Originally Posted by

**Ackbeet** So, as I understand it, your current problem is this:

$\displaystyle \ddot{x}=u_{1}$

$\displaystyle \ddot{\theta}=u_{2}$

$\displaystyle \ddot{y}=u_{1}\tan(\theta)+\dot{x}\tan(\theta)-\dot{y},$

and you want to convert to

$\displaystyle \ddot{y}_{1}=v_{1}$

$\displaystyle \ddot{y}_{2}=v_{2}$

$\displaystyle \ddot{y}_{3}=y_{2}v_{1}+\dots$

Is that correct? If so, can I ask if you have the freedom to assign the $\displaystyle y_{k}$ and $\displaystyle v_{k}?$ If so, I would go with the following assignments:

$\displaystyle y_{1}=x$

$\displaystyle v_{1}=u_{1}$

$\displaystyle y_{2}=\tan(\theta).$

Differentiating $\displaystyle y_{2}$ yields the following:

$\displaystyle \dot{y}_{2}=\sec^{2}(\theta)\,\dot{\theta},$ and

$\displaystyle \ddot{y}_{2}=\sec^{2}(\theta)(\ddot{\theta}+2 \dot{\theta}^{2}\tan(\theta))=\sec^{2}(\theta)(u_{ 2}+2\dot{\theta}^{2}\tan(\theta)).$

So, let

$\displaystyle v_{2}=\sec^{2}(\theta)(u_{2}+2\dot{\theta}^{2}\tan (\theta)),$ and

$\displaystyle y_{3}=y.$

We would thus have the following system:

$\displaystyle \ddot{y}_{1}=v_{1}$

$\displaystyle \ddot{y}_{2}=v_{2}$

$\displaystyle \ddot{y}_{3}=v_{1}y_{2}+\dot{y}_{1}y_{2}-\dot{y}_{3}.$

Is that what you were seeking?