Angular Velocity via Extrinsic Euler Angles

Question

I am wondering if the angular velocity of a rotating coordinate system, if expressed through extrinsic Euler angles, is $(\dot{\alpha},\dot{\beta}, \dot{\gamma})$ since extrinsic Euler angles are rotations about fixed axes so the rates should be orthogonal to each other.

Answer 1

$\newcommand{\bl}[1]{\boldsymbol{#1}} \newcommand{\e}{\bl=} \newcommand{\p}{\bl+} \newcommand{\m}{\bl-} \newcommand{\gr}{\bl>} \newcommand{\les}{\bl<} \newcommand{\greq}{\bl\ge} \newcommand{\leseq}{\bl\le} \newcommand{\plr}[1]{\left(#1\right)} \newcommand{\blr}[1]{\left[#1\right]} \newcommand{\lara}[1]{\langle#1\rangle} \newcommand{\lav}[1]{\langle#1|} \newcommand{\vra}[1]{|#1\rangle} \newcommand{\lavra}[2]{\langle#1|#2\rangle} \newcommand{\lavvra}[3]{\langle#1|\,#2\,|#3\rangle} \newcommand{\vp}{\vphantom{\dfrac{a}{b}}} \newcommand{\hp}[1]{\hphantom{#1}} \newcommand{\x}{\bl\times} \newcommand{\qqlraqq}{\qquad\bl{-\!\!\!-\!\!\!-\!\!\!\longrightarrow}\qquad}$

$\color{blue}{\bl\S\textbf{A. Rotation of coordinate axes via Euler angles}}$

Reference : $''$Classical Mechanics$''$, H.Goldstein-C.Poole-J.Safko, 3rd Edition, 4.4 $\blacksquare$ THE EULER ANGLES.

A rotation of coordinate axes via the Euler angles $\:\psi,\theta,\phi \:$ is represented by the matrix ^{\begin{equation} \!\!\!\! \mathrm A\plr{\phi,\theta,\psi}\e \begin{bmatrix} \hp\m\cos\psi\cos\phi\m\cos\theta\sin\phi\sin\psi& \hphantom\p\cos\psi\sin\phi\p\cos\theta \cos\phi\sin\psi &\sin\psi\sin\theta \vp\\ \,\m\sin\psi\cos\phi\m\cos\theta \sin\phi \cos\psi&\,\m\sin\psi\sin\phi\p\cos\theta\cos\phi \cos\psi&\cos\psi\sin\theta \vp\\ \sin\theta \sin\phi&\m\sin\theta \cos\phi&\cos\theta\vp \end{bmatrix} \tag{A-01}\label{A-01} \end{equation}} according to the following scheme \begin{equation} \bl{xyz} \quad \stackrel{\bl{\phi,\,z}}{\bl\Longrightarrow} \quad {\color{red}{\bl{\xi \eta \zeta}} } \quad \stackrel{\bl{\theta,\,\xi}}{\bl\Longrightarrow} \quad {\color{green}{\bl{\xi' \eta' \zeta'}}} \quad \stackrel{\bl{\psi,\,\zeta'}}{\bl\Longrightarrow} \quad {\color{blue}{\bl{x'y'z'}}} \tag{A-02}\label{A-02} \end{equation} as shown in Figure-01.

The first rotation $\:\mathrm D\:$ is around axis $\:\bl z\:$ by angle $\:\phi$, so $\:\bl{\zeta\equiv z}$. The second rotation $\:\mathrm C\:$ is around axis $\:\bl\xi\:$ (the $''$new $\:\bl x\:$-axis$''$) by angle $\:\theta$, so $\bl{\xi'\equiv\xi}$. The third rotation $\:\mathrm B\:$ is around axis $\:\bl{\zeta'}\:$ by angle $\:\psi\:$, so $\:\bl{z'\equiv\zeta'}$.

Consequently we have \begin{equation} \mathrm A\plr{\phi,\theta,\psi}\e\mathrm B\plr\psi\mathrm C\plr{\theta}\mathrm D\plr{\phi} \tag{A-03}\label{A-03} \end{equation} where \begin{equation} \begin{split} \mathrm D\plr{\phi}&\e \begin{bmatrix} \hphantom\m\cos\phi&\sin\phi&\:\:0\:\:\vp\\ \m\sin\phi&\cos\phi&\:\:0\:\:\vp\\ \hphantom\m0&0&\:\:1\:\:\vp \end{bmatrix}\\ \mathrm C\plr{\theta}&\e \begin{bmatrix} \:\:1\:\:&\hphantom\m0&0\:\vp\\ \:\:0\:\:&\hphantom\m \cos\theta &\sin\theta\:\vp\\ \:\:0\:\:&\m\sin\theta&\cos\theta\:\vp \end{bmatrix}\\ \mathrm B\plr{\psi}&\e \begin{bmatrix} \hphantom\m\cos\psi &\sin\psi&\:0\:\:\vp\\ \m\sin\psi &\cos\psi &\:0\:\:\vp\\ \hphantom\m0&\hphantom\m0&\:1\:\:\vp \end{bmatrix}\\ \end{split} \tag{A-04}\label{A-04} \end{equation} Each one of the rotation matrices $\:\mathrm{A,B,C,D}\:$ is orthonormal with inverse its transpose.

Note that the transformation matrix $\:\mathrm A\:$ of equation \eqref{A-01} represents a passive rotation, that is a change of the coordinates between the frames $\:\bl{xyz}\:$ and $\:\bl{x'y'z'}\:$ due to the rotation of the former to the latter. It's not an active transformation that is it doesn't represent rotation of a rigid body.

$=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!$

$\color{blue}{\bl\S\textbf{B. Rotation of a rigid body via Euler angles}}$

To express the angular velocity of a rotating rigid body via the Euler angles in the $\underline{\texttt{space frame}\:\:\bl{xyz}}\:$ we attach to this body the $\underline{\texttt{body frame}\:\:\bl{x'y'z'}}\:$ shown in Figure-02.

