I am currently studying dynamics and trying to understand the relation between angular velocity $\omega$ of a rotating frame and the eulerian rotation matrix $\mathbf{R=\mathbf{R}\mathrm{(\psi)\mathbf{R}(\theta)\mathbf{R}(\phi)}}$, which accomplishes the rotation. I found a derivation at MIT:
MIT Courseware – Kinematics of moving frames
They basically try to find the derivative of $\overrightarrow{x}(t)=\overrightarrow{x}_{0}(t)+\mathbf{R}^{T}\overrightarrow{x}_{b}(t)$, where
$\overrightarrow{x}(t)$ is a vector in the inertial frame
$\overrightarrow{x}_{b}(t)$ is a vector in the moving frame and
$\mathbf{R}^{T}=\left[\begin{array}{ccc}
\cos\theta\cos\phi & -\cos\psi\sin\phi+\sin\psi\sin\theta\cos\phi & \sin\psi\sin\phi+\cos\psi\sin\theta\cos\phi\\
\cos\theta\sin\phi & \cos\psi\cos\phi+\sin\psi\sin\theta\sin\phi & -\sin\psi\cos\phi+\cos\psi\sin\theta\sin\phi\\
-\sin\theta & \sin\psi\cos\theta & \cos\psi\cos\theta
\end{array}\right]$
(the above vectors shall be defined as the triple of projections of directed line segments along the coordinate axes)
Now, if the angles of rotation are small, $\mathbf{R}^{T}$ can be approximated as:
$\mathbf{R}^{T}\simeq\left[\begin{array}{ccc}
1 & -\delta\phi & \delta\theta\\
\delta\phi & 1 & -\delta\psi\\
-\delta\theta & \delta\psi & 1
\end{array}\right]=\underbrace{\left[\begin{array}{ccc}
0 & -\delta\phi & \delta\theta\\
\delta\phi & 0 & -\delta\psi\\
-\delta\theta & \delta\psi & 0
\end{array}\right]}_{\textrm{cross product operator}}+\boldsymbol{I}_{3×3}=\boldsymbol{I}_{3×3}+\delta\overrightarrow{E}\times$
where $\delta\overrightarrow{E}=\left[\begin{array}{c}
\delta\psi\\
\delta\theta\\
\delta\phi
\end{array}\right]$
Now the derivative of vector $\overrightarrow{x}(t)$ would be:
$\begin{eqnarray*}
\overrightarrow{x}(t) & = & \overrightarrow{x}_{0}(t)+\overrightarrow{x}_{b}(t)\\
\overrightarrow{x}(t+\delta t) & = & \overrightarrow{x}_{0}(t)+\delta\overrightarrow{x}_{0}(t)+\mathbf{R}^{T}\overrightarrow{x}_{b}(t)+\delta\overrightarrow{x}_{b}(t)\\
& = & \overrightarrow{x}_{0}(t)+\delta\overrightarrow{x}_{0}(t)+\overrightarrow{x}_{b}(t)+\delta\overrightarrow{E}\times\overrightarrow{x}_{b}(t)+\delta\overrightarrow{x}_{b}(t)\\
\frac{\delta\overrightarrow{x}(t)}{\delta t} & = & \frac{\delta\overrightarrow{x}_{0}(t)}{\delta t}+\frac{\delta\overrightarrow{E}}{\delta t}\times\overrightarrow{x}_{b}(t)+\frac{\delta\overrightarrow{x}_{b}(t)}{\delta t}\\
& = & \frac{\delta\overrightarrow{x}_{0}(t)}{\delta t}+\overrightarrow{\omega}\times\overrightarrow{x}_{b}+\frac{\delta\overrightarrow{x}_{b}(t)}{\delta t}
\end{eqnarray*}$
Now my question: Why can this small angle approximation be made? Isn't that approximation only valid for small rotations and therefore the derived formula only valid in that case?
I know that there are quite some other people who had problems with this, sadly I didnt understand the explanations I found in the web.
edit: I just started to doubt the general validity of the above derivation,
because I compared
$\frac{\delta}{\delta t}\overrightarrow{x}(t)=\frac{\delta}{\delta t}\overrightarrow{x}_{0}(t)+\overrightarrow{\omega}\times\overrightarrow{x}_{b}(t)+\frac{\delta}{\delta t}\overrightarrow{x}_{b}(t)$ (last equation from above)
to
$\frac{\delta}{\delta t}\overrightarrow{x}(t)=\frac{\delta}{\delta t}\overrightarrow{x}_{0}(t)+\frac{\delta}{\delta t}\mathbf{R\cdot}\overrightarrow{x}_{b}(t)+\mathbf{R}\cdot\frac{\delta}{\delta t}\overrightarrow{x}_{b}(t)$,
last of which was derived from $\overrightarrow{x}(t)=\overrightarrow{x}_{0}(t)+\mathbf{R}\cdot\overrightarrow{x}_{b}(t)$
by applying product rule.
That would mean, that $\frac{\delta}{\delta t}\mathbf{R}=\overrightarrow{\omega}\times$
and $\mathbf{R}=\boldsymbol{I}_{3×3}$, which is wrong, in general.
Applying the above derivation to a real problem I got two different
vectors for $\overrightarrow{\omega}$, one from inspection and another
one from:
$\overrightarrow{\omega}=\frac{\delta}{\delta t}\overrightarrow{E}=\left[\begin{array}{c}
\frac{\delta}{\delta t}\psi\\
\frac{\delta}{\delta t}\theta\\
\frac{\delta}{\delta t}\phi
\end{array}\right]$
That leads to my second question, how can $\overrightarrow{\omega}$
be expressed in terms of the rotation matrix $\mathbf{R}$ in the
general case?
I found a not so general solution to that question here (the second answer) which
basically says that $\frac{\delta}{\delta t}\overrightarrow{E}\times=\overrightarrow{\omega}\times=\dot{\mathbf{R}}\cdot\mathbf{R}^{T}$.
So now we have three different terms that should be related as following
(but aren't): $\frac{\delta}{\delta t}\overrightarrow{E}\times=\dot{\mathbf{R}}\cdot\mathbf{R}^{T}=\left[\begin{array}{c}
\frac{\delta}{\delta t}\psi\\
\frac{\delta}{\delta t}\theta\\
\frac{\delta}{\delta t}\phi
\end{array}\right]\times=\dot{\mathbf{R}}$
I would love to derive an equation for $\overrightarrow{\omega}$ in
a similar manner, only for the situation of shifted rotating
and inertial frame coordinate vectors: $\overrightarrow{x}(t)=\overrightarrow{x}_{0}(t)+\mathbf{R}\cdot\overrightarrow{x}_{b}(t)$.
Sadly my linear algebra knowledge is somewhat limited. That is why
I would be happy about some further help. Many thanks in advance!
Best Answer
The derivation you quoted assumes the angles $\psi,\theta,\phi$ depend on time, i.e.
$$\begin{cases}\psi=\psi(t)\\\theta=\theta(t)\\\phi=\phi(t)\end{cases}$$
In this case, at the moment $t+\delta t$ you can approximate
$$\begin{cases}\psi(t+\delta t)=\psi(t)+\delta\psi(t)\\\theta(t+\delta t)=\theta(t)+\delta\theta(t)\\\phi(t+\delta t)=\phi(t)+\delta\phi(t)\end{cases}$$
In other words, an infinitesimal increment in time leads to infinitesimal increments of the angles captured by the vector
$$\delta\boldsymbol{E}=\begin{pmatrix}\delta\psi\\\delta\theta\\\delta\phi\end{pmatrix}$$