# Angular velocity relative to some frame

In "Introduction to Robotics" by John Craig, we have the following statement: The vector $${}^A \Omega_B$$ describes the angular velocity of $$B$$ with respect to $$A$$, and $${}^C({}^A \Omega_B)$$ additionally specifies a coordinate frame $$C$$.

I have a really hard time understanding why frame $$A$$ matters.

In my head the axis of rotation that $$B$$ spins around is a geometric object that does not depend on any other frame $$A$$.

Can you give an example where the angular velocity vector is different for two "respect to" frames $$A_1$$ and $$A_2$$ and intuition what the relative frame means?

EDIT: I can visualize in 2D case that the relative frame matters for the $$\dot{\theta}$$ part of the angular velocity if the relative frame is moving. But at least the rotation axis seems unaffected... or its hard to intuit what the relationship is between the rotation axis and relative frame.

EDIT 2: There is a statement in Proposition 3.9, p77 of Modern Robotics: https://hades.mech.northwestern.edu/images/2/2e/MR-largefont-v2.pdf

It says that the angular velocity $$\omega_b$$ is "relative to" the stationary frame {b} that is instantaneously coincident with a frame attached to the moving body. Is this true just because $$\omega_b = R_{sb}^{-1} \omega_s$$? I can't argue it rigorously or thoroughly. Honestly I'm not even sure what the mathematical definition of "relative" is.

EDIT 3: Also, I'm not sure why later on in Proposition 3.9 they can claim that $$\dot{R} R^{-1}$$ is independent of {b} and similarly for $$R{-1} \dot{R}$$. I don't see any proof.

Consider the case where A is itself a rotating frame. In such case, the angular velocity with respect to frame A will be more than just the angular velocity of B in an inertial frame, it must also include the effects of the rotation of frame A such that, in frame A, the angular velocity describes the rate of change.

One thing to consider, which may help: consider that there is a rotation matrix $$(^AR_B)$$ which rotates from frame A to frame B. We can consider the derivative of this matrix, $$\frac{d(^AR_B)}{dt}$$. With some math we can show that $$\frac{d(^AR_B)}{dt}=\left [\begin{matrix}0 & -z & y \\ z & 0 & -x \\ -y & x & 0\end{matrix}\right ]\cdot (^AR_B) = [\omega]_\times\cdot (^AR_B)$$. That skew-symmetric matrix $$[\omega]_\times$$ is where the angular velocity came from. If I use a transform to get to some other frame, such as $$(^CR_A)$$, I can write $$\frac{d(^AR_B)}{dt}\cdot(^CR_A)=[\omega]_\times\cdot (^AR_B)\cdot (^CR_A)=[\omega]_\times\cdot (^CR_B)$$. Now if $$\omega$$ was independent of frame A, and was valid for all frames, then that last bit, $$[\omega]_\times\cdot (^CR_B)$$ would need to be equal to $$\frac{d(^CR_B)}{dt}$$ which would mean $$\frac{d(^CR_B)}{dt} \stackrel ? = \frac{d(^AR_B)}{dt}\cdot(^CR_A)$$ for all frames A B and C. I mark that as $$\stackrel ? =$$ because its using an assumption which we're still trying to determine if it is true (which is that angular velocity did not depend on a reference frame)

But we quickly find that this is not true. For any rotating frame, we find that there is an extra term, which the Wikipedia page on Rotating Frames denotes with $$\Omega \times f$$: $$\frac{d}{dt}f = (\frac{df}{dt})_r+\Omega \times f$$ This does not match with the above simpler equation, which has no additional $$\Omega \times f$$ term, we must conclude that the assumption we started from is false. We must assume that "the angular velocity vector is independent of the frame of reference" is false.

• Thanks for your response. I have a followup to my question if you are able to answer. Commented Apr 4 at 5:32

$$\def \b {\mathbf}$$

you want to obtain the components of vector $$~\vec r_A~$$ , where $$~A~$$ is the coordinate system that the components are given, in coordinate system $$~B~$$ .

I use this notation:

$$\vec r_B= \left[_A^B \,\b R\right]\,\vec r_A$$

notice that the index $$~A~$$ canceled .

the transformation matrix $$~\left[_A^B \,\b R\right]~$$ is from coordinate $$~A~$$ to coordinate $$~B~$$ systems.

now take the time derivative

$$\dot{\vec{r}}_B= \left[_A^B \,\dot{\b{R}}\right]\,\vec r_A+ \left[_A^B \,{\b{R}}\right]\,\dot{\vec{r}}_A\tag 1$$

with

$$~I~$$

$$\left[_A^B \,\dot{\b{R}}\right]=\left[_A^B \,{\b{R}}\right] \left[ \begin {array}{ccc} 0&-\omega_{{z}}&\omega_{{y}} \\ \omega_{{z}}&0&-\omega_{{x}}\\ -\omega_{{y}}&\omega_{{x}}&0\end {array} \right]_A \quad,\Rightarrow\quad\vec\omega_A$$

thus equation (1)

$$\dot{\vec{r}}_B= \left[_A^B \,{\b{R}}\right]\,\left(\vec\omega_A\,\times\vec r_A\right)+ \left[_A^B \,{\b{R}}\right]\,\dot{\vec{r}}_A$$

again the index $$~A~$$ canceled

II

the angular velocity vector is given in $$~B~$$ system

the velocity is then

$$\dot{\vec{r}}_B= \,\left(\left[_B^A \,{\b{R}}\right]\vec\omega_B\,\times\vec r_A\right)+ \left[_A^B \,{\b{R}}\right]\,\dot{\vec{r}}_A$$

where $$\left[_B^A \,{\b{R}}\right]=\left[_A^B \,{\b{R}}\right]^T$$

and $$\left[_A^B \,\dot{\b{R}}\right]= \left[ \begin {array}{ccc} 0&-\omega_{{z}}&\omega_{{y}} \\ \omega_{{z}}&0&-\omega_{{x}}\\ -\omega_{{y}}&\omega_{{x}}&0\end {array} \right]_B\ \left[_A^B \,{\b{R}}\right] \quad,\Rightarrow\quad\vec\omega_B$$