## 1 Introduction
### Kinematic Analysis of the Spatial RRURR Electric Releasing Mechanism
As shown in Figure 10, the instantaneous motion of each joint can be represented by corresponding motion screws \(\Figure 8:0_{B},\Figure 8:1_{C^{\prime}},\Figure 8:2_{E}\). 2. Calculate for the input motion screw, motion screw of each joint axis and corresponding transmission force screw in sub-chain 1. Solve the input motion/force transmission performance index of sub-chain 1. The motion screw space of sub-chain 1 in the base coordinate system of \(O-x_{0}y_{0}z_{0}\) is  
\[\$_{1}=\left\{ \begin{aligned} \$_{A}&=(0,0,1; \overset{\rightarrow}{OA}\times(0,0,1))\\ &=(0,0,1;0,0,0)\\ \$_{B}&=(0,0,1;\overset{\rightarrow}{OB}\times(0,0,1))\\ &=(0,0,1;a_{1}sin\theta_{1},-a_{1}cos\theta_{1},0)\\ \$_{C}&=(0,0,1;\overset{\rightarrow}{OC}\times(0,0,1))\\ &=(0,0,1;a_{2}sin\theta_{3}+a_{1}sin\theta_{1},a_{2}cos\theta_{3} -a_{1}cos\theta_{1},0)\end{aligned} \right. \tag{8}\]  
where  
\[\left\{ \begin{aligned} \overset{\rightarrow}{OA}&=(0,0,d_{1})\\ \overset{\rightarrow}{OB}&=(a_{1}cos\theta_{1},a_{1}sin \theta_{1},d_{1})\\ \overset{\rightarrow}{OC}&=(-a_{2}cos\theta_{3}+a_{1} cos\theta_{1},a_{2}sin\theta_{3}+a_{1}sin\theta_{1},d_{1})\end{aligned} \right.\]  
The input motion screw of sub-chain 1 is  
\[\$_{I_{1}}=\$_{A}=(0,0,1;0,0,0) \tag{9}\]  
The transmission-force screw, expressed as \((p_{1},p_{2},p_{3};p_{4},p_{5},p_{6})\), of sub-chain 1 is reciprocal to all the passive joint screws in sub-chain 1. The following equations are obtained:  
\[\left\{ \begin{aligned} \$_{T_{1}}\circ\$_{B}=0& \Rightarrow a_{1}sin\theta_{1}p_{1}+(-a_{1}cos\theta_{1})p_{2}+p_{6}=0\\ \$_{T_{1}}\circ\$_{C}=0&\Rightarrow(a_{2}sin\theta_ {3}+a_{1}sin\theta_{1})p_{1}+(a_{2}cos\theta_{3}-a_{1}cos\theta_{1})p_{2}+p_{6 }=0\end{aligned} \right. \tag{10}\]  
Figure 10: Illustrated diagram of the motion screws of the spatial RRURR mechanism. The following equations can be obtained according to Equation (10):  
\[ \begin{cases}-a_{2}sin\theta_{3}p_{1}=a_{2}cos\theta_{3}p_{2}\\ p_{6}=a_{2}cos\theta_{3}p_{2}-a_{2}sin\theta_{3}p_{1}\end{cases}  \tag{11}\]  
The necessary and sufficient condition for the reciprocal product of two lines to be zero is that they are coplanar [24; 25]. Also, the transmission-force screw should act on the joints [26]. It can be deduced that the transmission-force screw of sub-chain 1 passes through the center of joint C, is perpendicular to the axis of \(\$_{B}\) and \(\$_{C}\), and can be expressed as  
\[\$_{T_{1}}=(a_{2}cos\theta_{3},-a_{2}sin\theta_{3},0;d_{1}a_{2}sin\theta_{3}, d_{1}a_{2}cos\theta_{3},-a_{1}a_{2}sin\theta_{1}cos\theta_{3}-a_{1}a_{2}cos \theta_{1}sin\theta_{3}) \tag{12}\]  
where \((a_{2}cos\theta_{3},-a_{2}sin\theta_{3},0)\) is the direction of vector BC. Substituting Equations (5), (9) and (12) into Equation (6), the input motion/force transmission index of sub-chain 1 is calculated as follows:  
\[ \begin{array}{l}\gamma_{I_{1}}=\frac{|\$_{T_{1}}\circ\$_{S_{I_{1}}}|}{|\$_{S_ {T_{1}}\circ\$_{S_{I_{1}}}|max}}\\ =\frac{|-a_{1}a_{2}sin\theta_{1}cos\theta_{3}-a_{1}a_{2}cos\theta_{1}sin\theta _{3}|}{|-a_{1}a_{2}sin\theta_{1}cos\theta_{3}-a_{1}a_{2}cos\theta_{1}sin\theta _{3}|max}\\ =|sin(\theta_{2})|\end{array}  \tag{13}\]  
From Equations (5) and (13), we can observe that \(\gamma_{I_{1}}\) is a function of \(\theta_{1}\). 3. Invert the moving/fixed platforms and transfer the drive. Calculate for the input motion screw, motion screw of each joint axis and corresponding transmission force screw in-verted sub-chain 2. Solve the input motion/force transmission performance index of inverted sub-chain 2. Driving transference is achieved by transferring the motion/force of joint \(R_{A}\) along sub-chain 1 and the moving platform to joint \(R_{C^{\prime}}\) using Equation (5): \[\theta_{4}=n\pi+(-1)^{n}arcsin\frac{K_{3}}{K_{2}}-\phi_{1}=f(\theta_{1})\]  
The motion screw space of inverted sub-chain 2 in the base coordinate system is  
\[\$_{2}=