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 &amp; -\omega_{3} &amp; \omega_{2}\\<br /> \omega_{3} &amp; 0 &amp; -\omega_{1}\\<br /> -\omega_{2} &amp; \omega_{1} &amp; 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 &amp; -\omega_z(t) &amp; \omega_y(t) \\<br /> \omega_z(t) &amp; 0 &amp; -\omega_x(t) \\<br /> -\omega_y(t) &amp; \omega_x(t) &amp; 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}.