Angular velocity and rotation matrix

[srpvx]

Hello. Sorry for my English

There are $R$ - rotation matrix (that performs transformation from associated coordinate system $IE$ to static coordinate system $OI$) and $\omega$ - angular velocity. The matrix $R$ depends on parameters $\xi$ (for example, Euler angles). I need to express $\omega$ as function of $\xi$.

Let $r^e$ - components of vector $r$ in the associated coordinate system: $r=Rr^{e}$ and $r^{e}=R^{T}r$. Than variation of vector $r$:
$\delta r=\sum_{i}\frac{\partial Rr^{e}}{\partial\xi_{i}}\delta\xi_{i}=$ $\sum_{i}\left(\frac{\partial R}{\partial\xi_{i}}\delta\xi_{i}\right)r^{e}=$ $\sum_{i}\left(\frac{\partial R}{\partial\xi_{i}}\delta\xi_{i}\right)R^{T}r$

On the other hand if I rotate $r$ about $l$ on angle $\Delta\varphi$ then variation of $r$ equals: $\delta r=\Delta\varphi\left[l\times r\right]=\delta\varphi\times r$, where $\delta\varphi = \Delta \varphi l$.

Comparing $\delta r=\delta\varphi\times r$ with $\delta r=\sum_{i}\left(\frac{\partial R}{\partial\xi_{i}}\delta\xi_{i}\right)R^{T}r$, we get the following expression for omega:
$\omega = \frac{\delta \varphi}{d t}$
$\left[\omega\times\right]=\left(\begin{array}{ccc}0 & -\omega_{3} & \omega_{2}\\<br /> \omega_{3} & 0 & -\omega_{1}\\<br /> -\omega_{2} & \omega_{1} & 0\end{array}\right)=\sum_{i}\left(\frac{\partial R}{\partial\xi_{i}}\dot{\xi_{i}}\right)R^{T}$

Is it right?

 

[A.T.]

Try this:
http://www.kwon3d.com/theory/euler/avel.html

 

[srpvx]

I found the answer on my question at http://en.wikipedia.org/wiki/Angular_velocity" :

It can be introduced from rotation matrices. Any vector $\vec r$ that rotates around an axis with an angular speed vector $\vec \omega$ (as defined before) satisfies:

$\frac {d \vec r(t)} {dt} = \vec{\omega} \times\vec{r}$

We can introduce here the '''angular velocity tensor''' associated to the angular speed $\omega$:

$W(t) = \begin{pmatrix}<br /> 0 & -\omega_z(t) & \omega_y(t) \\<br /> \omega_z(t) & 0 & -\omega_x(t) \\<br /> -\omega_y(t) & \omega_x(t) & 0 \\<br /> \end{pmatrix}$

This tensor $W(t)$ will act as if it were a $(\vec \omega \times)$ operator :

$\vec \omega(t) \times \vec{r}(t) = W(t) \vec{r}(t)$

Given the orientation matrix $A(t)$ of a frame, we can obtain its instant '''angular velocity tensor''' W as follows. We know that:

$\frac {d \vec r(t)} {dt} = W \cdot \vec{r}$

As angular speed must be the same for the three vectors of a rotating frame $A(t)$, we can write for all the three:

$\frac {dA(t)} {dt} = W \cdot A (t)$

And therefore the angular velocity tensor we are looking for is:

$W = \frac {dA(t)} {dt} \cdot A^{-1}(t)$

But $W$ is not a tensor, $W$ is a pseudotensor: $W_{ij} = e_{iwj} \omega_{w}$.