An analysis of grasp quality measures for the application of sheet metal parts grasping

Abstract

Acquiring a qualitative force-closure grasp requires the determination of feasible contact points on the object based on a defined criterion. The determination must be fast in order to implement feasible synthesis algorithms. Moreover, grasping of sheet metal parts has further requirements derived from its geometry and clamping tools. This paper presents a grasp quality analysis for the application of sheet metal parts. Moreover, a novel grasp quality measure approach is proposed based on standard deviation computation of the contact’s coordinates. The proposed measure is frame invariant, simple to implement, and has low computational complexity. A comparative analysis over other measures is presented. Further, a stress analysis was performed to show that the proposed criterion yields low stress on the sheet metal part compared to other criteria. Simulations show advantage to grasp synthesis with the proposed quality measure.

Appendix

The appendix contains proofs of the results stated in Sect. 4.

Lemma 1

Given two sets $\mathcal {P}_1=\left\{ \mathbf {a}_{\mathbf{1}},\ldots ,\mathbf {a}_{\mathbf{n}}\right\}$ and $\mathcal {P}_2=\left\{ \mathbf {b}_{\mathbf{1}},\ldots ,\mathbf {b}_{\mathbf{n}}\right\}$, where there exists a rotation matrix $R\in SO(3)$ such that $\mathbf {b}_{\mathbf{k}}=R\cdot \mathbf {a}_{\mathbf{k}}$ for all $k=1,\ldots ,n$. The rotation matrices $R_{SD}^{\mathbf {a}}$ and $R_{SD}^{\mathbf {b}}$ are computed by the PCA function such that $\mathbf {a}_{\mathbf{k}}'=R_{SD}^{\mathbf {a}}\mathbf {a}_{\mathbf{k}}$ and $\mathbf {b}_{\mathbf{k}}'=R_{SD}^{\mathbf {b}}\mathbf {b}_{\mathbf{k}}$. By that, the respected vectors are equal; that is, $\mathbf {a}_{\mathbf{k}}'=\mathbf {b}_{\mathbf{k}}'$ for all $k=1,\ldots ,n$.

Proof

We define the mean vector of $\mathcal {P}_1$ to be

$$\begin{aligned} \bar{\mathbf{a}}=\frac{1}{n}\sum _{k=1}^n\mathbf {a}_{\mathbf{k}}. \end{aligned}$$

(28)

By definition, there exists a rotation matrix R such that $\mathbf {b}_{\mathbf{k}}=R\cdot \mathbf {a}_{\mathbf{k}}$ for all $k=1,\ldots ,n$. Therefore, the mean vector of $\mathcal {P}_2$ with respect to $\bar{\mathbf{a}}$ is shown to be

$$\begin{aligned} \bar{\mathbf{b}}=\frac{1}{n}\sum _{k=1}^n\mathbf {b}_{\mathbf{k}}=\frac{1}{n}\sum _{k=1}^nR\mathbf {a}_{\mathbf{k}}=R\bar{\mathbf{a}}. \end{aligned}$$

(29)

Let $\mathbf {a}$ and $\mathbf {B}$ be matrices concatenating the vectors in $\mathcal {P}_1$ and $\mathcal {P}_2$, respectively, such that

(30)

and

(31)

The covariance matrices of $\mathcal {P}_1$ and $\mathcal {P}_2$ are given by

$$\begin{aligned} M_{\mathbf {a}}=\frac{1}{n}\mathbf {a}\mathbf {a}^T-\bar{\mathbf{a}} \bar{\mathbf{a}}^T \end{aligned}$$

(32)

and

$$\begin{aligned} M_{\mathbf {b}}=\frac{1}{n}\mathbf {B}\mathbf {B}^T-\bar{\mathbf{b}}\bar{\mathbf{b}}^T, \end{aligned}$$

(33)

respectively. Applying (29) and (31) on (33) yields

$$\begin{aligned} M_{\mathbf {b}}= & {} \frac{1}{n}R\mathbf {a}\mathbf {a}^TR^T-R\bar{\mathbf{a}}\bar{\mathbf{a}}^TR^T\nonumber \\= & {} R\left( \frac{1}{n}\mathbf {a}\mathbf {a}^T-\bar{\mathbf{a}}\bar{\mathbf{a}}^T\right) R^T. \end{aligned}$$

(34)

According to (32) we acquire the relation between both covariance matrices to be

$$\begin{aligned} M_{\mathbf {b}}=RM_{\mathbf {a}}R^T. \end{aligned}$$

(35)

Next, the eigenvalues $\lambda _{\mathbf {a}_i},~i=1,2,3$ of $M_{\mathbf {a}}$ are the solution of

$$\begin{aligned} det\left( M_{\mathbf {a}}-\lambda _{\mathbf {a}}I\right) =0. \end{aligned}$$

(36)