Now we have an active rotation so the transformation matrix $\:\mathcal A\:$ which represents this body rotation is the inverse of that representing the passive one $\:\mathrm A$ \begin{equation} \mathcal A\e\mathrm A^{\m 1}\e\mathrm A^{\bl\top} \tag{B-01}\label{B-01} \end{equation} So \begin{equation} \mathcal A\e\mathrm D^{\m 1}\mathrm C^{\m 1}\mathrm B^{\m 1}\e\mathrm D^{\bl\top}\mathrm C^{\bl\top}\mathrm B^{\bl\top} \tag{B-02}\label{B-02} \end{equation} or \begin{equation} \mathcal A\e\mathcal D\,\mathcal C\,\mathcal B \tag{B-03}\label{B-03} \end{equation} where \begin{equation} \begin{split} \mathcal D&\e \begin{bmatrix} \hp\m\cos\phi&\m\sin\phi&\hp\m 0\vp\\ \hp\m\sin\phi&\hp\m\cos\phi&\hp\m 0\vp\\ \hp\m 0&\hp\m 0&\hp\m 1\vp \end{bmatrix}\e\mathrm D^{\m 1}\e\mathrm D^{\bl\top}\\ \mathcal C & \e \begin{bmatrix} \hp\m 1&\hp\m 0&\hp\m 0\:\vp\\ \hp\m 0&\hp\m\cos\theta&\m\sin\theta\:\vp\\ \hp\m 0&\hp\m\sin\theta&\hp\m\cos\theta\:\vp \end{bmatrix}\e \mathrm C^{\m 1}\e \mathrm C^{\bl\top}\\ \mathcal B & \e \begin{bmatrix} \hp\m\cos\psi&\m\sin\psi&\hp\m 0\vp\\ \hp\m\sin\psi&\hp\m\cos\psi&\hp\m 0 \vp\\ \hp\m 0&\hp\m 0&\hp\m 1\vp \end{bmatrix}\e\mathrm B^{\m 1}\e\mathrm B^{\bl\top}\\ \end{split} \tag{B-04}\label{B-04} \end{equation} Note that in order to express the body rotation via the Euler angles we apply a first rotation $\:\mathcal B\:$ which is a rotation around the $\underline{\text{space}}$ $\:\bl z\m$axis by an angle $\:\psi$, then a second rotation $\:\mathcal C\:$ which is a rotation around the $\underline{\text{space}}$ $\:\bl x\m$axis by an angle $\:\theta\:$ and finally a third rotation $\:\mathcal D\:$ which is a rotation around the $\underline{\text{space}}$ $\:\bl z\m$axis by an angle $\:\phi$. In the end the two frames, space and body, are found in the arrangement of Figure-01.

It's important to note here that neither $\:\mathcal C\:$ is around the new $\:\bl x-$axis after application of the$\:\mathcal B\m$rotation nor $\:\mathcal D\:$ is around the new $\:\bl z\m$axis after application of the combined $\:\mathcal{C\,B}\m$rotation. This is necessary in order to achieve finally the arrangement of Figure-01.

In case of rotational motion of the rigid body the Euler angles are functions of time $\:t\:$ and the same is valid for all rotation matrices $\:\mathcal{A ,B,C,D}$. We suppose that at time $\:t\e 0\:$ the body frame $\:\bl{x'y'z'}\:$ coincides with the space frame $\:\bl{xyz}$.

As shown in $\color{blue}{\textbf{Appendix-01}}$ if a rotation matrix $\:\mathcal S\plr t\:$ represents the rotational motion of a rigid body in a space frame $\:\bl{xyz}\:$ then the components of the instantaneous angular velocity $\:\bl\omega\e\plr{\omega_x,\omega_y,\omega_z}\:$ are determined as elements of an antisymmetric matrix \begin{equation} \dot{\mathcal S}\,\mathcal S^{\bl\top}\e \begin{bmatrix} \hp\m 0&\m\omega_{\bl z}&\hp\m\omega_{\bl y}\vp\\ \hp\m \omega_{\bl z}&\hp\m 0&\m\omega_{\bl x}\vp\\ \m\omega_{\bl y}&\hp\m\omega_{\bl x}&\hp\m 0\vp \end{bmatrix}\e\bl\omega\x \tag{B-05}\label{B-05} \end{equation}

The angular velocity $\:\bl\omega\:$ of our case will be determined by the antisymmetric matrix $\:\dot{\mathcal A}\,\mathcal A^{\bl\top}$. For this we take advantage of the composition $\:\mathcal A \e \mathcal{ D\, C\, B}\:$ so we have \begin{equation} \begin{split} \dot{\mathcal A}\mathcal A^{\bl\top}&\e\plr{\dot{\mathcal D}\mathcal C\mathcal B\p\mathcal D\dot{\mathcal C}\mathcal B\p\mathcal D\mathcal C\dot{\mathcal B}}\plr{\mathcal B^{\bl\top}\mathcal C^{\bl\top}\mathcal D^{\bl\top}}\\ & \e \dot{\mathrm D}\mathcal C\mathcal B\mathcal B^{\bl\top}\mathcal C^{\bl\top}\mathcal D^{\bl\top}\p\mathcal D\dot{\mathcal C}\mathcal B\mathcal B^{\bl\top}\mathcal C^{\bl\top}\mathcal D^{\bl\top}\p \mathcal D\mathcal C\dot{\mathcal B}\mathcal B^{\bl\top}\mathcal C^{\bl\top}\mathcal D^{\bl\top}\bl\implies\\ \end{split} \nonumber \end{equation}

\begin{equation} \underbrace{\dot{\mathcal A}\mathcal A^{\bl\top}\vp}_ {\boxed{\bl\omega\x\vphantom{_\psi}}}\e\underbrace{\dot{\mathcal D}\mathcal D^{\bl\top}\vphantom{\dfrac{a}{b}}}_{\boxed{\bl\omega_\phi\x\vphantom{_\psi}}}\p\underbrace{\mathcal D\plr{\dot{\mathcal C}\mathcal C^{\bl\top}}\mathcal D^{\bl\top}}_{\boxed{\bl\omega_\theta\x\vphantom{_\psi}}}\p \underbrace{\plr{\mathcal D\mathcal C\vphantom{\dfrac{a}{b}}}\plr{\dot{\mathcal B}\mathcal B^{\bl\top}}\plr{\mathcal D\mathcal C}^{\bl\top}}_{\boxed{\bl\omega_\psi\x}}\\ \tag{B-06}\label{B-06} \end{equation} that is \begin{equation} \boxed{\:\bl\omega\e\bl\omega_\phi\p\bl\omega_\theta\p\bl\omega_\psi\vp\:} \tag{B-07}\label{B-07} \end{equation}

