Rotation Matrix Time Derivative

In Coordinate Systems, I found the challenge of “how do we find the time derivative of the rotation matrix?”. So because I didn’t want to load that page with explanations, I am doing it here (the mathematical argument)

Basics

So we know that the rotation matrix is orthogonal: $R^{T}R=R{R}^T=I$

• So we can write:

$$R_{\mathbf{x}}(\theta )R_{\mathbf{x}}^T(\theta )=I$$

• If we derivate this wr.t. time:

$$\left[\begin{array}{c}\frac{d}{d\theta }{R}_{\mathbf{x}}(\theta )\end{array}\right]R_{\mathbf{x}}^T(\theta )+R_{\mathbf{x}}(\theta )\frac{d}{d\theta }{R}_{\mathbf{x}}^T(\theta )=0$$

• We know that $A^T{B}^T=(BA{)}^T$, thus:

$$\left[\begin{array}{c}\frac{d}{d\theta }{R}_{\mathbf{x}}(\theta )\end{array}\right]R_{\mathbf{x}}^T(\theta )+{\left[\begin{array}{c}\frac{d}{d\theta }{R}_{\mathbf{x}}(\theta )R_{\mathbf{x}}^T(\theta )\end{array}\right]}^T=0$$

• We define the first part as S: $\left[\begin{array}{c}\frac{d}{d\theta }{R}_{\mathbf{x}}(\theta )\end{array}\right]R_{\mathbf{x}}^T(\theta )$
• And we know that S is a skew-symmetric matrix! ($S+S^T=0$) and ($-S=S^T$)
• In 3 dimensions, the skew-symmetric matrix has a definite form. It’s also an alternative way to express the cross product $a\times b=S(a)b$

$$S_{3D}(\mathbf{u})=\left(\begin{array}{ccc}0& -u_z& u_y\\ u_z& 0& -u_x\\ -u_y& u_x& 0\end{array}\right),\mathbf{u}=[u_x,u_y,u_z]$$

Now here comes some generalization

For one dimension:

• $S=\frac{d}{d\theta }{R}_{\mathbf{x}}(\theta )R_{\mathbf{x}}^T(\theta )$ AND $R_{\mathbf{x}}(\theta )=\left[\begin{array}{ccc}1& 0& 0\\ 0& \cos \theta & -\sin \theta \\ 0& \sin \theta & \cos \theta \end{array}\right]$.
• If we solve this, we will get that $S=\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& -1\\ 0& 1& 0\end{array}\right]=S([1,0,0])$
• Therefore, $\frac{d}{d\theta }{R}_{\mathbf{x}}(\theta )=S([1,0,0])R_{\mathbf{x}}(\theta )$

THIS HOLDS FOR ALL OF THE EXAMPLES IN 1D!

$\frac{d}{d\theta }{R}_{\mathbf{y}}(\theta )=S([0,1,0])R_{\mathbf{y}}(\theta )$ $\frac{d}{d\theta }{R}_{\mathbf{z}}(\theta )=S([0,0,1])R_{\mathbf{z}}(\theta )$

• So we get the general form: $\frac{d}{d\theta }{R}_{\mathbf{l}}(\theta )=S(l)R_{\mathbf{l}}(\theta )$

Going further..

• Now we look at the time derivative of the rotation.

$$\frac{d\theta }{dt}\frac{d}{d\theta }{R}_l(\theta )=\frac{d\theta }{dt}S(l)R_l(\theta )$$ $$\frac{\cancel{d\theta }}{dt}\frac{d}{\cancel{d\theta }}{R}_l(\theta )=\frac{d\theta }{dt}S(l)R_l(\theta )$$ $$\frac{d}{dt}{R}_l(\theta )=S(\omega )R_l(\theta ),\omega =\dot{\theta }l$$

• The angular velocities (${\omega }_x,{\omega }_y,{\omega }_z$) come directly from the IMU gyroscopes. $S(\omega )=\left(\begin{array}{ccc}0& -{\omega }_z& {\omega }_y\\ {\omega }_z& 0& -{\omega }_x\\ -{\omega }_y& {\omega }_x& 0\end{array}\right)$

Based on Taylor Series, we can write

• The expansion of a Rotation Matrix around a point C: $R(\alpha )\approx R(c)+\frac{R^{\prime }(c)}{1!}(\alpha -c)+\frac{R^{\prime \prime }(c)}{2!}(\alpha -c{)}^2+...$
• We know that $R_{\mathbf{x}}(\alpha )=\left[\begin{array}{ccc}1& 0& 0\\ 0& \cos \alpha & -\sin \alpha \\ 0& \sin \alpha & \cos \alpha \end{array}\right]$ and $\frac{d{R}_x}{d\alpha }=\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& -1\\ 0& 1& 0\end{array}\right]$
• If we evaluate at point $c=0$ and keep only the first 2 terms from the expansion above, we get: $R_{\mathbf{x}}(\alpha )\approx R(0)+\frac{R^{\prime }(0)}{1!}(\alpha -0)=\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)+\left(\begin{array}{ccc}0& 0& 0\\ 0& 0& -1\\ 0& 1& 0\end{array}\right)\alpha =I_3+S([1,0,0]\alpha$

So if we go back to the first question from Coordinate Systems, ${\dot{r}}^i={\dot{R}}_b^i{r}^b+R_b^i{\dot{r}}^b$, we can write:

$${\dot{r}}^i=(R_b^i{\Omega }_{ib}^b)r^b+R_b^i{\dot{r}}^b$$ $${\dot{r}}^i=R_b^i({\Omega }_{ib}^b{r}^b+{\dot{r}}^b),{\Omega }_{ib}^b=S(\omega )=\left(\begin{array}{ccc}0& -{\omega }_z& {\omega }_y\\ {\omega }_z& 0& -{\omega }_x\\ -{\omega }_y& {\omega }_x& 0\end{array}\right)$$

Second Derivative (Time Derivative of Velocity)

$${\dot{r}}^i=R_b^i({\Omega }_{ib}^b{r}^b+{\dot{r}}^b)$$ $$\ddot{r}={\dot{R}}_b^i({\Omega }_{ib}^b{r}^b+{\dot{r}}^b)+R_b^i\frac{d}{dt}({\Omega }_{ib}^b{r}^b+{\dot{r}}^b)$$ $$\frac{d}{dt}({\Omega }_{ib}^b{r}^b+{\dot{r}}^b)={\dot{\Omega }}_{ib}^b{r}^b+{\Omega }_{ib}^b{\dot{r}}^b+{\ddot{r}}^b$$ $${\ddot{r}}^i=R_b^i({\ddot{r}}^b+2{\Omega }_{ib}^b{\dot{r}}^b+{\dot{\Omega }}_{ib}^b{r}^b+{\Omega }_{ib}^b{\Omega }_{ib}^b{r}^b)$$

The last part, ${\Omega }_{ib}^b{\Omega }_{ib}^b{r}^b$ simply means the skew-symmetric matrix is applied twice. This is equivalent to the cross product applied twice: $\omega \times (\omega \times r^b)$. This double cross product represents the centrifugal acceleration, which points radially outward from the rotation axis.