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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:

$\mathbf{R}^{T}\simeq\left[\begin{array}{ccc} 1 & -\delta\phi & \delta\theta\\ \delta\phi & 1 & -\delta\psi\\ -\delta\theta & \delta\psi & 1 \end{array}\right]=\underbrace{\left[\begin{array}{ccc} 0 & -\delta\phi & \delta\theta\\ \delta\phi & 0 & -\delta\psi\\ -\delta\theta & \delta\psi & 0 \end{array}\right]}_{\textrm{cross product operator}}+\boldsymbol{I}_{3x3}=\boldsymbol{I}_{3x3}+\delta\overrightarrow{E}\times$

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 $\mathbf{R}=\boldsymbol{I}_{3x3}$, 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!

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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

$$\delta\boldsymbol{E}=\begin{pmatrix}\delta\psi\\\delta\theta\\\delta\phi\end{pmatrix}$$

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  • $\begingroup$ Hmm, I dont think I get it... $\endgroup$ – B. Preiss Dec 17 '17 at 21:42
  • $\begingroup$ @B.Preiss Do you agree that $\delta\boldsymbol{x}=\delta\boldsymbol{E}\times\boldsymbol{x}$? $\endgroup$ – eranreches Dec 17 '17 at 21:45
  • $\begingroup$ @B.Preiss For small rotation? $\endgroup$ – eranreches Dec 17 '17 at 21:50
  • $\begingroup$ Hmm, I dont think I get it... if I apply what you have written above to my sines and cosines: $\sin(\psi(t+\delta t))=\sin(\psi(t)+\delta\psi(t))=\sin(\psi(t))\cdot\cos(\delta\psi(t))+\cos(\psi(t))\cdot\sin(\delta\psi(t))$. Now, when I apply: $\cos(\delta\psi(t))=0$ still $\cos(\psi(t))\cdot\sin(\delta\psi(t))$ remains. Isn't that term only equal to $\delta\psi(t)$ if $\psi(t)$ is close to $0$? $\endgroup$ – B. Preiss Dec 17 '17 at 21:53
  • $\begingroup$ "for small rotation": can we assume a small rotation in a general case? Couldn't the angles of a general rotation be significantly bigger than 0? $\endgroup$ – B. Preiss Dec 17 '17 at 21:57

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