# Dynamics Modelling and Simulation for Deployment Characteristics of Mesh Reflector Antennas
## 3 Deployment Modelling of Truss Antenna Using Torsional Spring  
To improve the efficiency of deployment, a torsional spring is usually assembled in the synchronous joint to deploy the hoop truss to a certain value of deploy angle. A simplified mechanics model of torsion spring was given in the literature [24], in which the torque varies linearly from 4 to 2 N \(\cdot\) m during the whole deployment. Assuming that the torsional angle acted by the torsional spring ranges with \([0,\,\alpha_{0}]\), then the torque produced by the torsional spring can be expressed as  
\[\tau=K(\alpha_{0}-\alpha)+c\dot{\alpha} \tag{18}\]  
where \(K=128/31\pi\) Nm/rad and \(c=270\) Nms/rad are the stiffness coefficient and damping coefficient of the torsional spring, respectively. As shown in Figure 4, \(\alpha\) is the angle between the horizontal rods and perpendicular rods. Defining \(F\) as the driving force equivalent to the torque of the torsional spring, as shown in Figure 4, using the principle of virtual work, we have  
\[F\delta l+2\tau\delta\alpha=0 \tag{19}\]  
The virtual change in \(l\) can be obtained by variation of \(l\) with respect to \(\alpha\)  
\[\delta l=\frac{-L_{1}L_{2}\sin\alpha}{\sqrt{L_{1}^{2}+L_{2}^{2}+2L_{1}L_{2} \cos\alpha}}\delta\alpha \tag{20}\]  
Figure 4: Equivalence of torsional spring. Substituting Equation (20) into Equation (19), the torque in Equation (19) is transformed into an equivalent driving force applied on the diagonal rod, expressed as  
\[F=\frac{2\tau\sqrt{L_{1}^{2}+L_{2}^{2}+2L_{1}L_{2}\cos\alpha}}{L_{1}L_{2}\sin\alpha} \tag{21}\]  
In order to consider the attenuation of pulling force caused by friction of bearing, Figure 5 schematically depicts a diagram of the pulley. As illustrated in Figure 5, the frictional moment generated by the bearing can be obtained by the formula of Palmgren [24]. \[M=fF_{\lambda} \tag{22}\]  
where \(f=ud\) denotes the frictional coefficient, in which \(u\) is the frictional factor and \(d\) is the pitch diameter of the bearing. \(F_{\lambda}\) is the synthetic moment load acted on the bearing. As for the ball bearing, in this case, \(F_{\lambda}\) equals to the radial load \(F_{r}\)[6], thus Equation (22) can be written as  
\[M=fF_{r} \tag{23}\]  
Dynamic equations of the pulley can be obtained using the law of motion of the center of mass and theorem of moment of momentum, respectively. \[\left\{ \begin{array}{l}F_{n}+F_{n+1}+F_{r}-m_{p}a_{p}=0\\ R_{n}\times F_{n}+R_{n+1}\times F_{n+1}+M-\frac{1}{2}m_{p}\Big{|}R_{i}\Big{|} \frac{d\omega}{dt}=0\end{array} \right. \tag{24}\]  
where \(m_{p},a_{p}\), and \(\omega\) are the mass, acceleration, and angular velocity of the pulley. \(R_{n}\) denotes the vector from the center of the pulley towards \(F_{n}\). Considering \(a_{p},|R_{i}|\) and \(d\omega/dt\) are very small in Equation (24), which can be transformed as  
\[\left\{ \begin{array}{l}F_{n}+F_{n+1}+F_{r}=0\\ R_{n}\times F_{n}+R_{n+1}\times F_{n+1}+M=0\end{array} \right. \tag{25}\]  
According to Equations (24) and (25), one can obtain the following expressions  
\[\left\{ \begin{array}{l}F_{r}^{2}=F_{n}^{2}+F_{n+1}^{2}+2F_{n}F_{n+1}\cos \beta\\ (F_{n}-F_{n+1})R=fF_{r}\end{array} \right. \tag{26}\]  
Further, Equation (26) can be transformed as  
\[(F_{n}-F_{n+1})R-f\sqrt{F_{n}^{2}+F_{n+1}^{2}+2F_{n}F_{n+1}\cos\beta}=0 \tag{27}\]  
Figure 5: Schematic diagram of the pulley. The open square item in Equation (27) can be transformed as  
\[ \begin{array}{ll}F_{n}^{2}+F_{n+1}^{2}+2F_{n}F_{n+1}\cos\beta=&F_{n}^{2}/2+F_{n} ^{2}/2+F_{n+1}^{2}/2+F_{n+1}^{2}/2+\\ &F_{n}F_{n+1}\cos\beta/2+F_{n}F_{n+1}\cos\beta/2+F_{n}F_{n+1}\cos\beta\end{array}  \tag{28}\]  
Considering the fact that the driving force on each side of a pulley is nearly the same, that is \(F_{n}\approx F_{n+1}\), one can obtain  
\[ \begin{array}{ll}F_{n}^{2}+F_{n+1}^{2}+2F_{n}F_{n+1}\cos\beta\approx&F_{n}^{ 2}/2+F_{n}F_{n+1}/2+F_{n+1}^{2}/2+F_{n}F_{n+1}/2+\\ &F_{n}^{2}\cos\beta/2+F_{n+1}^{2}\cos\beta/2+F_{n}F_{n+1}\cos\beta\end{array}