Let us say you have a sequence of rotations about two axes, $\hat{x}$ and $\hat{y}$ by the angles $\varphi$ and $\theta$. You use rotation matrices to find the final orientation

$$ \mathtt{E} = {\rm Rot}(\hat{x},\varphi)\,{\rm Rot}(\hat{y},\theta) $$

Now lets add some motion, and give the angles some speed

$$ \begin{align} \dot{\varphi} & = \frac{{\rm d}\varphi}{{\rm d}t} \\ \dot{\theta} & = \frac{{\rm d}\theta}{{\rm d}t} \end{align} $$

The rotational velocity of this body is thus defined as

$$ \vec{\omega} = \hat{x} \dot{\varphi} + {\rm Rot}(\hat{x},\varphi) \hat{y} \dot{\theta} $$

This is because $\hat{y}$ rotates about $\hat{x}$.

This motion is decomposed onto a rotation axis $\hat{z}$ and a rotation magnitude $\omega$

$$ \omega = \| \hat{x} \dot{\varphi} + {\rm Rot}(\hat{x},\varphi) \hat{y} \dot{\theta} \| \\ \hat{z} = \frac{\vec{\omega}}{\omega} $$

Appendix I

To derive this you use the fact that the time derivative of a rotation matrix $\mathtt{E}$ is $$\dot{\mathtt{E}} = \vec{\omega} \times \mathtt{E}$$

Use the product rule to evaluate the left hand side (with $\mathtt{E} = \mathtt{R}_x \mathtt{R}_y$) as $$\dot{\mathtt{E}} = \left( \frac{{\rm d}}{{\rm d}t} \mathtt{R}_x \right) \mathtt{R}_y + \mathtt{R}_x \left(\frac{{\rm d}}{{\rm d}t} \mathtt{R}_y \right) $$ $$ = ((\hat{x} \dot{\varphi}) \times \mathtt{R}_x ) \mathtt{R}_y +\mathtt{R}_x ( (\hat{y} \dot{\theta})\times \mathtt{R}_y) $$ $$ = (\hat{x} \dot{\varphi}) \times (\mathtt{R}_x \mathtt{R}_y) + (\mathtt{R}_x \hat{y} \dot{\theta})\times (\mathtt{R}_x \mathtt{R}_y) $$ $$ = (\hat{x} \dot{\varphi} + \mathtt{R}_x \hat{y} \dot{\theta}) \times (\mathtt{R}_x \mathtt{R}_y) $$ $$ = \vec{\omega} \times \mathtt{E} $$

Let us say you have a sequence of rotations about two axes, $\hat{x}$ and $\hat{y}$ by the angles $\varphi$ and $\theta$. You use rotation matrices to find the final orientation

$$ \mathtt{E} = {\rm Rot}(\hat{x},\varphi)\,{\rm Rot}(\hat{y},\theta) $$

Now lets add some motion, and give the angles some speed

$$ \begin{align} \dot{\varphi} & = \frac{{\rm d}\varphi}{{\rm d}t} \\ \dot{\theta} & = \frac{{\rm d}\theta}{{\rm d}t} \end{align} $$

The rotational velocity of this body is thus defined as

$$ \vec{\omega} = \hat{x} \dot{\varphi} + {\rm Rot}(\hat{x},\varphi) \hat{y} \dot{\theta} $$

This is because $\hat{y}$ rotates about $\hat{x}$.

This motion is decomposed onto a rotation axis $\hat{z}$ and a rotation magnitude $\omega$

$$ \omega = \| \hat{x} \dot{\varphi} + {\rm Rot}(\hat{x},\varphi) \hat{y} \dot{\theta} \| \\ \hat{z} = \frac{\vec{\omega}}{\omega} $$

Appendix I

To derive this you use the fact that the time derivative of a rotation matrix $\mathtt{E}$ is $$\dot{\mathtt{E}} = \vec{\omega} \times \mathtt{E}$$