Similarly, the eigenvalues $\lambda _{\mathbf {B}_i},~i=1,2,3$ of $M_{\mathbf {b}}$ are the solution of

$$\begin{aligned} det\left( M_{\mathbf {b}}-\lambda _{\mathbf {b}}I\right) =0. \end{aligned}$$

(37)

Equivalently, from (35), the left hand side of (37) can be written as

$$\begin{aligned} det\left( RM_{\mathbf {a}}R^T-\lambda _{\mathbf {b}}I\right)= & {} det\left( RM_{\mathbf {a}}R^T-\lambda _{\mathbf {b}}RR^T\right) \nonumber \\= & {} det(R)det\left( M_{\mathbf {a}}-\lambda _{\mathbf {b}}I\right) det\left( R^T\right) .\nonumber \\ \end{aligned}$$

(38)

By definition, matrix R is an orthogonal matrix such that $det(R)=det(R^T)=1$ and therefore

$$\begin{aligned} det\left( M_{\mathbf {b}}-\lambda _{\mathbf {b}}I\right) =det\left( M_{\mathbf {a}}-\lambda _{\mathbf {b}}I\right) =0. \end{aligned}$$

(39)

From (36) and (39) we can conclude that the eigenvalues of $M_{\mathbf {a}}$ and $M_{\mathbf {b}}$ are equal. That is,

$$\begin{aligned} \lambda _{\mathbf {a}_i}=\lambda _{\mathbf {B}_i}=\lambda _i,~~\forall i=1,2,3. \end{aligned}$$

(40)

To find the rotation matrices $R_{SD}^{\mathbf {a}}$ and $R_{SD}^{\mathbf {b}}$ we need to compute the eigenvectors of $M_{\mathbf {a}}$ and $M_{\mathbf {b}}$ by solving the following equations

$$\begin{aligned} (M_{\mathbf {a}}-\lambda _iI)\mathbf {v}_{\mathbf {a}_{\mathbf{i}}}=0 \end{aligned}$$

(41)

and

$$\begin{aligned} (M_{\mathbf {b}}-\lambda _iI)\mathbf {v}_{\mathbf {B}_{\mathbf{i}}}=0 \end{aligned}$$

(42)

for $i=1,2,3$ where $\mathbf {v}_{\mathbf{a}_{\mathbf{i}}}$ and $\mathbf {v}_{\mathbf{b}_{\mathbf{i}}}$ are the respective eigenvectors. Using (35) we can rewrite (42)

$$\begin{aligned} (RM_{\mathbf {a}}R^T-\lambda _iI)\mathbf {v}_{\mathbf {B}_{\mathbf{i}}}= & {} \left( RM_{\mathbf {a}}R^T-\lambda _iRR^T\right) \mathbf {v}_{\mathbf {B}_{\mathbf{i}}}\nonumber \\= & {} R((M_{\mathbf {a}}-\lambda _iI)R^T\mathbf {v}_{\mathbf {B}_{\mathbf{i}}}\nonumber \\= & {} ((M_{\mathbf {a}}-\lambda _iI)R^T\mathbf {v}_{\mathbf {B}_{\mathbf{i}}}=0 \end{aligned}$$

(43)

and from (41) we show that

$$\begin{aligned} \mathbf {v}_{\mathbf{a}_{\mathbf{i}}}=R^T\mathbf {v}_{\mathbf{b}_{\mathbf{i}}}, \forall i=1,2,3. \end{aligned}$$

(44)

From its definition, matrix $R_{SD}^{\mathbf {a}}$ is formed by the eigenvectors as follows

(45)

and from (44)

$$\begin{aligned} {R_{SD}^{\mathbf {a}}}^T=R^T \begin{bmatrix} \mathbf {v}_{\mathbf{b}_{\mathbf{1}}}&\mathbf {v}_{\mathbf{b}_{\mathbf{2}}}&\mathbf {v}_{\mathbf{b}_{\mathbf{3}}} \end{bmatrix}=R^T {R_{SD}^{\mathbf {b}}}^T \end{aligned}$$

(46)

or

$$\begin{aligned} R_{SD}^{\mathbf {a}}=R_{SD}^{\mathbf {b}}R. \end{aligned}$$

(47)

Applying (47) to $\mathbf {a}_{\mathbf{k}}'$ yields

$$\begin{aligned} \mathbf {a}_{\mathbf{k}}'=R_{SD}^{\mathbf {a}}\mathbf {a}_{\mathbf{k}}=R_{SD}^{\mathbf {b}}R\mathbf {a}_{\mathbf{k}}=R_{SD}^{\mathbf {b}}\mathbf {b}_{\mathbf{k}}=\mathbf {b}_{\mathbf{k}}'. \end{aligned}$$

(48)

That is, we have shown that

$$\begin{aligned} \mathbf {a}_{\mathbf{k}}'=\mathbf {b}_{\mathbf{k}}',~\forall k=1,\ldots ,n. \end{aligned}$$

(49)

We have shown that two sets represented in two rotated reference frames are equal after a PCA rotation. $\square$

