## 3 Launch Trajectory Planning  
A mass is launched at the launch instant \(t=t_{L}\) from the EE and we require it to pass through a given target point \(P_{T}\), which has position \(\mathbf{p}_{T}=\left[x_{T},y_{T},z_{T}\right]^{T}\) (see Figure 1). The position \(\mathbf{p}_{L}\) at launch point \(P_{L}\) of the EE and its launch velocity \(\dot{\mathbf{p}}_{L}\) can be obtained from Equation (6) as:\[\mathbf{p}_{L}= \begin{bmatrix}x_{L}\\ y_{L}\\ z_{L}\end{bmatrix} =\mathbf{p}_{C}+\mathbf{b} \begin{bmatrix}\sin\alpha_{L}\\ \cos\alpha_{L}\end{bmatrix} ,\quad\dot{\mathbf{p}}_{L}= \begin{bmatrix}\dot{x}_{L }\\ \dot{y}_{L}\\ \dot{z}_{L}\end{bmatrix} =\omega\mathbf{b} \begin{bmatrix}\cos\alpha_{L}\\ -\sin\alpha_{L}\end{bmatrix}  \tag{7}\]  
where \(\alpha_{L}=\omega t_{L}\) is the launch angle (using \(\alpha\) as the path coordinate along \(\Gamma\)); note that \(\mathbf{p}_{L}\) and \(\dot{\mathbf{p}}_{L}\) define the initial conditions for the motion of the mass after launch. The trajectory planning of the launched mass requires some simplifying assumptions. Most authors [13; 14; 15; 16; 17; 19; 21; 25; 26; 27; 21; 25; 27; 23; 24; 25; 27; 26; 28; 29; 30; 31; 32; 33; 34; 36; 37; 38; 39] consider a simplified model where, during the free-flight phase, the only force acting is gravity; under these assumptions, it can be readily shown that the center of mass of the launched object will move along a parabolic arc in a vertical plane. Considering the effect of air drag leads to a more realistic model; however, the dynamic problem including the air-drag model does not have a closed-form solution. Therefore, we will disregard this effect, which is acceptable if the object being launched is sufficiently small and heavy. With these assumptions, gravity is the only external force acting during the free-flight motion and the equation of ballistic flight can be written as:  
\[\mathbf{p}(t)=\frac{1}{2}\mathbf{g}\left(t-t_{L}\right)^{2}+\dot{\mathbf{p}}_{ L}\left(t-t_{L}\right)+\mathbf{p}_{L} \tag{8}\]  
If we denote by \(\Delta t_{LT}\) the flight time in which the mass goes from \(P_{L}\) to \(P_{T}\), then \(\mathbf{p}_{T}=\mathbf{p}\left(t_{L}+\Delta t_{LT}\right)\), and thus:  
\[x_{T} =\dot{x}_{L}\Delta t_{LT}+x_{L} \tag{9}\] \[y_{T} =\dot{y}_{L}\Delta t_{LT}+y_{L}\] (10) \[z_{T} =\dot{z}_{L}\Delta t_{LT}-\frac{1}{2}g\Delta t_{LT}^{2}+z_{L} \tag{11}\]  
Equations (9) and (10) yield  
\[\frac{x_{T}-x_{L}}{y_{T}-y_{L}}=\frac{\dot{x}_{L}}{\dot{y}_{L}} \tag{12}\]  
By substituting Equation (7) into Equation (12) and simplifying, we obtain  
\[ \begin{split}&\left[\left(x_{T}-x_{C}\right)y_{rs}-\left(y_{T}-y_{ C}\right)x_{rs}\right]\cos\alpha_{L}\\ &-\left[\left(x_{T}-x_{C}\right)y_{rc}-\left(y_{T}-y_{C}\right)x_{ rc}\right]\sin\alpha_{L}+x_{rs}y_{rc}-x_{rc}y_{rs}=0\end{split}  \tag{13}\]  
which, using the tangent half-angle substitution, can be rewritten as a 2 nd-degree equation in \(t_{n}=\tan\left(\frac{\alpha_{L}}{2}\right)\), namely:  
\[Gt_{n}^{2}+Ht_{n}+I=0 \tag{14}\]  
where \(G\), \(H\) and \(I\) are known coefficients that depend on the geometry of ellipse \(\Gamma\) and the coordinates of the target point \(P_{T}\), namely:  
\[ \begin{split} G&=x_{rs}y_{rc}-x_{rc}y_{rs}-x_{T}y_{ rs}+x_{C}y_{rs}+x_{rs}y_{T}-x_{rs}y_{C}\\ H&=2\left(-x_{T}y_{rc}+x_{C}y_{rc}+x_{rc}y_{T}-x_{ rc}y_{C}\right)\\ I&=x_{rs}y_{rc}-x_{rc}y_{rs}+x_{T}y_{rs}-x_{C}y_{ rs}-x_{rs}y_{T}+x_{rs}y_{C}\end{split}  \tag{15}\]  
Solving Equation (14) yields two values for \(t_{n}\):  
\[t_{n,1}=\frac{-H+\sqrt{H^{2}-4 GI}}{2G},\quad t_{n,2}=\frac{-H-\sqrt{H^{2}-4 GI}}{ 2G} \tag{16}\]and thus two values for \(\alpha_{L}\):  
\[\alpha_{L1}=2\arctan\left(t_{n,1}\right),\quad\alpha_{L2}=2\arctan\left(t_{n,2}\right) \tag{17}\]  
Two possible launch positions \(\mathbf{p}_{L1}\) and \(\mathbf{p}_{L2}\) are obtained substituting the two values of \(\alpha_{Li}\) in Equation (7); it is worth pointing out that neither \(\alpha_{L}\) nor \(\mathbf{p}_{L}\) depend on \(\omega\).