## 4 Simulation Results
### ROPOS Vehicle  
The parameters of the selected vehicle are given in Table 1[67]. The ROPOS vehicle has rectangular shape with length \(L\), width \(W\), and height \(H\). The use of this parameters set means, that \(m_{11}=6648\) kg, \(m_{22}=11\),786 kg, \(m_{33}=7459\) kg\(\cdot\)m\({}^{2}\) and consequently \(N_{1}=6648\) kg, and \(N_{2}=11\),786 kg, \(N_{3}=7348\) kg\(\cdot\)m\({}^{2}\). For the point \(h\) value \(l=1.3\) m was chosen, where \(l\) is the distance between the pivot point in the NED frame described by \((x,y)\) (Figure 1) and the hand position point \(h\). In [67], the matrix \(M\) was a diagonal one. Here, in order to consider the matrix \(M\) with off-diagonal elements, it is assumed that \(m_{23}=mx_{g}=-1134\) kg\(\cdot\)m, which corresponds to a center of gravity shift of \(-0.5\) m.  
For this vehicle \(\left\|Y^{-1}\right\|=1.0493\), which means about 10% couplings (1 is equivalent to 0%, whereas 1.1008 to 20%). This value means that the couplings are weak. Of course, we can take the values of the parameters so that the couplings in the system are larger, but for practical reasons it seems unrealistic. **Linear trajectory**. At first, the linear trajectory (48) and using (43) was tested. The gains for the controller given by (14) and (15) were selected to ensure acceptable errors convergence:  
\[k_{p_{x}}=k_{p_{y}}=4,\ \ k_{p_{x}}=k_{p_{y}}=1,\ \ k_{I_{x}}=k_{I_{y}}=0.25. \tag{50}\]  
Moreover, due to the dynamics of the ROPOS vehicle \(s_{f}=10^{-3}\), \(s_{f}N_{1}=6.648\), and \(s_{f}N_{3}=7.348\). Results of simulation are presented in Figure 2. As it is observed from Figure 2a, the actual trajectory reaches the desired trajectory. From Figure 2b,c one can see that all error states \(\Delta\zeta_{i}\) tend to zero after about 40 s, which confirms that the algorithm is working properly. The applied force \(\tau_{u}\) and torque \(\tau_{r}\) (Figure 2d) have big values only if the vehicle starts and then their values are small. In Figure 2e, the kinetic energy corresponding to each variable is given (they are accessible only for the QV controller). The most of the energy is consuming by the \(\zeta_{2}\), which is related to the lateral movement of the vehicle, which means that the lateral velocity \(v_{r}\) plays the crucial role in dissipating this energy. The error \(\Delta\zeta_{2}\) in Figure 2f represents velocity deformation caused by couplings at the beginning of the movement is up to 0.02 m/s is almost an insignificant change. The kinetic energy average values are \(K_{m}=4911\) J (and \(K_{m1}=1503\), \(K_{m2}=3401\), \(K_{m3}=7\)). In the next case the gains for the original controller (14) and (15) were assumed the same as in [56] due to a lack of information as to their choice:  
\[k_{p_{x}}=k_{p_{y}}=50,\ \ k_{p_{x}}=k_{p_{y}}=5,\ \ k_{I_{x}}=k_{I_{y}}=0.5. \tag{51}\]  
\begin{table}
\begin{tabular}{c c c} \hline \hline  & **XX AUV Model** & \\ \hline
**Symbol** & **Value** & **Unit** \\ \hline \(L\) & approx. 1.2 & m \\ \(m\) & approx. 45 & kg \\ \(m_{11}\) & 47.5 & kg \\ \(m_{22}\) & 94.1 & kg \\ \(m_{23}\) & 5.2 & kg\(\cdot\)m \\ \(m_{32}\) & 5.2 & kg\(\cdot\)m \\ \(m_{33}\) & 13.6 & kg\(\cdot\)m\({}^{2}\) \\ \(d_{11}\) & 13.5 & kg/s \\ \(d_{22}\) & 50.2 & kg/s \\ \(dAs it follows from Figure 3a the trajectory is reached in a longer time than in the case of the QV tracking algorithm. Additionally, the position errors shown in Figure 3b,c are approaching zero in a longer time (after about 80 s). The applied force and torque have similar values if we compare Figures 3d and 2d. Therefore, it can be concluded that the values of gains have a significant impact on the performance of the control algorithm. Figure 3: Simulation results for ROPOS original controller (CL) and linear trajectory: (**a**) desired and realized trajectory; (**b**) position error states; (**c**) velocity error states; (**d**) applied force and torque.