## 4 Results
### Transverse Motion  
Tourist submarines are equipped with transverse thrusters to enable transverse motion and maneuvering. Additionally, they are used to prevent the transverse motion by acting in the opposite direction to the sea current. Therefore, there is an interest in studying the transverse resistance of the submarine as well as the transverse motion of the submarine, and to suggest equations to estimate the resistance coefficient in the transverse direction. To accomplish this, 90 different cases are modeled in OpenFOAM using the developed code for 10 different transverse speeds (0.05, 0.10, 0.15, 0.20, 0.25, 0.30, 0.35, 0.4, 0.45 and 0.5 m/s). Therefore, 90 cases listed in Appendix B were analyzed and are shown in Figure 12. OpenFOAM computation for each of the 90 cases lasted for about 18,000 s, so the total running time was about 27,000 min. Figure 12 shows that the resistance coefficient in the transverse direction (\(C_{D\_t}\)) is linearly related to the \(L\)/\(D_{S}\). Nevertheless, for all \(L\)/\(D_{S}\), the relationship between \(C_{D\_t}\) and transverse speed (\(V_{t}\)) is nonlinear. \(NLLS\) and \(ANN\)s were applied to fit the best curve to the obtained results shown in Figure 13 and to find a nonlinear equation to estimate the resistance coefficient of the bare hull based on two dimensionless parameters, \(L\)/\(D_{S}\) and Reynolds number (\(R_{el}\)). The following equations were derived for the resistance coefficient of the submarine with transverse speed using the code in MATLAB. In \(ANN\), 70%, 15%, and 15% of data (corresponding to 64, 13, and 13 data) were also used in training, testing, and evaluation steps, respectively:  
Figure 12: Variation of the resistance coefficient versus the dimensionless parameters, \(L\)/\(D_{S}\) and log (\(R_{el}\)) for the transverse directions. Figure 13: The \(SSE\) of \(ANN\) versus the \(NON\) for transverse motion. NLS:  
\[C_{R-t}=232.71-5.211889\frac{L}{D_{5}}-105.35826(log(R_{et}))+0.083 091\left(\frac{L}{D_{5}}\right)^{2}+1.46633\frac{L}{D_{5}}(log(R_{et}))\] \[+16.000585(log(R_{et}))^{2}-0.0133027\left(\frac{L}{D_{5}}\right)^ {2}log(R_{et})-0.101747\frac{L}{D_{5}}(log(R_{et}))^{2}\] \[-0.815011(log(R_{et}))^{3} \tag{20}\]  
ANN (Figure 3):  
\[C_{R-t}=1.772+1.513 tanh(-0.093\frac{L}{D_{5}}-1.92 log(R_{et})+10.361) \tag{21}\]  
ANN (Figure 4):  
\[C_{R-t}=1.516+1.22 tanh\left(-0.1093\frac{L}{D_{5}}-2.278 log(R_{et})+12.466 \right)+0.0374 tanh(-0.534\frac{L}{D_{5}}\] \[-3.695 log(R_{et})+26.177) \tag{22}\]  
The data in Appendix B has also been used to train _ANN_s with more than two neurons in the hidden layer (see Figure 8). The resistance coefficient was estimated at 0.02 m/s as a test of the ability of the proposed models to estimate it at different speeds. This speed is lower and outside the range used for the extraction of relations. In addition, the resistance coefficient of the submarine at transverse speeds of 0.125 and 0.375 was estimated, which was included in the range of speeds used in the extraction of relationships, but not in the training. In a similar manner, the resistance coefficient was estimated for a speed of 0.6, which was higher than the range of speeds used to extract the relationships. Training _ANN_s with different _NON_s were used to predict the coefficient of resistance of the submarine at speeds of 0.02 m/s, 0.125, 0.375, and 0.6, and _SSE_ is shown in Figure 13.  
It is shown in Figure 13 that _ANN_ with seven neurons in the hidden layer (_NON_ = 7) performs the best at predicting resistance coefficients.