here's my question:

I have position and velocity vectors of a body in the inertial frame. Now i need to switch the reference system to body frame.

So i have

$\bar{x}_b = \hat{R}\bar{x}_I$


$\hat{R}=\hat{R}( \phi(t),\theta(t),\psi(t)) = \left[ \begin{matrix}\cos{(\theta)}\cos{(\psi)} & \cos{(\theta)}\sin{(\psi)} & -\sin{(\theta)}\\ \sin{(\phi)}\sin{(\theta)}\cos{(\psi)}-\cos{(\phi)}\sin{(\psi)} & \sin{(\phi)}\sin{(\theta)}\sin{(\psi)}+\cos{(\phi)}\cos{(\psi)} & \sin(\phi)\cos(\theta) \\ \cos{(\phi)}\sin{(\theta)}\cos{(\psi)}+\sin{(\phi)}\cos{(\psi)} & \sin{(\phi)}\sin{(\theta)}\sin{(\psi)}-\sin{(\phi)}\cos{(\psi)} & \cos(\phi)\sin(\theta) \end{matrix} \right]$

so now i shoud do:

$\bar{v}_b = \dot{\bar{x}}_b = \frac{\partial{\hat{R}}}{\partial{t}}\cdot \bar{x}_I +\hat{R} \cdot\frac{\partial{\bar{x}_I}}{\partial{t}} $

is this correct? My doubt is about rotational matrix, should it be derived? Or am i missing something?

Thanks in advance!

  • $\begingroup$ Or just can i do: $\bar{v}_b = \hat{R}\bar{v}_I$ ? $\endgroup$ – DiTTiD Sep 27 '12 at 23:54

I can say quite surely that the right answer is that:

$$ V_\text{b} = \hat{R}_{\text{i} \rightarrow \text{b}} \cdot V_\text{i}$$

where $V_\text{b}$ is $\dot{x}_\text{b}$ (velocity in body frame) and $V_\text{i}$ the velocity in inertial frame.

  • $\begingroup$ Some explanation would be appropriate, but I suppose you are answering your own question... $\endgroup$ – Austin Mohr Nov 16 '12 at 7:30
  • $\begingroup$ It simply that a rotation matrix transforms an anykind vector expressed in a reference frame to another. So there's no need to derive itself. I'm no so mathematical expert so probably my answer could not be fully useful. $\endgroup$ – DiTTiD Nov 18 '12 at 20:18

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