Dynamics and input–output feedback linearization control of a wheeled mobile cable-driven parallel robot

Abstract

In comparison to the conventional parallel robots, cable-driven parallel robots (CDPRs) have generally superior features such as simple production technology, low energy consumption, large workspace, high payload to moving weight ratio, and also low cost. On the other hand, a wheeled mobile robot (WMR) which is capable of covering a vast area can be used when no specific space is designated for the stationary accessories of a robot. In this paper, the integration of a CDPR with a WMR is proposed to overcome some of the issues related to each of these robots. The kinematic equations of the robot are presented. To derive the dynamic equations, Gibbs–Appel (G–A) formulation is used, which in contrary to the Lagrange formulation benefits from advantages of quasi-velocities over generalized coordinates as well as not requiring Lagrange multipliers. The dynamic equations of the two parts are coupled, and the interacting effects are observable from the governing equations. By considering non-holonomic wheels for the robot, internal dynamics appears in the equations. However, based on some conditions, the equations are input–output linearizable via a static feedback. The platform trajectory is designed based on the given end-effector trajectory. The effectiveness of the controller is shown through simulations and experimental tests.

Appendices

Appendix A: Kinematic equations of the wheeled mobile CDPR

The vector ${}^{{A}}\mathbf{x}_{{BA}}$ is the relative position of the end-effector with respect to the platform expressed in the platform frame. It can be expressed in terms of absolute positions of the platform ${}^{{N}}\mathbf{x}_{{AN}}$ and the end-effector ${}^{{N}}\mathbf{x}_{{BN}}$ as

$$ {}^{{A}}\mathbf{x}_{{BA}} = {}^{{N}} \mathbf{R}_{{A}}^{{T}} \bigl({}^{{N}} \mathbf{x}_{{BN}} - {}^{{N}}\mathbf{x}_{{AN}} \bigr). $$

(23)

The first derivative of Eq. (23) is as follows:

$$ {}^{{A}}\dot{\mathbf{x}}_{{BA}} = {}^{{N}} \mathbf{R}_{{A}}^{{T}} \bigl({}^{{N}}\dot{\mathbf{x}}_{{BN}} - {}^{{N}}\dot{\mathbf{x}}_{{AN}} \bigr) + {}^{{A}}\bar{\mathbf{x}}_{{BA}}{}^{{A}} \boldsymbol{\omega}_{{AN}} $$

(24)

where ${}^{{A}}\mathbf{x}_{{BA}}$ is dual vector of the anti-symmetric matrix of ${}^{{A}}\bar{\mathbf{x}}_{{BA}}$. The relative vector of the end-effector angular velocity with respect to the platform in the frame attached to the end-effector can be obtained as

$$ {}^{{B}}\boldsymbol{\omega}_{{BA}} = {}^{{B}} \boldsymbol{\omega}_{{BN}} - {}^{{N}}\mathbf{R}_{{B}}^{{T}}{}^{{N}} \mathbf{R}_{{A}}{}^{{A}}\boldsymbol{\omega}_{{AN}}. $$

(25)

By arranging Eqs. (24) and (25) into the more compact form, the inverse kinematic equation of the CDPR part of the robot can be obtained as

$$ \dot{\boldsymbol{\beta}} = \frac{1}{r}\boldsymbol{\varLambda}^{{T}}{}^{{A}} \bar{\mathbf{R}}_{{B}} \bigl[{}^{{A}}\dot{\mathbf{x}}_{{BA}}^{{T}},{}^{{B}}\boldsymbol{\omega}_{{BA}}^{{T}} \bigr]^{{T}} = \frac{1}{r} \boldsymbol{\varLambda}^{{T}}{}^{{A}}\bar{\mathbf{R}}_{{B}}(\mathbf{C}_{{1}} \dot{\mathbf{x}}_{{A}} + \mathbf{C}_{{2}} \dot{\tilde{\mathbf{x}}}_{{B}}) $$

(26)

where

$$\mathbf{C}_{{1}} = \left [ \textstyle\begin{array}{c@{\quad}c} - {}^{{N}}\mathbf{R}_{{A}}^{{T}} & {}^{{A}}\bar{\mathbf{X}}_{{BA}} \\ \mathbf{0}_{{3}} & - {}^{{N}}\mathbf{R}_{{B}}^{{T}}{}^{{N}}\mathbf{R}_{{A}} \end{array}\displaystyle \right ], \quad\quad \mathbf{C}_{{2}} = \left [ \textstyle\begin{array}{c@{\quad}c} {}^{{N}}\mathbf{R}_{{A}}^{{T}} & \mathbf{0}_{{3}} \\ \mathbf{0}_{{3}} & \mathbf{I}_{{3}} \end{array}\displaystyle \right ]. $$

