Relating angular velocity between rotating frames

Question

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 $\omega^N_{BN}$.

Answer 1

@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.