2点を補間した位置への回転操作のための四元数
位置ベクトル\(\mathbf{p}\)への四元数\(\mathbf{q_1}\), \(\mathbf{q_2}\)による回転操作を行った結果を\(\mathbf{p_1}\)，\(\mathbf{p_2}\)とする． また\(\mathbf{p_1}\)と\(\mathbf{p_2}\)との間を補間した位置を\(\mathbf{p_t}\)とした時，\(\mathbf{p}\)に対する回転操作で\(\mathbf{p_t}\)となる四元数\(\mathbf{q_{t}}\)を考える． $$ \begin{eqnarray} \mathbf{q}&=&w+x\mathbf{i}+y\mathbf{j}+z\mathbf{k} \\&=&(w, x\mathbf{i}+y\mathbf{j}+z\mathbf{k}) \;\cdots\;\left(実部(スカラー),ijk部(ベクトル)\right) \\&=&\left(w,\mathbf{V}\right) \;\cdots\;\mathbf{V}=x\mathbf{i}+y\mathbf{j}+z\mathbf{k} \\\mathbf{\overline{q}} &=&w-x\mathbf{i}-y\mathbf{j}-z\mathbf{k} \\&=&(w, -x\mathbf{i}-y\mathbf{j}-z\mathbf{k}) \\&=&\left(w,-\mathbf{V}\right) \\\mathbf{q_{回転}}&=&\left(\cos{\left(\frac{\theta}{2}\right)}, \sin{\left(\frac{\theta}{2}\right)}\mathbf{V}\right) \\&&\;\cdots\;特に位置ベクトルを回転軸\mathbf{V}(ただし\mathbf{V}は単位ベクトル.\;\left|\mathbf{V}\right|=1),回転角\thetaで回す際の四元数 \\&&\href{https://shikitenkai.blogspot.com/2020/05/blog-post_41.html}{\mathbf{q_{回転}}と\mathbf{q_{回転}^{-1}}で位置ベクトル\mathbf{p}へ左右から作用させ\mathbf{q_{回転}}\mathbf{p}\mathbf{q_{回転}^{-1}}として用いる.} \\\mathbf{q_1} &=&\left(\cos{\left(\frac{\theta_1}{2}\right)}, \sin{\left(\frac{\theta_1}{2}\right)}\mathbf{V_1}\right) \\&&\;\cdots\;\left|q_1\right|=1, \left|\mathbf{V_1}\right|=1,\; 位置ベクトルpをp_1に向けて回転させる際の軸 \\&=&\left(C_1, S_1\mathbf{V_1}\right) \\\mathbf{q_2} &=&\left(\cos{\left(\frac{\theta_2}{2}\right)}, \sin{\left(\frac{\theta_2}{2}\right)}\mathbf{V_2}\right) \\&&\;\cdots\;\left|q_2\right|=1, \left|\mathbf{V_2}\right|=1位置ベクトルpをp_2に向けて回転させる際の軸 \\&=&\left(C_2, S_2\mathbf{V_2}\right) \\\mathbf{q_{1 \rightarrow2}} &=&\left(\cos{\left(\frac{\theta_{1 \rightarrow2}}{2}\right)}, \sin{\left(\frac{\theta_{1 \rightarrow2}}{2}\right)}\mathbf{V_{1 \rightarrow2}}\right) \\&&\;\cdots\;\left|q_{1 \rightarrow2}\right|=1, \left|\mathbf{V_{1 \rightarrow2}}\right|=1,\; 位置ベクトルpをp_1からp_2に向けて回転させる際の軸 \\&=&\left(C_{1 \rightarrow2}, S_{1 \rightarrow2}\mathbf{V_{1 \rightarrow2}}\right) \\\mathbf{q_{1 \rightarrow t}} &=&\left(\cos{\left(t\frac{\theta_{1 \rightarrow2}}{2}\right)}, \sin{\left(t\frac{\theta_{1 \rightarrow2}}{2}\right)}\mathbf{V_{1 \rightarrow2}}\right) \\&&\;\cdots\;0\leq t \leq 1,\; 位置ベクトルpをp_1からp_2に向けて回転させる途中のp_tまでの際の軸なので\mathbf{V_{1 \rightarrow2}}で変わらない.