## 3 Cam Wave Generators with Free Deformation in Non-Working Area
### Sample Design with Double Eccentric Arc Cam Wave Generator  
A double eccentric arc is usually used for cam wave generator design, which mainly follows the basic idea of a double disc wave generator, but eliminating the additional force couple effect in the general double disc wave generator by setting up double arcs on both sides in one plane, as shown in Figure 4b. In the design of a double eccentric arc cam, if the flexspline's deformation coincides exactly with the cam's contour, the stress mutation will occur at the junction of the arc and the straight line, and the stress will be very large. If the flexspline is deformed only fitting the cam in the working section, there would still be free deformation in the non-working section, and the rigidity of the flexspline and the shape remaining during rotating are all problems. Based on the new deformation model, a double eccentric arc wave generator would be designed and optimized in this paper. The convex curve of the double eccentric circular arc cam wave generator based on this deformation model is an eccentric circular arc curve. According to the thin-walled ring theory, the change of the ring shape caused by tangential load is consistent with that caused by radial load. The radial load is equal to the integral of the tangential load and the effect of the circumferential load \(\mathbf{q_{t}}\) is equivalent to that of the equivalent radial load \(\mathbf{q_{rt}}\):  
\[\mathbf{q_{rt}}=\int\mathbf{q_{t}}\mathbf{d\varphi} \tag{5}\]  
The total load on the outer ring is the sum of the equivalent radial load \(\mathbf{q_{rt}}\) and the radial load \(\mathbf{q_{r}}\). Here, taking cosine deformation as an example, the tangential load of the equivalent load of circular arc shape's forced deformation is set as \(q(\varphi)\)\(=0\), and radial load is set as  
\[p(\varphi)=-(A\cos\varphi+B) \tag{6}\]  
According to Equation (1), the differential equation is  
\[u=-\frac{AR^{4}\varphi\sin\varphi}{EI}+C_{1}+C_{2}\cos\varphi+C_{3}\varphi\sin \varphi+C_{4}\varphi\cos\varphi+C_{5}\sin\varphi \tag{7}\]  
Here, \(C_{1},\ C_{2},\ C_{3},C_{4}\) and \(C_{5}\) are the undetermined coefficients of the general solution. Due to symmetry of deformation, that is, \(u(\varphi)=u(-\varphi)\), the odd function in the general solution is omitted, and the deflection equation of the ring is simplified as  
\[u=C_{1}+C_{2}\cos\varphi+C_{3}\varphi\sin\varphi \tag{8}\]According the known deformation elongation of the ring with \(\varphi\)= 0, the rotation angle [26] of the ring at \(\varphi=\frac{\pi}{2}\) is 0, and the total tangential displacement of the ring is zero with \(\varphi\in[0,\alpha]\). Then the constraint equation is developed as following:  
\[\left\{\begin{array}{l} \begin{array}{l}\int_{q}^{\frac{\pi}{2}}-(C_{1}+C_{2} \cos\phi+C_{3}\phi\sin\phi)d\phi+\int_{0}^{\delta}-(\sqrt{r^{2}+e^{2}+2 re\cos \theta}-R)d\theta=0\\ u\big{|}_{\phi=\alpha}=C_{1}+C_{2}\cos\alpha+C_{3}\alpha\sin\alpha=\sqrt{r^{2}+ e^{2}+2 re\cos\delta}-R\\ \frac{du}{d\varphi}\big{|}_{\phi=\frac{\pi}{2}}=(-C_{2}\sin\phi+C_{3}\sin\phi +C_{3}\phi\cos\phi)\big{|}_{\phi=\frac{\pi}{2}}=0\end{array} \right. \tag{9}\]  
Then, the undetermined coefficients of \(C_{1}\), \(C_{2}\) and \(C_{3}\) can be gained by solving the Equation (9) as:  
\[\left\{ \begin{array}{l}C_{2}=C_{3}=\frac{\int_{0}^{\delta}-(\sqrt{r^{2}+e^{2 }+2 re\cos\theta}-R)d\theta-(\frac{\pi}{2}-a)(\sqrt{r^{2}+e^{2}+2 re\cos\delta}- R)}{2(1-\sin a)-(\frac{\pi}{2}-2a)\cos a-(\frac{\pi}{2}-a)\sin\alpha}\\ C_{1}=(\sqrt{r^{2}+e^{2}+2 re\cos\delta}-R)-(\cos\alpha+\alpha\sin\alpha)C_{3} \end{array} \right. \tag{10}\]  
As shown in Figure 4b, the polar diameter \(\rho(\varphi)\) of the flexspline's neutral layer after deformation in the working area is:  
\[\rho(\varphi)=\sqrt{r^{2}+e^{2}+2 re\cos\theta} \tag{11}\]  
\[\theta=\frac{\phi r}{R} \tag{12}\]  
Here, \(r\) is radius of the arc, \(\theta\) is the center angle of the arc and \(R\) is the radius of the neutral layer of the flexspline.