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 Aug 27 '20 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 Aug 27 '20 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 Aug 27 '20 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 Aug 27 '20 at 19:32
  • $\begingroup$ The go between is the Jacobian matrix which converts angle speed vectors to cartesian rotational velocity. $\endgroup$ – JAlex Aug 28 '20 at 19:46

@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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.