$\blacksquare$ The angular velocity $\bl\omega_\phi\:$ : \begin{equation} \begin{split} \bl\omega_\phi\x\e\dot{\mathcal D}\mathcal D^{\bl\top}&\e \begin{bmatrix} \m\sin\phi&\m\cos\phi&\hp\m 0\vp\\ \hp\m\cos\phi&\m\sin\phi&\hp\m 0\vp\\ \hp\m 0&\hp\m 0&\hp\m 0\vp \end{bmatrix}\:\dot{\!\!\phi} \begin{bmatrix} \hp\m\cos\phi&\hp\m\sin\phi&\hp\m 0\vp\\ \m\sin\phi&\hp\m\cos\phi&\hp\m 0\vp\\ \hp\m 0&\hp\m 0&\hp\m 1\vp \end{bmatrix}\\ &\e \:\dot{\!\!\phi} \begin{bmatrix} \hp\m 0&\m 1&\hp\m 0\vp\\ \p 1&\hp\m 0&\hp\m 0\vp\\ \hp\m 0&\hp\m 0&\hp\m 0\vp \end{bmatrix}\e \plr{\:\dot{\!\!\phi}\,\mathbf{\hat{\mathbf z}}}\x \boldsymbol \implies\\ \end{split} \nonumber \end{equation}

\begin{equation} \boxed{\:\bl\omega_\phi\e\:\dot{\!\!\phi}\,\mathbf{\hat{\mathbf z}}\vp\:} \tag{B-08}\label{B-08} \end{equation}

$\blacksquare$ The angular velocity $\bl\omega_\theta\:$ :

First \begin{equation} \begin{split} \dot{\mathcal C}\mathcal C^{\bl\top}&\e \begin{bmatrix} \hp\m 0&\hp\m 0&\hp\m 0\:\vp\\ \hp\m 0&\m\sin\theta&\m\cos\theta\:\vp\\ \hp\m 0&\hp\m\cos\theta&\m\sin\theta\:\vp \end{bmatrix} \:\dot{\!\!\theta} \begin{bmatrix} \hp\m 1&\hp\m 0&\hp\m 0\:\vp\\ \hp\m 0&\hp\m\cos\theta&\hp\m\sin\theta\:\vp\\ \hp\m 0&\m\sin\theta&\hp\m\cos\theta\:\vp \end{bmatrix}\bl\implies\\ \end{split} \nonumber \end{equation} \begin{equation} \dot{\mathcal C}\mathcal C^{\bl\top}\e \:\dot{\!\!\theta} \begin{bmatrix} \hp\m 0&\hp\m 0&\hp\m 0\vp\\ \hp\m 0&\hp\m 0&\m 1\vp\\ \hp\m 0&\p 1&\hp\m 0\vp \end{bmatrix}\e \plr{\:\dot{\!\!\theta}\,\mathbf{\hat{\mathbf x}}}\x \tag{B-09}\label{B-09} \end{equation} so \begin{equation} \begin{split} \bl\omega_\theta\x&\e\mathcal D\plr{\dot{\mathcal C}\mathcal C^{\bl\top}}\mathcal D^{\bl\top}\\ &\e\hp{\dot{\theta}} \begin{bmatrix} \hp\m\cos\phi&\m\sin\phi&\hp\m 0\vp\\ \hp\m\sin\phi&\hp\m\cos\phi&\hp\m 0\vp\\ \hp\m 0&\hp\m 0&\hp\m 1\vp \end{bmatrix} \:\dot{\!\!\theta} \begin{bmatrix} \hp\m 0&\hp\m 0&\hp\m 0\vp\\ \hp\m 0&\hp\m 0&\m 1\vp\\ \hp\m 0&\p 1&\hp\m 0\vp \end{bmatrix} \begin{bmatrix} \hp\m\cos\phi&\hp\m\sin\phi&\hp\m 0\vp\\ \m\sin\phi&\hp\m\cos\phi&\hp\m 0\vp\\ \hp\m 0&\hp\m 0&\hp\m 1\vp \end{bmatrix}\\ & \e \:\dot{\!\!\theta} \begin{bmatrix} \hp\m 0&\hp\m 0&\hp\m\sin\phi\vp\\ \hp\m 0&\hp\m 0&\m\cos\phi\vp\\ \m\sin\phi&\hp\m\cos\phi&\hp\m 0\vp \end{bmatrix}\e\plr{\:\dot{\!\!\theta}\cos\phi\,\mathbf{\hat{\mathbf x}}\p\:\dot{\!\!\theta}\sin\phi\,\mathbf{\hat{\mathbf y}}}\x \e \plr{\:\dot{\!\!\theta}\, \boldsymbol{\hat{\boldsymbol \xi}}}\x \\ \end{split} \nonumber \end{equation} that is \begin{equation} \boxed{\:\bl\omega_\theta\e\:\dot{\!\!\theta}\cos\phi\,\mathbf{\hat{\mathbf x}}\p\:\dot{\!\!\theta}\sin\phi\,\mathbf{\hat{\mathbf y}}\e\: \dot{\!\!\theta}\, \bl{\hat{\bl \xi}}\vp\:} \tag{B-10}\label{B-10} \end{equation} where \begin{equation} \bl{\hat{\bl\xi}}\e\cos\phi\,\mathbf{\hat{\mathbf x}}\p\sin\phi\,\mathbf{\hat{\mathbf y}} \tag{B-11}\label{B-11} \end{equation}

$\blacksquare$ The angular velocity $\bl\omega_\psi\:$ :

