# Flap and Wing Dynamics for a Light Sport Aircraft Analysis Using a Topological Model
## 4 Dynamic Analysis  
In Section 2, a mechanism with joints and a planar motion was associated with a planar mechanism consisting only of bars. From a mechanical point of view, the two mechanisms  
Figure 14: The command mechanism.  
are equivalent. To perform a dynamic analysis of the mechanism, the motion equations of each bar that compose the equivalent mechanism will be written, and then, taking into account the liaison conditions, the motion of the entire assembly will be obtained. Figure 15 shows a bar with joints at both ends, and its motion equations will be written in the center of mass of the bar.  
By considering the free body diagram for one element, it is possible to write the motion equations for a single bar element, numbered \(k\):  
\[\left\{\begin{array}{l}m_{k}a_{Ckx}=X_{ki}+X_{kj}+X_{k}^{ext}\ ;\\ m_{k}a_{Cky}=Y_{ki}+Y_{kj}+Y_{k}^{ext}\ ;\\ J_{k}\epsilon_{k}=X_{ki}a_{k}\sin a_{k}-X_{kj}(L_{k}-a_{k})\sin a_{k}-Y_{ki}a_{k }\cos a_{k}+Y_{kj}(L_{k}-a_{k})\cos a_{k}+M_{k}^{ext}\end{array}\right., \tag{29}\]  
or  
\[\begin{bmatrix}m_{k}&0&0\\ 0&m_{k}&0\\ 0&0&J_{k}\end{bmatrix}\begin{Bmatrix}a_{kx}\\ a_{ky}\\ \epsilon_{k}\end{Bmatrix}=\left\{\begin{array}{l}X_{ki}+X_{kj}\\ Y_{ki}+Y_{kj}\\ X_{ki}a_{k}\sin a_{k}-X_{kj}(L_{k}-a_{k})\sin a_{k}-Y_{ki}a_{k}\cos a_{k}+Y_{kj} (L_{k}-a_{k})\cos a_{k}\end{array}\right\}+\left\{\begin{array}{l}X_{k}^{ ext}\\ Y_{k}^{ext}\\ M_{k}^{ext}\end{array}\right\}; \tag{30}\]  
or in a symbolic form  
\[\left[m\right]_{k}\{a\}_{k}=\{Q\}_{k}^{\mathit{liaison}}+\{Q\}_{k}^{ext}, \tag{31}\]  
where the following is denoted:  
\[\left[m\right]_{k}=\begin{bmatrix}m_{k}&0&0\\ 0&m_{k}&0\\ 0&0&J_{k}\end{bmatrix}\ ;\ \left\{a\right\}_{k}=\begin{Bmatrix}a_{kx}\\ a_{ky}\\ \epsilon_{k}\end{Bmatrix}\ ;\ \left\{Q\right\}_{k}^{ext}=\left\{\begin{array}{l}X_{k}^{ ext}\\ Y_{k}^{ext}\\ M_{k}^{ext}\end{array}\right\}, \tag{32}\]  
\[\{Q\}_{k}^{\mathit{liaison}}=\left\{\begin{array}{l}X_{ki}+X_{kj}\\ Y_{ki}+Y_{kj}\\ X_{ki}a_{k}\sin a_{k}-X_{kj}(L_{k}-a_{k})\sin a_{k}-Y_{ki}a_{k}\cos a_{k}+Y_{kj} (L_{k}-a_{k})\cos a_{k}\end{array}\right\}. \tag{33}\]  
The mass center accelerations are obtained by one of the following relations:  
\[\begin{array}{l}a_{Ckx}=a_{ix}-\epsilon_{k}b_{ky}-\omega_{k}^{2}b_{kx};&a_{ Ckx}=a_{jx}-\epsilon_{k}\Big{(}L_{k}-b_{ky}\Big{)}-\omega_{k}^{2}\Big{(}L_{k}-b_{ ky}\Big{)};\\ \text{or}\\ a_{Cky}=a_{iy}+\epsilon_{k}b_{kx}-\omega_{k}^{2}b_{ky};&a_{Cky}=a_{iy}+\epsilon_{ k}\Big{(}L_{k}-b_{ky}\Big{)}-\omega_{k}^{2}\Big{(}L_{k}-b_{ky}\Big{)}.\end{array} \tag{34}\]  
For the mechanism considered in this paper, the decomposition into component bars is presented in Figure 16.  
Figure 15: A free body diagram for an element \(k\) defined by the nodes \(i\) and \(j\).  
When considering Equation (37) from Equation (36), we obtain  
\[[m]\Big{(}\{A_{1}\}\varepsilon_{2}+\{A_{2}\}\omega_{2}^{2}\Big{)}=\{Q\}^{liaison}+ \{Q\}^{ext}, \tag{38}\]  
In previous papers, the possibility of eliminating the internal liaison forces by a simple matrix multiplication (based on the fact that in the absence of friction, the mechanical work of the liaison forces is zero) was demonstrated [61; 62; 63]:  
\[0=\{A_{1}\}^{T}\{Q\}^{liaison}. \tag{39}\]  
Premultiplying Equation (38) with \(\{A_{1}\}^{T}\) results in  
\[\{A_{1}\}^{T}[m]\{A_{1}\}\varepsilon_{2}+\{A_{1}\}^{T}[m]\{A_{2}\}\omega_{2}^ {2}=\{A_{1}\}^{T}\Big{(}\{q^{ext}\}+\Big{\{}q^{\text{\it flasons}}\Big{\}} \Big{)}. \tag{40}\]  
and using Equation (39), the final form of the equations is obtained:  
\[\{A_{1}\}^{T}[m]\{A_{1}\}\varepsilon_{2}+\{A_{1}\}^{T}[m]\{A_{2}\}\omega_{2}^ {2}=\{A_{1}\}^{T}\{q^{ext}\}. \tag{41}\]  
Or, finally,  
\[\alpha\varepsilon_{2}+\beta\omega_{2}^{2}=Q^{ext} \tag{42}\]  
where  
\[\alpha=\{A_{1}\}^{T}[m]\{A_{1}\}\ ;\beta=\{A_{1}\}^{T}[m]\{A_{2}\}\ ;Q^{ext}=\{A_{1}\}^{T}\{q^{ext}\}. \tag{43}\]  
To study the dynamic of the system, the differential equation remains to be solved:  
\[\alpha\frac{d^{2}\alpha_{2}}{dt^{2}}+\beta\bigg{(}\frac{d\alpha_{2}}{dt}\bigg{)} ^{2}=Q^{ext}. \tag{44}\]