1
$\begingroup$

I have to study the controllability of the kinematic model of a Cycab: enter image description here

$\dot{q}=g_1(q)v+g_2(q)\omega_R+g_3(q)\omega_L$

where

$\dot{q}=\begin{bmatrix}\dot{x}\\\dot{y}\\\dot{\theta}\\\dot{\gamma}\\\dot{\phi}\end{bmatrix}$ $g_1(q)=\begin{bmatrix}cos(\theta+\gamma)\\sin(\theta+\gamma)\\\frac{sin(\phi-\gamma)}{lcos(\phi)}\\0\\0\end{bmatrix}$ $g_2(q)=\begin{bmatrix}0\\0\\0\\1\\0\end{bmatrix}$ $g_3(q)=\begin{bmatrix}0\\0\\0\\0\\1\end{bmatrix}$

with $x$ and $y$ the Cartesian coordinates of the midpoint of the rear segment joining the two rear wheels and $\theta$ the direction of the midpoints of the two segments joining the wheels centers with respect to the axis $x$. So, I have to study the accessibility distribution $\{g_1,g_2,g_3.[g_1,g_2],[g_2,[g_1,g_2]],...\}$ so I computed

$[g_1,g_2]=\begin{bmatrix}sin(\theta+\gamma)\\-cos(\theta+\gamma)\\\frac{cos(\phi-\gamma)}{lcos(\phi)}\\0\\0\end{bmatrix}$ $[g_2,[g_1,g_2]]=\begin{bmatrix}-cos(\theta+\gamma)\\-sin(\theta+\gamma)\\\frac{-sin(\phi-\gamma)}{lcos(\phi)}\\0\\0\end{bmatrix}$

so the rank of $[g_1,g_2,g_3,[g_1,g_2],[g_2,[g_1,g_2]]]$ is equal to 5 so we can say that the system is controllable.

Now, is it correct to study the accessibility distribution without using the vector field $g_3$, so is it correct to say that the system is controllable without using the vector field $g_3$?

Improve this question
$\endgroup$
4
  • $\begingroup$ Would you include a reference for this material (preferably a textbook)? There are a couple of potential issues that I am worried about. First, you want to determine controllability of the system, but you then use accessibility--is it true that the system is controllable if the system is accessible (i.e. if the accessibility distribution is full rank)? Second, something seems off with the accessibility distribution--what is $g4(q)$? $\endgroup$
    – JSycamore
    Commented Apr 4, 2017 at 13:09
  • $\begingroup$ The book is "Robotics: Modelling, Planning and Control" for which a system is controllable if its accessibility distribution is full rank. $g_2$ and $g_3$ are vector fields and the $g_4$ was a mistake of mine, now it's all correct. $\endgroup$
    – B.junior
    Commented Apr 4, 2017 at 13:11
  • $\begingroup$ We also need to know what $\theta, \gamma,$ and $ \phi$ are. If they are states, then we should express them in terms of the generalized coordinates. $\endgroup$
    – JSycamore
    Commented Apr 4, 2017 at 13:29
  • $\begingroup$ I've added a pic of the wheeled robot $\endgroup$
    – B.junior
    Commented Apr 4, 2017 at 13:57

2 Answers 2

Reset to default
0
$\begingroup$

Chow's Theorem states that:

The driftless system $$\dot q=g_1(q)u_1+...+g_m(q)u_m$$ is locally controllable at $q_0\in\mathbb{R}^n$ if $\text{rank}\bar\Delta(q_0)=n.$

Here, $\bar\Delta$ is the involutive closure of the distribution $\Delta=\text{span}\{g_1,...,g_m\}$ (i.e. the smallest involutive distribution containing $\Delta$). $\bar\Delta,$ also known as the Control Lie Algebra, is found by forming successively larger distributions by repeated computation of the Lie Brackets until an involutive distribution is found.

To answer your question, there is nothing about $\bar \Delta$ that requires the use of $g_3$ as long as $\bar \Delta$ contains $\Delta,$ so we can make statements about the local controllability without including $g_3$ in the distribution. Because this is a local result, you will have to prove that the system is controllable to any point in $\mathbb{R}^n$ in order to say that the system is controllable.

Improve this answer
$\endgroup$
6
  • $\begingroup$ Just one more question about the involutive closure: if I find $n$ linearly independent, they represent my involutive closure? there's no need to keep on computing other Lie brackets because the involutive closure has reached a rank equal to dimension of the system, right? $\endgroup$
    – B.junior
    Commented Apr 5, 2017 at 13:07
  • $\begingroup$ *$n$ linearly independent vectors $\endgroup$
    – B.junior
    Commented Apr 5, 2017 at 14:57
  • $\begingroup$ I'm not sure that is correct. The definition of $\bar \Delta$ states that it is the smallest involutive distribution containing $\Delta;$ this means that the vectors of $\bar \Delta$ must be a basis for $\Delta$ (i.e. the vectors are a minimal generating set of $\Delta$). Once you find this, however, you can stop computing Lie Brackets. $\endgroup$
    – JSycamore
    Commented Apr 6, 2017 at 11:54
  • $\begingroup$ But the dimension of the basis of $\Delta$ should be equal to or less than the dimension of the system we are studying so if we find $n$ linearly independent Lie Brackets we can stop computing Lie Brackets $\endgroup$
    – B.junior
    Commented Apr 6, 2017 at 12:22
  • $\begingroup$ Close--you can find $n$ linearly independent vectors, a subset of which forms a basis, so finding $n$ linearly independent vectors is not sufficient; you must find $n$ linearly independent vectors that form a basis. $\endgroup$
    – JSycamore
    Commented Apr 6, 2017 at 12:27
0
$\begingroup$

I noticed that $g_1$ and $[g_2,[g_1,g_2]]$ are not linearly independent so I computed the Lie bracket $[g_3,[g_1,g_2]]$ that is equal to $\begin{bmatrix}0\\0\\\frac{-cos(\gamma)}{lcos(\phi)}\\0\\0\end{bmatrix}$ and the rank of $[g_1,g_2,g_3,[g_1,g_2],[g_3,[g_1,g_2]]]$ is equal to 5 and the system should be controllable

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.