## 3 Kinematic Analysis
\begin{array}{l}X^{\prime\prime}=x\cos\theta_{y}+y\sin\theta_{x}\sin \theta_{y}-z\cos\theta_{x}\sin\theta_{y}-\rho\cos\theta\sin 3\varphi_{0}\\ Y^{\prime\prime}=y\cos\theta_{x}+z\sin\theta_{x}-\rho\cos\theta\cos 3 \varphi_{0}\\ Z^{\prime\prime}=x\sin\theta_{y}-y\sin\theta_{x}\cos\theta_{y}+z\cos\theta_{x} \cos\theta_{y}-\rho\sin\theta\end{array} \right. \tag{8}\]  
The angles \(\theta_{x}\) and \(\theta_{y}\) can be determined by the intersection point S of axes \(oz\) and \(O^{\prime\prime}Z^{\prime\prime}\). By substituting the coordinates of point S in both coordinate systems, \(\left(0,0,\frac{1}{cos\theta}\right)\) and \(\left(0,0,l\right)\), into the above equation, \(\theta_{x}\) and \(\theta_{y}\) can be solved. \[\theta_{x}=arcsin\left(\frac{\rho}{l}\cos^{2}\theta\cos 3\varphi_{0}\right) \tag{9}\]  
\[\theta_{y}=arccos\left(\frac{\rho\sin\theta\cos\theta}{\sqrt{l^{2}-\rho^{2} \cos^{4}\theta\cos^{2}(3\varphi_{0})}}\right) \tag{10}\]  
Substitute Equation (8) into Equation (4), and then, after simplification, the equation of the axis of the three rods (Rod \(n_{1}\), Rod \(n_{2}\), Rod \(n_{3}\)) in the \(oxyz\) coordinate system can be obtained. \[\left\{ \begin{array}{l}x\cos\theta_{y}tg\varphi_{0}+y\left(\sin\theta_{x} \sin\theta_{y}tg\varphi_{0}-\cos\theta_{y}\right)-z\left(\cos\theta_{x}\sin \theta_{y}tg\varphi_{0}+\sin\theta_{x}\right)\\ =\rho\cos\theta\sin 3\varphi_{0}tg\varphi_{0}-\rho\cos\theta\cos 3\varphi_{0} \\ x\sin\theta_{y}-y\sin\theta_{x}\cos\theta_{y}+z\cos\theta_{x}\cos\theta_{y}= \rho\sin\theta\end{array} \right. \tag{11}\]  
\[\left\{ \begin{array}{l}x\cos\theta_{y}tg\left(\varphi_{0}+\frac{2}{3}\pi \right)+y\left[\sin\theta_{x}\sin\theta_{y}tg\left(\varphi_{0}+\frac{2}{3}\pi \right)-\cos\theta_{y}\right]\\ -z\left[\cos\theta_{x}\sin\theta_{y}tg\left(\varphi_{0}+\frac{2}{3}\pi\right)+ \sin\theta_{x}\right]\\ =\rho\cos\theta\sin 3\left(\varphi_{0}+\frac{2}{3}\pi\right)tg\left(\varphi_{0}+ \frac{2}{3}\pi\right)-\rho\cos\theta\cos 3\left(\varphi_{0}+\frac{2}{3}\pi\right)\\ x\sin\theta_{y}-y\sin\theta_{x}\cos\theta_{y}+z\cos\theta_{x}\cos\theta_{y}= \rho\sin\theta\end{array} \right. \tag{12}\]  
\[\left\{ \begin{array}{l}x\cos\theta_{y}tg\left(\varphi_{0}+\frac{4}{3}\pi \right)+y\left[\sin\theta_{x}\sin\theta_{y}tg\left(\varphi_{0}+\frac{4}{3} \pi\right)-\cos\theta_{y}\right]\\ -z\left[\cos\theta_{x}\sin\theta_{y}tg\left(\varphi_{0}+\frac{4}{3}\pi\right)+ \sin\theta_{x}\right]\\ =\rho\cos\theta\sin 3\left(\varphi_{0}+\frac{4}{3}\pi\right)tg\left(\varphi_{0}+ \frac{4}{3}\pi\right)-\rho\cos\theta\cos 3\left(\varphi_{0}+\frac{4}{3}\pi\right)\\ x\sin\theta_{y}-y\sin\theta_{x}\cos\theta_{y}+z\cos\theta_{x}\cos\theta_{y}= \rho\sin\theta\end{array} \right. \tag{13}\]  
To determine the relationship between the input shaft angle \(\varphi_{1}\) and the output shaft angle \(\varphi_{0}\), we can utilize the characteristic that the axis of the plunger slot \(m_{1}\) should intersect with the rod \(n_{1}\). Therefore, we can express this by utilizing Equations (3) and (11). \[\left\{ \begin{array}{l}A_{1}x+B_{1}y+C_{1}z+D_{1}=0\\ A_{1}^{\prime}x+B_{1}^{\prime}y+C_{1}^{\prime}z+D_{1}^{\prime}=0\\ A_{1}^{\prime\prime}x+B_{1}^{\prime}y+C_{1}^{\prime\prime}z+D_{1}^{\prime \prime}=0\\ A_{1}^{\prime\prime\prime}x+B_{1}^{\prime\prime\prime}y+C_{1}^{\prime\prime \prime}z+D_{1}^{\prime\prime\prime}=0\end{array} \right. \tag{14}\]  
Among which  
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