## 4 Calculation Method of Equal Load Factor for Dual-Input Counter-Rotating Gearbox  
### Calculation of the Elastic Deflection Angle of the Gear Shaft System  
For the dual-input counter-rotating transmission system, when the two input torques are the same, the load difference between G\({}_{\rm f1}\), G\({}_{\rm r1}\), and G\({}_{\rm o2}\) is neglected because the cumulative error between G\({}_{\rm f1}\), G\({}_{\rm r1}\), and G\({}_{\rm o2}\) is small, the two gears are very close to the input motor, and the deformation of the shaft system is small. During the loading process, one of the two idler pulleys enters into engagement first. Due to the influence of manufacturing and installation errors, there is a gap between the idler pulley that enters into engagement later and G\({}_{\rm o1}\), and with the increase in the input torque T, the gap is gradually eliminated through the elastic deformation of the engagement sub and input shaft system. Then, all idler pulleys enter into engagement. The rotation angle of G\({}_{\rm f1}\) and G\({}_{\rm r1}\) is \(\theta_{\rm GII}=\theta_{Gr1}\). The rotation angles of G\({}_{\rm f2}\) and G\({}_{\rm r2}\) are \(\theta_{\rm GIZ}\) and \(\theta_{Gr2}\), respectively, while the G\({}_{\rm o1}\) and G\({}_{\rm o2}\) rotation angles are \(\theta_{\rm G01}\)and\(\theta_{\rm G02}\), respectively. The left and right input shaft torsion angles are \(\theta_{\rm nll}\)and\(\theta_{\rm nrr}\), respectively. The gear meshing deformation and rotation angle were calculated as shown in Equation (18). The torsion angle formed by the torsional deformation of the shaft was calculated as shown in Equation (19). The equations for the static balance of the input shaft, idler  
\begin{table}
\begin{tabular}{c c c c c} \hline \hline \multirow{2}{*}{**N**} & \multicolumn{2}{c}{**Closed-Loop z-Translation Coordinates/mm**} & \multicolumn{2}{c}{**Closed-Loop z-Rotation Coordinates/mm**} & \multirow{2}{*}{_jn/mm_} \\  & \(FR_{1w}\) & \(FR_{2w}\) & \(FR_{1\delta}\) & \(FR_{2\delta}\) & \\ \hline
1000 & \((-0.405,\,0.391)\) & \((-0.047,\,0.048)\) & \((-0.0059,\,0.0062)\) & \((-  
shaft, and output shaft 1 are shown in Equation (20). The theoretical normal backlash could be converted into a circumferential backlash with Equation (21). \[\left\{ \begin{array}{l}T_{\text{G1I}}=(\theta_{\text{G1I}}-\theta_{\text{G02}} )K_{1}\cdot r_{\text{G1I}}{}^{2}\\ T_{\text{G1I2}}=(\theta_{\text{G12}}-\theta_{\text{G01}}-\theta_{nl}-j_{lt})K_{2} \cdot r_{\text{B}\text{G12}}{}^{2}\\ T_{\text{G1r1}}=(\theta_{\text{G1r1}}-\theta_{\text{G02}})K_{1}\cdot r_{\text{B }\text{G1r1}}{}^{2}\\ T_{\text{G72}}=(\theta_{\text{G72}}-\theta_{\text{G01}}-\theta_{mr}-j_{lr})K_{2} \cdot r_{\text{B}\text{G72}}{}^{2}\\ \theta_{\text{G01}}=\theta_{\text{G12}}/i;\theta_{\text{G02}}=\theta_{\text{G11 }}/i\\ \end{array} \right. \tag{18}\]  
\[\theta_{n}=\frac{32T_{\text{G1I}}{}^{2}l}{\pi Gd^{4}} \tag{19}\]  
\[\left\{ \begin{array}{l}T_{3}=T_{\text{G11}}+T_{\text{G71}}\\ T_{\text{G11}}=T_{\text{G02}}\cdot i\cdot\text{$T_{\text{G71}}$}=T_{\text{G02}} \cdot i\\ T_{1}=T_{\text{G01}}+T_{\text{G02}}\\ T_{\text{G4}}=T_{\text{G12}}+T_{\text{G72}}\\ T_{\text{G12}}=T_{\text{G01}}\cdot i\cdot\text{$T_{\text{G72}}$}=T_{\text{G01}} \cdot i\\ \end{array} \right. \tag{20}\]  
\[jt=\frac{jn}{\cos(\beta_{b})\cdot\cos(\alpha)\cdot r_{bG01}} \tag{21}\]  
where \(i\) is the transmission ratio, \(K_{1}\) is the average meshing stiffness of gears G\({}_{o2}\) and G\({}_{l1}\), and \(K_{2}\) is the average meshing stiffness of gears G\({}_{o1}\) and G\({}_{l2}\). \(T_{\text{G11}}\) is the torque applied to gear G\({}_{l1}\), and \(r_{b\text{G11}}\) is the radius of the base circle of the gear. \(T_{\text{G12}}\) is the torque applied to gear G\({}_{l2}\) and \(r_{b\text{G12}}\) is the radius of the base circle of the G\({}_{l2}\).