## 2 Cable-Driven Parallel-Series Mechanism
### Inverse Kinematics and Static Solution  
The lengths of four cables \(I=[l_{1},l_{2},l_{3},l_{4}]\) are calculated in the inverse kinematics model based on the angles and the length of \(d_{2}\):  
\[I=f(x),f:\mathbf{R}^{4}\to\mathbf{R}^{4}, \tag{11}\]  
where \(x=[\theta_{m},\,\theta_{b},\,\theta_{r},d_{2}]^{\mathrm{T}}\in\mathbf{R}^{4}\). The length of the cable is calculated by the following distance Formula (12) according to the transformation matrix:  
\[I=\|^{O}\mathbf{p}-{}^{O}\mathbf{T}_{o}^{o}\mathbf{q}\|. \tag{12}\]  
Transformation matrix \({}^{o}\mathbf{T}_{O}\) from the global coordinate system to the local coordinate system is shown in Label (13):  
\[{}^{o}\mathbf{T}_{O}=\left[ \begin{array}{cccc}s_{11}&s_{21}&s_{31}&-a\\ s_{12}&s_{22}&s_{32}&-b\\ s_{13}&s_{23}&s_{33}&-c\\ 0&0&0&1\end{array} \right]. \tag{13}\]  
The position of the cable holes on the lower platform in the local coordinate system is as follows:  
\[{}^{o}\mathbf{p}_{1}= {}^{o}\mathbf{T}_{O}\cdot{}^{O}\mathbf{p}_{1} \tag{14}\] \[= [-p\cdot s_{21}-d_{1}\cdot s_{31}-a,-p\cdot s_{22}-d_{1}\cdot s_{3 2}-b,\] \[-p\cdot s_{23}-d_{1}\cdot s_{33}-c,1]^{T}\]  
\[{}^{o}\mathbf{p}_{2}={}^{o}\mathbf{T}_{O}\cdot{}^{O}\mathbf{p}_{2}=(p\cdot s_{21}-d_{1} \cdot s_{31}-a,p\cdot s_{22}-d_{1}\cdot s_{32}-b, \tag{15}\] \[\qquad\qquad\qquad\qquad\qquad p\cdot s_{23}-d_{1}\cdot s_{33}-c,1]^{T}\]  
\[{}^{o}\mathbf{q}_{1}=(-q,0,0,1)^{T},{}^{o}\mathbf{q}_{2}=(q,0,0,1)^{T}. \tag{16}\]  
Due to the angular limitation of the mechanism bending, the spring provides a support force \(F_{k}\) along the direction of the upper shaft and perpendicular to the moving platform. The force of the spring is calculated by [PERSON]'s law,  
\[F_{k}=K(l-l_{0}), \tag{17}\]where \(K\) is the elastic coefficient of the spring, \(l\) is the length of the spring, and \(l_{0}\) is the initial length of the spring. For the upper platform, the combined force of the four driving cables is decomposed into component forces \(F_{1}\) and \(F_{2}\) in the direction of the upper shaft and the plane of the upper platform, respectively (Figure 3). Figure 3 shows that \(F_{1}\) cancels the force of the spring \(F_{k}\). \(F_{2}\) is the force subjected from the upper platform to the upper shaft. Therefore, the torque balance equation of the upper shaft is achieved between \(F_{2}\) and the gravity of the upper platform:  
\[Mg\sin\theta_{b}\cdot d_{2}=F_{2}d_{2}, \tag{18}\]  
where \(M\) is the mass of the upper platform. The torque and force balance equations of the upper platform are shown in Labels (20) and (21):  
\[\sum\
olimits_{i=1}^{4}T_{i}{}^{o}\mathbf{u}=[-F_{2}\cos(\theta_{r}-\theta_{m}),F _{2}\sin(\theta_{r}-\theta_{m}),F_{k}]^{T}, \tag{19}\]  
\[\sum\
olimits_{i=1}^{4}{}^{o}\mathbf{r}\times T^{o}\mathbf{u}=\Big{[}m_{x},m_{y},m_{z }\Big{]}^{T}, \tag{20}\]  
where \(m_{x}\), \(m_{y}\), and \(m_{z}\) are the component of the combined torque along the \(x\)-axis, \(y\)-axis, and \(z\)-axis, respectively. \(T_{i}\) represents the force on each cable:  
\[= [-p\cdot s_{21}-d_{1}\cdot s_{31}-a,-p\cdot s_{22}-d_{1}\cdot s_{3 2}-b,\]0  
\[= [-p\cdot s_{21}-d_{1}\cdot s_{31}-a,-p\cdot s_{22}-d_{1}\cdot s_{3 2}-b,\]1  
\[= [-p\cdot s_{21}-d_{1}\cdot s_{31}-a,-p\cdot s_{22}-d_{1}\cdot s_{3 2}-b,\]2  
According to the characteristics of the spring-shaft structure, the torque balance equation must be achieved along the \(z\)-axis because the combined torque on the \(oxy\) plane is zero when the mechanism is in the static state. Therefore, three force balance equations and one torque balance equation along the \(z\)-axis are obtained because the torques \(m_{x}\) and \(m_{y}\) are canceled by the interaction between the shaft and the upper platform:  
\[= [-p\cdot s_{21}-d_{1}\cdot s_{31}-a,-p\cdot s_{22}-d_{1}\cdot s_{3 2}-b,\]3 Force analysis of the mechanism. \[\left[ \begin{array}{cccc}M_{1}&M_{2}&M_{3}&M_{4}\end{array} \right]\cdot\left[ \begin{array}{c}T_{1}\\ T_{2}\\ T_{3}\\ T_{4}\end{array} \right]=\left[