# Research on the Flight Performance of Biomimetic Moth Based on Flapping Function Control
## 3 Calculation and Simulation
### Lift Calculation of Rigid Wing Flapping Motion
\tag{5}\]  
For a differential element of length \(dr\), according to Equation (3), the lift \(L\) perpendicular to the wing plane produced by this differential element can be derived as shown in Equation (6), where the subscripts \(d\) and \(u\) represent the downstroke and upstroke processes, respectively. \[\left\{ \begin{array}{ll}dL(t,r)=dL_{d}(t,r)-dL_{u}(t,r)\\ dL_{d}(t,r)=\frac{1}{2}\rho C_{Ld}U_{d}^{2}c_{r}(t,r)dr\\ dL_{u}(t,r)=\frac{1}{2}\rho C_{Lu}U_{d}^{2}c_{r}(t,r)dr\end{array} \right. \tag{6}\]  
It is clear that even when simplified, the flapping motion cannot be equated to a straightforward translational movement of the wing. The flapping speed \(U\) continues to depend on both \(t\) and \(r\). [PERSON] and others [27] analyzed and believed that compared with the flapping speed, the influence of the aircraft's forward speed on force generation is secondary, and this view has been verified by experiments. Therefore, ignoring the forward speed of the aircraft, it can be considered that the flapping speed satisfies Equation (7):  
\[U=U(t,r)=r\omega=r\dot{\phi}(t) \tag{7}\]Taking the derivative of the flapping function equation, the angular velocity \(\omega\) at any time can be obtained as Equation (8):  
\[\omega=\frac{d\phi}{dt}=\dot{\phi}(t)=2\pi f\phi_{max}\text{cos}(2\pi ft+\psi) \tag{8}\]  
Due to the wing being a rigid wing and undergoing rigid-body motion, meaning the leading and trailing edges of the airfoil do not change over time, the time-averaged function of the trailing edge is as shown in Equation (9). \[\bar{c}_{r}(r)=c_{r}(r)=-4.5r^{2}+1.485r+0.126 \tag{9}\]  
Substituting Equation (7) into Equation (6) and then integrating it, the lift perpendicular to the wing plane during the downward and upward strokes is obtained as Equation (10):  
\[ \begin{cases}L_{d}(t)=\int_{0}^{l_{w}}dL_{d}(t,r)=\frac{1}{2}\rho C_{Ld}\dot{ \phi}_{d}^{2}(t)\int_{0}^{l_{w}}r^{2}\bar{c}_{r}(r)dr\\ L_{u}(t)=\int_{0}^{l_{w}}dL_{u}(t,r)=\frac{1}{2}\rho C_{Ld}\dot{\phi}_{u}^{2}( t)\int_{0}^{l_{w}}r^{2}\bar{c}_{r}(r)dr\end{cases}  \tag{10}\]  
Substituting Equations (8) and (9) into Equation (10), Equation (11) is obtained:  
\[ \begin{cases}L_{d}(t)=\left[2\rho C_{Ld}\tau^{2}f^{2}\phi_{max}^{2}\int_{0}^{ l_{w}}r^{2}\bar{c}_{r}(r)dr\right]\text{-}\text{cos}^{2}(2\pi ft+\psi)\\ L_{u}(t)=\left[2\rho C_{Lu}\tau^{2}f^{2}\phi_{max}^{2}\int_{0}^{l_{w}}r^{2}\bar {c}_{r}(r)dr\right]\text{-}\text{cos}^{2}(2\pi ft+\psi)\end{cases}  \tag{11}\]  
Then, the average lift calculation formula perpendicular to the wing plane is derived as in Equation (12):  
\[ \begin{cases}\bar{L}_{d}=f\int_{0}^{\frac{1}{2}}L_{d}(t)dt=4\rho C_{Ld}\tau^{ 3}f^{4}\phi_{max}^{2}\int_{0}^{l_{w}}r^{2}\bar{c}_{r}(r)dr\\ \bar{L}_{u}=f\int_{0}^{\frac{1}{2}}L_{u}(t)dt=4\rho C_{Lu}\tau^{3}f^{4}\phi_{ max}^{2}\int_{0}^{l_{w}}r^{2}\bar{c}_{r}(r)dr\end{cases}  \tag{12}\]  
Further, the lift calculation formula in the vertical direction \(F_{L}\) is derived as in Equation (13):  
\[ \begin{cases}\bar{F}_{Ld}=\frac{1}{2\phi_{max}}\int_{-\phi_{max}+\phi_{0}}^{ \phi_{max}+\phi_{0}}\bar{L}_{d}cos\phi d\phi=\frac{L_{d}}{2\phi_{max}}\int_{- \phi_{max}+\phi_{0}}^{\phi_{max}+\phi_{0}}cos\phi d\phi\\ \bar{F}_{Lu}=\frac{1}{2\phi_{max}}\int_{-\phi_{max}+\phi_{0}}^{\phi_{max}+\phi _{0}}\bar{L}_{u}cos\phi d\phi=\frac{L_{u}}{2\phi_{max}}\int_{-\phi_{max}+\phi _{0}}^{\phi_{max}+\phi_{0}}cos\phi d\phi\end{cases}  \tag{13}\]  
Finally, the lift difference generated by the upward and downward strokes within a period is in Equation (14):  
\[\bar{F}_{L}=\bar{F}_{Ld}-\bar{F}_{Lu}=4\rho(C_{Ld}-C_{Lu})\pi^{3}f^{4}\phi_{ max}sin\phi_{max}cos\phi_{0}\int_{0}^{l_{w}}r^{2}\bar{c}_{r}(r)dr \tag{14}\]  
Analyzing Equation (14), the air density \(\rho=1.29\) kg/m\({}^{3}\) at room temperature, which is a constant. In the simplified model, it is assumed that \(C_{Ld}=0.9\) and \(C_{Lu}=0.6\) are constants; in the simplified model, it is assumed that the shape of the wing remains unchanged, then the definite integral \(\int_{0}^{l_{w}}r^{2}\bar{c}_{r}(r)dr\) is a constant.