Existing controllers for sailboat robots are usually developed for speed performances and for long straight lines. In this context, the accuracy is not the main concern. In this paper, we consider the tight slalom problem which requires accuracy. We propose a feedback-linearization based method combined with a vector field approach to control the sailboat. Some simulations show that the robot is able to perform the slalom without missing any gate.
We consider a mobile robot described by the state equations
The approach we propose is to find a controller so that the vector p˙ be collinear (instead of equal) to the
required field. This is illustrated in this paper in the case where the mobile robot is a sailboat
Let us choose as the control output the variable
and let us find a classical feedback linearization based controller
In order to facilitate the understanding of our approach, we will deal with a Dubins car, which is much simpler than a sailboat. The extension to other type of mobile robots is straightforward. 2.1
To introduce our approach, we consider a robot (here a Dubins car) moving on a plane and described by the following state equations: Note that we do not have any singularity. As illustrated by the simulation depicted on Figure 1, the associated vector field makes the car attracted by the line x2 = 0. where α is a small positive coefficient, e.g., α = 0.1. In such a case, the robot will go to the line in a finite time (and not asymptotically, as previously). Moreover, it will remains exactly on the line even if some small uncertainties occur. 2.2
We want our robot to follow the field ψ(p), more precisely, we want that ψ(p) and p˙ point toward the same direction. This condition can be translated into the form ϕ (ψ(p), p˙ ) = 0, where ϕ is a collinearity function which satisfies
ϕ (r, s) = 0 ⇔ ∃λ > 0, λr = s.
Typically, this function corresponds to one angle (the heading) if m = 1 and two angles (heading, elevation) for m = 2. Note that the function ϕ cannot be expressed with a determinant since r, s should not point toward opposite directions. We define the output y = ϕ (ψ(p), p˙ ) = ϕ ψ(g (x)), ∂g ∂x (x) · f (x, u) .
Since y ∈ Rm, we can apply a feedback linearization method and we get y → 0. This means that the robot will follows the required field. Note that we have no control on the speed, which is not our main concern in this paper.
2D case. Consider for instance the case where m = 1. We have
We take as an output y, the angle between the actual heading vector p˙ = ∂∂xg (x) · f (x, u) and the desired heading vector given by ψ(p). Denote by θ (x) the argument of the vector p˙ . We have
The sawtooth function is given by: y (=8) = angle ψ(g (x)), ∂∂xg (x) · f (x, u) = angle ψ(g (x)), scionsθθ = sawtooth(θ−atan2(ψ2(g (x)), ψ1(g (x))))
| {bz } | {az } sawtooth(θe) = 2atan tan θ2e = mod(θe + π, 2π) − π As illustrated in Figure 2, the function corresponds to an error in heading. The interest in taking an error θe filtered by the sawtooth function is to avoid the problem of the 2kπ modulus: we would like a 2kπ to be considered non-zero.
(7) (8) (9) (10) (11) (12) if we assume that the input u corresponds to the desired angular velocity. We propose a feedback linearization based control based on the required equation y˙ = −y. We have u (12) = We thus have the guarantee that after some time, the error angle y is 0 and that we follow exactly the vector field. Take g (x) = (x1, x2)T which means that we want to build the paths in the (x1, x2)-space. We have and Thus and where ρ1 = 0.003, ρ2 = 0.2, ρ3 = 3. In this equation u1, u2 correspond to the tuning of the rudder and the sail, respectively. We would like our robot to follow a path which makes a tight slalom through doors that have to be passed. We assume that we have a Cartesian equation for our path. For instance, we consider that the path is described by e (p) = 10 sin
− p2 = 0 where e (p) corresponds to an error. This path corresponds to a path that should be possible for a normal sailboat robot for crosswind conditions. We take a vector field which corresponds to a pole placement strategy. For instance, we want the error satisfies e˙ = −0.1 e, so that it will converge to zero in about 10 sec. Thus cos | p1 10 {z e˙(p)
1 p˙1 − p˙2 = − 10 10 sin } | p1 10 {z e(p) − p2 } We take p˙1 = 1, to go to the right. As a consequence, we get the following field: which is attracted by the curve p2 = 10 sin p101 .
We have (20) (21) (22) (23) (24) and
and
From (13), we get that the desired angular velocity should be Now, since the true angular velocity is θ˙ = −ρ2v sin 2u1, we take ∂g ∂x a = b =
. The saturation function tanh is needed since the rudder cannot respond to any required ωˆ. Indeed, if our ωˆ controller ask to turn too fast for the boat, ρ2v will be more than 1, and the rudder can only do its best. The behavior of our controller is illustrated by Figure 4, where the sailboat has to slalom tightly between doors. We can see that the trajectory follows exactly the sine path (magenta).
The Python source codes associated to the simulation can be found at:
https://www.ensta-bretagne.fr/jaulin/slalompy.zip