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I am stuck with understanding how can I make my robot move along planned path. For instance, if we have a grid map of an environment and applied, for example, A* to plan a path then after that we have to make our robot move through each cell in our path. Assuming that we know center coordinates of cells, the task is to generate control commands which will lead robot along the trajectory.

I have two differential wheeled robot so equations of motion are going to be like these, where b is a distance between wheels:

$v = \frac{1}{2}(v_{1}+v_{2})\\ \dot\theta=\frac{1}{b}(v_{2}-v_{1})\\ \theta = \frac{\delta t}{b}(v_{2}-v_{1}) + \theta_{0}\\ \dot x = v\cos(\theta)\\ \dot y = v\sin(\theta)\\ x = x_{0} + \frac{b(v_{1}+v_{2})}{2(v_{2}-v_{1})}(\sin(\theta)-\sin(\theta_{0}))\\ y = y_{0} - \frac{b(v_{1}+v_{2})}{2(v_{2}-v_{1})}(\cos(\theta)-\cos(\theta_{0})) $

Suppose that we can control speeds of both wheels therefore we are able to set any possible angular and linear velocities. So what actually I have to do with these equations to make robot move through each cell?

Moreover there may be different constraints like moving with constant linear speed etc. I understand that I have to solve these equations somehow. Will appreciate practical advises, certain names of algorithms and etc. Thanks!

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The model you have given is called the unicycle model and is widely used n robotics. In general the model is given by \begin{eqnarray} \dot{x} &= v\cos \theta \\ \dot{y} &= v\sin \theta \\ \dot{\theta} &= \omega \end{eqnarray} where $v = 1/2(v_1+v_2)$ and $\omega = 1/b(v_2-v_1)$ are the controls of the robot.

The method that you have mentioned is a waypoint following method, where the destination would be updated as you reach a point along a path. The problem is, due to the Lyapunov design, the destination is approached exponentially, i.e, as $t\rightarrow \infty$. Thus, the robot would slow down exponentially as it approaches the target.

One way to overcome this is to use design process that uses feedback linearization. But first you need to parameterize the path you get from A* with time to get a desired trajectory $\mathbf{x_d}(t)$. This is easy to do by selecting a speed for the robot to traverse the path.

Next select a point $p$ that is a distance $d$ away from the point of rotation of the robot, along the central axis of the robot. Assume that this point $p$ has a fully actuated kinematic model given by, $$ \mathbf{\dot{x}_p} = \mathbf{v_p}$$ where $\mathbf{v_p}$ is the control for this point. If we set, $$ \mathbf{v_p} = k_v(\mathbf{x_d}(t) - \mathbf{x_p}(t) ) + \mathbf{\dot{x}_d}(t), $$ then $\mathbf{x_p} \rightarrow \mathbf{x_d}$ exponentially, i.e., the point $p$ will follow the path. Note that its the point $p$ that follows the path and not the rotation center of the robot.

Control Design,

Next you should set the controls $v$ and $\omega$ of the robot so that point $p$ has this desired control $\mathbf{v_p}$. The position of the point $p$ can be written as, $$ \mathbf{x_p} = \mathbf{x_0} + d[\cos \theta, \sin \theta ]^T.$$ Thus the velocity of the point $p$ is, $$ \mathbf{\dot{x}_p} = \mathbf{\dot{x}_0} + d\dot{\theta}[-\sin \theta, \cos \theta ]^T.$$ Using the kinematic model of the robot to substitute for $\mathbf{x_0}$ and $\dot{\theta}$, this can be written as, \begin{equation} \mathbf{\dot{x}_p} = \left[ \begin{array}[c c] g\cos\theta & -d\sin \theta\\ \sin\theta & d\cos \theta \end{array} \right] \left[ \begin{array}[c] sv\\ \omega \end{array} \right]. \end{equation}

In order to get $p$ to follow the trajectory, we need $\mathbf{\dot{x}_p} = \mathbf{v_p} = k_v(\mathbf{x_d}(t) - \mathbf{x_p}(t) ) + \mathbf{\dot{x}_d}(t)$. Thus, the required controls are given by, \begin{equation} \left[ \begin{array}[c] sv\\ \omega \end{array} \right] = 1/d\left[ \begin{array}[c c] gd\cos\theta & d\sin \theta\\ -\sin\theta & \cos \theta \end{array} \right]\bigg(k_v(\mathbf{x_d}(t) - \mathbf{x_p}(t) ) + \mathbf{\dot{x}_d}(t)\bigg). \end{equation}

Once the controls are set, the wheel speeds can be set as, \begin{eqnarray} v_1 = \frac{2v +b\omega}{2}\\ v_2 = \frac{2v -b\omega}{2} \end{eqnarray}

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  • $\begingroup$ Thanks! It looks like a great explanation and example of how control is being constructed! $\endgroup$ – Long Smith Jun 5 '17 at 21:55
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I found what I was looking for. It is called tangential escape. In general it is used to move from point A to point B avoiding obstacles, but as we have already planned path we can omit the last part.

The idea is to define the model in polar coordinates which describes distance to a target and the angle between current position and target position, therefore our aim is to set $\rho = 0, \alpha=0$. After that use this model with Lyapunov function:

$V(\rho, \alpha) = \frac{1}{2}\rho^{2} + \frac{1}{2}\alpha^2$

It is necessary that derivative of $V$ must be negative in order to make the system stable. Then inserting our model into $\dot V$ we can find such linear and angular velocities functions that will make our derivative negative. Finally we will get these equations for them:

$\begin{cases} v=v_{max}\tanh\rho\cos(\alpha)\\ w=k_{w}\alpha+v_{max}\frac{\tanh\rho}{\rho}\sin(\alpha)\cos(\alpha),k_{w}>0 \end{cases}$

It may seem like a hard math but actually just read the original article carefully.

Here you can find example of code at the very end of the article. This article is written in Russian but nevertheless code examples might appear to be very useful.

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