\;回転量だけtで調整される. \\&=&\left(C_{1 \rightarrow t}, S_{1 \rightarrow t}\mathbf{V_{1 \rightarrow2}}\right) \\\mathbf{q_{t}} &=&\left(\cos{\left(\frac{\theta_{t}}{2}\right)}, \sin{\left(\frac{\theta_{t}}{2}\right)}\mathbf{V_{t}}\right) \\&&\;\cdots\;\left|q_{t}\right|=1, \left|\mathbf{V_{t}}\right|=1,\; 位置ベクトルpをp_tに向けて回転させる際の軸 \\&=&\left(C_{t}, S_{t}\mathbf{V_{t}}\right) \end{eqnarray} $$\(\mathbf{p}\)，\(\mathbf{p_1}\)，\(\mathbf{p_2}\)，\(\mathbf{q_1}\)，\(\mathbf{q_2}\)の関係性から\(\mathbf{q_{1 \rightarrow2}}=\left(C_{1 \rightarrow2}, S_{1 \rightarrow2}\mathbf{V_{1 \rightarrow2}}\right)\)を求める． $$ \begin{eqnarray} \mathbf{p_1}&=&\mathbf{q_1}\mathbf{p}\mathbf{q_1^{-1}} \\ \\\mathbf{p_2}&=&\mathbf{q_2}\mathbf{p}\mathbf{q_2^{-1}} \\ \\\mathbf{p_2}&=&\mathbf{q_{1\rightarrow2}}\mathbf{p_1}\mathbf{q_{1\rightarrow2}^{-1}} \\&=&\mathbf{q_{1 \rightarrow2}}\left(\mathbf{q_1}\mathbf{p}\mathbf{q_1^{-1}}\right)\mathbf{q_{1 \rightarrow2}^{-1}} \;\cdots\;\mathbf{p_1}=\mathbf{q_1}\mathbf{p}\mathbf{q_1^{-1}} \\&=&\left(\mathbf{q_{1 \rightarrow2}}\mathbf{q_1}\right)\mathbf{p}\left(\mathbf{q_1^{-1}}\mathbf{q_{1 \rightarrow2}^{-1}}\right) \\ \\\mathbf{q_2}&=&\mathbf{q_{1 \rightarrow2}}\mathbf{q_1} \\ \\\mathbf{q_{1 \rightarrow2}}&=&\mathbf{q_2}\mathbf{q_1^{-1}} \\&=&\mathbf{q_2}\frac{\mathbf{\overline{q_1}}}{|\mathbf{q_1}|^2} \;\cdots\;\href{https://shikitenkai.blogspot.com/2020/05/blog-post_58.html}{\mathbf{q^{-1}}=\frac{\mathbf{\overline{q}}}{|\mathbf{q}|^2}} \\&=&\mathbf{q_2}\mathbf{\overline{q_1}} \;\cdots\;|\mathbf{q_1}|^2=1^2=1 \\&=&\left(C_2, S_2\mathbf{V_2}\right)\left(C_1, -S_1\mathbf{V_1}\right) \\&=&\left( C_1C_2-\{-S_1\}S_2\mathbf{V_2}\cdot\mathbf{V_1} ,\; -C_2S_1\mathbf{V_1}+C_1S_2\mathbf{V_2}+\{-S_1\}S_2\mathbf{V_2}\times\mathbf{V_1} \right) \\&&\;\cdots\;\href{https://shikitenkai.blogspot.com/2020/05/quaternion.html}{\mathbf{q}_1\mathbf{q}_2=\left(w_1,\mathbf{V_1}\right)\left(w_2,\mathbf{V_2}\right)=\left(w_1w_2-\mathbf{V}_1\cdot\mathbf{V}_2,\;w_1\mathbf{V}_2+w_2\mathbf{V}_1+\mathbf{V}_1\times\mathbf{V}_2\right)} \\&=&\left( C_1C_2+S_1S_2\mathbf{V_1}\cdot\mathbf{V_2} ,\; -C_2S_1\mathbf{V_1}+C_1S_2\mathbf{V_2}+S_1S_2\mathbf{V_1}\times\mathbf{V_2} \right) \;\cdots\;\mathbf{A}\cdot\mathbf{B}=\mathbf{B}\cdot\mathbf{A},\;\mathbf{A}\times\mathbf{B}=-\mathbf{B}\times\mathbf{A} \\ \\C_{1 \rightarrow2}&=&C_1C_2+S_1S_2\mathbf{V_1}\cdot\mathbf{V_2} \\S_{1 \rightarrow2}\mathbf{V_{1 \rightarrow2}}&=&-C_2S_1\mathbf{V_1}+C_1S_2\mathbf{V_2}+S_1S_2\mathbf{V_1}\times\mathbf{V_2} \end{eqnarray} $$
