## 3 Modeling of the Cable Tension Distribution Problem Oriented the Enhanced Stiffness of the 3 DJM  
For a given pose, the Jacobian matrix of the 3 DJM is constant. The stiffness of the 3 DJM only depends on the cable tensions according to (22). That means the stiffness of the 3 DJM can be adjusted by regulating the cable tensions. Given a desired feasible stiffness matrix \(\mathbf{K}_{\text{des}}\in\mathbb{R}^{3\times 3}\) for the 3 DJM at a given pose \(\mathbf{R}\) with a load \(\mathbf{\tau}\), the corresponding cable tensions should be solved from the following equation  
\[\mathbf{D}+\mathbf{S}\mathbf{K}_{\text{diag}}\mathbf{S}^{T}-\widehat{\mathbf{S}T}=\mathbf{K}_{\text{ des}}. \tag{23}\]  
Due to the complexity of the stiffness formulation of the 3 DJM, it is difficult to solve the cable tensions \(\mathbf{T}\in\mathbb{R}^{6}\) from (23) analytically. As the stiffness \(\mathbf{K}_{\text{des}}\) is a \(3\times 3\) real symmetric matrix in the frame \(\{\mathbf{A}\}\), there exists a real orthogonal matrix \(\mathbf{Q}\in\mathbb{R}^{3\times 3}\) to transfer the stiffness matrix from the frame \(\{\mathbf{A}\}\) to another frame \(\{\mathbf{E}\}\), in which the symmetric stiffness matrix can be represented by a diagonal matrix [34], i.e.,  
\[\mathbf{Q}^{T}\mathbf{K}_{\text{des}}\mathbf{Q}={}^{E}\mathbf{K}_{\text{des}}=\mathbf{diag}\{K_ {d1},K_{d2},K_{d3}\}, \tag{24}\]  
where the orthogonal matrix \(\mathbf{Q}\in SO(3)\) satisfies  
\[\mathbf{Q}^{T}\mathbf{Q}=\mathbf{Q}\mathbf{Q}^{T}=\mathbf{I}_{3\times 3}. \tag{25}\]  
and \({}^{E}\mathbf{K}_{\text{des}}\) represents the desired stiffness in the frame \(\{\mathbf{E}\}\). Here, \(K_{di}\)\((i=1,2,3)\) is the diagonal component of the diagonal matrix \({}^{E}\mathbf{K}_{\text{des}}\). Since the desired stiffness is a diagonal matrix in the frame \(\{\mathbf{E}\}\), we can discuss the SCTD problem in the frame \(\{\mathbf{E}\}\) to make it simple. On the other hand, as the actual stiffness matrix \(\mathbf{K}_{\text{act}}\) may not be exactly symmetric, a symmetric stiffness matrix \(\mathbf{\overline{K}}_{\text{act}}\) can be obtained as below  
\[\mathbf{\overline{K}}_{\text{act}}=\frac{1}{2}(\mathbf{K}_{\text{act}}+\mathbf{K}_{\text{ act}}^{T}). \tag{26}\]  
In the frame \(\{\mathbf{E}\}\), the symmetric actual stiffness \(\mathbf{\overline{K}}_{\text{act}}\) is represented by \({}^{E}\mathbf{\overline{K}}_{\text{act}}\), which yields  
\[{}^{E}\mathbf{\overline{K}}_{\text{act}}=\mathbf{Q}^{T}\mathbf{\overline{K}}_{\text{act} }\mathbf{Q}=\mathbf{Q}^{T}(\mathbf{D}+\mathbf{S}\mathbf{K}_{\text{diag}}\mathbf{S}^{T}-\mathbf{\widehat{ S}}\mathbf{T})\mathbf{Q}. \tag{27}\]  
Denote \(K_{ai}\)\((i=1,2,3)\) as the diagonal elements of \({}^{E}\mathbf{\overline{K}}_{\text{act}}\), the SCTD problem can be formulated as an optimization model  
\[ \begin{array}{ll}\text{Minimize}&g(\mathbf{T})=\sqrt{\sum_{i=1}^{3}(K_{ai}( \mathbf{T})-K_{di})^{2}},\\ \text{Subject to}&\mathbf{S}\mathbf{T}=\mathbf{\tau},\\ &\mathbf{T}_{min}\preceq\mathbf{T}\preceq\mathbf{T}_{max},\end{array}  \tag{28}\]  
where \(\mathbf{T}_{min}\in\mathbb{R}^{6}\) represents the minimum of the tension vector \(\mathbf{T}\) to avoid the cable be slack and \(\mathbf{T}_{max}\in\mathbb{R}^{6}\) represents the maximum of the tension vector \(\mathbf{T}\) to avoid the cable tensions exceeding the capability of the cable driving units. Here, \(\mathbf{X}\preceq\mathbf{Y}\)\((\mathbf{X},\mathbf{Y}\in\mathbb{R}^{n})\) represents that each element of \(\mathbf{X}\) is no more than the corresponding element of \(\mathbf{Y}\). According to analysis of the rank of the Jacobian matrix of the 3 DJM in the Appendix A, we have \(\mathbf{rank}(\mathbf{S})=3\). Then, the column vectors of \(\mathbf{S}\in\mathbb{R}^{3\times 6}\) can be divided into two parts, \(\mathbf{S}_{p},\mathbf{S}_{s}\in\mathbb{R}^{3\times 3}\). Here \(\mathbf{S}_{p}\) yields \(\mathbf{rank}(\mathbf{S}_{p})=3\) and it is called a basis of matrix \(\mathbf{S}\).