The derivative of Eq. (26) leads to the angular acceleration vector of the output shaft of the CDPR motors

$$\begin{aligned} \begin{aligned} &\ddot{\boldsymbol{\beta}} = \mathbf{C}_{{3}} \ddot{\mathbf{x}}_{{A}} + \mathbf{C}_{{4}} \ddot{\tilde{\mathbf{x}}}_{{B}} + \mathbf{C}_{{5}}, \\ &\mathbf{C}_{{3}} = \frac{1}{r}\boldsymbol{\varLambda}^{{T}}\left [ \textstyle\begin{array}{c@{\quad}c} - {}^{{N}}\mathbf{R}_{{A}}^{{T}} & {}^{{A}}\bar{\mathbf{x}}_{{BA}} \\ \mathbf{0}_{{3}} & - \mathbf{I}_{{3}} \end{array}\displaystyle \right ], \quad\quad \mathbf{C}_{{4}} = \frac{1}{r}\boldsymbol{\varLambda}^{{T}}{}^{{A}}\bar{\mathbf{R}}_{{B}} \mathbf{C}_{{2}},\\ & \mathbf{C}_{{5}} = \frac{1}{r}\boldsymbol{\varLambda}^{{T}}\left [ \textstyle\begin{array}{c} - {}^{{A}}\bar{\mathbf{W}}_{{AN}}^{{2}}{}^{{N}}\mathbf{R}_{{A}}^{{T}}({}^{{N}}\mathbf{x}_{{BN}} - {}^{{N}}\mathbf{x}_{{AN}}) - {}^{{A}}\bar{\mathbf{W}}_{{AN}}({}^{{N}}\mathbf{R}_{{A}}^{{T}}({}^{{N}}\dot{\mathbf{x}}_{{B}} - {}^{{N}}\dot{\mathbf{x}}_{{A}}) + {}^{{A}}\bar{\mathbf{x}}_{{BA}}{}^{{A}}\boldsymbol{\omega}_{{AN}}) \\ - {}^{{A}}\boldsymbol{\omega}_{{AN}} \times {}^{{N}}\mathbf{R}_{{A}}^{{T}}{}^{{N}}\mathbf{R}_{{B}}{}^{{B}}\boldsymbol{\omega}_{{BN}} \end{array}\displaystyle \right ] \\ &\phantom{\mathbf{C}_{{5}} =\,\,}{} + \frac{1}{r}\dot{\boldsymbol{\varLambda}}^{{T}}{}^{{A}} \bar{\mathbf{R}}_{{B}}(\mathbf{C}_{{1}} \dot{\mathbf{x}}_{{A}} + \mathbf{C}_{{2}} \dot{\tilde{\mathbf{x}}}_{{B}}) \end{aligned} \end{aligned}$$

(27)

where ${}^{{A}}\bar{\mathbf{W}}_{{AN}}$ is the anti-symmetric matrix of the dual vector ${}^{{A}}\boldsymbol{\omega}_{{AN}}$. Considering the relation between angular velocity and Euler angle rates, the acceleration vector of the end-effector can be expressed as

$$ \ddot{\tilde{\mathbf{x}}}_{{B}} = \left [ \textstyle\begin{array}{c}^{{B}}\ddot{\mathbf{x}}_{{BN}} \\ {}^{{B}}\dot{\boldsymbol{\omega}}_{{BN}} \end{array}\displaystyle \right ] = \left [ \textstyle\begin{array}{c@{\quad}c} \mathbf{I}_{{3}} & \mathbf{0}_{{3}} \\ \mathbf{0}_{{3}} & \mathbf{P}_{{B}} \end{array}\displaystyle \right ]\left [ \textstyle\begin{array}{c}^{{B}}\ddot{\mathbf{x}}_{{BN}} \\ \ddot{\boldsymbol{\varPsi}}_{{B}} \end{array}\displaystyle \right ] + \left [ \textstyle\begin{array}{c} \mathbf{0}_{{3 \times 1}} \\ \dot{\mathbf{P}}_{{B}}\dot{\boldsymbol{\varPsi}}_{{B}} \end{array}\displaystyle \right ] = \mathbf{C}_{{8}}\ddot{\mathbf{x}}_{{B}} + \mathbf{C}_{{9}}. $$

(28)

