## 5 Cam Mechanism
Through simulation, it can be determined that both rollers simultaneously reach extremities at approximately \(\alpha_{1}=-7.2^{\circ}\) and \(\alpha_{2}=172.8^{\circ}\), which correspond to the angles of the central gear when the left and right roller switch between active and passive stretching. In the context of this switching process, there exists a transition angle \(\alpha_{0}\) between the non-circular and circular profiles, thus ensuring that one of the left or right roller is stretched along a circular or non-circular path while the other is stretched along a non-circular or circular path by the synchronous string. This angle, \(\alpha_{0}\), should result in equal rotation angles for the left and right rollers; hence, it is the midpoint between \(\alpha_{1}\) and \(\alpha_{2}\), i.e.,  
\[\alpha_{0}=\frac{1}{2}(\alpha_{1}+\alpha_{2})=82.8^{\circ} \tag{16}\]  
Figure 13: Motion schematic of the cam mechanism. From (**a**–**h**), the double-layer bearing change from 45\({}^{\circ}\) to 360\({}^{\circ}\) with step of 45\({}^{\circ}\). When the central gear is at this angle, it remains at a tangent to the circular paths on both the left and right rollers. Upon a counterclockwise rotation, the left roller retracts the string along the non-circular path while the right roller releases it along the circular path. Conversely, during clockwise rotation, the left roller releases the string along the circular path while the right roller retracts it along the non-circular path. Furthermore, considering the structural symmetry, it becomes evident that the simulation results for \(\alpha_{1}\) to \(\alpha_{2}\) and \(\alpha_{2}\) to \(\alpha_{1}+360^{\circ}\) yield identical non-circular cam profiles. Therefore, for ease of analysis, this paper will focus on the analysis of the non-circular cam profile as the central gear rotates from \(\alpha_{0}\) to \(\alpha_{2}\). Figure 15 shows a representation diagram of the cam mechanism, in which the direction of the connection between the center and the intersection point of the circular trajectory and the cam trajectory is taken as the positive direction of the polar axis, and where the direction of the cam trajectory is taken as the positive direction of the angle. A polar coordinate system is then established, and the cam trajectory can be expressed as Equation (17). \[\rho=\rho(\theta),\rho(0)=r_{1} \tag{17}\]  
where \(\theta\) is the angle between the positive direction of the polar axis and the trajectory. As shown in Figure 16, the initial position was set as the central gear rotation angle \(\alpha_{0}\). At that moment, the intersection point \(B_{0}\) of the left roller's circular trajectory and the cam trajectory was at a tangent to the line \(B_{0}C_{0}\). Similarly, the intersection point of the right roller was also at a tangent. At that time, the initial angles \(\theta_{0}\) and \(\delta_{0}\) of the two cams could be solved through the tangential equation of the circle, and the initial length \(m_{0}\) of the string could also be calculated. The values of the above parameters are shown in Table 4.  
Figure 14: Flapping angle of both rollers when \(r_{3}=0\). Figure 15: Polar coordinate representation of the cam mechanism. Due to the clockwise rotation that followed, the right roller was directly pulled by the central gear, so the flapping angle of the entire mechanism as a function of the central gear rotation angle \(\alpha\) was the same as that generated by the circular trajectory of the right roller. Therefore, the positional mapping relationship between the rotation angle of the left roller and the central gear angle \(\theta=f(\alpha)\) could be obtained. The Curve Fitting toolbox in MATLAB was used to fit the simulated data. The mapping equation could be fitted in the form of first-order Fourier series, which can be expressed as Equation (18). Figure 12.0 (18) \[=69.32-81.68\text{cos}  
Consider Point A, which moves along a known trajectory with Point D as its center and \(r_{1}\) as its radius.