## 4 Dynamics Analysis
\begin{array}{l}R_{x^{\prime\prime}}+F_{1x^{\prime\prime}}+F_{2x^{ \prime\prime}}+F_{3x^{\prime\prime}}=0\\ R_{y^{\prime\prime}}+F_{1y^{\prime\prime}}+F_{2y^{\prime\prime}}+F_{3y^{ \prime\prime}}=0\\ R_{z^{\prime\prime}}+F_{1z^{\prime\prime}}+F_{2z^{\prime\prime}}+F_{3z^{ \prime\prime}}=0\end{array} \right\} \tag{25}\]  
The support reaction force of the output shaft bearing can be obtained from Equation (23). \[\left. \begin{array}{l}R_{x^{\prime\prime}}=0\\ R_{y^{\prime\prime}}=0\\ R_{z^{\prime\prime}}=-F_{1z^{\prime\prime}}-F_{2z^{\prime\prime}}-F_{3z^{ \prime\prime}}\end{array} \right\} \tag{26}\]  
The results show that the restraining forces \(R_{x^{\prime\prime}},R_{y^{\prime\prime}}\) on axes \(X^{\prime\prime}\) and \(Y^{\prime\prime}\) at the output end of the spherical roller bearing are both zero, leaving only the restraining force \(R_{z^{\prime\prime}}\) on axis \(Z^{\prime\prime}\) unknown. According to the principle of action and reaction, the forces \(F_{x^{\prime\prime}}\), \(F_{y^{\prime\prime}}\), \(F_{z^{\prime\prime}}\) acting on the triple rod are transformed into the \(OX^{\prime}Y^{\prime\prime}Z^{\prime}\) coordinate system, and should be equal in magnitude but opposite in direction to the forces \(F_{x^{\prime}}\)\(F_{y^{\prime}}\)\(F_{z^{\prime}}\) acting on the ball joints connected to the piston. Consequently:  
Figure 9: A pictorial drawing of the twisting forces on the three-pronged. \[ \begin{bmatrix}F_{x^{\prime}}\\ F_{y^{\prime}}\\ F_{z^{\prime}}\end{bmatrix} =-[C_{o^{\prime}o^{\prime\prime}}] \begin{bmatrix}F_{x^{ \prime\prime}}\\ F_{y^{\prime\prime}}\\ F_{z^{\prime\prime}}\end{bmatrix}  \tag{27}\]  
In the above equation, the following are known:  
\[F_{z^{\prime}}=ma_{i},F_{x^{\prime\prime}}=-F\sin\left(\varphi+i\frac{2\pi}{3} \right),F_{y^{\prime\prime}}=F\cos\left(\varphi+i\frac{2\pi}{3}\right)\quad(i= 0,1,2) \tag{28}\]  
\(m\) represents the mass of the piston rod, and \(a_{i}\) represents the acceleration of the piston. The unknown parameters are \(F_{x^{\prime}},F_{y^{\prime}},F_{z^{\prime}}\). Expanding the equation results in three equations with three unknowns, therefore, Equation (27) can be solved. Among them:  
\[[C_{o^{\prime}o^{\prime\prime}}]= \begin{bmatrix}A_{1}&A_{2}&A_{3}\\ B_{1}&B_{2}&B_{3}\\ C_{1}&C_{2}&C_{3}\end{bmatrix} \]  
Among them [17]:  
\[A_{2}=\cos\beta\cos\theta_{y}-\frac{\sin\beta\sin 2\theta\cos 3\varphi_{0}}{2 \cos\theta_{y}}\;\;A_{2}=-\frac{\sin\beta\sin\theta\sin 3\varphi_{0}}{\cos \theta_{y}}\]  
\[A_{3}=-\cos\beta\sin\theta\cos 3\varphi_{0}-\sin\beta\cos\theta\]  
\[B_{1}=-\frac{\sin^{2}\theta\sin 6\varphi_{0}}{2\cos\theta_{y}}\;\;B_{2}= \frac{\cos\theta}{\cos\theta_{y}}\;\;B_{3}=-\sin\theta\sin 3\varphi_{0}\]  
\[C_{1}=\sin\beta\cos\theta_{y}+\frac{\cos\beta\sin 2\theta\cos 3\varphi_{0}}{2 \cos\theta_{y}}\;\;C_{2}=\frac{\cos\beta\sin\theta\sin 3\varphi_{0}}{\cos \theta_{y}}\]  
\[C_{3}=-\sin\beta\sin\theta\cos 3\varphi_{0}+\cos\beta\cos\theta\]  
Expanding Equation (27), we obtain:  
\[ \begin{cases}F_{x^{\prime}}=-A_{1}F_{x^{\prime\prime}}-A_{2}F_{y^{\prime\prime }}-A_{3}F_{z^{\prime\prime}}\\ F_{y^{\prime}}=-B_{1}F_{x^{\prime\prime}}-B_{2}F_{y^{\prime\prime}}-B_{3}F_{z^ {\prime\prime}}\\ F_{z^{\prime}}=-C_{1}F_{x^{\prime\prime}}-C_{2}F_{y^{\prime\prime}}-C_{3}F_{z^ {\prime\prime}}\end{cases}