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 $B$: $dot{eta}= H(eta) cdotomega^N_{BN} = H(eta) cdot A^{BN}cdot omega^B_{BN}$… but I’m not sure why $dot{eta}$ is not equal to $omega^N_{BN}$.