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At a given time t, the rotation matrix R has the value:

$$R= \begin{pmatrix} 0.675 & −0.1724 &0.7174\\0.2474 & 0.9689& 0 &\\−0.6951& 0.1775&0.6967. \end{pmatrix}$$

The angular velocity in a body fixed frame $\hat ω^b$ at that same time t is:

$$\hat ω^b=\begin{pmatrix}0& −1 &0.9689&\\1 & 0 &−0.2474\\−0.9689&0.2474&0. \end{pmatrix}$$ What is $\hatω^s$, the angular velocity in an inertial frame?

I know that \begin{align*}\omega^s:\dot q &= \dot RR^Tq\\&=R\hat\omega^b\end{align*}

Therfore I did on Matlab:

>> R = [0.675 -0.1724 0.7174; 0.2474 0.9689 0; -0.6951 0.1775 0.6967];
>> wb= [0 -1 0.9689; 1 0 -0.2474; -0.9689 0.2747 0];
>> R*wb

ans =

   -0.8675   -0.4779    0.6967
    0.9689   -0.2474         0
   -0.4975    0.8865   -0.7174

Which is none of the provided answer:

enter image description here

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  • $\begingroup$ Hi! Did you find the solution to this? $\endgroup$ – octavian Apr 3 '17 at 16:49
  • $\begingroup$ @octavian SE works by providing an answer if you have one. In this case the problem is not well-defined enough to answer. $\endgroup$ – Mortified Through Math Apr 3 '17 at 17:00
  • $\begingroup$ @ALB if you don't understand it, it doesn't necessarily mean it's not well defined. $\endgroup$ – octavian Apr 3 '17 at 17:15
  • $\begingroup$ @Octavian given I work professionally on attitude control, I'm fairly sure I understand it. If you think you have an answer however, provide it. $\endgroup$ – Mortified Through Math Apr 3 '17 at 17:24
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    $\begingroup$ @Octavian I don't see the point of your combative attitude. For what it's worth, since you mention you need an answer to this question, OP posted the matrix form of $\omega \times$. This doesn't transform like a vector under a change of basis, but $\omega$ will. If you extract the components of $\omega$ from this representation of $\omega\times$ then you can use the rotation matrix (presuming it rotates correctly between the frames) to find the components in the desired frame and then recalculate $\omega\times$ using the transformed components. $\endgroup$ – Mortified Through Math Apr 3 '17 at 17:48
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The angular velocity in a body fixed frame $\hat{\omega}^b$ at that same time $t$ is: [insert matrix here]

No, that's not the angular velocity, that's the matrix version of $\omega\times$, and while we're at it the angular velocity of what in the body frame of what? Can you please explain the setup of the problem? How many bodies are there? What is rotating and what isn't? How is the inertial frame defined?

From what you wrote I would just read off the components of $\omega^b$ (a vector, not a matrix) and multiply them be $R$, trying to see if it agrees with any one of those answers, but without any further details it's tough to give you good advice here.

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This is down to the relationship between the angular velocity in the fixed body frame ($w^b$) of a rigid body and in the inertial frame ($w^s$) given a rotation matrix.

You have the wrong equations above for $w^s$ and $w^b$ The points don't come into it.

It should be as below: ($R^T$ is the transpose, $\dot R$ differentiated against time ($t$))

$$ωs=R\dot R^T$$ $$wb=R^T \dot R$$

The missing piece needed to calculate $w^s$ is $\dot R$.

Take the $w^b$ equation and pre-multiply each side of the equation by $inv(RT)$

$$inv(R^T)R^T\dot R= inv(R^T)wb$$

$$\dot R=inv(RT)wb$$

Now you know $R$ and $\dot R$ So you can calculate $ws=\dot RR^T$

You can confirm this by seeing that the answer is one of the 4 options above.

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