The derivative of Eq. (1) which is used in dynamic equation derivation can be expressed as

$$ \ddot{\mathbf{x}}_{{A}} = \mathbf{C}_{{6}} \ddot{\boldsymbol{\theta}} + \mathbf{C}_{{7}} \dot{\varphi} \dot{\boldsymbol{\theta}} $$

(29)

where

$$\mathbf{C}_{{7}} = \frac{r_{\mathit{wh}}}{2b}\left [ \textstyle\begin{array}{c@{\quad}c} d\cos \varphi - b\sin \varphi & - d\cos \varphi - b\sin \varphi \\ d\sin \varphi + b\cos \varphi & - d\sin \varphi + b\cos \varphi \\ \mathbf{0}_{{4 \times 1}} & \mathbf{0}_{{4 \times 1}} \end{array}\displaystyle \right ]. $$

Appendix B: Dynamic equations of the wheeled mobile CDPR

In order to derive the equations of motion, the Gibbs function and generalized power are extended in terms of $\dot{\tilde{\mathbf{x}}}_{{B}}$ and $\dot{\boldsymbol{\theta}}$ as follows:

$$ \left \{ \textstyle\begin{array}{l} \biggl[ \frac{\partial \ddot{\boldsymbol{\beta}}}{\partial \ddot{\tilde{\mathbf{x}}}_{{B}}} \biggr]^{T} \biggl[ \frac{\partial S_{\beta}}{\partial \ddot{\boldsymbol{\beta}}} \biggr]^{T} + \biggl[ \frac{\partial S_{B}}{\partial \ddot{\tilde{\mathbf{x}}}_{{B}}} \biggr]^{T} = \biggl[ \frac{\partial \dot{\boldsymbol{\beta}}}{\partial \dot{\tilde{\mathbf{x}}}_{{B}}} \biggr]^{T}\boldsymbol{\tau}_{\boldsymbol{\beta}} + [ \mathbf{M}_{{B}} ]g\left [ \textstyle\begin{array}{c@{\quad}c@{\quad}c} \mathbf{0}_{{1 \times 2}} & g & \mathbf{0}_{{1 \times 3}} \end{array}\displaystyle \right ]^{T} \\ \biggl[ \frac{\partial \ddot{\mathbf{x}}_{{A}}}{\partial \ddot{\boldsymbol{\theta}}} \biggr]^{T} \biggl[ \frac{\partial S_{A}}{\partial \ddot{\mathbf{x}}_{{A}}} \biggr]^{T} + \biggl[ \frac{\partial \ddot{\mathbf{x}}_{{A}}}{\partial \ddot{\boldsymbol{\theta}}} \biggr]^{T} \biggl[ \frac{\partial S_{\beta}}{ \partial \ddot{\mathbf{x}}_{{A}}} \biggr]^{T} + \mathbf{I}_{{w}} \ddot{\boldsymbol{\theta}} = \biggl[ \frac{\partial \dot{\mathbf{x}}_{{A}}}{\partial \dot{\boldsymbol{\theta}}} \biggr]^{T} \biggl[ \frac{\partial \dot{\boldsymbol{\beta}}}{\partial \dot{\mathbf{x}}_{{A}}} \biggr]^{T}\boldsymbol{\tau}_{\boldsymbol{\beta}} + \boldsymbol{\tau}_{\boldsymbol{\theta}} .\end{array}\displaystyle \right . $$

(30)

By substituting the kinematic equations (1), (3), (27)–(29), and the dynamic equations of (6) and (7) into Eq. (30), the combined dynamic equations of the end-effector and platform can be obtained as

$$ \mathbf{M}\left [ \textstyle\begin{array}{c} \ddot{\mathbf{x}}_{{B}} \\ \ddot{\boldsymbol{\theta}} \end{array}\displaystyle \right ] + \mathbf{C} + \mathbf{G} = \mathbf{F} \boldsymbol{\tau} $$

(31)

