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.


  • 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


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.

  • 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 $\omega^N_{BN}$.
  • $\begingroup$ 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. $\endgroup$
    – JAlex
    Commented Aug 27, 2020 at 17:21
  • $\begingroup$ 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 $\endgroup$
    – FemtoComm
    Commented Aug 27, 2020 at 17:51
  • 1
    $\begingroup$ 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. $\endgroup$
    – JAlex
    Commented Aug 27, 2020 at 18:53
  • $\begingroup$ @JAlex So angle vectors are never to be mixed with cartesian vectors? Or is that a product of the specific treatment done by Hahn? $\endgroup$
    – FemtoComm
    Commented Aug 27, 2020 at 19:32
  • $\begingroup$ The go between is the Jacobian matrix which converts angle speed vectors to cartesian rotational velocity. $\endgroup$
    – JAlex
    Commented Aug 28, 2020 at 19:46

1 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.


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