# Design and Verification of a Large-Scaled Flapping-Wing Aircraft Named \"Cloud Owl\"
## Materials and Methods
### Power System Design  
A flapping-wing aircraft's flapping wing produces lift and thrust, and the drive mechanism, which is directly attached to the flapping wing and controls its movement, including its frequency, amplitude, and other characteristics, does the same. The drive mechanism must output adequate torque to the flapping wing in order for the flapping wing to generate enough lift for the aircraft to fly normally. In the conceptual design stage, there are no detailed design parameters of the driving mechanism, so it is challenging to directly acquire the instantaneous flapping angular velocity in the conceptual design stage in order to calculate the average periodic angular velocity. The common flapping-wing drive mechanism is the crank-connecting rod mechanism, and the motion equation of its output is close to the sine function, so it is advisable to set the motion equation of the instantaneous flapping angular displacement of the output end of the drive mechanism as the form of the sine function shown in the equation. The corresponding instantaneous flapping angular velocity is obtained by taking the derivative. \[\theta_{flap}=\frac{A}{2}sin(2\pi ft) \tag{5}\]  
\[\omega_{flap}=\overset{\bullet}{\theta}_{flap}=\pi Afcos(2\pi ft) \tag{6}\]  
In the formula, \(\theta_{\text{flap}}\) stands for the instantaneous flapping angular displacement; \(\omega_{\text{flap}}\) is the instantaneous flapping angular velocity; A is the flapping amplitude; and f is the flapping frequency. In the cruise state, the weight of the aircraft is assumed to be the lift produced by the flapping wings, and the maximum cruise power and output torque of the drive mechanism are determined on the assumption that the wingspan is known [23]. \[F_{L}=W_{to}\bullet g \tag{7}\]  
\[T_{wingmax}=F_{L}\bullet\frac{b}{4} \tag{8}\]  
\[P_{wing\ max}=T_{wing\ max}\bullet\omega_{flapmax} \tag{9}\]  
Figure 9: Curve of lift and thrust with velocity. \[P_{motor}=\frac{P_{wingmax}}{\eta_{trans}} \tag{10}\]  
\[P_{motormax}=\frac{P_{motor}}{\eta_{motor}} \tag{11}\]  
In the formula, \(\eta_{trans}\) is the mechanical efficiency of the driving system and \(\eta_{motor}\) is the motor efficiency. The drive system is driven by a micro brushless motor. For the two stages, the mechanical transmission efficiency is about 81%. The maximum output power of the motor is 104 W. At the same time, the battery capacity is 10.8 AH according to the designed target endurance of 100 min after getting the maximum cruising power. Table 4 lists the optional motor models. In each level, the motor with the lower KV value should be chosen in consideration of the lower flapping frequency. After taking the installation mode, weight, and output power requirements into account, the 2312 motor (KV1400) is chosen. The AT2312 motor is tested, and the relationship between motor speed n and motor output torque T is determined by fitting data points in Figure 10.  
\[n_{m}=-39,704T_{m}+16,771 \tag{12}\]  
The reduction ratio must be satisfied in order to meet the maximum flapping frequency requirements of f = 6 Hz. \[\omega_{flap}=\overset{\bullet}{\theta}_{flap}=\pi Afcos(2\pi ft) \tag{6}\]0  
The torque necessary to propel the flapping wing, according to a conservative estimate, which is the reduction ratio must meet the requirement of a factor of safety k = 1.5. \[\omega_{flap}=\overset{\bullet}{\theta}_{flap}=\pi Afcos(2\pi ft) \tag{6}\]1 \begin{table}
\begin{tabular}{c c c c} \hline \hline
**Motor Type** & **KV of Motor** & **Maximum Continuous Input Power** & **Weight** \\ \hline
2308 & 1450 & 250 W, 11.1 V & 47 g \\  & 2600 & 300 W, 11.1 V & \\ \hline
2312 & 1400 & 320 W, 11.1 V & 60 g \\  & 1150 & 350 W, 14.8 V & 60 g \\ \hline
2317 & 880 & 340 W, 14.8 V & 80 g \\  & 1250 & 520 W, 14.8 V & 80 g \\ \hline \hline \end{tabular}
\end{table} 
Table 4: List of selectable motors. Figure 10: The curve of torque and RPM.