where

$$\begin{aligned} & \mathbf{M} = \left [ \textstyle\begin{array}{c@{\quad}c} \mathbf{C}_{{8}}^{{T}} ( \mathbf{C}_{{4}}^{{T}}\mathbf{I}_{{m}}\mathbf{C}_{{4}} + [ \mathbf{M}_{{B}} ] )\mathbf{C}_{{8}} & \mathbf{C}_{{8}}^{{T}} ( \mathbf{C}_{{4}}^{{T}}\mathbf{I}_{{m}}(\mathbf{C}_{{3}} + \mathbf{C}_{{10}}) \mathbf{C}_{{6}} ) \\ ( \mathbf{C}_{{8}}^{{T}} ( \mathbf{C}_{{4}}^{{T}}\mathbf{I}_{{m}}(\mathbf{C}_{{3}} + \mathbf{C}_{{10}}) \mathbf{C}_{{6}} ) )^{{T}} & \mathbf{C}_{{6}}^{{T}} [ \mathbf{M}_{{A}} ]\mathbf{C}_{{6}} + \mathbf{C}_{{6}}^{{T}}(\mathbf{C}_{{3}} + \mathbf{C}_{{10}})^{{T}}\mathbf{I}_{{m}}(\mathbf{C}_{{3}} + \mathbf{C}_{{10}})\mathbf{C}_{{6}} + \mathbf{I}_{{w}} \end{array}\displaystyle \right ], \\ &\mathbf{C} = \left [ \textstyle\begin{array}{c} \mathbf{C}_{{8}}^{{T}} ( ( \mathbf{C}_{{4}}^{{T}}\mathbf{I}_{{m}}\mathbf{C}_{{4}} + [ \mathbf{M}_{{B}} ] )\mathbf{C}_{{9}} + \mathbf{C}_{{4}}^{{T}}\mathbf{I}_{{m}} ( (\mathbf{C}_{{3}} + \mathbf{C}_{{10}}) \mathbf{C}_{{7}}\dot{\varphi} \dot{\boldsymbol{\theta}} + \mathbf{C}_{{5}} ) ) \\ ( \mathbf{C}_{{6}}^{{T}}(\mathbf{C}_{{3}} + \mathbf{C}_{{10}})^{{T}}\mathbf{I}_{{m}}\mathbf{C}_{{4}} )\mathbf{C}_{{9}} + \mathbf{C}_{{6}}^{{T}} [ \mathbf{M}_{{A}} ]\mathbf{C}_{{7}}\dot{\varphi} \dot{\boldsymbol{\theta}} + \mathbf{C}_{{6}}^{{T}}(\mathbf{C}_{{3}} + \mathbf{C}_{{10}})^{{T}}\mathbf{I}_{{m}} ( (\mathbf{C}_{{3}} + \mathbf{C}_{{10}}) \mathbf{C}_{{7}}\dot{\varphi} \dot{\boldsymbol{\theta}} + \mathbf{C}_{{5}} ) \end{array}\displaystyle \right ] \\ &\phantom{\mathbf{C} =\,\,}{}+ \left [ \textstyle\begin{array}{c} \left [ \textstyle\begin{array}{c} \mathbf{0}_{{3 \times 1}} \\ \mathbf{P}_{{BN}}^{{T}} ( \mathbf{P}_{{BN}}\dot{\boldsymbol{\varPsi}}_{{BN}} \times \mathbf{I}_{{B}}\mathbf{P}_{{BN}}\dot{\boldsymbol{\varPsi}}_{{BN}} ) \end{array}\displaystyle \right ] \\ \mathbf{C}_{{6}}^{{T}}\left [ \textstyle\begin{array}{c} \mathbf{0}_{{3 \times 1}} \\ \mathbf{P}_{{AN}}\dot{\boldsymbol{\varPsi}}_{{AN}} \times \mathbf{I}_{{A}}\mathbf{P}_{{AN}}\dot{\boldsymbol{\varPsi}}_{{AN}} \end{array}\displaystyle \right ] \end{array}\displaystyle \right ], \\ &\mathbf{G} = [ \mathbf{M}_{{B}} ]\left [ \textstyle\begin{array}{c} \mathbf{0}_{{2 \times 1}} \\ - g \\ \mathbf{0}_{{5 \times 1}} \end{array}\displaystyle \right ],\quad\quad \mathbf{F} = \left [ \textstyle\begin{array}{c@{\quad}c} \frac{1}{r}\mathbf{C}_{{8}}^{{T}} \mathbf{C}_{{2}}^{{T}}{}^{{A}}\bar{\mathbf{R}}_{{B}}^{{T}}\boldsymbol{\varLambda} & \mathbf{0}_{{2}} \\ \frac{1}{r}\mathbf{C}_{{6}}^{{T}}\mathbf{C}_{{1}}^{{T}}{}^{{A}}\bar{\mathbf{R}}_{{B}}^{{T}}\boldsymbol{\varLambda} & \mathbf{I}_{{2}} \end{array}\displaystyle \right ], \quad\quad \boldsymbol{\tau} = \left [ \textstyle\begin{array}{c} \boldsymbol{\tau}_{\boldsymbol{\beta}} \\ \boldsymbol{\tau}_{\boldsymbol{\theta}} \end{array}\displaystyle \right ]. \end{aligned}$$