直升机飞行控制

直升机飞行控制中坐标转换问题

关于直升机飞行控制中坐标转换问题

• 欧拉角 $\Theta$：(roll: $\phi$, pitch: $\theta$, yaw: $\psi$)
• 姿态角速率 $\Omega$：(roll: $p$, pitch: $q$, yaw: $r$)
• 转换矩阵：
  $$R = \left[ {\begin{array}{*{20}{c}} {C\theta C\psi }&{S\phi S\theta C\psi - C\phi S\psi }&{C\phi S\theta C\psi + S\phi S\psi }\\ {C\theta S\psi }&{S\phi S\theta S\psi + C\phi C\psi }&{C\phi S\theta S\psi - S\phi C\psi }\\ { - S\theta }&{S\phi C\theta }&{C\phi C\theta } \end{array}} \right]$$

$$H\left( \Omega \right) = \left[ {\begin{array}{*{20}{c}} 1&{S\phi S\theta /C\theta }&{C\phi S\theta /C\theta }\\ 0&{C\phi }&{ - S\phi }\\ 0&{S\phi /C\theta }&{C\phi /C\theta } \end{array}} \right]$$

$$sk\left( \Omega \right)=\left[ {\begin{array}{*{20}{c}} 0&{ - r}&q\\ r&0&{ - p}\\ { - q}&p&0 \end{array}} \right]$$

$$\dot R\left( \Theta \right) = R\left( \Theta \right)sk\left( {{\Omega }} \right)\\ \dot\Theta=R(\Theta)\Omega$$

$$H\left( {\dot \Omega } \right) = \left[ {\begin{array}{*{20}{c}} 0&{C\phi T\theta \cdot \dot \phi + S\phi /{C^2}\theta \cdot \dot \theta }&{ - S\phi T\theta \cdot \dot \phi + C\phi /{C^2}\theta \cdot \dot \theta }\\ 0&{ - S\phi \cdot \dot \phi }&{ - C\phi \cdot \dot \phi }\\ 0&{C\phi \cdot \dot \phi /C\theta + S\phi S\theta /{C^2}\theta \cdot \dot \theta }&{ - S\phi \cdot \dot \phi /C\theta + C\phi S\theta /{C^2}\theta \cdot \dot \theta } \end{array}} \right]$$

---