\(\mathbf{p}\)，\(\mathbf{p_1}\)，\(\mathbf{p_t}\)，\(\mathbf{q_1}\)，\(\mathbf{q_{1 \rightarrow t}}\)の関係性から\(\mathbf{q_t}=\left(C_t, S_t\mathbf{V_t}\right)\)を求める． $$ \begin{eqnarray} \mathbf{p_t}&=&\mathbf{q_{t}}\mathbf{p}\mathbf{q_{t}^{-1}} \\ \\\mathbf{p_t}&=&\mathbf{q_{1\rightarrow t}}\mathbf{p_1}\mathbf{q_{1\rightarrow t}^{-1}} \\&=&\mathbf{q_{1\rightarrow t}}\left(\mathbf{q_1}\mathbf{p}\mathbf{q_1^{-1}}\right)\mathbf{q_{1\rightarrow t}^{-1}} \;\cdots\;\mathbf{p_1}=\mathbf{q_1}\mathbf{p}\mathbf{q_1^{-1}} \\&=&\left(\mathbf{q_{1\rightarrow t}}\mathbf{q_1}\right)\mathbf{p}\left(\mathbf{q_1^{-1}}\mathbf{q_{1\rightarrow t}^{-1}}\right) \\ \\\mathbf{q_{t}}&=&\mathbf{q_{1 \rightarrow t}}\mathbf{q_1} \\&=&\left(C_{1 \rightarrow t}, S_{1 \rightarrow t}\mathbf{V_{1 \rightarrow 2}}\right)\left(C_1, S_1\mathbf{V_1}\right) \\&=&\left( C_1C_{1 \rightarrow t}-S_1S_{1 \rightarrow t}\mathbf{V_{1 \rightarrow 2}}\cdot\mathbf{V_1} ,\; C_{1 \rightarrow t}S_{1}\mathbf{V_1}+C_1S_{1 \rightarrow t}\mathbf{V_{1 \rightarrow 2}}+S_1S_{1 \rightarrow t}\mathbf{V_{1 \rightarrow 2}}\times\mathbf{V_1} \right) \\&=&\left( C_1C_{1 \rightarrow t}-S_1S_{1 \rightarrow t}\mathbf{V_1}\cdot\mathbf{V_{1 \rightarrow 2}} ,\; C_{1 \rightarrow t}S_{1}\mathbf{V_1}+C_1S_{1 \rightarrow t}\mathbf{V_{1 \rightarrow 2}}+S_1S_{1 \rightarrow t}\mathbf{V_{1 \rightarrow 2}}\times\mathbf{V_1} \right) \;\cdots\;\mathbf{A}\cdot\mathbf{B}=\mathbf{B}\cdot\mathbf{A} \\ \\C_{t}&=&C_1C_{1 \rightarrow t}-S_1S_{1 \rightarrow t}\mathbf{V_1}\cdot\mathbf{V_{1 \rightarrow 2}} \\S_{t}\mathbf{V_{t}}&=&C_{1 \rightarrow t}S_{1}\mathbf{V_1}+C_1S_{1 \rightarrow t}\mathbf{V_{1 \rightarrow 2}}+S_1S_{1 \rightarrow t}\mathbf{V_{1 \rightarrow 2}}\times\mathbf{V_1} \end{eqnarray} $$
上記で求めた\(S_{1 \rightarrow2}\mathbf{V_{1 \rightarrow2}}\)を用いて，\(\mathbf{V_1}\cdot\mathbf{V_{1 \rightarrow2}}\)を求めておく． $$ \begin{eqnarray} S_{1 \rightarrow2}\mathbf{V_{1 \rightarrow2}}&=&-C_2S_1\mathbf{V_1}+C_1S_2\mathbf{V_2}+S_1S_2\mathbf{V_1}\times\mathbf{V_2} \\ \\\mathbf{V_{1 \rightarrow2}} &=&\frac{1}{S_{1 \rightarrow2}}\left\{-C_2S_1\mathbf{V_1}+C_1S_2\mathbf{V_2}+S_1S_2\mathbf{V_1}\times\mathbf{V_2}\right\} \\ \\\mathbf{V_1}\cdot\mathbf{V_{1 \rightarrow2}} &=&\mathbf{V_1}\cdot\left[ \frac{1}{S_{1 \rightarrow2}}\left\{ -C_2S_1\mathbf{V_1} +C_1S_2\mathbf{V_2} +S_1S_2\left(\mathbf{V_1}\times\mathbf{V_2}\right) \right\} \right] \\&=&\frac{1}{S_{1 \rightarrow2}}\left\{ -C_2S_1\mathbf{V_1}\cdot\mathbf{V_1} +C_1S_2\mathbf{V_1}\cdot\mathbf{V_2} +S_1S_2\mathbf{V_1}\cdot\left(\mathbf{V_1}\times\mathbf{V_2}\right) \right\} \\&=&\frac{1}{S_{1 \rightarrow2}}\left\{ -C_2S_1 +C_1S_2\mathbf{V_1}\cdot\mathbf{V_2} +S_1S_2\mathbf{V_2}\cdot\left(\mathbf{V_1}\times\mathbf{V_1}\right)\right\} \;\cdots\;\mathbf{V_1}\cdot\mathbf{V_1}=\left|V_1\right|^2=1^2=1 ,\;\href{https://shikitenkai.blogspot.com/2020/06/blog-post_42.html}{\mathbf{A}\cdot\left(\mathbf{B}\times\mathbf{C}\right)=\mathbf{C}\cdot\left(\mathbf{A}\times\mathbf{B}\right)(スカラー三重積)} \\&=&\frac{1}{S_{1 \rightarrow2}}\left\{ -C_2S_1 +C_1S_2\mathbf{V_1}\cdot\mathbf{V_2} \right\} \;\cdots\;\mathbf{A}\times\mathbf{A}=0 \end{eqnarray} $$
上記で求めた\(\mathbf{V_1}\cdot\mathbf{V_{1 \rightarrow 2}}\)を用いて，\(C_{t}\)を変形する． $$ \begin{eqnarray} C_{t} &=&C_1C_{1 \rightarrow t}-S_1S_{1 \rightarrow t}\mathbf{V_1}\cdot\mathbf{V_{1 \rightarrow 2}} \\&=&C_1C_{1 \rightarrow t}-S_1S_{1 \rightarrow t}\left[\frac{1}{S_{1 \rightarrow2}}\left\{-C_2S_1+C_1S_2\mathbf{V_1}\cdot\mathbf{V_2}\right\}\right] \\&&\;\cdots\;\mathbf{V_1}\cdot\mathbf{V_{1 \rightarrow 2}}=\frac{1}{S_{1 \rightarrow2}}\left\{-C_2S_1+C_1S_2\mathbf{V_1}\cdot\mathbf{V_2}\right\} \\&=&C_1C_{1 \rightarrow t}-\frac{S_{1 \rightarrow t}}{S_{1 \rightarrow2}}\left\{-C_2S_1^2+C_1S_1S_2\mathbf{V_1}\cdot\mathbf{V_2}\right\} \\&=&C_1C_{1 \rightarrow t}-\frac{S_{1 \rightarrow t}}{S_{1 \rightarrow2}}\left\{-C_2\left(1-C_1^2\right)+C_1S_1S_2\mathbf{V_1}\cdot\mathbf{V_2}\right\} \;\cdots\;\sin^2{\left(\theta\right)}=1-\cos^2{\left(\theta\right)} \\&=&C_1C_{1 \rightarrow t}-\frac{S_{1 \rightarrow t}}{S_{1 \rightarrow2}}\left\{-C_2+C_1^2C_2+C_1S_1S_2\mathbf{V_1}\cdot\mathbf{V_2}\right\} \\&=&C_1C_{1 \rightarrow t}-\frac{S_{1 \rightarrow t}}{S_{1 \rightarrow2}}\left\{-C_2+C_1\left(C_1C_2+S_1S_2\mathbf{V_1}\cdot\mathbf{V_2}\right)\right\} \\&=&C_1C_{1 \rightarrow t}-\frac{S_{1 \rightarrow t}}{S_{1 \rightarrow2}}\left\{-C_2+C_1C_{1 \rightarrow 2}\right\} \;\cdots\;C_{1 \rightarrow 2}=C_1C_2+ S_1S_2\mathbf{V_1}\cdot\mathbf{V_2} \\&=&C_1C_{1 \rightarrow t} +\frac{S_{1 \rightarrow t}}{S_{1 \rightarrow2}}C_2 -\frac{S_{1 \rightarrow t}}{S_{1 \rightarrow2}}C_1C_{1 \rightarrow 2} \\&=&\frac{S_{1 \rightarrow 2}C_{1 \rightarrow t}-C_{1 \rightarrow 2}S_{1 \rightarrow t}} {S_{1 \rightarrow2}}C_1 +\frac{S_{1 \rightarrow t}}{S_{1 \rightarrow2}}C_2 \\&=&\frac{ \sin{\left(\frac{\theta_{1 \rightarrow 2}}{2}\right)}\cos{\left(t\frac{\theta_{1 \rightarrow 2}}{2}\right)} -\cos{\left(\frac{\theta_{1 \rightarrow 2}}{2}\right)}\sin{\left(t\frac{\theta_{1 \rightarrow 2}}{2}\right)} }{S_{1 \rightarrow2}}C_1 +\frac{S_{1 \rightarrow t}}{S_{1 \rightarrow2}}C_2 \\&=&\frac{\sin{\left(\frac{\theta_{1 \rightarrow 2}}{2}-t\frac{\theta_{1 \rightarrow 2}}{2}\right)}}{S_{1 \rightarrow2}}C_1 +\frac{S_{1 \rightarrow t}}{S_{1 \rightarrow2}}C_2 \;\cdots\;\sin{\left(\alpha\right)}\cos{\left(\beta\right)}-\cos{\left(\alpha\right)}\sin{\left(\beta\right)}=\sin{\left(\alpha-\beta\right)} \\&=& \frac{\sin{\left(\left(1-t\right)\frac{\theta_{1 \rightarrow 2}}{2}\right)}}{S_{1 \rightarrow2}} C_1 +\frac{S_{1 \rightarrow t}}{S_{1 \rightarrow2}} C_2 \\&=& \frac{\sin{\left(\left(1-t\right)\frac{\theta_{1 \rightarrow 2}}{2}\right)}}{\sin{\left(\frac{\theta_{1 \rightarrow 2}}{2}\right)}} C_1 +\frac{ \sin{\left( t\frac{\theta_{1 \rightarrow 2}}{2} \right)} } { \sin{\left( \frac{\theta_{1 \rightarrow 2}}{2} \right)} } C_2 \end{eqnarray} $$
上記で求めた\(\mathbf{V_{1 \rightarrow 2}}\)を用いて，\(S_{t}\mathbf{V_{t}}\)を変形する． $$ \begin{eqnarray} S_{t}\mathbf{V_{t}} &=&C_{1 \rightarrow t}S_{1}\mathbf{V_1} +C_1S_{1 \rightarrow t}\mathbf{V_{1 \rightarrow 2}} +S_1S_{1 \rightarrow t}\mathbf{V_{1 \rightarrow 2}}\times\mathbf{V_1} \\&=&C_{1 \rightarrow t}S_{1}\mathbf{V_1} +C_1S_{1 \rightarrow t}\left[\frac{1}{S_{1 \rightarrow2}}\left\{-C_2S_1\mathbf{V_1}+C_1S_2\mathbf{V_2}+S_1S_2\mathbf{V_1}\times\mathbf{V_2}\right\}\right] +S_1S_{1 \rightarrow t}\left[\frac{1}{S_{1 \rightarrow2}}\left\{-C_2S_1\mathbf{V_1}+C_1S_2\mathbf{V_2}+S_1S_2\mathbf{V_1}\times\mathbf{V_2}\right\}\right]\times\mathbf{V_1} \\&&\;\cdots\;\mathbf{V_{1 \rightarrow 2}}=\frac{1}{S_{1 \rightarrow2}}\left\{-C_2S_1\mathbf{V_1}+C_1S_2\mathbf{V_2}+S_1S_2\mathbf{V_1}\times\mathbf{V_2}\right\} \\&=&C_{1 \rightarrow t}S_{1}\mathbf{V_1} +\frac{S_{1 \rightarrow t}}{S_{1 \rightarrow2}}\left\{-C_1C_2S_1\mathbf{V_1}+C_1^2S_2\mathbf{V_2}+C_1S_1S_2\mathbf{V_1}\times\mathbf{V_2}\right\} +\frac{S_{1 \rightarrow t}}{S_{1 \rightarrow2}}\left\{-C_2S_1^2\mathbf{V_1}\times\mathbf{V_1}+C_1S_1S_2\mathbf{V_2}\times\mathbf{V_1}+S_1^2S_2\mathbf{V_1}\times\mathbf{V_2}\times\mathbf{V_1}\right\} \\&=&C_{1 \rightarrow t}S_{1}\mathbf{V_1} +\frac{S_{1 \rightarrow t}}{S_{1 \rightarrow2}}\left\{ -C_1C_2S_1\mathbf{V_1} +C_1^2S_2\mathbf{V_2} +C_1S_1S_2\mathbf{V_1}\times\mathbf{V_2} -C_2S_1^2\mathbf{V_1}\times\mathbf{V_1} +C_1S_1S_2\mathbf{V_1}\times\mathbf{V_2} +S_1^2S_2\mathbf{V_1}\times\mathbf{V_2}\times\mathbf{V_1} \right\} \\&=&C_{1 \rightarrow t}S_{1}\mathbf{V_1} +\frac{S_{1 \rightarrow t}}{S_{1 \rightarrow2}}\left[ -C_1C_2S_1\mathbf{V_1} +C_1^2S_2\mathbf{V_2} +S_1^2S_2\left\{ \left(\mathbf{V_1}\cdot\mathbf{V_1}\right)\mathbf{V_2}-\left(\mathbf{V_2}\cdot\mathbf{V_1}\right)\mathbf{V_1} \right\} \right] \;\cdots\;\mathbf{A}\times\mathbf{A}=0,\; \href{https://shikitenkai.blogspot.com/2020/06/blog-post_35.html}{\mathbf{A}\times\mathbf{B}\times\mathbf{C}=\left(\mathbf{A}\cdot\mathbf{C}\right)\mathbf{B}-\left(\mathbf{B}\cdot\mathbf{C}\right)\mathbf{A}} \\&=&C_{1 \rightarrow t}S_{1}\mathbf{V_1} +\frac{S_{1 \rightarrow t}}{S_{1 \rightarrow2}}\left[ -C_1C_2S_1\mathbf{V_1} +C_1^2S_2\mathbf{V_2} +S_1^2S_2\left\{ \left|\mathbf{V_1}\right|^2\mathbf{V_2}-\left|\mathbf{V_2}\right|\left|\mathbf{V_1}\right|\cos{\left(\theta_{1 \rightarrow2}\right)}\mathbf{V_1} \right\} \right] \;\cdots\;\mathbf{A}\cdot\mathbf{A}=\left|\mathbf{A}\right|^2 ,\;\mathbf{A}\cdot\mathbf{B}=\left|\mathbf{A}\right|\left|\mathbf{B}\right|\cos{\left(\theta\right)} \\&=&C_{1 \rightarrow t}S_{1}\mathbf{V_1} +\frac{S_{1 \rightarrow t}}{S_{1 \rightarrow2}}\left\{ -C_1C_2S_1\mathbf{V_1} +C_1^2S_2\mathbf{V_2} +S_1^2S_2\mathbf{V_2}-S_1^2S_2\cos{\left(\theta_{1 \rightarrow2}\right)}\mathbf{V_1} \right\} \;\cdots\;\left|\mathbf{V_1}\right|=1 ,\;\left|\mathbf{V_2}\right|=1 \\&=&C_{1 \rightarrow t}S_{1}\mathbf{V_1} +\frac{S_{1 \rightarrow t}}{S_{1 \rightarrow2}}\left[ \left\{-C_1C_2S_1-S_1^2S_2\cos{\left(\theta_{1 \rightarrow2}\right)}\right\} \mathbf{V_1} +\left\{C_1^2S_2+S_1^2S_2\right\} \mathbf{V_2} \right] \\&=&C_{1 \rightarrow t}S_{1}\mathbf{V_1} +\frac{S_{1 \rightarrow t}}{S_{1 \rightarrow2}}\left[ -S_1\left\{C_1C_2+S_1S_2\cos{\left(\theta_{1 \rightarrow2}\right)}\right\} \mathbf{V_1} +S_2\left\{C_1^2+S_1^2\right\} \mathbf{V_2} \right] \\&=&C_{1 \rightarrow t}S_{1}\mathbf{V_1} +\frac{S_{1 \rightarrow t}}{S_{1 \rightarrow2}}\left\{ -S_1C_{1 \rightarrow2}\mathbf{V_1} +S_2\mathbf{V_2} \right\} \;\cdots\;C_{1 \rightarrow2}=C_1C_2+S_1S_2\cos{\left(\theta_{1 \rightarrow2}\right)} ,\;\cos^2{\left(\theta\right)}+\sin^2{\left(\theta\right)}=1 \\&=& \frac{S_{1 \rightarrow2}C_{1 \rightarrow t}-C_{1 \rightarrow2}S_{1 \rightarrow t}}{S_{1 \rightarrow2}} S_{1}\mathbf{V_1} +\frac{S_{1 \rightarrow t}}{S_{1 \rightarrow2}}S_2 \mathbf{V_2} \\&=& \frac{ \sin{ \left(\frac{\theta_{1 \rightarrow 2}}{2}\right) } \cos{ \left(t\frac{\theta_{1 \rightarrow 2}}{2}\right) } -\cos{ \left(\frac{\theta_{1 \rightarrow 2}}{2}\right) } \sin{ \left(t\frac{\theta_{1 \rightarrow 2}}{2}\right) } } { S_{1 \rightarrow2} } S_{1}\mathbf{V_1} +\frac{S_{1 \rightarrow t}}{S_{1 \rightarrow2}} S_2\mathbf{V_2} \\&=& \frac{ \sin{ \left(\frac{\theta_{1 \rightarrow 2}}{2}-t\frac{\theta_{1 \rightarrow 2}}{2}\right) } } { S_{1 \rightarrow2} } S_{1}\mathbf{V_1} +\frac{S_{1 \rightarrow t}}{S_{1 \rightarrow2}} S_2\mathbf{V_2} \;\cdots\;\sin{\left(\alpha\right)}\cos{\left(\beta\right)}-\cos{\left(\alpha\right)}\sin{\left(\beta\right)}=\sin{\left(\alpha-\beta\right)} \\&=& \frac{ \sin{ \left(\left(1-t\right)\frac{\theta_{1 \rightarrow 2}}{2}\right) } } { S_{1 \rightarrow2} } S_{1}\mathbf{V_1} +\frac{S_{1 \rightarrow t}}{S_{1 \rightarrow2}} S_2\mathbf{V_2} \\&=& \frac{\sin{\left(\left(1-t\right)\frac{\theta_{1 \rightarrow 2}}{2}\right)}}{\sin{\left(\frac{\theta_{1 \rightarrow 2}}{2}\right)}} S_{1}\mathbf{V_1} +\frac{\sin{\left(t\frac{\theta_{1 \rightarrow 2}}{2}\right)}}{\sin{\left(\frac{\theta_{1 \rightarrow 2}}{2}\right)}} S_2\mathbf{V_2} \end{eqnarray} $$
