## 2 Kinematic and Dynamic Model  
In our work, we consider a 3-DoF robot suspended by three cables such as the one shown in Figures 1 and 2; the cables are modeled as inextensible, massless and having fixed cable exit points \(A_{i}\). The points \(A_{i}\) have position vectors \(\mathbf{a}_{i}=A_{i}-O\) with respect to the fixed reference frame (on the robot base) having its origin in \(O\). The EE is modeled as a point \(P\) having mass \(m\) and position \(\mathbf{p}=P-O=\left[x,y,z\right]^{T}\). One can write the inverse kinematic equations (which provide the cable lengths \(\rho_{i}\) as a function of \(\mathbf{p}\)) as  
\[\rho_{i}=\sqrt{\left(\mathbf{p}-\mathbf{a}_{i}\right)^{T}\left(\mathbf{p}- \mathbf{a}_{i}\right)},\quad i=1,2,3. \tag{1}\]  
Figure 1: A 3-DoF (Degrees-of-Freedom) spatial Cable-Suspended Parallel Robot (CSPR) launching an object from the launch point \(P_{L}\) to the target point \(P_{T}\). Figure 2: An example 3-DoF, 3-cable spatial CSPR (image taken with permission from [43]). We define the unit vectors \(\mathbf{e}_{i}\) directed as the \(i\)th cable and oriented towards the EE as  
\[\mathbf{e}_{i}=\frac{\mathbf{p}-\mathbf{a}_{i}}{\rho_{i}},\quad i=1,2,3. \tag{2}\]  
We obtain the dynamic model of the robot by writing the force equilibrium equation for the EE:  
\[mg-\sum_{i=1}^{3}\tau_{i}\mathbf{e}_{i}=m\ddot{\mathbf{p}} \tag{3}\]  
In Equation (3), \(\tau_{i}\) denotes the tensile force in the \(i\)th cable, \(\mathbf{g}=-\left[0,0,g\right]^{T}\) is the gravity vector, and \(g\) is the gravitational acceleration. It is convenient to refer to the cable tensions per unit mass \(\mu_{i}=\tau_{i}/m\), as the EE mass varies before and after launch. We now define the vector \(\boldsymbol{\mu}\) and the structure matrix \(\mathbf{M}\) as  
\[\boldsymbol{\mu}= \begin{bmatrix}\mu_{1}\\ \mu_{2}\\ \mu_{3}\end{bmatrix} =\frac{1}{m} \begin{bmatrix}\tau_{1}\\ \tau_{2}\\ \tau_{3}\end{bmatrix} =\frac{1}{m}\boldsymbol{\tau},\quad\mathbf{M}= \begin{bmatrix} \mathbf{e}_{1}&\mathbf{e}_{2}&\mathbf{e}_{3}\end{bmatrix}  \tag{4}\]  
After some algebraic manipulation, Equation (3) can be solved (in matrix form) for \(\boldsymbol{\mu}\) as  
\[\boldsymbol{\mu}=\mathbf{M}^{-1}\left(\mathbf{g}-\ddot{\mathbf{p}}\right) \tag{5}\]  
where matrix \(\mathbf{M}\) is invertible if and only if \(P\) does not lie on the plane through points \(A_{1}\), \(A_{2}\) and \(A_{3}\)[4]. In the following, for simplicity we assume that the EE mass varies instantaneously from \(m_{1}\) (before the launch) to \(m_{2}\) (after the launch) and that this happens by simply releasing the launched object, without introducing external forces on the EE; with the previous assumptions of point-mass EE and inextensible cables, this implies that the tensions \(\boldsymbol{\tau}\) instantaneously vary at launch. For a more comprehensive treatment of variable-mass systems, we refer the reader to [44]. The robot is fully actuated, that is, it has as many controlled DoFs as actuated cables: therefore, any launch trajectory could in principle be used in order to reach a desired launch condition (defined by a launch point and launch velocity), provided that said trajectory is feasible. We assume that \(P\) moves along an elliptical path \(\Gamma\), so that we may usefully apply many results from previous works [5; 7; 8], especially in regards to determining trajectory feasibility by means of analytical methods. The position \(\mathbf{p}\) of the EE is then defined by  
\[\mathbf{p}=\mathbf{p}_{C}+\mathbf{b} \begin{bmatrix}\sin(\omega t)\\ \cos(\omega t)\end{bmatrix} ,\quad\mathbf{p}_{C}= \begin{bmatrix}x_{C}\\ y_{C}\\ z_{C}\end{bmatrix} ,\quad\mathbf{b}= \begin{bmatrix}x_{rs}&x_{rc}\\ y_{rs}&y_{rc}\\ z_{rs}&z_{rc}\end{bmatrix}  \tag{6}\]  
where \(\mathbf{p}_{C}=C-O\) denotes the position of the center \(C\) of \(\Gamma\), while the entries in matrix \(\mathbf{b}\) are path parameters that define the shape and orientation of the trajectory.