How to derive the relation between Euler angles and angular velocity and get the same form as mentioned in the bellow figure

Question

How to derive the relation between Euler angles and angular velocity and get this form:

$$ \left. \begin{cases}{} P \\ Q\\ R \\ \end{cases} \right\}= \left[ \begin{array}{c} 1&0&-\sin\Theta\\ 0&\cos\Phi&\cos\Theta\sin\Phi\\ 0&-\sin\Phi&\cos\Theta\cos\Phi \end{array} \right] \left. \begin{cases}{} \dot{\Phi} \\ \dot{\Theta}\\ \dot{\Psi} \\ \end{cases} \right\} $$

$$ \left. \begin{cases}{} \dot{\Phi} \\ \dot{\Theta}\\ \dot{\Psi} \\ \end{cases} \right\}= \left[ \begin{array}{c} 1&\sin\Phi\tan\Theta&\cos\Phi\tan\Theta\\ 0&\cos\Phi&-\sin\Phi\\ 0&\sin\Phi\sec\Theta&\cos\Phi\sec\Theta \end{array} \right] \left. \begin{cases}{} P \\ Q\\ R \\ \end{cases} \right\} $$

Answer 1

How to derive the relation between euler angles and angular velocity \begin{align*} &\text{The equations to calculate the angular velocity $\vec{\omega}$ for a given transformation matrix $S$ are: } \\\\ &\left[_B^I\dot{S}\right]=\left[\tilde{\vec{\omega}}_I\right]\left[_B^I S\right]\,\quad \Rightarrow \left[\tilde{\vec{\omega}}_I\right]=\left[_B^I\dot{S}\right]\left[_I^B S\right]\\ &\text{or}\\ &\left[_B^I\dot{S}\right]=\left[_B^I S\right]\left[\tilde{\vec{\omega}}_B\right]\,\quad \Rightarrow \left[\tilde{\vec{\omega}}_B\right]=\left[_I^B S\right]\left[_B^I\dot{S}\right]\\ &\text{with}\\ &\left[_B^I S\right]\,\left[_I^B S \right]= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix}\\\\ &\text{$\left[_B^I{S}\right]$ Transformation matrix between B- System and I-System }\\ &\text{$\left[\vec{\omega}\right]_B$ Vector components B-System}\\ &\text{$\left[\vec{\omega}\right]_I$ Vector components I-System and}\\ &\tilde{\vec{\omega}}= \begin{bmatrix} 0 & -\omega_z &\omega_y \\ \omega_z & 0 & -\omega_x \\ -\omega_y & \omega_x & 0 \\ \end{bmatrix}\\\\ &\textbf{Example: Transformation matrix Euler angle}\\ &\left[_B^I{S}\right]=S_z(\psi)\,S_y(\vartheta)\,S_z(\varphi)\\ &S_z(\psi)=\left[ \begin {array}{ccc} \cos \left( \psi \right) &-\sin \left( \psi \right) &0\\ \sin \left( \psi \right) &\cos \left( \psi \right) &0\\ 0&0&1\end {array} \right]\\ &S_y(\vartheta)=\left[ \begin {array}{ccc} \cos \left( \vartheta \right) &0&\sin \left( \vartheta \right) \\ 0&1&0 \\ -\sin \left( \vartheta \right) &0&\cos \left( \vartheta \right) \end {array} \right]\\ &S_z(\varphi)=\left[ \begin {array}{ccc} \cos \left( \varphi \right) &-\sin \left( \varphi \right) &0\\ \sin \left( \varphi \right) &\cos \left( \varphi \right) &0\\ 0&0&1\end {array} \right]\\ &\Rightarrow\\ &\vec{\omega}_B= \left[ \begin {array}{ccc} 0&\sin \left( \varphi \right) &-\cos \left( \varphi \right) \sin \left( \vartheta \right) \\ 0&\cos \left( \varphi \right) &\sin \left( \varphi \right) \sin \left( \vartheta \right) \\ 1 &0&\cos \left( \vartheta \right) \end {array} \right] \begin{bmatrix} \dot{\varphi} \\ \dot{\vartheta}\\ \dot{\psi} \\ \end{bmatrix}\\ &\begin{bmatrix} \dot{\varphi} \\ \dot{\vartheta}\\ \dot{\psi} \\ \end{bmatrix}= \left[ \begin {array}{ccc} {\frac {\cos \left( \varphi \right) \cos \left( \vartheta \right) }{\sin \left( \vartheta \right) }}&-{ \frac {\sin \left( \varphi \right) \cos \left( \vartheta \right) }{ \sin \left( \vartheta \right) }}&1\\ \sin \left( \varphi \right) &\cos \left( \varphi \right) &0\\ -{\frac {\cos \left( \varphi \right) }{\sin \left( \vartheta \right) }}&{\frac {\sin \left( \varphi \right) }{\sin \left( \vartheta \right) }}&0\end {array} \right] \begin{bmatrix} \omega_x \\ \omega_y\\ \omega_z \\ \end{bmatrix}_B\\ &\vec{\omega}_I= \left[ \begin {array}{ccc} \cos \left( \psi \right) \sin \left( \vartheta \right) &-\sin \left( \psi \right) &0\\ \sin \left( \psi \right) \sin \left( \vartheta \right) &\cos \left( \psi \right) &0\\ \cos \left( \vartheta \right) &0& 1\end {array} \right] \begin{bmatrix} \dot{\varphi} \\ \dot{\vartheta}\\ \dot{\psi} \\ \end{bmatrix}\\ &\begin{bmatrix} \dot{\varphi} \\ \dot{\vartheta}\\ \dot{\psi} \\ \end{bmatrix}= \left[ \begin {array}{ccc} {\frac {\cos \left( \psi \right) }{\sin \left( \vartheta \right) }}&{\frac {\sin \left( \psi \right) }{\sin \left( \vartheta \right) }}&0\\ -\sin \left( \psi \right) &\cos \left( \psi \right) &0\\ -{\frac { \cos \left( \vartheta \right) \cos \left( \psi \right) }{\sin \left( \vartheta \right) }}&-{\frac {\cos \left( \vartheta \right) \sin \left( \psi \right) }{\sin \left( \vartheta \right) }}&1\end {array} \right] \begin{bmatrix} \omega_x \\ \omega_y\\ \omega_z \\ \end{bmatrix}_I \end{align*}