Use the product rule to evaluate the left hand side (with $\mathtt{E} = \mathtt{R}_x \mathtt{R}_y$) as $$\begin{align}\dot{\mathtt{E}} &= \left( \frac{{\rm d}}{{\rm d}t} \mathtt{R}_x \right) \mathtt{R}_y + \mathtt{R}_x \left(\frac{{\rm d}}{{\rm d}t} \mathtt{R}_y \right)\\ &= ((\hat{x} \dot{\varphi}) \times \mathtt{R}_x ) \mathtt{R}_y +\mathtt{R}_x ( (\hat{y} \dot{\theta})\times \mathtt{R}_y) \\ &= (\hat{x} \dot{\varphi}) \times (\mathtt{R}_x \mathtt{R}_y) + (\mathtt{R}_x \hat{y} \dot{\theta})\times (\mathtt{R}_x \mathtt{R}_y) \\ &= (\hat{x} \dot{\varphi} + \mathtt{R}_x \hat{y} \dot{\theta}) \times (\mathtt{R}_x \mathtt{R}_y) \\ &= \vec{\omega} \times \mathtt{E} \,.\end{align}$$

Let us say you have a sequence of rotations about two axes, $\hat{x}$ and $\hat{y}$ by the angles $\varphi$ and $\theta$. You use rotation matrices to find the final orientation

$$ E = {\rm Rot}(\hat{x},\varphi)\,{\rm Rot}(\hat{y},\theta) $$

Now lets add some motion, and give the angles some speed

$$ \begin{align} \dot{\varphi} & = \frac{{\rm d}\varphi}{{\rm d}t} \\ \dot{\theta} & = \frac{{\rm d}\theta}{{\rm d}t} \end{align} $$

The rotational velocity of this body is thus defined as

$$ \vec{\omega} = \hat{x} \dot{\varphi} + {\rm Rot}(\hat{x},\varphi) \hat{y} \dot{\theta} $$

This is because $\hat{y}$ rotates about $\hat{x}$.

This motion is decomposed onto a rotation axis $\hat{z}$ and a rotation magnitude $\omega$

$$ \omega = \| \hat{x} \dot{\varphi} + {\rm Rot}(\hat{x},\varphi) \hat{y} \dot{\theta} \| \\ \hat{z} = \frac{\vec{\omega}}{\omega} $$

Let us say you have a sequence of rotations about two axes, $\hat{x}$ and $\hat{y}$ by the angles $\varphi$ and $\theta$. You use rotation matrices to find the final orientation

$$ \mathtt{E} = {\rm Rot}(\hat{x},\varphi)\,{\rm Rot}(\hat{y},\theta) $$

Now lets add some motion, and give the angles some speed

$$ \begin{align} \dot{\varphi} & = \frac{{\rm d}\varphi}{{\rm d}t} \\ \dot{\theta} & = \frac{{\rm d}\theta}{{\rm d}t} \end{align} $$

The rotational velocity of this body is thus defined as

$$ \vec{\omega} = \hat{x} \dot{\varphi} + {\rm Rot}(\hat{x},\varphi) \hat{y} \dot{\theta} $$

This is because $\hat{y}$ rotates about $\hat{x}$.

This motion is decomposed onto a rotation axis $\hat{z}$ and a rotation magnitude $\omega$

$$ \omega = \| \hat{x} \dot{\varphi} + {\rm Rot}(\hat{x},\varphi) \hat{y} \dot{\theta} \| \\ \hat{z} = \frac{\vec{\omega}}{\omega} $$

Appendix I

To derive this you use the fact that the time derivative of a rotation matrix $\mathtt{E}$ is $$\dot{\mathtt{E}} = \vec{\omega} \times \mathtt{E}$$

Use the product rule to evaluate the left hand side (with $\mathtt{E} = \mathtt{R}_x \mathtt{R}_y$) as $$\dot{\mathtt{E}} = \left( \frac{{\rm d}}{{\rm d}t} \mathtt{R}_x \right) \mathtt{R}_y + \mathtt{R}_x \left(\frac{{\rm d}}{{\rm d}t} \mathtt{R}_y \right) $$ $$ = ((\hat{x} \dot{\varphi}) \times \mathtt{R}_x ) \mathtt{R}_y +\mathtt{R}_x ( (\hat{y} \dot{\theta})\times \mathtt{R}_y) $$ $$ = (\hat{x} \dot{\varphi}) \times (\mathtt{R}_x \mathtt{R}_y) + (\mathtt{R}_x \hat{y} \dot{\theta})\times (\mathtt{R}_x \mathtt{R}_y) $$ $$ = (\hat{x} \dot{\varphi} + \mathtt{R}_x \hat{y} \dot{\theta}) \times (\mathtt{R}_x \mathtt{R}_y) $$ $$ = \vec{\omega} \times \mathtt{E} $$