以上により\(\mathbf{p_{t}}\)へ\(\mathbf{p}\)を回転させる四元数\(\mathbf{q_{t}}\)が\(\mathbf{q_{1}}\)，\(\mathbf{q_{2}}\)(及び\(\theta_{1 \rightarrow 2}\)と\(t\))によって表すことができた． $$ \begin{eqnarray} \mathbf{q_{t}}&=&\left(C_{t},\;S_{t}\mathbf{V_{t}}\right) \\&=&\left( \frac{\sin{\left(\left(1-t\right)\frac{\theta_{1 \rightarrow 2}}{2}\right)}}{\sin{\left(\frac{\theta_{1 \rightarrow 2}}{2}\right)}} C_1 +\frac{\sin{\left(t\frac{\theta_{1 \rightarrow 2}}{2}\right)}}{\sin{\left(\frac{\theta_{1 \rightarrow 2}}{2}\right)}} C_2 ,\; \frac{\sin{\left(\left(1-t\right)\frac{\theta_{1 \rightarrow 2}}{2}\right)}}{\sin{\left(\frac{\theta_{1 \rightarrow 2}}{2}\right)}} S_{1}\mathbf{V_1} +\frac{\sin{\left(t\frac{\theta_{1 \rightarrow 2}}{2}\right)}}{\sin{\left(\frac{\theta_{1 \rightarrow 2}}{2}\right)}} S_2\mathbf{V_2} \right) \\&=&\frac{\sin{\left(\left(1-t\right)\frac{\theta_{1 \rightarrow 2}}{2}\right)}}{\sin{\left(\frac{\theta_{1 \rightarrow 2}}{2}\right)}} C_1 +\frac{\sin{\left(t\frac{\theta_{1 \rightarrow 2}}{2}\right)}}{\sin{\left(\frac{\theta_{1 \rightarrow 2}}{2}\right)}} C_2 +\frac{\sin{\left(\left(1-t\right)\frac{\theta_{1 \rightarrow 2}}{2}\right)}}{\sin{\left(\frac{\theta_{1 \rightarrow 2}}{2}\right)}} S_{1}\mathbf{V_1} +\frac{\sin{\left(t\frac{\theta_{1 \rightarrow 2}}{2}\right)}}{\sin{\left(\frac{\theta_{1 \rightarrow 2}}{2}\right)}} S_2\mathbf{V_2} \\&=&\frac{\sin{\left(\left(1-t\right)\frac{\theta_{1 \rightarrow 2}}{2}\right)}}{\sin{\left(\frac{\theta_{1 \rightarrow 2}}{2}\right)}} C_1 +\frac{\sin{\left(\left(1-t\right)\frac{\theta_{1 \rightarrow 2}}{2}\right)}}{\sin{\left(\frac{\theta_{1 \rightarrow 2}}{2}\right)}} S_{1}\mathbf{V_1} +\frac{\sin{\left(t\frac{\theta_{1 \rightarrow 2}}{2}\right)}}{\sin{\left(\frac{\theta_{1 \rightarrow 2}}{2}\right)}} C_2 +\frac{\sin{\left(t\frac{\theta_{1 \rightarrow 2}}{2}\right)}}{\sin{\left(\frac{\theta_{1 \rightarrow 2}}{2}\right)}} S_2\mathbf{V_2} \\&=&\frac{\sin{\left(\left(1-t\right)\frac{\theta_{1 \rightarrow 2}}{2}\right)}}{\sin{\left(\frac{\theta_{1 \rightarrow 2}}{2}\right)}} \left(C_1,S_{1}\mathbf{V_1}\right) +\frac{\sin{\left(t\frac{\theta_{1 \rightarrow 2}}{2}\right)}}{\sin{\left(\frac{\theta_{1 \rightarrow 2}}{2}\right)}} \left(C_2,S_{2}\mathbf{V_2}\right) \\&=&\frac{\sin{\left(\left(1-t\right)\frac{\theta_{1 \rightarrow 2}}{2}\right)}}{\sin{\left(\frac{\theta_{1 \rightarrow 2}}{2}\right)}} \mathbf{q_1} +\frac{\sin{\left(t\frac{\theta_{1 \rightarrow 2}}{2}\right)}}{\sin{\left(\frac{\theta_{1 \rightarrow 2}}{2}\right)}} \mathbf{q_2} \end{eqnarray} $$ これは“ 2点を補間した位置への回転操作のための三角凾数”と同様の式の形をしている．
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