First \begin{equation} \dot{\mathcal B}\mathcal B^{\bl\top}\e \begin{bmatrix} \m\sin\psi&\m\cos\psi&\hp\m 0\vp\\ \hp\m\cos\psi&\m\sin\psi&\hp\m 0\vp\\ \hp\m 0&\hp\m 0&\hp\m 0\vp \end{bmatrix}\:\dot{\!\!\psi} \begin{bmatrix} \hp\m\cos\psi&\hp\m\sin\psi&\hp\m 0\vp\\ \m\sin\psi&\hp\m\cos\psi&\hp\m 0\vp\\ \hp\m 0&\hp\m 0&\hp\m 1\vp \end{bmatrix} \nonumber \end{equation} so \begin{equation} \dot{\mathcal B}\mathcal B^{\bl\top}\e\:\dot{\!\!\psi} \begin{bmatrix} \hp\m 0&\m 1&\hp\m 0\vp\\ \p 1&\hp\m 0&\hp\m 0\vp\\ \hp\m 0&\hp\m 0&\hp\m 0\vp \end{bmatrix} \e\plr{\:\dot{\!\!\psi}\,\mathbf{\hat{\mathbf z}}}\x \tag{B-12}\label{B-12} \end{equation} Also \begin{equation} \mathcal D\mathcal C\e \begin{bmatrix} \hp\m\cos\phi&\m\sin\phi&\hp\m 0\vp\\ \hp\m\sin\phi&\hp\m\cos\phi&\hp\m 0\vp\\ \hp\m 0&\hp\m 0&\hp\m 1\vp \end{bmatrix} \begin{bmatrix} \hp\m 1&\hp\m 0&\hp\m 0\:\vp\\ \hp\m 0&\hp\m\cos\theta&\m\sin\theta\:\vp\\ \hp\m 0&\hp\m\sin\theta&\hp\m\cos\theta\:\vp \end{bmatrix} \nonumber \end{equation} that is \begin{equation} \hp{^{\bl\top}}\plr{\mathcal D\mathcal C}\e \begin{bmatrix} \cos\phi&\m\sin\phi\cos\theta&\hp\m\sin\phi\sin\theta\vp\\ \sin\phi&\hp\m\cos\phi\cos\theta&\m\cos\phi\sin\theta \vp\\ 0&\hp\m\sin\theta&\hp\m\cos\theta\vp \end{bmatrix} \tag{B-13}\label{B-13} \end{equation} and consequently \begin{equation} \plr{\mathcal D\mathcal C}^{\bl\top}\e \begin{bmatrix} \hp\m\cos\phi&\hp\m\sin\phi&\hp\m 0\vp\\ \m\sin\phi\cos\theta&\hp\m\cos\phi\cos\theta&\hp\m\sin\theta\vp\\ \hp\m\sin\phi\sin\theta&\m\cos\phi\sin\theta&\hp\m\cos\theta\vp \end{bmatrix} \tag{B-14}\label{B-14} \end{equation} Secondly \begin{equation} \plr{\mathcal D\mathcal C\vp}\plr{\dot{\mathcal B}\mathcal B^{\bl\top}}\e \begin{bmatrix} \cos\phi&\m\sin\phi\cos\theta&\hp\m\sin\phi\sin\theta\vp\\ \sin\phi&\hp\m\cos\phi\cos\theta&\m\cos\phi\sin\theta\vp\\ 0&\hp\m\sin\theta&\hp\m\cos\theta\vp \end{bmatrix} \:\dot{\!\!\psi} \begin{bmatrix} \hp\m 0 &\m 1&\hp\m 0\vp\\ \p 1&\hp\m 0&\hp\m 0\vp\\ \hp\m 0&\hp\m 0&\hp\m 0\vp \end{bmatrix} \nonumber \end{equation} so \begin{equation} \plr{\mathcal D\mathcal C\vp}\plr{\dot{\mathcal B}\mathcal B^{\bl\top}}\e\:\dot{\!\!\psi} \begin{bmatrix} \m\sin\phi\cos\theta&\m\cos\phi&\hp\m 0\vp\\ \hp\m\cos\phi\cos\theta&\m\sin\phi&\hp\m 0\vp\\ \hp\m\sin\theta&\hp\m 0&\hp\m 0\vp \end{bmatrix} \tag{B-15}\label{B-15} \end{equation} and

