I'm struggling to understand the relation between the angles used to compose a rotation matrix and the angular velocity vector of the body expressed in the body frame.

Assume there is no translation between the body frame and the world frame. If I have a 3D vector specified in the body coordinate system, $v_b$, and I want to express it in the world coordinate system, $v_w$, I use a rotation matrix like this: $$ v_w=R\cdot v_b $$

where $R$ is formed by multiplying simpler rotation matrices, like this: $$ R = R_x(\theta_r).R_y(\theta_p).R_z(\theta_y) $$

where $R_i(\theta)$ is the rotation matrix that rotates a vector around the coordinate $i$, by $\theta$ radians.

Now, I would like to calculate the angular velocity vector (i.e. the vector, $\omega$, specified in the body coordinate system, which is aligned with the axis of rotation and has the magnitude equal to the angular speed which the frame is rotating). At first I thought this should simply be:

$$ \omega = [\dot{\theta}_r, \dot{\theta}_p, \dot{\theta}_y]^T $$

but now I'm really starting to doubt that. Can someone provide some insight into how I can compute this vector?


1 Answer 1


To get the angular velocity from a time-dependent rotation matrix $R(t)\in \mathrm{SO}(3)$, all you need to do is differentiate.

More specifically, since you know that $R(t)^TR(t)\equiv \mathbb 1$, differentiating with respect to time you get $R(t)^TR'(t)+R'(t)^TR(t) = 0$, and w.l.o.g. taking $t=0$ and therefore $R(t)=\mathbb 1$, you get $R'(t) = -R'(t)^T$, i.e. that the derivative of the rotation matrix is skew symmetric. This means that you can write it as $$ \frac{\mathrm dR}{\mathrm dt} = \begin{pmatrix}0 & -\omega_z & \omega_y \\ \omega_z & 0 & -\omega_x \\ -\omega_y & \omega_x & 0\end{pmatrix} $$ for some triplet of numbers $\omega_x,\omega_y,\omega_z$. These turn out to be the components of the angular velocity, at least at those times when $R(t)=\mathbb 1$. In general you need to do some playing around with frames of reference, which are duly explained in Wikipedia, but the idea remains unchanged.

The mechanism you propose, i.e. getting the angular velocity directly from the Euler angles, is rather more complicated - for the details see a previous question on this site, Angular Velocity expressed via Euler Angles.


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