## 2 The Principle and Kinematics Analysis of the Drive Mechanism of
### Kinematics Analysis of Drive Mechanism of Die-Cutting Device  
When operating at high speed, it is necessary to obtain the inertia force and moment of inertia of the drive rod through the kinematic analysis of the drive mechanism, which provides a prerequisite for the subsequent structural parameter optimization and dynamic simulation verification. Considering that the drive mechanism of the moving platform is a symmetrical structure, only one side of the drive mechanism is taken as the analysis object in the kinematics analysis. According to the research results of other scholars [21; 22; 23; 24], the lower drive rod of BC in Figure 1 is the most stressed part. Due to assembly restrictions, the upper drive rod of BD and the lower drive rod of BC must be the same size. Therefore, it is sufficient to analyze the drive rod with the greatest stress. In this paper, the lower drive rod of BC on one side of the drive mechanism is selected as the final analysis object. To determine the inertial force of the drive rod, it is necessary to analyze the motion of the connecting rod and calculate the speed and acceleration of the connecting rod. According to the assembly relationship of crank, power rod and drive rod, its motion diagram can be established, and the force vector diagram of drive rod can be drawn according to the diagram (as shown in Figure 2). From the closed vector polygon OABCD, it can be obtained [25]:  
\[ \begin{split}\overset{\rightharpoonup}{l_{1}}+\overset{ \rightharpoonup}{l_{2}}=\overset{\rightharpoonup}{l_{5}}+\overset{ \rightharpoonup}{l_{4}}+\overset{\rightharpoonup}{l_{3}}\end{split}  \tag{1}\]  
Among them: \(l_{1},l_{2},l_{3},l_{4},l_{5}\) are the lengths of \(OA,AB,BC,CD,DO\). Figure 1: Schematic diagram of drive mechanism of die-cutting device. 1-crankshaft; 2-power rod; 3-drive rod; 4-moving platform. Figure 2: Vector relationship of the drive rod. By projecting Formula (1) onto the coordinate axis, the position equation of the part can be obtained:  
\[ \begin{array}{l}l_{1}cos\varphi_{1}+l_{2}cos\varphi_{2}=l_{4}+l_{3}cos\varphi_{ 3}\\ l_{1}sin\varphi_{1}+l_{2}sin\varphi_{2}=-l_{5}+l_{3}cos\varphi_{3}\end{array}  \tag{2}\]  
The solution can be obtained:  
\[\varphi_{2}=arcsin\frac{C_{1}}{\sqrt{A_{1}^{2}+B_{1}^{2}}}-arctan\frac{B_{1}}{ A_{1}}\]  
\[\varphi_{3}=arcsin\frac{C_{2}}{\sqrt{A_{2}^{2}+B_{2}^{2}}}-arctan\frac{B_{2}}{ A_{2}}\]  
In the formula:  
\[A_{1}=-2l_{5}l_{3}-2l_{1}l_{3}sin\varphi_{1}\]  
\[A_{2}=2l_{1}l_{2}sin\varphi_{1}+2l_{2}l_{5}\]  
\[B_{1}=2l_{4}l_{3}-2l_{1}l_{3}cos\varphi_{1}\]  
\[B_{2}=2l_{1}l_{2}cos\varphi_{1}-2l_{1}l_{4}\]  
\[C_{1}=l_{2}^{2}-l_{3}^{2}-l_{5}^{2}-l_{4}^{2}-l_{1}^{2}-2l_{1}l_{5}sin\varphi _{1}+2l_{1}l_{4}cos\varphi_{1}\]  
\[C_{2}=l_{3}^{2}-l_{1}^{2}-l_{2}^{2}-l_{4}^{2}-l_{5}^{2}-2l_{1}l_{5}sin\varphi _{1}+2l_{1}l_{4}cos\varphi_{1}\]  
Here, take \(l_{1}=45\) mm, \(l_{2}=217\) mm, \(l_{3}=178\) mm, \(l_{4}=280\) mm, \(l_{5}=185.3\) mm. The position change information of the drive rod can be calculated by substituting the rod length parameter into Formula (2) (see Figure 3):  
Among them: the abscissa represents the position of the drive rod vertex in the \"X\" axis direction on the operation plane, and the ordinate represents the position of the drive rod vertex in the \"Y\" axis. Figure 3: Position of drive rod. By taking the first derivative of Formula (2) with respect to time, the angular velocity equation can be obtained:  
\[ \begin{array}{l}l_{1}\omega_{1}cos\varphi_{1}+l_{2}\omega_{2}cos\varphi_{2}=l_ {3}\omega_{3}cos\varphi_{3}\\ l_{1}\omega_{1}sin\varphi_{1}+l_{2}\omega_{2}sin\varphi_{2}=l_{3}\omega_{3}sin \varphi_{3}\end{array}  \tag{3}\]  
The solution can be obtained:  
\[\omega_{2}=-\frac{l_{1}\omega_{1}sin(\varphi_{3}-\varphi_{1})}{l_{2}sin( \varphi_{3}-\varphi_{2})}\]  
\[\omega_{3}=\frac{l_{1}\omega_{1}sin(\varphi_{1}-\varphi_{2})}{l_{2}sin( \varphi_{3}-\varphi_{2})}\]  
In the formula: \(\omega_{1}\), \(\omega_{2}\), \(\omega_{3}\) are the angular velocities of \(OA\), \(AB\) and \(BC\), respectively. Here take \(\omega_{1}=125\) r/min.