A compliant self-adaptive gripper with proprioceptive haptic feedback

Abstract

Grippers and robotic hands are an important field in robotics. Recently, the combination of grasping devices and haptic feedback has been a promising avenue for many applications such as laparoscopic surgery and spatial telemanipulation. This paper presents the work behind a new self-adaptive, a.k.a. underactuated, gripper with a proprioceptive haptic feedback in which the apparent stiffness of the gripper as seen by its actuator is used to estimate contact location. This system combines many technologies and concepts in an integrated mechatronic tool. Among them, underactuated grasping, haptic feedback, compliant joints and a differential seesaw mechanism are used. Following a theoretical modeling of the gripper based on the virtual work principle, the authors present numerical data used to validate this model. Then, a presentation of the practical prototype is given, discussing the sensors, controllers, and mechanical architecture. Finally, the control law and the experimental validation of the haptic feedback are presented.

Appendices

Appendix 1: Kinematic loop-closure equations

$$\begin{aligned} 0&= L_1\cos {\alpha _{k1}}+L_2\cos {\alpha _{k2}}+c\cos {\alpha _{k3}} +b\cos {\alpha _{k4}}\nonumber \\&\quad +a\cos {\alpha _{k5}}-d\cos {\lambda },\end{aligned}$$

(14)

$$\begin{aligned} 0&= L_1\sin {\alpha _{k1}}+L_2\sin {\alpha _{k2}}+c\sin {\alpha _{k3}} +b\sin {\alpha _{k4}}\nonumber \\&\quad +a\sin {\alpha _{k5}}-d\sin {\lambda }, \end{aligned}$$

(15)

$$\begin{aligned} 0&= m\cos ({\psi +\zeta _k\beta _{2}})+h\cos {(\beta _{k1})}-g\cos {\alpha _{k6}}-f,\end{aligned}$$

(16)

$$\begin{aligned} 0&= m\sin ({\psi +\zeta _k\beta _{2}})+h\sin {(\beta _{k1})} -g\sin {\alpha _{k6}}-Y_a,\end{aligned}$$

(17)

$$\begin{aligned} 0&= (\zeta _k +1) f + g \sin (\alpha _{R6}+(\zeta _k+1)\frac{\pi }{4})\nonumber \\&\quad -h \sin (\beta _{R1}+(\zeta _k+1)\frac{\pi }{4})- n \sin (\beta _{2}+(\zeta _k+1)\frac{\pi }{4})\nonumber \\&\quad -\zeta _k h \sin (\beta _{L1}\!+\!(\zeta _k\!+\!1)\frac{\pi }{4})\!+\! \zeta _k g \sin (\alpha _{L6}\!+\!(\zeta _k\!+\!1)\frac{\pi }{4}).\nonumber \\ \end{aligned}$$

(18)

Please note that the gripper is designed to be symmetrical, i.e. all lengths are the same on both sides.

Appendix 2: Virtual work matrices

The virtual work matrices for each side of the gripper are:

$$\begin{aligned} \mathbf M _k&= \begin{bmatrix} \mathbf m _{k1}&\ldots&\mathbf m _{k15} \end{bmatrix}, \end{aligned}$$

(19)

$$\begin{aligned} \mathbf m _{k1}&= \begin{bmatrix} 1&0&0 \end{bmatrix}^T, \end{aligned}$$

(20)

$$\begin{aligned} \mathbf m _{k2}&= \begin{bmatrix} -1&1&0 \end{bmatrix}^T, \end{aligned}$$

(21)

$$\begin{aligned} \mathbf m _{k3}&= \begin{bmatrix} 0&-1&1 \end{bmatrix}^T, \end{aligned}$$

(22)

$$\begin{aligned} \mathbf m _{k4}&= \begin{bmatrix} \frac{\delta \alpha _{k4}}{\delta \alpha _{k1}}&\frac{\delta \alpha _{k4}}{\delta \alpha _{k2}}&\frac{\delta \alpha _{k4}}{\delta \alpha _{k3}}-1 \end{bmatrix}^T, \end{aligned}$$

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$$\begin{aligned} \mathbf m _{k5}&= \begin{bmatrix} \frac{\delta \alpha _{k5}}{\delta \alpha _{k1}} -\frac{\delta \alpha _{k4}}{\delta \alpha _{k1}}\\ \frac{\delta \alpha _{k5}}{\delta \alpha _{k2}} -\frac{\delta \alpha _{k4}}{\delta \alpha _{k2}} \\ \frac{\delta \alpha _{k5}}{\delta \alpha _{k3}} -\frac{\delta \alpha _{k4}}{\delta \alpha _{k3}} \end{bmatrix}, \end{aligned}$$

