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I am having some troubles understanding a concept regarding the path planning for a unicycle. Suppose I want to plan a path that brings a unicycle from the origin to the point $\begin{pmatrix} 1\\ 1\\ \frac{\pi }{2} \end{pmatrix}$.

The way I operate is the following:

Since we are considering a path planning, we can consider the geometric kinematic model of a unicycle:

${x}'= cos\theta \cdot \tilde{v}$

${y}'= sin\theta \cdot \tilde{v}$

${\theta }'=\tilde{\omega }$

and to solve this problem we exploit the fact that x and y are flat outputs. So we can use third order polynomials to define the path with the parameter $s\in [0,1]$. After, we define the boundary conditions:

$x(0) = x_{i} = 0$

$x(1) = x_{f} = 1$

${x}'(0)= cos\theta _{i}=k$

${x}'(1)= cos\theta _{f}=0$

and the same for y:

$y(0) = y_{i} = 0$

$y(1) = y_{f} = 1$

${y}'(0)= cos\theta _{i}=0$

${y}'(1)= cos\theta _{f}=k$

and now we can define the polynimials:

$x(s)=(k-2)s^{3}+(3-2k)s^{2}+ks$

$y(s)=(k-2)s^{3}+(3-k)s^{2}$

the first derivatives:

${x}'(s)=3(k-2)s^{2}+2(3-2k)s+k$

${y}'(s)=3(k-2)s^{2}+2(3-k)s$

the second derivatives:

${x}''(s)=6(k-2)s+2(3-2k)$

${y}''(s)=6(k-2)s+2(3-k)$

at this point to find the evolution of $\theta $ is :

$\theta = atan2\left \{ {y}(s)',{x}(s)' \right \}$

Now, my question is : What values do we have to put inside the $atan2$ ? So, in particular, how do I compute ${y}(s)'$ and ${x}(s)'$ manually without using a program?

Can somebody please help me? Thank's in advance.

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    $\begingroup$ You have your equation for $x'$ and $y'$, And I think that you mean $\theta ' (s) = atan2(x'(s) ,y'(s))$ and you calculate $x', y'$ for a specific $s \in[0,1] $ $\endgroup$ – nionios Oct 31 '19 at 12:30

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