Theorem 2

Given two sets $\mathcal {P}_1=\left\{ \mathbf {a}_{\mathbf{1}},\ldots ,\mathbf {a}_{\mathbf{n}}\right\}$ and $\mathcal {P}_2=\left\{ \mathbf {b}_{\mathbf{1}},\ldots ,\mathbf {b}_{\mathbf{n}}\right\}$, where there exist a rotation matrix $R\in SO(3)$ and a translation vector $\mathbf {d}\in {\mathbb {R}}^3$ such that $\mathbf {b}_{\mathbf{k}}=R\cdot \mathbf {a}_{\mathbf{k}}+\mathbf {d}$ for all $k=1,\ldots ,n$. Let $\tau _{\mathbf {a}}'$ and $\tau _{\mathbf {b}}'$ be the PCA-SD vectors of $\mathcal {P}_1$ and $\mathcal {P}_2$, respectively. Therefore, both PCA-SD vectors are invariant to any arbitrary rotation R and translation d such that $\tau _{\mathbf {a}}'=\tau _{\mathbf {b}}'$.

Proof

The mean vector of $\mathcal {P}_1$ is given in (28) and therefore the mean vector of $\mathcal {P}_2$ can be written as

$$\begin{aligned} \bar{\mathbf{b}}= & {} \frac{1}{n}\sum _{k=1}^n\mathbf {b}_{\mathbf{k}}=\frac{1}{n}\sum _{k=1}^n(R\mathbf {a}_{\mathbf{k}}+\mathbf {d})\nonumber \\= & {} \frac{1}{n}\sum _{k=1}^nR\mathbf {a}_{\mathbf{k}}+\frac{1}{n}\sum _{k=1}^n\mathbf {d}=R\bar{\mathbf{a}}+\mathbf {d}. \end{aligned}$$

(50)

The SD vector for $\mathcal {P}_2$ is given by

$$\begin{aligned} \tau _{\mathbf {b}}=\sqrt{\frac{1}{n}\sum _{k=1}^n\left( \mathbf {b}_{\mathbf{k}}-\bar{\mathbf{b}}\right) ^2}. \end{aligned}$$

(51)

According to the definition of $\mathbf {b}_{\mathbf{k}}=R\cdot \mathbf {a}_{\mathbf{k}}+\mathbf {d}$ and using (50), $\tau _{\mathbf {b}}$ with respect to $\tau _{\mathbf {a}}$ has the form

$$\begin{aligned} \tau _{\mathbf {b}}=\sqrt{\frac{1}{n}\sum _{k=1}^n(R\mathbf {a}_{\mathbf{k}}+\mathbf {d}-(R\bar{\mathbf{a}}+\mathbf {d}))^2} =R\tau _{\mathbf {a}}. \end{aligned}$$

(52)

Recall $\mathbf {a}_{\mathbf{k}}'$ and $\mathbf {b}_{\mathbf{k}}'$ to be the PCA rotated vectors of $\mathbf {a}_{\mathbf{k}}$ and $\mathbf {b}_{\mathbf{k}}$, respectively, such that $\mathbf {a}_{\mathbf{k}}'=R_{SD}^{\mathbf {a}}\mathbf {a}_{\mathbf{k}}$ and $\mathbf {b}_{\mathbf{k}}'=R_{SD}^{\mathbf {b}}\mathbf {b}_{\mathbf{k}}$. Therefore, the PCA-SD vector of $\mathcal {P}_1$ is

$$\begin{aligned} \tau _{\mathbf {a}}'= & {} \sqrt{\frac{1}{n}\sum _{k=1}^n\left( \mathbf {a}_{\mathbf{k}}'-\bar{\mathbf {a}}'\right) ^2}\nonumber \\= & {} \sqrt{\frac{1}{n}\sum _{k=1}^n\left( R_{SD}^{\mathbf {a}}\mathbf {a}_{\mathbf{k}}-R_{SD}^{\mathbf {a}}\bar{\mathbf {a}}'\right) ^2}=R_{SD}^{\mathbf {a}}\tau _{\mathbf {a}} \end{aligned}$$

(53)

and similarly, the PCA-SD vector for $\mathcal {P}_2$ is $\tau _{\mathbf {b}}'=R_{SD}^{\mathbf {b}}\tau _{\mathbf {b}}$. The PCA-SD vector for $\mathcal {P}_2$ could be rewritten using (52) as

$$\begin{aligned} \tau _{\mathbf {b}}'=R_{SD}^{\mathbf {b}}R\tau _{\mathbf {a}} \end{aligned}$$

(54)

or according to (47) of Lemma 1 and (53) as

$$\begin{aligned} \tau _{\mathbf {b}}'=R_{SD}^{\mathbf {a}}\tau _{\mathbf {a}}=\tau _{\mathbf {a}}'. \end{aligned}$$

(55)

That is, the PCA-SD vector is invariant to any arbitrary rotation R and translation. $\square$