## 3 Simulations  
### Two-Satellite Problem  
In this study, simulations were conducted with two satellites, so the problem in (36) becomes as in (37). \[\left\{ \begin{aligned} &\min_{(\Omega_{1},\Omega_{2},\
u_{1},\
u_{ 2})\in[0,2\pi]^{4}}\big{(}\sigma_{\mathrm{P}}\big{)}\\ & s.t.\quad\Omega_{1}^{2}+\
u_{1}^{2}\leq\delta^{2}\\ & s.t.\quad\Omega_{2}^{2}+\
u_{2}^{2}\leq\delta^{2}\\ \end{aligned} \right. \tag{37}\]  
Recalling the hypothesis of target pointing and the equation in (28), with a constant value of \(\sigma\), representative of the fact that every satellite has the same receiver, the minimum of \(\sigma_{\mathrm{P}}\) occurs for the same set of variables for the minimum of \(PDOP\). As observed in [26], the problem in (37) can also be expressed in term of differences, as shown in (38). In particular, once the angular position of the first satellite is fixed, \(\Omega_{1}\) and \(\
u_{1}\), then the angular position of the second satellite, \(\Omega_{2}\) and \(\
u_{2}\), is calculated considering the difference with respect to the first satellite with \(\Delta\Omega=\Omega_{2}-\Omega_{1}\) and \(\Delta\
u=\
u_{2}-\
u_{1}\). \[\left\{\begin{aligned} &\min_{(\Delta\Omega,\Delta\
u)\in[0,2\pi]^{2}} \big{(}\sigma_{\mathrm{P}}\big{)}\\ & s.t.\quad\Delta\Omega^{2}+\Delta\
u^{2}\leq\delta_{1,2}^{2}\\ &\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \quad\The meaning of (39) is that the maximum displacement of the first satellite \(S_{1}\) plus the relative displacement of the second satellite \(S_{2}\) with respect to the first one, expressed by \(\delta_{1,2}\), must be lower than or equal to the absolute angular condition \(\delta\).  
A convenient hypothesis, which is more conservative for simulation purposes, is to consider \(\max|\Omega_{1}|=\max|\
u_{1}|=\delta_{1,2}=\delta_{\mathrm{r}}\).  
With this assumption, the condition (39) becomes as in (40).  
\[\delta\geq\delta_{\mathrm{r}}+\sqrt{\delta_{\mathrm{r}}^{2}+\delta_{\mathrm{ r}}^{2}}=\delta_{\mathrm{r}}+\sqrt{2\delta_{\mathrm{r}}^{2}}=\delta_{ \mathrm{r}}+\sqrt{2}\delta_{\mathrm{r}}=\left(1+\sqrt{2}\right)\delta_{ \mathrm{r}} \tag{40}\]  
The condition (40) can also be expressed as in (41). \[\delta_{\mathrm{r}}\leq\frac{\delta}{1+\sqrt{2}} \tag{41}\]  
From (41), the ranges for \(\Omega_{1}\) and \(\
u_{1}\) in which the problem in (38) has to be iterated become as in (42) and guarantee the LOS condition in the case of \(\Omega_{1,0}=\
u_{1,0}=0\) (i.e., the LOS condition is written with respect to the Nadir alignment between the first satellite and the emitter).