# Relating angular velocity between rotating frames

### Notation

I'll be using Hubert Hahn's notation for my question. Hahn has an algebraic treatment of all values.

• $$\omega_{GN}^{G}$$ is the angular velocity of frame $$G$$ with respect to frame $$N$$, represented in frame $$G$$, that is to say $$\omega_{GN}^{G} = \omega_1.\hat{g}_1 +\omega_2.\hat{g}_2 +\omega_3.\hat{g}_3$$
• $$A^{BN}$$ shall be the transformation matrix that transforms an orthogonal vector represented in frame $$N$$ to a vector represented in frame $$B$$, i.e. $$\omega^G_{GN} = A^{GN} \cdot \omega^{N}_{GN}$$, where $$\cdot$$ is algebraic multiplication.

### Details

• Rotations using Bryant angles a.k.a Cardan Angles, euler angles.
• I have a space-fixed frame with no rotation $$N$$
• a body-fixed frame on a rotating body $$B$$ whose $$\dot{\eta}=\omega_{BN}^{N}$$ I know (Angular velocity of frame $$B$$ with respect to $$N$$, represented in frame $$N$$. My absolute angles $$\eta$$ represents this body.)
• Another frame $$G$$ which rotates about a fixed point on the first body (body with frame $$B$$). I have information on $$G$$'s rotation with respect to $$B$$: $$\omega_{GB}^{G}$$ known.
• 6dof in play

### Problem

How would I go about calculating $$G$$'s rotation relative to space-fixed frame $$N$$ ($$\omega_{GN}^{N}$$)?

### Attempt at a solution

Since $$G$$'s rotation is defined with respect to $$B$$ I'd argue we split $$\omega_{GN}^G$$ like so $$\omega_{GN}^G = \omega_{GB}^G + \omega_{BN}^G =\omega_{GB}^G + A^{GB}\omega_{BN}^B$$

I worry I'm missing out on the kinematic attitude treatment.

According to Hahn: $$\dot{\eta} = H(\eta)\cdot \omega^R_{LR} = H(\eta)\cdot A^{RL} \cdot \omega^L_{LR}$$, where $$H(\eta)$$ is the kinematic attitude matrix.
thus:

• We can calculate space-fixed angular velocity of frame $$B$$: $$\dot{\eta}= H(\eta) \cdot\omega^N_{BN} = H(\eta) \cdot A^{BN}\cdot \omega^B_{BN}$$... but I'm not sure why $$\dot{\eta}$$ is not equal to $$\omega^N_{BN}$$.
• Can you elaborate on $H(\eta)$ a bit. I am not immediately familiar with the kinematic attitude matrix, and why it would be different from change of basis $A$ matrix. – JAlex Aug 27 '20 at 17:21
• Right, so the physical interpretation of the vector $\omega_{LR}^L$ is complicated by the fact it defines an angular rate in a frame which is described by a change of basis matrix which does not get along well with angle vectors (or so I understand from Hahn's extended proofs). imgur.com/a/MCMvcbP – FemtoComm Aug 27 '20 at 17:51
• That is pretty stanard, where $\tfrac{{\rm d}}{{\rm d}t} {A}^{BN} = \omega_{BN}^B \times A^{BN}$. But the angle vectors belong in configuration space and are not to be mixed with cartesian vectors. – JAlex Aug 27 '20 at 18:53
• @JAlex So angle vectors are never to be mixed with cartesian vectors? Or is that a product of the specific treatment done by Hahn? – FemtoComm Aug 27 '20 at 19:32
• The go between is the Jacobian matrix which converts angle speed vectors to cartesian rotational velocity. – JAlex Aug 28 '20 at 19:46

## 1 Answer

@JAlex Answered the question in the comments. $$\eta$$ is NOT a cartesian vector. The attitude matrix converts a cartesian angular rate of the rotating frame (say $$\omega_{BN}^{N}$$) to the rigid-body-orientation-parameter-representation rate of change $$\dot{\eta}$$! I call them parameters because their derivative ($$\dot{\eta}$$) is not to be confused with an angular velocity. It is more related to the derivative of the transformation matrix, like JAlex points out:

$$\dot{A}^{BN} = \omega_{BN}^{B} \times A^{BN}$$

My mind is blown. I had read many rigid body related documents but none were clear on this matter.