## 2 Materials and Methods
### Optimal Synthesis Problem and Its Formulation  
A planar cable-driven parallel robot with three degrees of freedom (DOFs) is considered in this paper. Since at least one more cable than the DOFs was needed to fully constrain the robot, four cables were used [24]. Their mass and elasticity were neglected. They were modeled as straight lines. The design process consisted of finding the optimal position of each actuator and the end-effector size satisfying a set of criteria and constraints. The motors' positions were defined using the parameters \(\mathfrak{a}_{i}\) and \(\mathfrak{b}_{i}\), which represent the coordinates of the center of the pulley \(\mathcal{P}_{i}\), fixed on the \(i^{th}\) actuator. The mobile platform was considered as a square of side \(\mathfrak{c}\) (see Figure 5). The design vector is given by Equation (1). The global and the local frames \((\mathbf{X},\mathbf{Y})\) and \((\mathbf{x},\mathbf{y})\), respectively, matched those represented in Figure 2, used for the motion capture analysis. \[\mathbf{I}=[\mathfrak{a}_{1},\mathfrak{b}_{1},\ \mathfrak{a}_{2},\mathfrak{b}_{2},\ \mathfrak{a}_{3},\mathfrak{b}_{3},\ \mathfrak{a}_{4},\mathfrak{b}_{4},\ \mathfrak{c}], \tag{1}\]  
\(\mathfrak{n}_{i}\) denotes the unit vector along the \(i^{th}\) cable. It is defined as follows, where \(\psi_{i}\) and \(\theta_{i}\) are expressed as given in Equation (2). \(L_{i}\) is the length of the \(i^{th}\) cable and \(R_{p}\) is the pulleys' radius. \(R_{p}\) is considered to be fixed, since the cable radius and the coiling effects are neglected. Figure 4: (**a**) Hand trajectory (composed of 331 equidistant points); (**b**) hand orientation. “+” denotes the starting and the ending position. Figure 3: Local and global frames. \[ \begin{array}{c}\text{n}_{1}=\left[\begin{array}{c}-\sin\left(\psi_{1}+\theta_{1 }\right)\\ -\cos\left(\psi_{1}+\theta_{1}\right)\end{array}\right]\text{,n}_{2}=\left[ \begin{array}{c}\sin\left(\psi_{2}-\theta_{2}\right)\\ -\cos\left(\psi_{2}-\theta_{2}\right)\end{array}\right]\text{,n}_{3}=\left[ \begin{array}{c}\sin\left(\psi_{3}+\theta_{3}\right)\\ \cos\left(\psi_{3}+\theta_{3}\right)\end{array}\right]\text{, n}_{4}=\left[ \begin{array}{c}-\sin\left(\psi_{4}-\theta_{4}\right)\\ \cos\left(\psi_{4}-\theta_{4}\right)\end{array}\right]\text{,}\end{array}  \tag{2}\]  
\[\psi_{i}=\tan^{-1}(\frac{\left|\mathcal{P}_{ix}-\mathcal{B}_{i_{x}}\right|}{ \left|\mathcal{P}_{iy}-\mathcal{B}_{i_{y}}\right|})\text{,} \tag{3}\]  
\[\theta_{i}=\tan^{-1}(\frac{R_{p}}{L_{i}})\text{,} \tag{4}\]  
The end-effector dynamic model was obtained using Newton-Euler formulation with the assumption of neglecting the cables mass and the dynamics of the pulleys. Its expression is given by Equation (5). \[\left[ \begin{array}{cc}\sum\mathbf{F}\\ \sum\mathbf{M}\end{array} \right]=\mathbf{M}\ddot{\mathbf{x}}+\mathbf{C}\dot{ \mathbf{x}}=\mathbf{J}^{\mathbf{T}}\mathbf{T}+\mathbf{f}_{\mathbf{g}}+\mathbf{ F}_{\mathbf{ext}/\mathbf{EE}}\text{,} \tag{5}\]  
where \(\mathbf{M}=\left[ \begin{array}{cc}m_{p}\mathbb{I}_{3\times 3}&0_{3\times 3}\\ 0_{3\times 3}&\mathbf{R}_{\Phi}\ \mathbb{I}_{p}\ \mathbf{R}_{\Phi}^{\top}\end{array} \right]\) and \(\mathbf{C}=\left[ \begin{array}{cc}0_{3\times 1}\\ \boldsymbol{\omega}\times\mathbf{R}_{\Phi}\mathbb{I}_{p}\ \mathbf{R}_{\Phi}^{\top}\mathbf{\omega}\end{array} \right]\) are the mass and the Coriolis matrices, respectively. \(m_{p}\) and \(\boldsymbol{\omega}\) are the mass and the angular velocity of the end-effector, \(\mathbf{R}_{\Phi}\) is the rotation matrix, \(\mathbb{I}_{\mathbf{p}}\) denotes the inertial matrix of the mobile platform written in its center of mass, \(\mathbf{x}=\left[x\ y\ \Phi\right]^{\text{T}}\) is the pose vector of the end-effector, \(\mathbf{J}^{\mathbf{T}}\) is the transpose of the Jacobian matrix given in Equation (6), and \(\mathbf{T}\), \(\mathbf{f}_{\mathbf{g}}\), and \(\mathbf{F}_{\mathbf{ext}/\mathbf{EE}}\) are the cable tensions vector, the gravity force and the external forces applied on the end-effector, respectively. \[\mathbf{J}^{\mathbf{T}}=\left[