^{\begin{equation} \begin{split} \bl\omega_\psi\x&\e\plr{\mathcal D\mathcal C\vp}\plr{\dot{\mathcal B}\mathcal B^{\bl\top}}\plr{\mathcal D\mathcal C\vp}^{\bl\top}\\ & \e\:\dot{\!\!\psi} \begin{bmatrix} \m\sin\phi\cos\theta&\m\cos\phi&\hp\m 0\vp\\ \hp\m\cos\phi\cos\theta&\m\sin\phi&\hp\m 0\vp\\ \hp\m\sin\theta&\hp\m 0&\hp\m 0 \vp \end{bmatrix} \begin{bmatrix} \hp\m\cos\phi&\hp\m\sin\phi&\hp\m 0\vp\\ \m\sin\phi\cos\theta &\hp\m\cos\phi\cos\theta&\hp\m\sin\theta\vp\\ \hp\m\sin\phi\sin\theta&\m\cos\phi\sin\theta&\hp\m\cos\theta\vp \end{bmatrix}\\ &\e\:\dot{\!\!\psi} \begin{bmatrix} 0&\m\cos\theta&\m\cos\phi\sin\theta\vp\\ \cos\theta&\hp\m 0&\m\sin\phi\sin\theta\vp\\ \cos\phi\sin\theta&\hp\m\sin\phi\sin\theta&\hp\m 0\vp \end{bmatrix}\\ &\e\plr{\:\dot{\!\!\psi}\sin\theta\sin\phi\,\mathbf{\hat{\mathbf x}}\m\:\dot{\!\!\psi}\sin\theta\cos\phi\,\mathbf{\hat{\mathbf y}}\p\:\dot{\!\!\psi}\cos\theta\,\mathbf{\hat{\mathbf z}}}\x\e\plr{\:\dot{\!\!\psi}\,\mathbf{\hat{\mathbf z}}'}\x\\ \end{split} \tag{B-16}\label{B-16} \end{equation}} that is \begin{equation} \boxed{\:\bl\omega_\psi\e\:\dot{\!\!\psi}\sin\theta\sin\phi\,\mathbf{\hat{\mathbf x}}\m\:\dot{\!\!\psi}\sin\theta\cos\phi\,\mathbf{\hat{\mathbf y}}\p\:\dot{\!\!\psi}\cos\theta\,\mathbf{\hat{\mathbf z}}\e\:\dot{\!\!\psi}\,\mathbf{\hat{\mathbf z}}'\vp\:} \tag{B-17}\label{B-17} \end{equation} where \begin{equation} \mathbf{\hat{\mathbf z}}'\e\sin\theta\sin\phi\,\mathbf{\hat{\mathbf x}}\m\sin\theta\cos\phi\,\mathbf{\hat{\mathbf y}}\p\cos\theta\,\mathbf{\hat{\mathbf z}} \tag{B-18}\label{B-18} \end{equation} Using equations \eqref{B-07},\eqref{B-08},\eqref{B-10},\eqref{B-11},\eqref{B-17} and \eqref{B-18} we construct the following tables \eqref{Table-01} and \eqref{Table-02}. ^{\begin{equation} \begin{array}{||c|c|c|c||} \hline \textbf{angular}&\mathbf{\hat{\mathbf x}}&\mathbf{\hat{\mathbf y}}&\mathbf{\hat{\mathbf z}}\\ \textbf{velocity}&\textbf{component}&\textbf{component}&\textbf{component}\\ \hline \bl\omega_\phi\e\:\dot{\!\!\phi}\,\mathbf{\hat{\mathbf z}}&0&0&\:\dot{\!\!\phi}\vphantom{\tfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ \hline \bl\omega_\theta\e\:\dot{\!\!\theta}\,\bl{\hat{\bl\xi}}&\:\dot{\!\!\theta}\cos\phi&\:\dot{\!\!\theta}\sin\phi&0\vphantom{\tfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ \hline \bl\omega_\psi\e\:\dot{\!\!\psi}\,\mathbf{\hat{\mathbf z}}'&\:\dot{\!\!\psi}\sin\theta\sin\phi&\m \:\dot{\!\!\psi}\sin\theta\cos\phi&\:\dot{\!\!\psi}\cos\theta \vphantom{\tfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ \hline \bl\omega\e\underbrace{\bl\omega_\phi\p\bl\omega_\theta \p\bl\omega_\psi}_{\:\dot{\!\!\phi}\,\mathbf{\hat{\mathbf z}}\p\:\dot{\!\!\theta}\,\bl{\hat{\bl\xi}}\p\:\dot{\!\!\psi}\,\mathbf{\hat{\mathbf z}}'}& \underbrace{\:\dot{\!\!\theta}\cos\phi\p\:\dot{\!\!\psi}\sin\theta\sin\phi\vphantom{\dfrac{\tfrac{a}{b}}{b}}}_{\omega_x} & \underbrace{\:\dot{\!\!\theta}\sin\phi\m \:\dot{\!\!\psi}\sin\theta\cos\phi\vphantom{\dfrac{\tfrac{a}{b}}{b}}}_{\omega_y} & \underbrace{\:\dot{\!\!\psi}\cos\theta\p\:\dot{\!\!\phi}\vphantom{\dfrac{\tfrac{a}{b}}{b}}}_{\omega_z}\\ &&&\\ \hline \end{array} \tag{Table-01}\label{Table-01} \end{equation} \begin{equation} \begin{array}{||c|c|c|c||} \hline \textbf{angular}&\mathbf{\hat{\mathbf x}'}&\mathbf{\hat{\mathbf y}'}&\mathbf{\hat{\mathbf z}'}\\ \textbf{velocity}&\textbf{component}&\textbf{component}&\textbf{component} \\ \hline \bl\omega_\phi\e\:\dot{\!\!\phi}\,\mathbf{\hat{\mathbf z}}&\:\dot{\!\!\phi}\sin\theta\sin\psi&\:\dot{\!\!\phi}\sin\theta\cos\psi &\:\dot{\!\!\phi}\cos\theta\vphantom{\tfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ \hline \bl\omega_\theta\e\:\dot{\!\!\theta}\,\bl{\hat{\bl\xi}}&\:\dot{\!\!\theta}\cos\psi&\m\:\dot{\!\!\theta}\sin\psi&0\vphantom{\tfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ \hline \bl\omega_\psi\e\:\dot{\!\!\psi}\,\mathbf{\hat{\mathbf z}}'&0&0&\:\dot{\!\!\psi}\vphantom{\tfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ \hline \bl\omega\e\underbrace{\bl\omega_\phi\p\bl\omega_\theta \p\bl\omega_\psi}_{\:\dot{\!\!\phi}\,\mathbf{\hat{\mathbf z}}\p\:\dot{\!\!\theta}\,\bl{\hat{\bl\xi}}\p\:\dot{\!\!\psi}\,\mathbf{\hat{\mathbf z}}'} & \underbrace{\:\dot{\!\!\phi}\sin\theta\sin\psi\p\:\dot{\!\!\theta }\cos\psi}_{\omega_{x'}} & \underbrace{\:\:\dot{\!\!\phi}\sin\theta\cos\psi\m\:\dot{\!\!\theta }\sin\psi}_{\omega_{y'}} & \underbrace{\:\dot{\!\!\phi}\cos\theta\p\:\dot{\!\!\psi}}_{\omega_{z'}} \vphantom{\tfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}\\ &&&\\ \hline \end{array} \tag{Table-02}\label{Table-02} \end{equation}} In both tables we see the analysis of the total angular velocity $\:\bl\omega\:$ in terms of the partial angular velocities $\:\bl\omega_\phi,\bl\omega_\theta,\bl\omega_\psi$ \begin{equation} \bl\omega\e\bl\omega_\phi\p\bl\omega_\theta\p\bl\omega_\psi\e\:\dot{\!\!\phi}\,\mathbf{\hat{\mathbf z}}\p\:\dot{\!\!\theta}\,\bl{\hat{\bl\xi}}\p\:\dot{\!\!\psi}\,\mathbf{\hat{\mathbf z}}' \tag{B-19}\label{B-19} \end{equation} with graphical representation in Figure-03.

