# Development of an Autonomous Underwater Helicopter with High Maneuverability
## 3 Dynamic Analysis of the AUH
### Analysis of Motion in the Horizontal Plane  
The main three special functions of AUH's working mode are short-range flexible motion, fixed-point hovering, and landing and take-off. This means that the AUH has a need for 6 DOF control. The force analysis of horizontal motion is shown in Figure 3. The equation of linear motion is:  
\[F_{1}+F_{2}-f_{p}=m\frac{du}{dt} \tag{4}\]  
Figure 3: Force analysis of horizontal motion. (**a**) Linear motion; (**b**) rotation motion. Figure 2: Fixed coordinate system and moving coordinate system. The equation of rotation motion is:  
\[F_{1}R+F_{2}R-M_{fr}=J_{r}\frac{d\omega}{dt} \tag{5}\]  
where \(F_{1}\) and \(F_{2}\) are the force of the horizontal thrusters. \(f_{p}\) is the resistance of flow. \(M_{fp}\) is the resistance moment of the \(z\)-axis. \(u\) is the velocity of the \(p\)-axis. \(\omega\) is the angular velocity about the \(r\)-axis. \(R\) is the radius of AUH, which is effectively the moment arm for each thruster. The resistance of AUH is mainly inertial fluid force and viscous fluid force. Since inertial fluid force can be regarded as additional mass, we mainly discuss viscous fluid force. Viscous fluid force can be divided into viscous resistance related to the velocity and contact area of the submarine and differential pressure resistance related to the shape of the submarine. Because the AUH presents a remarkable non-streamlined line, this may make the viscous resistance in translation direction negligible for differential pressure resistance [18]. We can get  
\[f_{p}=C_{u|u|}|u|+C_{u}Su\approx C_{u|u|}|u| \tag{6}\]  
where \(C_{u|u|}\) is the quadratic differential pressure coefficient, \(C_{u}\) is translational viscous drag coefficient, S is the surface area of AUH. In the rotary motion, due to the rotatory characteristics of the AUH, there is only viscous resistance and no differential pressure resistance. So  
\[M_{fr}=\int\limits_{S}C_{\omega}\omega rdr \tag{7}\]  
where \(C_{u}\) is the rotary viscous drag coefficient. Since \(C_{u}\) and \(C_{\omega}\) are only related to fluid properties and shell material, they are of the same order of magnitude. When \(u\) is the same as \(\omega r\), \(M_{fr}\) is the same order of magnitude as \(C_{u}SuR\). The AUH design translation speed is 0.8 m/s. Assuming that the AUH edge has a rotational linear velocity of 0.8 m/s, its angular velocity is about 90 deg/s, which is sufficient for high maneuverability. The thrust of the propeller selected according to the translational design speed is at least \(f_{p}\), and the minimum torque \(M_{fr}\) required to maintain the design rotational speed is much smaller than the torque \(2f_{p}R\) that the propeller can provide. Therefore, \(M_{fr}\) can be ignored in the analysis. Higher maneuverability means that linear motion and rotary motion have faster responses. From Equations (4) and (5), it can be seen that higher \(F_{1}\), \(F_{2}\), \(R\) and smaller \(f_{p}\), \(M_{fr}\), \(m\), and \(J_{r}\) mean higher response speed. It can be seen from the above that \(M_{fr}\) is negligible. So the response speed of rotation is directly correlated with \(J_{r}\). Because of the special shape of AUH, \(J_{r}\) can be quite small when the mass distribution is concentrated near the center as much as possible. Therefore, AUH has good horizontal plane rotation ability. In addition, AUH can achieve zero radius rotation by providing a set of symmetrical opposite driving forces. The response of plane motion is similar to that of rotation motion, but the difference is that \(f_{x}\) is not zero and the situation is more complex. The analysis is as follows;  
When AUH advances under the action of a horizontal propeller, the force acting on it is shown in Figure 4. Considering stability, AUH's center of gravity is lower than the center of buoyancy. In addition, the special requirement of positioning and navigation system leads to the change of body shape.