## 4 Key Characteristics Analysis
### Traction Characteristic  
EMPU and ERSD will be subject to a very high load from the probe rod when towing the helicopter. The force analysis was performed on the dynamic system consisting of Helicopter-ERSD-Ship to determine the maximum thrust force required by EMPU and ERSD or the maximum lateral and longitudinal external load on the claw.  
As shown in Figure 12, the Cartesian coordinate system is established. Assume that the deck rolls to the negative \(X\)-axis direction, while the maximum lateral acceleration acts along the negative \(X\)-axis direction, and pitches to the negative direction of the \(Y\)-axis, while the maximum longitudinal acceleration acts along the negative direction of the \(Y\)-axis. The maximum vertical acceleration acts in the positive direction of the \(Z\)-axis.  
Figure 12: Schematic diagram of rolling and pitching of the deck.  
As shown in Figure 13, assume that the center of mass of the helicopter (\(W_{\rm H}\)), the force point of the claw (\(E\)), and the action center of the wind force (\(W_{\rm W}\)) are all located on the central axis of the helicopter. Suppose that the ground points of the three tires of the helicopter are 1, 2, and 3, and the rolling friction coefficient is \(f\).  
Through simple force analysis, the following conclusions can be drawn.  
1. In Figure 13, lateral resistance is maximum when the following three conditions are met: * The card-axis of the front wheel is deflected by 90\({}^{\circ}\); * The wind is blowing along the negative direction of the X-axis; * The claw pushes the helicopter in the direction of the yellow arrow. Assume that the thrust force required by the claw is \(P_{\rm X}\).
2. In Figure 13, the longitudinal resistance is maximum when the following three conditions are met: * The card-axis of the front wheel is deflected 0\({}^{\circ}\); * The wind is blowing along the negative direction of the Y-axis; * The claw pushes the helicopter forward in the direction of the yellow arrow. * Assume that the thrust force required by the claw is \(P_{\rm Y}\).  
In Figure 13, let the roll angle be \(\theta\), and the force acting on the center of mass is  
\[\left\{\begin{array}{l}F_{\rm X}=M_{\rm H}(a_{x}+g\sin\theta)\\ F_{\rm Z}=M_{\rm H}(a_{z}+g\cos\theta)\end{array}\right.. \tag{14}\]  
The rolling resistance of the front wheel is  
\[F_{1X}=W_{1}f=\frac{F_{Z}L_{\rm H}}{L}f. \tag{15}\]  
The rolling resistance of the rear wheel is  
\[F_{2Y}+F_{3Y}=(W_{2}+W_{3})f=(F_{Z}-W_{1})f. \tag{16}\]  
The wind force on the helicopter is  
\[F_{\rm WX}=\frac{1}{2}\rho A_{X}C_{\rm W}v_{\rm W}^{2}. \tag{17}\]  
Figure 13: Force analysis of the helicopter.  
If we take the torque equilibrium on the \(Z\)-axis, we can get  
\[P_{X}=\frac{2(F_{\rm WX}L_{\rm W}+F_{X}L_{\rm H}+F_{\rm IX}L)+(F_{ \rm ZY}+F_{\rm 3Y})B}{2L_{\rm E}}. \tag{18}\]  
In Figure 13, let the pitch angle be \(\lambda\), and the force acting on the center of mass is  
\[\left\{\begin{array}{l}F_{Y}=M_{H}(a_{Y}+g\sin\lambda)\\ F_{\rm Z}=M_{H}(a_{Z}+g\cos\lambda)\end{array}\right.. \tag{19}\]  
The sum of the rolling resistance of the front and rear wheels is  
\[F_{\rm 1Y}+F_{\rm 2Y}+F_{\rm 3Y}=(W_{1}+W_{2}+W_{3})f=F_{\rm Z}f. \tag{20}\]  
The wind force on the helicopter is  
\[F_{\rm WY}=\frac{1}{2}\rho A_{\rm Y}C_{\rm W}v_{\rm W}^{2}. \tag{21}\]  
If we take the force equilibrium on the \(Y\)-axis, we can get  
\[P_{Y}=F_{\rm WY}+F_{Y}+(F_{\rm 1Y}+F_{\rm 2Y}+F_{\rm 3Y}). \tag{22}\]  
The parameters of a specific type of helicopter and ship are shown in Table 1, in level-3 sea conditions. Taking the data in Table 1 into Equations (18) and (22), it can be calculated that the maximum lateral and longitudinal load of the claw in level-3 sea conditions are \(P_{\rm X}=34,543\) N and \(P_{\rm Y}=11,789\) N.