Moreover in \eqref{Table-01} we see the coordinates of the angular velocity $\:\bl\omega\:$ in the space frame $\:\bl{xyz}$ \begin{equation} \begin{split} \bl\omega&\e\omega_x\,\mathbf{\hat{\mathbf x}}\p\omega_y\,\mathbf{\hat{\mathbf y}}\p\omega_z\,\mathbf{\hat{\mathbf z}}\\ \omega_x&\e\:\dot{\!\!\theta}\cos\phi\p\:\dot{\!\!\psi}\sin\theta\sin\phi\vphantom{\dfrac{\tfrac{a}{b}}{b}} \\ \omega_y&\e\:\dot{\!\!\theta}\sin\phi\m \:\dot{\!\!\psi}\sin\theta\cos\phi \\ \omega_z&\e\:\dot{\!\!\psi}\cos\theta\p\:\dot{\!\!\phi}\\ \end{split} \tag{B-20}\label{B-20} \end{equation} shown in Figure-04. Note that equations \eqref{B-20} are identical to those in DERIVATION 15, page 182 in the Reference.

In \eqref{Table-02} we see also the coordinates of the angular velocity $\:\bl\omega\:$ in the body frame $\:\bl{x'y'z'}$ \begin{equation} \begin{split} \bl\omega&\e\omega_{x'}\,\mathbf{\hat{\mathbf x}'}\p\omega_{y'}\,\mathbf{\hat{\mathbf y}'}\p\omega_{z'}\,\mathbf{\hat{\mathbf z}'}\\ \omega_{x'}&\e\:\dot{\!\!\phi}\sin\theta\sin\psi\p\:\dot{\!\!\theta }\cos\psi \\ \omega_{y'}&\e\:\dot{\!\!\phi}\sin\theta\cos\psi\m\:\dot{\!\!\theta }\sin\psi \\ \omega_{z'}&\e\:\dot{\!\!\phi}\cos\theta\p\:\dot{\!\!\psi}\\ \end{split} \tag{B-21}\label{B-21} \end{equation} shown in Figure-05. Note that equations \eqref{B-21} are identical to those in equation (4.87), page 174 in the Reference.

$=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!$

$\color{blue}{\textbf{Appendix-01 : Angular velocity from rotation matrix}}$

Consider that a rotation matrix $\:\mathcal S\plr t\:$ represents the rotational motion of a rigid body around the origin of a space frame $\:\bl{xyz}$. Let $\:\mathrm P\:$ a given point on the rigid body with position vector $\:\mathbf r\plr 0\bl\equiv\mathbf r_0 \:$ at time $\:t\e 0$. At any instant $\:t\:$ its position vector $\:\mathbf r\plr t\:$ will be given by
\begin{equation} \mathbf r\plr t\e \mathcal S\plr t \mathbf r_0 \tag{01-01}\label{01-01} \end{equation} For the velocity $\:\mathbf v\plr t\:$ of point $\:\mathrm P\:$ at time $\:t\:$ we have \begin{equation} \mathbf v\e\mathbf{\dot{r}}\e\mathcal{\dot{S}}\mathbf r_0 \tag{01-02}\label{01-02} \end{equation} From \eqref{01-01} \begin{equation} \mathbf r_0\e\mathcal S^{\m 1}\mathbf r\e\mathcal S^{\bl\top}\mathbf r \tag{01-03}\label{01-03} \end{equation} so \begin{equation} \mathbf v\e\mathbf{\dot{r}}\e\mathcal{\dot{S}}\mathcal S^{\bl\top}\mathbf r \tag{01-04}\label{01-04} \end{equation} But \begin{equation} \mathcal S\mathcal S^{\bl\top}\e\mathcal I\quad \bl\implies\quad\mathcal{\dot{S}}\mathcal S^{\bl\top}\!\!\p\mathcal S\mathcal{\dot{S}}^{\bl\top}\!\!\e\mathcal O \tag{01-05}\label{01-05} \end{equation} yields \begin{equation} \boxed{\:\:\plr{\mathcal{\dot{S}}\mathcal S^{\bl\top}}\e\m\plr{\mathcal{\dot{S}}\mathcal S^{\bl\top}}^{\!\bl\top}\vp\:\:} \tag{01-06}\label{01-06} \end{equation} that is an antisymmetric matrix which could be expressed in terms of a real 3-vector $\:\bl\omega\e\plr{\omega_x,\omega_y,\omega_z}$, the instantaneous angular velocity of the rigid body \begin{equation} \dot{\mathcal S}\,\mathcal S^{\bl\top}\e \begin{bmatrix} \hp\m 0&\m\omega_{\bl z}&\hp\m\omega_{\bl y}\vp\\ \hp\m \omega_{\bl z}&\hp\m 0&\m\omega_{\bl x}\vp\\ \m\omega_{\bl y}&\hp\m\omega_{\bl x}&\hp\m 0\vp \end{bmatrix}\e\bl\omega\x \tag{01-07}\label{01-07} \end{equation} and equation \eqref{01-04} gives \begin{equation} \boxed{\:\:\mathbf v\e\mathbf{\dot{r}}\e\bl\omega\x \mathbf r\vp\:\:} \tag{01-08}\label{01-08} \end{equation}

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Answer 2

The Rotation matrix is created with those 3 matrices

Rotation about x-axes with the angle $~\alpha~$

$$\mathbf R_x= \left[ \begin {array}{ccc} 1&0&0\\0&\cos \left( \alpha \right) &-\sin \left( \alpha \right) \\ 0& \sin \left( \alpha \right) &\cos \left( \alpha \right) \end {array} \right] $$

Rotation about y-axes with the angle $~\beta~$

$$\mathbf R_y= \left[ \begin {array}{ccc} \cos \left( \beta \right) &0&\sin \left( \beta \right) \\ 0&1&0\\ -\sin \left( \beta \right) &0&\cos \left( \beta \right) \end {array} \right] $$

Rotation about z-axes with the angle $~\gamma~$ $$\mathbf R_z=\left[ \begin {array}{ccc} \cos \left( \gamma \right) &-\sin \left( \gamma \right) &0\\ \sin \left( \gamma \right) &\cos \left( \gamma \right) &0\\ 0&0&1\end {array} \right] $$

