# Analytic Derivation of Ascent Trajectories and Performance of Launch Vehicles
## 4 Analytic Derivation of Launcher Trajectories  
The above analytic formulas can be generalized to multistage launchers. Let us start with two stages with the following average parameters \((n_{1},Isp_{1},u_{1})\) and \((n_{2},Isp_{2},u_{2})\). The first stage has variables \(z_{1}\), \(V_{1}=A_{1}\,z_{1}^{n_{1}-1}(1+z_{1}^{2})\) in the time interval \([t_{10}=0,t_{1f}=t_{b_{1}}]\). The second stage has variables \(z_{2}\), \(V_{2}=A_{2}\,z_{2}^{n_{2}-1}(1+z_{2}^{2})\) in the time interval \([t_{20}=t_{b_{1}},t_{2f}=t_{b_{2}}+t_{b_{1}}]\) and the final value \(z_{2_{2}}\) is fixed (see Figure 5). Of course the two solutions\(z_{1}\), \(V_{1}\) and \(z_{2}\), \(V_{2}\) must match at the boundary point  
\[z_{1f}=z_{20}\] \[V_{1f}=V_{1}(z_{1f})=V_{2}(z_{20})=V_{20}\]  
The two matching conditions imply  
\[\frac{A_{1}}{A_{2}}=z_{20}^{n_{2}-n_{1}} \tag{18}\]  
Equation (11) gives at first-stage burn out  
\[t_{b1}=\frac{A_{1}}{g}\left(\frac{z_{20}^{\tilde{n}_{1}-1}}{\tilde{n}_{1}-1}+ \frac{z_{20}^{\tilde{n}_{1}+1}}{\tilde{n}_{1}+1}\right)=^{def}f_{1}(z_{20}) \tag{19}\]  
At the burn out of the second stage, one has \(z_{2f}\) and  
\[t_{b2}=\frac{A_{2}}{g}\left(\frac{z_{2f}^{n_{2}-1}}{\tilde{n}_{2}-1}+\frac{z_{ 2f}^{n_{2}+1}}{\tilde{n}_{1}+1}-\frac{z_{20}^{n_{2}-1}}{\tilde{n}_{2}-1}+ \frac{z_{20}^{n_{2}+1}}{\tilde{n}_{1}+1}\right)=^{def}f_{2}(z_{20}) \tag{20}\]  
We have a system of three Equations (18)-(20) in the three unknowns \(A_{1}\), \(A_{2}\) and \(z_{20}\). To solve it, let us consider the ratio  
\[\frac{t_{b2}}{t_{b1}}=\frac{f_{1}}{f_{2}}\]  
Because of (18), this ratio is a function of \(z_{20}\) only  
\[f(z_{20})=\frac{t_{b2}}{t_{b1}}-\frac{f_{1}}{f_{2}}=0\]  
The solution \(z_{20}\) of \(f(z_{20})=0\) (in the range [0,1]) provides the value of the flight-path angle at the end of the first stage. Moreover, by Equation (20), one finds \(A_{2}\) and by Equation (19) one finds \(A_{1}\). Hence, the velocity profile of the two-stage launcher with a prescribed flight-path angle \(z_{2f}\) at second-stage burn out is equal to  
\[ \begin{array}{l}V(z)=A_{1}z_{1}^{\tilde{n}_{1}-1}(1+z_{1}^{2})\ for\ z\in[0,z_{20}]\\ V(z)=A_{2}z_{2}^{\tilde{n}_{2}-1}(1+z_{2}^{2})\ for\ z\in[z_{20},z_{2f}]\end{array}  \tag{21}\]  
Figure 5: Two stages. With same arguments, we obtain the altitude and range  
\[ \begin{split} h(z)&=\frac{A_{1}^{2}}{g}\left[\frac{z_{1 }^{2n_{1}-2}}{2\bar{n}_{1}-2}-\frac{z_{1}^{2n_{1}+2}}{\bar{n}_{1}+2}\right]\,\ for\ z\in[0,z_{20}]\\ h(z)&=h(z_{20})+\frac{A_{2}^{2}}{g}\left[\frac{z_{2 }^{2n_{2}-2}}{2\bar{n}_{2}-2}-\frac{z_{2}^{2n_{2}+2}}{\bar{n}_{2}+2}\right]_{z_ {20}}^{z}\,\ for\ z\in[z_{20},z_{2f}]\end{split}  \tag{22}\]  
\[ \begin{split} s(z)&=2\frac{A_{1}^{2}}{g}\left[\frac{z ^{2n_{1}}}{2\bar{n}_{1}}+\frac{z^{2n_{1}+2}}{2\bar{n}_{1}+2}\right]\,\ for\ z\in[0,z_{20}]\\ s(z)&=R(z_{20})+2\frac{A_{2}^{2}}{g}\left[\frac{z ^{2n_{2}}}{2\bar{n}_{2}}+\frac{z^{2n_{2}+2}}{2\bar{n}_{2}+2}\right]_{z_{20}}^{ z}\,\ for\ z\in[z_{20},z_{2f}]\end{split}  \tag{23}\]  
Note that the gravity loss of the two-stage launcher is  
\[\Delta V_{g}=gIsp_{1}log(\frac{1}{U_{1}})+gIsp_{2}log(\frac{1}{U_{2}})-V(z_{2f})\]  
Any number of stage N can be handled: of course, more numerical work is needed to find out the unknown variables. The N-1-matching conditions on the flight-path angle generate the following N-1-matching conditions on the velocity  
\[ \begin{split} V_{1}(z_{20},A_{1})&=& V_{2}(z_{20},A_{2})\\ V_{2}(z_{30},A_{2})&=& V_{3}(z_{30},A_{3 })\\ &\vdots\\ V_{N-2}(z_{N-2\ 0},A_{N-2})&=& V_{N-1}(z_{N-1\ 0},A_{N-1})\end{split} \]  
These, together with the \(N\) relationships related to each burn time  
\[ \begin{split} t_{b_{1}}&=& t_{b_{1}}(A_{1 },z_{20})\\ t_{b_{2}}&=& t_{b_{2}}(A_{2},z_{20},z_{ 30})\\ &\vdots\\ t_{b_{N}}&=& t_{b_{N}}(A_{N},z_{20},\ldots,z_{ N\ 0})\end{split} \]  
generate \(2N-1\) equations to be solved with respect to the \(2N-1\) unknowns \(z_{20},z_{30},\ldots z_{N\ 0}\), \(A_{1},\ldots A_{N}\).