Compliance Modeling and Error Compensation of a 3-Parallelogram Lightweight Robotic Arm

Abstract

This paper presents compliance modeling and error compensation for lightweight robotic arms built with parallelogram linkages, i.e., $\varPi$ joints. The Cartesian stiffness matrix is derived using the virtual joint method. Based on the developed stiffness model, a method to compensate the compliance error is introduced, being illustrated with a 3-parallelogram robot in the application of pick-and-place operation. The results show that this compensation method can effectively improve the operation accuracy.

Appendix

The force Jacobian matrix is expressed as

$${\mathbf{J}}_{\theta } = \left[ {\begin{array}{*{20}c} {\hat{\vec{\$}}}_{\theta ,1} &{{\hat{\vec{{\$ }}}}_{\theta ,2} } & {{\hat{\vec{{\$}}}}_{21} } & \ldots & {{\hat{\vec{{\$ }}}}_{26} } & {{\hat{\vec{{\$ }}}}_{\theta ,3} } & {{\hat{\vec{{\$ }}}}_{31} } & \ldots & {{\hat{\vec{{\$}}}}_{36} } \\ \end{array} } \right] \in {\vec{R}}^{6 \times 15}$$

(21)

with the unit screws

$$\begin{aligned} & {\varvec {\mathbf{\hat{\$ }}}}_{\theta ,1} = \left[ {\begin{array}{*{20}c} {\mathbf{k}} \\ { - {\mathbf{q}} \times {\mathbf{k}}} \\ \end{array} } \right];{\kern 1pt} {\varvec{\mathbf{\hat{\$ }}}}_{\theta ,2} = \left[ {\begin{array}{*{20}c} {{\mathbf{z}}_{1} } \\ {{\mathbf{q}}_{1} \times {\mathbf{z}}_{1} } \\ \end{array} } \right];{\kern 1pt} {\varvec{\mathbf{\hat{\$ }}}}_{\theta ,3} = \left[ {\begin{array}{*{20}c} {{\mathbf{z}}_{2} } \\ {{\mathbf{q}}_{2} \times {\mathbf{z}}_{2} } \\ \end{array} } \right];{\kern 1pt} {\varvec{\mathbf{\hat{\$ }}}}_{n1} = \left[ {\begin{array}{*{20}c} {{\mathbf{x}}_{n - 1} } \\ {{\mathbf{q}}_{n} \times {\mathbf{x}}_{n - 1} } \\ \end{array} } \right] \hfill \\ & {\varvec{\mathbf{\hat{\$ }}}}_{n2} = \left[ {\begin{array}{*{20}c} {{\mathbf{y}}_{n - 1} } \\ {{\mathbf{q}}_{n} \times {\mathbf{y}}_{n - 1} } \\ \end{array} } \right];{\kern 1pt} {\varvec{\mathbf{\hat{\$ }}}}_{n3} = \left[ {\begin{array}{*{20}c} {{\mathbf{z}}_{n - 1} } \\ {{\mathbf{q}}_{n} \times {\mathbf{z}}_{n - 1} } \\ \end{array} } \right];{\kern 1pt} {\varvec{\mathbf{\hat{\$ }}}}_{n4} = \left[ {\begin{array}{*{20}c} {\mathbf{0}} \\ {{\mathbf{x}}_{n - 1} } \\ \end{array} } \right];{\kern 1pt} {\varvec{\mathbf{\hat{\$ }}}}_{n5} = \left[ {\begin{array}{*{20}c} {\mathbf{0}} \\ {{\mathbf{y}}_{n - 1} } \\ \end{array} } \right];{\kern 1pt} {\varvec{\mathbf{\hat{\$ }}}}_{n6} = \left[ {\begin{array}{*{20}c} {\mathbf{0}} \\ {{\mathbf{z}}_{n - 1} } \\ \end{array} } \right] \hfill \\ \end{aligned}$$

(22)

and

$${\mathbf{x}}_{n} = {\mathbf{R}}_{n} {\mathbf{i}};\quad {\mathbf{y}}_{n} = {\mathbf{R}}_{n} {\mathbf{j}};\quad {\mathbf{z}}_{n} = {\mathbf{R}}_{n} {\mathbf{k}};\quad {\mathbf{q}}_{n} = {\mathbf{p}}_{n} - {\mathbf{q}}$$

(23)

where ${\mathbf{R}}_{n}$ and ${\mathbf{p}}_{n}$, $n = 1,{\kern 1pt} 2,{\kern 1pt} 3$, respectively, are the rotation matrix and position vector extracted from $\prod\nolimits_{i = 1}^{n} {^{i - 1} {\mathbf{A}}_{i} }$ of Eq. (1). Moreover, vectors ${\mathbf{i}}$, ${\mathbf{j}}$ and ${\mathbf{k}}$ represent the unit vectors of $x$-, $y$- and $z$-axis, respectively.