Example

• first rotation about the z-axes $~\mathbf R_z(\gamma)$
• second rotation about the new axes y' $~\mathbf R_{y'}(\beta)$
• third rotation about the new axes z' $~\mathbf R_{z'}(\alpha)$

hence the rotation matrix $\mathbf R~$ is

$$\mathbf R=\mathbf R_z(\gamma)\,~\mathbf R_{y'}(\beta)\,\mathbf R_{z'}(\alpha)$$

from here you obtain that angular velocity $~\mathbf\omega$

$$\mathbf\omega=\mathbf A(\alpha~,\beta~,\gamma)\,\begin{bmatrix} \dot\alpha \\ \dot\beta\\ \dot\gamma\\ \end{bmatrix} \Rightarrow\quad \begin{bmatrix} \alpha \\ \beta\\ \gamma\\ \end{bmatrix}=\int\,\mathbf A^{-1}(\alpha~,\beta~,\gamma)\,\mathbf\omega\,dt $$

your question.

for a "small" rotation angle $~\varphi~,$ $~\cos(\varphi)=1~,\sin(\varphi)=\varphi~$

the rotation matrix is now: $$\mathbf R=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix}+ \left[ \begin {array}{ccc} 0&-\gamma&\beta\\ \gamma &0&-\alpha\\ -\beta&\alpha&0\end {array} \right] $$

and the angular velocity

$$\mathbf\omega=\begin{bmatrix} \dot\alpha \\ \dot\beta\\ \dot\gamma\\ \end{bmatrix} \Rightarrow\quad \underbrace{\begin{bmatrix} \alpha \\ \beta\\ \gamma\\ \end{bmatrix}}_{\mathbf \phi}=\int\mathbf\omega\,dt $$ $\mathbf \phi~$ is now a pseudo vector.

hence: only for a small angles $~\alpha~,\beta~,\gamma~$ the angles are rotation about the axes $~x~,y'~,z'~$

Edit

how to obtain the angular velocity from the rotation matrix $~\mathbf R$

with $$\mathbf{\dot{R}}=\mathbf R\, \left[ \begin {array}{ccc} 0&-\omega_{{z}}&\omega_{{y}} \\ \omega_{{z}}&0&-\omega_{{x}} \\ -\omega_{{y}}&\omega_{{x}}&0\end {array} \right] \quad \Rightarrow\quad \left[ \begin {array}{ccc} 0&-\omega_{{z}}&\omega_{{y}} \\ \omega_{{z}}&0&-\omega_{{x}} \\ -\omega_{{y}}&\omega_{{x}}&0\end {array} \right]=\mathbf R^T\,\mathbf{\dot{R}} $$

I use MAPLE program to obtain the result , for the example above ,you obtain

$$ \left[ \begin {array}{c} \omega_{{x}}\\ \omega_{{y} }\\\omega_{{z}}\end {array} \right] = \underbrace{\left[ \begin {array}{ccc} 1&0&-\sin \left( \beta \right) \\ 0&\cos \left( \alpha \right) &\sin \left( \alpha \right) \cos \left( \beta \right) \\ 0&-\sin \left( \alpha \right) &\cos \left( \alpha \right) \cos \left( \beta \right) \end {array} \right]}_{\mathbf A(\alpha,\beta)} \,\underbrace{\begin{bmatrix} \dot\alpha \\ \dot\beta \\ \dot\gamma \\ \end{bmatrix}}_{\mathbf{\dot{\phi}}}\\ \begin{bmatrix} \dot\alpha \\ \dot\beta \\ \dot\gamma \\ \end{bmatrix}=\left[ \begin {array}{ccc} 1&{\frac {\sin \left( \alpha \right) \sin \left( \beta \right) }{\cos \left( \beta \right) }}&{\frac {\cos \left( \alpha \right) \sin \left( \beta \right) }{\cos \left( \beta \right) }}\\ 0&\cos \left( \alpha \right) &-\sin \left( \alpha \right) \\ 0&{\frac {\sin \left( \alpha \right) }{\cos \left( \beta \right) }}&{\frac {\cos \left( \alpha \right) }{\cos \left( \beta \right) }}\end {array} \right] \begin{bmatrix} \omega_x \\ \omega_y \\ \omega_z \\ \end{bmatrix} $$

the components of the angular velocity are given in the rotating system, not in inertial system. the components of the angular velocity in inertial system are

$$\mathbf\omega_I=\mathbf R\,\mathbf\omega$$

notice the singularity at $~\beta=\pi/2~$ . each rotation matrix has singularity at some rotation angle.

Answer 3

So there is a formal way of deriving $\boldsymbol{\omega}$ from any sequence of rotations $\mathrm{R}_i$ given their angles and speeds.

1. Definitions

You have a sequence of three elementary rotations $\mathrm{R}_\alpha$, $\mathrm{R}_\beta$, $\mathrm{R}_\gamma$, each about their axis $\boldsymbol{z}_\alpha$, $\boldsymbol{z}_\beta$, $\boldsymbol{z}_\gamma$, and with angles $\alpha$, $\beta$ and $\gamma$. The final orientation is given by $$ \mathrm{R} = \mathrm{R}_\alpha \mathrm{R}_\beta \mathrm{R}_\gamma \tag{1}$$

2. Time Derivatives

The time derivative of each elementary rotation is given by the differentiation on a rotating frame formula, considering that each column of an $\mathrm{R}_i$ matrix represents a frame fixed basis vector.

$$ \begin{aligned} \dot{\mathrm{R}}_\alpha & = (\boldsymbol{z}_\alpha \dot{\alpha}) \times \mathrm{R}_\alpha \\ \dot{\mathrm{R}}_\beta & = (\boldsymbol{z}_\beta \dot{\beta}) \times \mathrm{R}_\beta \\ \dot{\mathrm{R}}_\gamma & = (\boldsymbol{z}_\gamma \dot{\gamma}) \times \mathrm{R}_\gamma \\ \end{aligned} \tag{2}$$

