Angular rates components order from time-derivative of rotation matrix I'm currently confused at the moment about the components order obtained from a well-known relationship between derivative of a rotation matrix and its angular velocity: $\dot{R} = R \hat{\Omega}$.
I constructed the rotation matrix $R$ from consecutive rotations around $Z,Y,X$ axis using the formulae given in https://en.wikipedia.org/wiki/Euler_angles. So is it true that the $\Omega$ vector (whose skew-symmetric form is $\hat{\Omega}$) will be $[\omega_z,\omega_y,\omega_x]^T$? Moreover, if $R$ is a rotation matrix from frame $B$ to frame $A$, is $\Omega$ angular rate written with respect to frame  $B$? 
I'd very grateful to hear from you. Thanks in advance.
 A: First of I will denote a rotation matrix from frame $B$ to frame $A$ as $R^{AB}$ and the angular velocity of frame $B$ with respect to frame $A$ expressed in frame $B$ as $^{BA}\vec{\omega}^B = \begin{bmatrix}^{BA}\omega^B_x & ^{BA}\omega^B_y & ^{BA}\omega^B_z\end{bmatrix}^\top$.
The way how I remember how to derive the time derivative of a rotation matrix is by considering the rotation that occurs of frame $B$ relative to frame $A$ in a really small time step. Namely this rotation can be found by assuming $^{BA}\vec{\omega}^B$ to be constant during that time step and to use an axis angle representation of that rotation from the current frame $B$ (relative to $A$) to frame $B$ (relative to $A$) a small time step into the future, denoted by $B^*$. Here the axis, $\begin{bmatrix}u_x & u_y & u_z\end{bmatrix}^\top$, is the normalized angular velocity vector and the angle, $\theta$, is its magnitude times the time step. In the limit of the time step to zero we can use that $\cos\theta = 1$ and $\sin\theta = \theta$, therefore
$$
R^{BB^*} = 
\begin{bmatrix}
\cos\theta + u_x^2(1-\cos\theta) & u_xu_y(1-\cos\theta)-u_z\sin\theta & u_xu_z(1-\cos\theta)+u_y\sin\theta \\
u_xu_y(1-\cos\theta)+u_z\sin\theta & \cos\theta + u_y^2(1-\cos\theta) & u_yu_z(1-\cos\theta)-u_x\sin\theta \\
u_xu_z(1-\cos\theta)-u_y\sin\theta & u_yu_z(1-\cos\theta)+u_x\sin\theta & \cos\theta + u_z^2(1-\cos\theta)
\end{bmatrix} = 
\begin{bmatrix}
1 & -u_z\,\theta & u_y\,\theta \\
u_z\,\theta & 1 & -u_x\,\theta \\
-u_y\,\theta & u_x\,\theta & 1
\end{bmatrix} = 
\begin{bmatrix}
1 & -^{BA}\omega^B_z\,dt & ^{BA}\omega^B_y\,dt \\
^{BA}\omega^B_z\,dt & 1 & -^{BA}\omega^B_x\,dt \\
-^{BA}\omega^B_y\,dt & ^{BA}\omega^B_x\,dt & 1
\end{bmatrix}
$$
which can also be written as
$$
R^{BB^*} = I + 
\begin{bmatrix}
0 & -^{BA}\omega^B_z & ^{BA}\omega^B_y \\
^{BA}\omega^B_z & 0 & -^{BA}\omega^B_x \\
-^{BA}\omega^B_y & ^{BA}\omega^B_x & 0
\end{bmatrix} dt = I + 
^{BA}\vec{\omega}^B_\times\,dt
$$
where $^{BA}\vec{\omega}^B_\times$ refers to what you called the skew-symmetric form $\hat{\Omega}$, however I prefer this notation since it better illustrates its connection to the cross product. So the combined rotation can then be written as
$$
R^{AB}(t+dt) = R^{AB^*}(t) = R^{AB}(t)\,R^{BB^*}(t) = R^{AB}(t)\,\left(I + 
^{BA}\vec{\omega}^B_\times(t)\,dt\right)
$$
which can be substituted into the definition of the time derivative of $R^{AB}$
$$
\begin{align}
\dot{R}^{AB}(t) &= \lim_{dt \to 0} \frac{R^{AB}(t+dt) - R^{AB}(t)}{dt} \\
&= \lim_{dt \to 0} \frac{R^{AB}(t)\,\left(I + 
^{BA}\vec{\omega}^B_\times(t)\,dt\right) - R^{AB}(t)}{dt} \\
&= R^{AB}(t)\,^{BA}\vec{\omega}^B_\times(t).
\end{align}
$$

If $^{BA}\vec{\omega}^A$, the angular velocity expressed in frame $A$, is given instead you would get
$$
R^{AA^*} = I - 
^{BA}\vec{\omega}^A_\times\,dt
$$
where the minus sign comes from the fact that $^{AB}\vec{\omega}^A = -^{BA}\vec{\omega}^A$. So the combined rotation can be written as
$$
R^{AB}(t+dt) = R^{A^*B}(t) = R^{A^*A}(t)\,R^{AB}(t)
$$
where $R^{A^*A} = {R^{AA^*}}^\top$. By combining this with $\vec{v}_\times^\top = -\vec{v}_\times$ then it can be shown that
$$
\dot{R}^{AB} = ^{BA}\vec{\omega}^A_\times\,R^{AB}.
$$

And if you have found $\dot{R}^{AB}$ but actually need $\dot{R}^{BA}$, then you can use that a rotation matrix multiplied by its transpose (${R^{AB}}^\top = R^{BA}$) is equal to the identity matrix. So taking the time derivative of that and applying the chain rule gives
$$
\frac{d}{dt}\left(R^{BA}\,R^{AB}\right) = \dot{R}^{BA}\,R^{AB} + R^{BA}\,\dot{R}^{AB} = 0
$$
Solving for $\dot{R}^{BA}$ gives
$$
\dot{R}^{BA} = -R^{BA}\,\dot{R}^{AB}\,R^{BA}.
$$
Substituting in the two previously derived expressions for $\dot{R}^{AB}$ gives
$$
\begin{align}
\dot{R}^{BA} &= -R^{BA}\,R^{AB}\,^{BA}\vec{\omega}^B_\times\,R^{BA} = -^{BA}\vec{\omega}^B_\times\,R^{BA} = ^{AB}\vec{\omega}^B_\times\,R^{BA} \\
&= -R^{BA}\,^{BA}\vec{\omega}^A_\times\,R^{AB}\,R^{BA} = -R^{BA}\,^{BA}\vec{\omega}^A_\times = R^{BA}\,^{AB}\vec{\omega}^A_\times.
\end{align}
$$
It can also be noted that if you have the rotation matrix $R^{AB}$ then it is only meaningful to pre-multiply it with things which are expressed in frame $A$ and post-multiply it with things which are expressed in frame $B$ (so basically the closest letter in the subscript of the rotation matrix). That is also why I used the notation of the rotation matrix with letters of the reference frames in its subscript, because it makes it easier to keep track of this.
