A high-performance control algorithm based on a curvature-dependent decoupled planning approach and flatness concepts for non-holonomic mobile robots

Abstract

This paper proposes a high-performance control strategy for an efficient manipulation of non-holonomic mobile robots in environments cluttered with static obstacles. Firstly, and based on the decoupled planning approach, a new algorithm for fast and safe motions planning is introduced. This algorithm defines the robot path as a sequence of smoothly interpolated functions using ($\eta ^3$) splines and then assigns a suitable curvature-dependent smooth motion profile to describe the robot velocity along such path. In order to achieve fast motions which fulfill all system constraints, the velocity profile is defined as a chain of local profiles smoothly linked together. Each of the local profiles is defined as a smooth limited-jerk function, which is obtained by applying a moving average FIR filter to a classic limited-acceleration profile. The appropriate bounds on velocities and accelerations of trapezoidal acceleration profiles are fixed according to the physical limits of the robot and the maximum bounds on the curvature in the corresponding path segment. The boundary conditions of the local profiles are assigned to ensure that the robot moves from its starting position without stopping until it reaches the goal configuration. Once the motion reference trajectories are obtained, a robust flatness-based feedback controller was defined to ensure the robust and the accurate execution of the planned tasks. Practical tests, using the P3DX model, have been reported to evaluate the performances of the proposed control strategy.

Appendices

Appendices

1.1 Appendix A: Proof of Proposition 1

From Eqs. 2 and 15, the left and right wheels velocities may be written as:

$$\begin{aligned} v_R= & {} v+lw = v(1+k(s)l) \\ v_L= & {} v-lw = v(1-k(s)l) \end{aligned}$$

where k denotes the curvature of the generated path. Then one can get:

$$\begin{aligned} v_R= & {} \psi (t)(1+k(s)l) \\ v_L= & {} \psi (t)(1-k(s)l) \end{aligned}$$

Thus, the velocity limits on the right wheel are given as:

$$\begin{aligned} |v_{R}|\le |\psi (t)| (1+|k(s)|l) \le v_{\max } \end{aligned}$$

which implies that:

$$\begin{aligned} |\psi (t)| (1+|k(s)|l) \le v_{\max } \end{aligned}$$

Note that $(1+|k(s)|l) > 0$ , then one can conclude that:

$$\begin{aligned} \dfrac{|\psi (t)| (1+|k(s)|l)}{(1+|k(s)|l)} \le \dfrac{v_{\max }}{(1+|k(s)|l)} \end{aligned}$$

Finally, one can conclude:

$$\begin{aligned} |\psi (t)| \le \psi _{\max } =\dfrac{v_{\max }}{(1+|k(s)|l)} \end{aligned}$$

The same conclusion may be obtained considering the equation that represents the linear velocity of the left wheel $v_L$.

1.2 Appendix B: $\eta ^3$ Interpolation parameters

Given any two sequential points A and B defined with the following extended representation:

$$\begin{aligned} A=[X_A \ Y_A \ \theta _A \ k_A \ k_A^\prime ] , B=[X_B \ Y_B \ \theta _B \ k_B \ k_B^\prime ] \end{aligned}$$

then the two points may be interpolated, while satisfying all boundary conditions, using the seventh-order polynomials as in Eq. 32, where the parameters $\alpha _i$ and $\beta _i$ are given as:

$$\begin{aligned} \alpha _0= & {} X_A \\ \alpha _1= & {} \eta _1 \cos (\theta _A)\\ \alpha _2= & {} \dfrac{1}{2}\eta _2 \cos (\theta _A) - \dfrac{1}{2} \eta _1^2 k_A \sin (\theta _A) \\ \alpha _3= & {} \dfrac{1}{6}\eta _5 \cos (\theta _A)-\dfrac{1}{6}(\eta _1^3 k_A^\prime + 3 \eta _1\eta _3 k_A)\sin (\theta _A)\\ \alpha _4= & {} 35(X_B-X_A)-\left( 20\eta _1+5\eta _3+ \dfrac{2}{3}\eta _5\right) \cos (\theta _A) \\&+\left( 5 \eta _1^2 k_A+\dfrac{2}{3} \eta _1^3k_A^\prime +2\eta _1\eta _3k_A\right) \sin (\theta _A)\\&-\left( 15\eta _2-\dfrac{5}{2}\eta _4 +\dfrac{1}{6}\eta _6\right) \cos (\theta _B)\\&-\left( \dfrac{5}{2} \eta _2^2 k_B -\dfrac{1}{6} \eta _2^3 k_B^\prime - \dfrac{1}{2} \eta _2 \eta _4 k_B\right) \sin (\theta _B)\\ \alpha _5= & {} -84 (X_B-X_A)+(45\eta _1+10\eta _3+\eta _5) \cos (\theta _A)\\&-(10 \eta _1^2k_A+ \eta _1^3 k_A^\prime +3\eta _1 \eta _3 k_A) \sin (\theta _A) \\&+\left( 39 \eta _2 - 7\eta _4+\dfrac{1}{2}\eta _6\right) \cos (\theta _B) \\&+\left( 7 \eta _2^2 k_B-\dfrac{1}{2}\eta _2^3 k_B^\prime - \dfrac{3}{2} \eta _2 \eta _4 k_B\right) \sin (\theta _B) \\ \alpha _6= & {} 70(X_B-X_A)-\left( 36\eta _1+\dfrac{15}{2}\eta _3+\dfrac{2}{3} \eta _5\right) \cos (\theta _A) \\&+\left( \dfrac{15}{2} \eta _1^2 k_A+\dfrac{2}{3} \eta _1^3 k_A^\prime +2 \eta _1 \eta _3 k_A\right) \sin (\theta _A) \\&-\left( 34 \eta _2-\dfrac{13}{2} \eta _4+\dfrac{1}{2} \eta _6\right) \cos (\theta _B)\\&-\left( \dfrac{13}{2} \eta _2^2 k_B-\dfrac{1}{2} \eta _2^3 k_B^\prime -\dfrac{3}{2} \eta _2 \eta _4 k_B\right) \sin (\theta _B) \\ \alpha _7= & {} -20(X_B-X_A)+\left( 10 \eta _1+2 \eta _3+\dfrac{1}{6}\eta _5\right) \cos (\theta _A) \\&-\left( 2 \eta _1^2 k_A+\dfrac{1}{6}\eta _1^3 k_A^\prime + \dfrac{1}{2}\eta _1 \eta _3 k_A\right) \sin (\theta _A)\\&+ \left( 10 \eta _2-2 \eta _4+\dfrac{1}{6} \eta 6\right) \cos (\theta _B) +2 \eta _2^2 k_B\\&-\dfrac{1}{6}\eta _2^3 k_B^\prime -\dfrac{1}{2}\eta _2 \eta _4 k_B)\sin (\theta _B) \end{aligned}$$