3. Rotational Velocity

Similarly, the time derivative of the rotation matrix is used to define the rotational velocity vector $\boldsymbol{\omega}$

$$ \dot{\mathrm{R}} = \boldsymbol{\omega} \times \mathrm{R} \tag{3} $$

Now take (3) and apply (1) and the product rule of differentiation

$$ \begin{aligned}\boldsymbol{\omega}\times\mathrm{R} & =\tfrac{{\rm d}}{{\rm d}t}\mathrm{R}\\ & =\tfrac{{\rm d}}{{\rm d}t}\left(\mathrm{R}_{\alpha}\mathrm{R}_{\beta}\mathrm{R}_{\gamma}\right)\\ & =\tfrac{{\rm d}}{{\rm d}t}\left(\mathrm{R}_{\alpha}\right)\mathrm{R}_{\beta}\mathrm{R}_{\gamma}+\mathrm{R}_{\alpha}\tfrac{{\rm d}}{{\rm d}t}\left(\mathrm{R}_{\beta}\mathrm{R}_{\gamma}\right)\\ & =\left(\boldsymbol{z}_{\alpha}\dot{\alpha}\right)\times\mathrm{R}_{\alpha}\mathrm{R}_{\beta}\mathrm{R}_{\gamma}+\mathrm{R}_{\alpha}\tfrac{{\rm d}}{{\rm d}t}\left(\mathrm{R}_{\beta}\mathrm{R}_{\gamma}\right) \end{aligned} \tag{4a}$$

Here we take a detour to deal with the last term. Note that we use the distributed property for rotation matrices $\mathrm{R} ( a \times b) = (\mathrm{R} a) \times (\mathrm{R} b)$.

$$\begin{aligned}\tfrac{{\rm d}}{{\rm d}t}\left(\mathrm{R}_{\beta}\mathrm{R}_{\gamma}\right) & =\tfrac{{\rm d}}{{\rm d}t}\left(\mathrm{R}_{\beta}\right)\mathrm{R}_{\gamma}+\mathrm{R}_{\beta}\tfrac{{\rm d}}{{\rm d}t}\left(\mathrm{R}_{\gamma}\right)\\ & =\left(\boldsymbol{z}_{\beta}\dot{\beta}\right)\times\left(\mathrm{R}_{\beta}\mathrm{R}_{\gamma}\right)+\mathrm{R}_{\beta}\left(\left(\boldsymbol{z}_{\gamma}\dot{\gamma}\right)\times\mathrm{R}_{\gamma}\right)\\ & =\left(\boldsymbol{z}_{\beta}\dot{\beta}\right)\times\left(\mathrm{R}_{\beta}\mathrm{R}_{\gamma}\right)+\left(\mathrm{R}_{\beta}\boldsymbol{z}_{\gamma}\dot{\gamma}\right)\times\left(\mathrm{R}_{\beta}\mathrm{R}_{\gamma}\right)\\ & =\left(\boldsymbol{z}_{\beta}\dot{\beta}+\mathrm{R}_{\beta}\boldsymbol{z}_{\gamma}\dot{\gamma}\right)\times\left(\mathrm{R}_{\beta}\mathrm{R}_{\gamma}\right) \end{aligned} \ \tag{5}$$

Now continuing from (4a)

$$ \begin{aligned}\boldsymbol{\omega}\times\mathrm{R} & =\left(\boldsymbol{z}_{\alpha}\dot{\alpha}\right)\times\mathrm{R}+\mathrm{R}_{\alpha}\tfrac{{\rm d}}{{\rm d}t}\left(\mathrm{R}_{\beta}\mathrm{R}_{\gamma}\right)\\ & =\left(\boldsymbol{z}_{\alpha}\dot{\alpha}\right)\times\mathrm{R}+\mathrm{R}_{\alpha}\left(\left(\boldsymbol{z}_{\beta}\dot{\beta}+\mathrm{R}_{\beta}\boldsymbol{z}_{\gamma}\dot{\gamma}\right)\times\left(\mathrm{R}_{\beta}\mathrm{R}_{\gamma}\right)\right)\\ & =\left(\boldsymbol{z}_{\alpha}\dot{\alpha}\right)\times\mathrm{R}+\left(\mathrm{R}_{\alpha}\boldsymbol{z}_{\beta}\dot{\beta}+\mathrm{R}_{\alpha}\mathrm{R}_{\beta}\boldsymbol{z}_{\gamma}\dot{\gamma}\right)\times\left(\mathrm{R}_{\alpha}\mathrm{R}_{\beta}\mathrm{R}_{\gamma}\right)\\ & =\underbrace{\left(\boldsymbol{z}_{\alpha}\dot{\alpha}+\mathrm{R}_{\alpha}\boldsymbol{z}_{\beta}\dot{\beta}+\mathrm{R}_{\alpha}\mathrm{R}_{\beta}\boldsymbol{z}_{\gamma}\dot{\gamma}\right)}_{\text{defn. }\boldsymbol{\omega}}\times \mathrm{R} \end{aligned} \tag{4b} $$

and since $\mathrm{R} = \mathrm{R}_{\alpha}\mathrm{R}_{\beta}\mathrm{R}_{\gamma}$ the above is used to define the rotational velocity vector and the jacobian matrix.

$$ \begin{aligned}\boldsymbol{\omega} & = \boldsymbol{z}_{\alpha}\dot{\alpha}+\mathrm{R}_{\alpha}\boldsymbol{z}_{\beta}\dot{\beta}+\mathrm{R}_{\alpha}\mathrm{R}_{\beta}\boldsymbol{z}_{\gamma}\dot{\gamma}\\ \boldsymbol{\omega} & =\underbrace{\begin{bmatrix}\vdots & \vdots & \vdots\\ \boldsymbol{z}_{\alpha} & \mathrm{R}_{\alpha}\boldsymbol{z}_{\beta} & \mathrm{R}_{\alpha}\mathrm{R}_{\beta}\boldsymbol{z}_{\gamma}\\ \vdots & \vdots & \vdots \end{bmatrix}}_\text{Jacobian Matrix}\begin{pmatrix}\dot{\alpha}\\ \dot{\beta}\\ \dot{\gamma} \end{pmatrix} \end{aligned} \tag{6} $$

4. Recursive Formula

I like to arrange (6) as follows

$$\boldsymbol{\omega}=\boldsymbol{z}_{\alpha}\dot{\alpha}+\mathrm{R}_{\alpha}\left(\boldsymbol{z}_{\beta}\dot{\beta}+\mathrm{R}_{\beta}\left(\boldsymbol{z}_{\gamma}\dot{\gamma}+\ldots\right)\right)$$

Indicating that each successive rotation provides a $\boldsymbol{z}_i \dot{q}_i$ term on the local (intermediate) coordinate frame of its previous operation. The formula would continue on in a similar fashion if there were more than 3 rotations, although that is only seen on >6 DOF robotic arms or serial chains of rigid bodies.