## 2 Cable-Driven Parallel Manipulators Model
### Quasi-Static and Inverse Kinematic Problem of Cable-Driven Parallel Manipulators  
The kinematic problem of CDPM that just describes the geometric relationship between a redefined pose of end-effector and the cables' length. However, each cable has its length depend on the tension which acts on it, i.e., the kinematic model alone no longer makes sense. Many works apply the static or dynamic problem to it and solve them simultaneously [14; 15; 16; 3; 19]. Meanwhile, the quasi-static  
Figure 1: Cable sag model in frame \(\{A\}\). problem of CDPM concerns finding the proper forces of the moving platform and forces of its constraints given a certain motion of it where the inertial effects are negligible. Once again, with the relationship of tension of each cable and its length, the inverse kinematic problem is solved. Let \(\{W\}\) be the reference frame attached to the fixed of CDPM and \(\{R\}=(\mathbf{P}(t),\mathbf{\theta}(t))\), illustrated in Figure 2, be the moving platform frame attached on its body, where \(\mathbf{P}=\left[ \begin{array}{cc}x_{\mathbf{P}}&y_{\mathbf{P}}&z_{\mathbf{P }}\end{array} \right]^{T}\), \(\mathbf{\theta}=\left[ \begin{array}{cc}\alpha_{\mathbf{P}}&\beta_{\mathbf{P }}&\gamma_{\mathbf{P}}\end{array} \right]^{T}\) are the coordinate of center of mass and the vector that represents Euler angle of the moving platform with respect to \(\{W\}\). From now on, any vector without a superscript related to any frame is implied to respect to \(\{W\}\). We also note that \(\mathbf{x}=\left[ \begin{array}{cc}\mathbf{P}&\mathbf{\theta}\end{array}  \right]^{T}\) represents the generalized coordinate of the moving platform. If the moving platform is isolated from the system, according to the Newton-Euler's equation, the resultant force \(\mathbf{F}\) and torque \(\mathbf{\tau}\) exert on its center of mass can be obtained via Equation (5)  
\[\left[ \begin{array}{cc}\mathbf{M}&\mathbf{0}\\ \mathbf{0}&\mathbf{\Gamma}\end{array} \right]\ddot{\mathbf{x}}+\left[ \begin{array} []{cc}\mathbf{M}\cdot\mathbf{W}&\mathbf{0}\\ \mathbf{0}&\mathbf{W}\cdot\mathbf{\Gamma}\end{array} \right]\dot{\mathbf{x}}+ \left[ \begin{array}{cc}\mathbf{M}\cdot\mathbf{g}\\ \mathbf{0}\end{array} \right]=\left[ \begin{array}{c}\mathbf{F}\\ \mathbf{\tau}\end{array} \right] \tag{5}\]  
where \(\mathbf{M}\) is simplified into the scalar matrix of mass due to the homogeneous density and symmetric geometry of the end-effector, \(\mathbf{\Gamma}\) is the inertia tensor respect to center of mass of the end-effector expressed in \(\{W\}\), \(\mathbf{g}\) is the gravitational acceleration vector and \(\mathbf{W}\) is the skew symmetric matrix of \(\dot{\mathbf{\theta}}\). On the other hand, the inertial forces of cables are trivial since the weights of these cables are negligible compared to the end-effector. Thus, if tensions of cables are treated as the external forces of the end-effector, then the resultant force \(\mathbf{F}\) and torque \(\mathbf{\tau}\) also can be expressed as in (6) and (7)  
\[F_{x}=\sum  
Naturally, in the ideal statistic condition, or mobile platform moving with speed and acceleration which are not too high, the ith cable profile that is fixed at its extremities \(\mathbf{A}_{i}\) and \(\mathbf{B}_{i}\) completely lies in the plane that contains these points. Similar to the sagging cable model section, let \(\{A_{i}\}\) be the 2D-frame that defines the considered plane and illustrated in Figure 2. We can define \(\mathbf{a}_{i}\) and \(\mathbf{T}_{i}\) respect to \(\{A_{i}\}\) as (10) and (11), respectively  
\[\mathbf{a}_{i}^{A_{i}}=\mathbf{R}_{A_{i}}^{W}\cdot\mathbf{a}_{i} \tag{10}\]  
\[\mathbf{T}_{i}^{A_{i}}=\mathbf{R}_{A_{i}}^{W}\cdot\mathbf{T}_{i} \tag{11}\]  
where  
\[\mathbf{R}_{A_{i}}^{W}=\left[ \begin{array}{ccc}C_{\theta_{i}}&S_{\theta_{i} }&0\\ 0&0&-1\end{array} \right] \tag{12}\]  
In the matrix (12), \(\theta_{i}\) can be given by (13). Vector \(\mathbf{a}_{i}\) in Equation (10) can be obtained by deploying Equation (14).