(24)

$$\begin{aligned} \mathbf m _{k6}&= \begin{bmatrix} \frac{\delta \alpha _{k5}}{\delta \alpha _{k1}}&\frac{\delta \alpha _{k5}}{\delta \alpha _{k2}}&\frac{\delta \alpha _{k5}}{\delta \alpha _{k3}} \end{bmatrix}^T, \end{aligned}$$

(25)

$$\begin{aligned} \mathbf m _{k7}&= \begin{bmatrix} \frac{\delta \beta _{k1}}{\delta \alpha _{k1}} -\frac{\delta \alpha _{k5}}{\delta \alpha _{k1}}\\ \frac{\delta \beta _{k1}}{\delta \alpha _{k2}} -\frac{\delta \alpha _{k5}}{\delta \alpha _{k2}} \\ \frac{\delta \beta _{k1}}{\delta \alpha _{k3}} -\frac{\delta \alpha _{k5}}{\delta \alpha _{k3}} \end{bmatrix}, \end{aligned}$$

(26)

$$\begin{aligned} \mathbf m _{k8}&= \begin{bmatrix} \frac{\delta \beta _{k1}}{\delta \alpha _{k1}} -\zeta _k\frac{\delta \beta _{2}}{\delta \alpha _{k1}}\\ \frac{\delta \beta _{k1}}{\delta \alpha _{k2}} -\zeta _k\frac{\delta \beta _{2}}{\delta \alpha _{k2}} \\ \frac{\delta \beta _{k1}}{\delta \alpha _{k3}} -\zeta _k\frac{\delta \beta _{2}}{\delta \alpha _{k3}} \end{bmatrix}, \end{aligned}$$

(27)

$$\begin{aligned} \mathbf m _{k9}&= \begin{bmatrix} \frac{\delta \beta _{2}}{\delta \alpha _{k1}}&\frac{\delta \beta _{2}}{\delta \alpha _{k2}}&\frac{\delta \beta _{2}}{\delta \alpha _{k3}} \end{bmatrix}^T, \end{aligned}$$

(28)

$$\begin{aligned} \mathbf m _{k10}&= \begin{bmatrix} \frac{\delta \beta _{k'1}}{\delta \alpha _{k1}}&\frac{\delta \beta _{k'1}}{\delta \alpha _{k2}}&\frac{\delta \beta _{k'1}}{\delta \alpha _{k3}} \end{bmatrix}^T, \end{aligned}$$

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$$\begin{aligned} \mathbf m _{k11}&= \begin{bmatrix} \frac{\delta \beta _{k'1}}{\delta \alpha _{k1}} +\zeta _k\frac{\delta \beta _{2}}{\delta \alpha _{k1}}\\ \frac{\delta \beta _{k'1}}{\delta \alpha _{k2}} +\zeta _k\frac{\delta \beta _{2}}{\delta \alpha _{k2}} \\ \frac{\delta \beta _{k'1}}{\delta \alpha _{k3}} +\zeta _k\frac{\delta \beta _{2}}{\delta \alpha _{k3}} \end{bmatrix}, \end{aligned}$$

(30)

$$\begin{aligned} \mathbf m _{k12}&= \begin{bmatrix} -k_{k1}&0&0 \end{bmatrix}^T, \end{aligned}$$

(31)

$$\begin{aligned} \mathbf m _{k13}&= \begin{bmatrix} -L_1 \cos (\alpha _{k2}-\alpha _{k1})&-k_{k2}&0 \end{bmatrix}^T, \end{aligned}$$

(32)

$$\begin{aligned} \mathbf m _{k14}&= \begin{bmatrix} -L_1 \cos (\alpha _{k3}-\alpha _{k1}) \\ -L_2 \cos (\alpha _{k3}-\alpha _{k2}) \\ -k_{k3} \end{bmatrix}, \end{aligned}$$

(33)

$$\begin{aligned} \mathbf m _{k15}&= \begin{bmatrix} -\frac{\delta Y_a}{\delta \alpha _{k1}}&-\frac{\delta Y_a}{\delta \alpha _{k2}}&-\frac{\delta Y_a}{\delta \alpha _{k3}} \end{bmatrix}^T. \end{aligned}$$

(34)

Finally, the coefficient of the stiffness matrix are :

$$\begin{aligned} K_{11,i}&= \frac{E w_i t_i^3}{3 l_i},\end{aligned}$$

(35)

$$\begin{aligned} K_{12,i}&= \frac{E w_i t_i^3}{2 l_i^2},\end{aligned}$$

(36)

$$\begin{aligned} K_{22,i}&= \frac{E w_i t_i^3}{l_i^3},\end{aligned}$$

(37)

$$\begin{aligned} K_{33,i}&= \frac{E w_i t_i}{l_i}. \end{aligned}$$

(38)