a &= \left\{
\begin{array}{ll}
-1 & (q_0 > 0) \; \cap \; (M_{21} - M_{12} < 0) \\
1 & else
\end{array}
\right.\\
b &= \left\{
\begin{array}{ll}
-1 & ( (q_0 > 0) \cap (M_{02} - M_{20} < 0) ) \; \cup \;
( (q_0 = 0) \cap (q_1 \neq 0) \cap (M_{10} + M_{02} < 0)) \\
1 & else
\end{array}
\right.\\
c &= \left\{
\begin{array}{ll}
-1 & ( (q_0 > 0) \cap (M_{10} - M_{01} < 0) ) \; \cup \;
( (q_0 = 0) \cap (q_1 \neq 0) \cap (M_{02} + M_{20} < 0)) \; \cup \;
( (q_0 = 0) \cap (q_1 = 0) \cap (M_{21} + M_{12} < 0)) \\
1 & else
\end{array}
\right.\\
\end{align}\\
※後で正規化が必要