## Appendix A
### Constraint conditions of the crank-rocker mechanism  
According to the conditions for the existence of a crank-rocker mechanism in the planar four-bar linkage \(P_{2}P_{4}P_{8}P_{6}\), the bars \(l_{2}\), \(l_{3}\), \(l_{4}\), and \(l_{5}\) have the following six linear constraints:  
\[\left\{ \begin{array}{l}l_{2}-l_{i}\leq 0\ (i=3,4,5)\\ l_{2}+l_{3}-l_{4}-l_{5}\leq 0\\ l_{2}+l_{4}-l_{3}-l_{5}\leq 0\\ l_{2}+l_{5}-l_{3}-l_{4}\leq 0\\ \end{array} \right. \tag{1}\]  
\(\mathbf{X}\) represents the 10 dimensions of the optimization design variable, and the 6 \(\times\) 10 dimensions matrix \(\mathbf{A_{1}}\) represents the coefficient matrix of the constraint Equation (1). The six dimensions of vector \(\mathbf{b_{1}}\) represent the constraint vector of Equation (1), and so Equation (1) can be changed into:  
\[\mathbf{A_{1}}\mathbf{X}\leq\mathbf{b_{1}} \tag{2}\]  
Linear constraints of geometric dimensions  
According to the requirements of the overall size of the robot, design variables should be restrained, that is:  
\[\left\{ \begin{array}{l}l_{\text{min}}\leq l_{i}\leq l_{\text{max}}\ (i=1,2,3,\ldots,8)\\ \beta_{\text{j/min}}\leq\beta_{\text{j}}\leq\beta_{\text{j/max}}\ (j=1,2)\\ \end{array} \right. \tag{3}\]  
The 10 dimensions of vector \(\mathbf{d_{\text{min}}}\) represent the lower-bound matrix of the design variable; the 10 dimensions of vector \(\mathbf{d_{\text{max}}}\) represent the upper-bound matrix of the design variable, then Equation (3) can be changed into:  
\[\left\{ \begin{array}{l}\mathbf{X}\geq\mathbf{d_{min}}\\ \mathbf{X}\leq\mathbf{d_{max}}\\ \end{array} \right. \tag{4}\]  
Considering the limitations of processing technology of the component, and the requirement for non-interference of the installation position, the linear constraints of the geometric dimensions are given as:  
\[\left\{ \begin{array}{l}l_{2}-l_{1}\leq d_{1}\\ l_{2}-l_{3}\leq d_{2}\\ \end{array} \right. \tag{5}\]The \(2\times 10\) dimensional matrix \(A_{2}\) represents the coefficient matrix of the constraint Equation (A5). The two dimensional vector \(b_{2}\) represents the constraint vector of Equation (A5), so that Equation (A5) can be changed into:  
\[A_{2}X\leq b_{2}\] (A6)  
3. Non-linear constraints of geometric dimensions  
Based on the actual processing requirements of the frame, the limitations of the original length, and the tensile deformation of the tension spring, the following nonlinear constraints exist:  
\[\Phi(X)\leq G\] (A7)  
where \(\Phi(X)\) is a 3-dimensional function vector and \(G\) is a 3-dimensional constant vector. \[\Phi(X)=\left( \begin{array}{c}\varphi_{1}(X)\\ \varphi_{2}(X)\\ \varphi_{3}(X)\end{array} \right)=\left( \begin{array}{c}P_{6x^{\prime}}\\ P_{9x}-P_{10x}\\ P_{6y^{\prime}}-P_{9y}\end{array} \right)\qquad G=\left( \begin{array}{c}g_{1 }\\ g_{2}\\ g_{3}\end{array} \right)=\left( \begin{array}{c}d_{3}\\ d_{4}\\ d_{5}\end{array} \right)\]  
4. Constraints on the kinematical chain  
The main chain and the secondary chain are coupled at the joint \(P_{8}\), which means:  
\[\left\{ \begin{array}{c}h_{1}(X)=l_{2}C\theta_{1}+l_{5}C(\theta_{1+2})-l_{3} C\theta_{1^{\prime}}-l_{4}C(\theta_{1^{\prime}-2^{\prime}})=0\\ h_{2}(X)=l_{2}S\theta_{1}+l_{5}S(\theta_{1+2})+l_{3}S\theta_{1^{\prime}}+l_{4} S(\theta_{1^{\prime}-2^{\prime}})=0\end{array} \right.\] (A8)  
Point \(P_{6}\), \(P_{8}\), and \(P_{10}\) form a link with three elements, i.e., the slope of the link \(P_{6}P_{8}\) and \(P_{8}P_{10}\) are equal:  
\[h_{3}(X)=\frac{S(\theta_{1+2+3})}{C(\theta_{1+2+3})}+\frac{S(\theta_{1^{\prime }-2^{\prime}})}{C(\theta_{1^{\prime}-2^{\prime}})}=0\] (A9)  
Since the relative distance between point \(P_{11}\) and \(P_{12}\) along the \(X_{0}\)-axis is constant, i.e., there is a constraint relationship \(P_{12x}-P_{11x}=l_{8}\), this means:  
\[h_{4}(X)=2[l_{2}C_{1}+l_{5}C_{2}+l_{6}C_{3}+l_{7}C_{4}]+\frac{1}{2}l_{1}-l_{8}\] (A10)  
where \(C_{1}=C(\theta_{1})\), \(C_{2}=C(\theta_{1+2})\), \(C_{3}=C(\theta_{1+2+3})\), \(C_{4}=C(\theta_{1+2+3+4})\).