# Unmanned Vehicles' Placement Optimisation for Internet of Things and Internet of Unmanned Vehicles
## 5 Towards Algorithm Development: Propagation Concepts, Path Loss and Communication Graph
### Overwater Channels (OW) and Ship-to-Shore Channels  
Overwater channels or ship-to-ship and ship-to-shore channels are more affected by physical phenomena than the air-to-ship channel, since both transmitter and receiver antennae are close to the water surface. If we would apply the same two-ray model (Equation (7)) and the same technical solution (transceiver in [57] for small USVs, the coverage distance would not be more than 120 m when both heights of the antennae only reach 0.5 m. The coverage reaches 373 m when it is possible to place the antenna in a higher position (4.5 m) for one of the USVs (Figure 6). The most problematic channel is related to the water environment. Consequently, we want to prove that the path loss characterising UAV-USV, USV-USV or USV-GW/WSN communications is minimum in terms of antenna height but zero in terms of distance. To this end, we handle Equation (7), and we suppose \(h_{t}\) is the height of the USV antenna, whereas \(h_{t}\) is the height of the UAV antenna. Because the USV cannot move vertically (on the z-axis), we state that \(h_{t}\) is constant, whereas the UAVs can increase or decrease their altitude and consequently, \(h_{t}\) is variable. In addition, the distance between UV increases or decreases as well. To conclude, both \(h_{t}\) and \(d\) are variables with respect to \(PL_{2-ray}\). We cannot find a minimum \(PL\) in terms of the distance because Equation (11) does not have solutions for \(d>0\). \[\frac{\partial PL_{2-ray}}{\partial d}=0 \tag{11}\] \begin{table}
\begin{tabular}{c c c} \hline \(h_{t}\)(m) & \(h_{r}\)(m) & Recommendations \\ \hline \multirow{5}{*}{5} & 5 & d \(<\) 1219 m \\  & 25 & d \(\
eq\) 235 m; d \(<\) 2777 m \\  & 45 & d \(\
eq\) 217 m; \(d\) \(\
eq\) 423 m; d \(<\) 3714 m \\ \cline{1-1}  & 65 & d \(\
eq\) 154 m; \(d\) \(\
eq\) 310 m; \(d\) \(\
eq\) 621 m; d \(<\) 4472 m \\ \cline{1-1}  & 85 & d \(\
eq\) 115 m; \(d\) \(\
eq\) 205 m; \(d\) \(\
eq\) 405 m; \(d\) \(\
otin\) [800,825] m; d \(<\) 4817 m \\ \cline{1-1}  & 105 & d \(\
eq\) 167 m; \(d\) \(\
eq\) 250 m; \(d\) \(\
eq\) 329–341 m; \(d\) \(\
eq\) 497 m; \(d\) \(\
otin\) [967,1053] m; d \(<\) 5520 m \\ \hline \end{tabular}
\end{table} 
Table 6: Examples of recommendations for the scenario when \(h_{t}=5\) m, \(LB=106\) dB. Figure 5: Path loss, link budget and critical distance for UAV–USV link, when USV antenna height is \(h_{t}=5\) m and the solution in [57] is used. **Proof. **  
\[\frac{\partial PL_{2-ray}}{\partial d}=\frac{2}{d\ln 10}\bigg{(}1+\frac{\beta}{d} \cot\frac{2\beta}{d}\bigg{)}=0 \tag{12}\]  
where \(\beta=\frac{h_{t}h_{t}f}{cd}>0\). By denoting \(\frac{\beta}{d}=\alpha>0\), we will solve Equation  
\[1+\alpha\cot\alpha=0 \tag{13}\]  
with the unique negative solution \(\alpha=-227.769\) rad which contradicts the afore-mentioned condition for \(\alpha\). Instead, we can find a minimum of \(PL\) with respect to \(h_{t}\) by solving Equation (14). \[\frac{\partial PL_{2-ray}}{\partial h_{t}}=0 \tag{14}\]  
\[\frac{\partial PL_{2-ray}}{\partial h_{t}}=-10\cdot\frac{\partial\bigg{[}log_ {10}\bigg{[}2\sin\bigg{(}\frac{2h_{t}h_{t}f}{cd}\bigg{)}\bigg{]}^{2}\bigg{]}}{ \partial h_{t}} \tag{15}\]  
\[\frac{\partial PL_{2-ray}}{\partial h_{t}}=-\frac{40}{\ln 10}\cdot\frac{h_{t}f} {cd}\cot\bigg{(}\frac{h_{t}h_{t}f}{cd}\bigg{)}=0 \tag{16}\]  
for  
\[h_{t}=\frac{(2k+1)\pi cd}{4h_{t}f} \tag{17}\]  
We replace Equation (17) in Equation (7) and we obtain Equation (18) and (19). \[PL_{2-ray}(d)=20\cdot log_{10}\bigg{(}\frac{4\pi fd}{c}\bigg{)}-10\cdot log_{1 0}\bigg{[}2\sin\bigg{(}\frac{(2k+1)\pi}{2}\bigg{)}\bigg{]}^{2} \tag{18}\]  
\[PL_{2-ray}(d)=20\cdot log_{10}\bigg{(}\frac{4\pi fd}{c}\bigg{)}-10\cdot log_{1 0}(4) \tag{19}\]  
Further, we must compute the distance, \(d\), for which \(PL_{2-ray}(d)\) is lower or equal than the link budget, \(LB\), expressed in dB. Figure 5:0  
By solving Equation (20), we found that the distance, \(d\), must follow Equation (21):  
Figure 5:1  
We introduce the following hypothesis: all USVs have the same antennae height.