## 1 Introduction
### Kinematic Analysis of the Spatial RRURR Electric Releasing Mechanism  
A kinematic analysis of the electric releasing RRURR mechanism of a vehicle side-door latch is performed to solve the relationship between the joint angle, \(\theta_{i}\), and the input angle, \(\theta_{1}\), laying the foundation for the calculation and analysis of the motion/force transmission performance index of the mechanism. In the electric releasing condition, the kinematic model of the RRURR mechanism is shown in Figure 7, where \(\theta_{1}\) is the input angle and the angles, \(\theta_{i}(i=2\dots 6)\), of each joint axis in the single-loop mechanism are to be calculated. This establishes the D-H coordinate system corresponding to each axis of the kinematic joints in the single-loop spatial RRURR mechanism, where \(Z_{0}\), \(Z_{1}\), and \(Z_{2}\) are parallel to each other; \(Z_{3}\), \(Z_{4}\), and \(Z_{5}\) are parallel to each other; and \(Z_{0}\) is perpendicular to \(Z_{5}\), i.e., \(\alpha_{1}=\alpha_{2}=\alpha_{4}=\alpha_{5}=0\), and \(\alpha_{3}=\alpha_{6}=-\pi/2\). The joint offsets are \(d_{2}=d_{3}=0\). The bar length is \(a_{3}=0\). To solve for the joint variables, the mechanism is disconnected at joint C; the equation of the closed-loop can be written using the D-H matrix:  
\[T_{0}^{1}T_{1}^{2}T_{2}^{3}=T_{0}^{5}T_{5}^{4}T_{4}^{3} \tag{3}\]  
From Equation (3), elements of the second, third, and fourth columns of the homogeneous transformation matrix are equal on both sides, and the RRURR kinematic equation system can be established:  
\[ \begin{cases}sin(\theta_{4}+\theta_{5}+\theta_{6})=0\\ cos(\theta_{4}+\theta_{5}+\theta_{6})=-1\\ sin(\theta_{1}+\theta_{2}+\theta_{3})=0\\ cos(\theta_{1}+\theta_{2}+\theta_{3})=-1\\ a_{2}cos(\theta_{1}+\theta_{2})+a_{1}cos\theta_{1}=-a_{6}-a_{5}cos\theta_{6}-a _{4}cos(\theta_{5}+\theta_{6})\\ a_{2}sin(\theta_{1}+\theta_{2})+a_{1}sin\theta_{1}=d_{6}+d_{5}+d_{4}\\ d_{1}=a_{4}sin(\theta_{5}+\theta_{6})+a_{5}sin\theta_{6}\\ \end{cases}  \tag{4}\]  
The solution is obtained:  
\[ \begin{cases}\theta_{3}=n\pi+(-1)^{n}arcsin\frac{-a_{1}sin\theta_{1}+d_{4}+d_ {5}+d_{6}}{a_{2}}\\ \theta_{2}=n\pi-\theta_{1}-\theta_{3}\\ \theta_{5}=2n\pi\pm arccos(\frac{K_{1}^{2}+d_{1}^{2}-d_{1}^{2}-d_{2}^{2}}{2a _{4}d_{5}})\\ \theta_{4}=n\pi+(-1)^{n}arcsin\frac{K_{3}}{K_{2}}-\phi_{1}\\ \theta_{6}=n\pi-\theta_{4}-\theta_{5}\\ \end{cases}  \tag{5}\]  
Figure 7: Schematic diagram of the spatial RRURR mechanism kinematic model. where  
\[ \begin{array}{l}K_{1}=-a_{1}cos\theta_{1}+a_{2}cos\theta_{3}\pm a_{6}\\ K_{2}=\left(U_{1}^{2}+V_{1}^{2}\right)^{\frac{1}{2}}\\ K_{3}=a_{1}cos\theta_{1}-a_{2}cos\theta_{3}\pm a_{6}\\ U_{1}=a_{5}cos\theta_{5}+a_{4}\\ V_{1}=-a_{5}sin\theta_{5}\\ \phi_{1}=arctan(U_{1}/V_{1})\end{array} \]  
The analytical expressions between the joint angles, \(\theta_{i}(i=2\dots 6)\), and the input angle, \(\theta_{1}\), are obtained by Equation (5). Construction and Application of Motion/Force Transmission Performance Index of the Single-Degree-of-Freedom Single-Loop Mechanism  
The equivalent matching methods for the moving/fixed platform inversion and chain-driving transference are proposed. Based on equivalent matching methods and the input motion/force transmission performance index of the parallel mechanism, the overall motion/force transmission performance index for single-degree-of-freedom single-loop mechanisms is constructed, and the corresponding calculation procedure is defined. Through the forward and inverse solutions of the mechanism kinematics, the overall motion/force transmission performance of the electric releasing RRURR mechanism is analyzed and a performance atlas is drawn. Construction of Overall Motion/Force Transmission Performance Index for Single-Degree-of-Freedom Single-Loop Mechanisms Based on Input Transmission Performance Index of the Parallel Mechanism  
The parallel mechanism transmits motion/force from the fixed platform to the moving platform through the drive of each chain, as shown in Figure 8a.