Answer 2

Here is how to evaluate the rotational kinematics of a rigid body from the Euler angles

$$\boldsymbol{\omega} = \hat{\imath} \dot{\Phi} + \mathrm{R}_X ( \hat{\jmath} \dot{\Theta} + \mathrm{R}_Y \hat{k} \dot{\Psi}) \tag{1} $$

Here is how to derive the above:

Consider the orientation to be defined as sequence of thre elementary rotations $$ \mathrm{R} = \mathrm{R}_X \mathrm{R}_Y \mathrm{R}_Z \tag{2}$$

where $\mathrm{R}_X = \mathrm{rot}(\hat{\imath},\,\Phi)$, $\mathrm{R}_Y = \mathrm{rot}(\hat{\jmath},\,\Theta)$ and $\mathrm{R}_Z = \mathrm{rot}(\hat{k},\,\Psi)$.

Now the derivative on a rotating frame dictates that $$ \begin{aligned} \dot{\mathrm{R}}_X & = (\hat{\imath} \dot{\Phi}) \times \mathrm{R}_X \\ \dot{\mathrm{R}}_Y & = (\hat{\jmath} \dot{\Theta}) \times \mathrm{R}_Y \\ \dot{\mathrm{R}}_Z & = (\hat{k} \dot{\Psi}) \times \mathrm{R}_Z \end{aligned} $$

and

$$\dot{\mathrm{R}} = \boldsymbol{\omega} \times \mathrm{R} \tag{3} $$ which is used to derive $\boldsymbol{\omega}$, the rotational velocity of the rigid body.

Starting from the product rule on the left-hand side of (3)

$$ \dot{\mathrm{R}} = \dot{\mathrm{R}}_X\mathrm{R}_Y \mathrm{R}_Z + \mathrm{R}_X\dot{\mathrm{R}}_Y \mathrm{R}_Z + \mathrm{R}_X \mathrm{R}_Y\dot{\mathrm{R}}_Z$$

and substitute the derivatives from rotating frames to equate to the right-hand side of (3)

$$ \boldsymbol{\omega} \times \mathrm{R} = \left((\hat{\imath} \dot{\Phi}) \times \mathrm{R}_X \right) (\mathrm{R}_Y \mathrm{R}_Z) + \mathrm{R}_X \left( (\hat{\jmath} \dot{\Theta}) \times \mathrm{R}_Y \right) \mathrm{R}_Z + (\mathrm{R}_X \mathrm{R}_Y) \left( (\hat{k} \dot{\Psi}) \times \mathrm{R}_Z \right) $$

now start grouping and distribute the rotations. Note that $\mathrm{R} (a \times b) = (\mathrm{R} a) \times (\mathrm{R} b)$ is used below.

$$\begin{aligned} \boldsymbol{\omega} \times \mathrm{R} & = (\hat{\imath} \dot{\Phi}) \times \mathrm{R} + (\mathrm{R}_X \hat{\jmath} \dot{\Theta}) \times \mathrm{R} + (\mathrm{R}_X \mathrm{R}_Y \hat{k} \dot{\Psi}) \times \mathrm{R} \\ & = \left( \hat{\imath} \dot{\Phi}+\mathrm{R}_X \hat{\jmath} \dot{\Theta}+\mathrm{R}_X \mathrm{R}_Y \hat{k} \dot{\Psi} \right) \times \mathrm{R} \end{aligned} $$

or

$$ \boxed{ \boldsymbol{\omega} = \hat{\imath} \dot{\Phi}+\mathrm{R}_X \hat{\jmath} \dot{\Theta}+\mathrm{R}_X \mathrm{R}_Y \hat{k} \dot{\Psi} } $$

Using linear algebra the above is

$$\boldsymbol{\omega} = \pmatrix{1 \\ 0 \\ 0} \dot{\Phi}+\pmatrix{0 \\ \cos\Phi \\ \sin\Phi } \dot{\Theta}+ \pmatrix{ \sin\Theta \\ -\sin\Phi\cos\Theta \\ \cos\Phi \cos\Theta } \dot{\Psi} $$