$$\begin{aligned} \beta _0= & {} Y_A \\ \beta _1= & {} \eta _1 \sin (\theta _A)\\ \beta _2= & {} \dfrac{1}{2}\eta _2 \sin (\theta _A) + \dfrac{1}{2} \eta _1^2 k_A \cos (\theta _A) \\ \beta _3= & {} \dfrac{1}{6}\eta _5 \sin (\theta _A)+\dfrac{1}{6}(\eta _1^3 k_A^\prime + 3 \eta _1\eta _3 k_A) \cos (\theta _A)\\ \beta _4= & {} 35(X_B-X_A)-\left( 20\eta _1+5\eta _3+ \dfrac{2}{3}\eta _5\right) \sin (\theta _A) \\&-\left( 5 \eta _1^2 k_A+\dfrac{2}{3} \eta _1^3k_A^\prime +2\eta _1\eta _3k_A\right) \cos (\theta _A)\\&-\left( 15\eta _2-\dfrac{5}{2}\eta _4 +\dfrac{1}{6}\eta _6\right) \sin (\theta _B)\\&+\left( \dfrac{5}{2} \eta _2^2 k_B -\dfrac{1}{6} \eta _2^3 k_B^\prime - \dfrac{1}{2} \eta _2 \eta _4 k_B\right) \cos (\theta _B)\\ \beta _5= & {} -84 (X_B-X_A)+(45\eta _1+10\eta _3+\eta _5) \sin (\theta _A)\\&+(10 \eta _1^2k_A+ \eta _1^3 k_A^\prime +3\eta _1 \eta _3 k_A) \cos (\theta _A) \\&+\left( 39 \eta _2 - 7\eta _4+\dfrac{1}{2}\eta _6\right) \sin (\theta _B) \\&+ \left( 7 \eta _2^2 k_B-\dfrac{1}{2}\eta _2^3 k_B^\prime - \dfrac{3}{2} \eta _2 \eta _4 k_B\right) \cos (\theta _B) \\ \beta _6= & {} 70(X_B-X_A)+\left( 36\eta _1+\dfrac{15}{2}\eta _3+\dfrac{2}{3} \eta _5\right) \sin (\theta _A) \\&+\left( \dfrac{15}{2} \eta _1^2 k_A+\dfrac{2}{3} \eta _1^3 k_A^\prime +2 \eta _1 \eta _3 k_A\right) \cos (\theta _A) \\&+\left( 34 \eta _2-\dfrac{13}{2} \eta _4+\dfrac{1}{2} \eta _6\right) \sin (\theta _B)\\&-\left( \dfrac{13}{2} \eta _2^2 k_B-\dfrac{1}{2} \eta _2^3 k_B^\prime -\dfrac{3}{2} \eta _2 \eta _4 k_B\right) \cos (\theta _B) \\ \beta _7= & {} -20(X_B-X_A)+\left( 10 \eta _1+2 \eta _3+\dfrac{1}{6}\eta _5\right) \sin (\theta _A) \\&+\left( 2 \eta _1^2 k_A+\dfrac{1}{6}\eta _1^3 k_A^\prime + 0.5\eta _1 \eta _3 k_A\right) \cos (\theta _A)\\&+ \left( 10 \eta _2 -2 \eta _4+\dfrac{1}{6} \eta _6\right) \sin (\theta _B) -\left( 2 \eta _2^2 k_B \right. \\&\left. -\dfrac{1}{6}\eta _2^3 k_B^\prime -\dfrac{1}{2}\eta _2 \eta _4 k_B\right) \cos (\theta _B) \end{aligned}$$