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

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

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

  • $\begingroup$ Hi, welcome to Physics SE! Please don't post formulae as pictures or plain text, but use MathJax instead. MathJax is easy for people on all devices to read, and can show up clearer on different screen sizes and resolutions. I've edited it here as an example. Look at this Math SE meta post for a quick tutorial. $\endgroup$
    – user191954
    Aug 2, 2018 at 11:54
  • $\begingroup$ your example for the angular velocity is not for Euler angles, but for cardan angles $\endgroup$
    – Eli
    Aug 2, 2018 at 21:57

2 Answers 2


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*}


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


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

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


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

Using linear algebra the above is

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

  • 1
    $\begingroup$ This is kind of a while ago, but what is a vector cross product with a matrix mean? I don't understand $\endgroup$ Jun 19, 2021 at 5:57
  • 2
    $\begingroup$ @FourierFlux - makes a new matrix where is each column is the cross of the vector to that column. Since $R = [ i\,j\,k]$ has directions on its columns, then $$\omega \times \mathrm{R} = \begin{bmatrix} \omega \times \hat{i} & \omega \times \hat{j} & \omega \times \hat{k} \end{bmatrix} $$ $\endgroup$ Jun 19, 2021 at 11:38
  • 2
    $\begingroup$ @FourierFlux - commonly the $\omega \times A$ operation is also represented as a matrix operation $$ \begin{bmatrix} 0 & -\omega_z & \omega_y \\ \omega_z & 0 & -\omega_x \\ -\omega_y & \omega_x & 0 \end{bmatrix} \begin{bmatrix} A_{11} & \cdots & A_{1m} \\ A_{21} & \cdots & A_{2m} \\ A_{31} & \cdots & A_{3m} \end{bmatrix}$$ and the dimensions work out for $A$ being matrix or vector. $\endgroup$ Jun 19, 2021 at 11:41

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