## 5 Analysis and Calculation of Power Loss and Efficiency of a Two-State Logic Transmission

Table 2: Calculation of speed and gear ratio of each component of a two-state logical planetary transmission. Among \(k_{2}=k_{21}k_{22}\), \(k_{4}=k_{41}k_{42}\) and \(k_{5}=k_{51}k_{52}\), it is also the characteristic parameter value of each single planet row. So, the dynamic transmission ratio is calculated by \(\widetilde{i}=\overline{k_{2}k_{4}}(1+\overline{k_{5}})\). \[\frac{\partial\ln i}{\partial k_{2}}=\frac{\partial}{\partial k_{2}}\ln\left[ \overline{k_{2}}\;\overline{k_{4}}\Big{(}1+\overline{k_{5}}\Big{)}\right]= \frac{\partial}{\partial k_{2}}[\ln k_{2}+\ln k_{4}+\ln(1+k_{5})]=\frac{1}{k_{2}} \tag{12}\]  
\[\frac{\partial\ln i}{\partial k_{4}}=\frac{\partial}{\partial k_{4}}\ln\left[ \overline{k_{2}}\;\overline{k_{4}}\Big{(}1+\overline{k_{5}}\Big{)}\right]= \frac{\partial}{\partial k_{4}}[\ln k_{2}+\ln k_{4}+\ln(1+k_{5})]=\frac{1}{k_{4}} \tag{13}\]  
\[\frac{\partial\ln i}{\partial k_{5}}=\frac{\partial}{\partial k_{5}}\ln\left[ \overline{k_{2}}\;\overline{k_{4}}\Big{(}1+\overline{k_{5}}\Big{)}\right]= \frac{\partial}{\partial k_{5}}[\ln k_{2}+\ln k_{4}+\ln(1+k_{5})]=\frac{1}{1+k _{5}} \tag{14}\]  
\[\widetilde{i}=k_{2}\Big{(}\eta^{2}\Big{)}^{x_{2}}k_{4}\Big{(}\eta^{2}\Big{)}^ {x_{4}}\big{(}1+k_{5}(\eta_{in}\eta_{out})^{x_{5}}\big{)}=k_{2}\eta^{2}k_{4} \eta^{2}(1+k_{5}\eta_{in}\eta_{out}) \tag{15}\]  
\[\eta_{1}=\frac{\widetilde{i}}{i}=\frac{k_{2}\eta^{4}k_{4}(1+k_{5}\eta_{in} \eta_{out})}{k_{2}k_{4}(1+k_{5})}=\frac{\eta^{4}(1+k_{5}\eta_{in}\eta_{out})} {(1+k_{5})} \tag{16}\]  
According to Equations (12)-(15), the partial derivative of \(k\) is greater than 0, and \(x_{2}\), \(x_{4}\) and \(x_{5}\) are equal to 1, and then, \(\widetilde{i}\) is obtained. Finally, the final expression is obtained according to the Cleinus efficiency formula. When the gear meshing efficiency is calculated using high coincidence double internal meshing gears, \(\eta\) = 0.985 (0.99), \(\eta_{in}\) = 0.98 (ordinary internal meshing) and \(\eta_{out}\) = 0.97 (ordinary external meshing). The efficiency of the calculated gears (the first five as an example) is shown in the Table 3 below. The efficiency of each gear of the aforementioned dual-state logic planetary gearbox is plotted as a general efficiency diagram as follows. In the Figure 5, the red curve is the gear efficiency of the ordinary planetary gear and the blue curve is the gear efficiency of the two-state logical planetary transmission, which improves the gear efficiency of the ordinary planetary gear transmission by an average of 2.6%. \begin{table}
\begin{tabular}{c c c c c} \hline \hline \multirow{2}{*}{**Gear**} & \multirow{2}{*}{**Transmission Ratio**} & \multirow{2}{*}{**Efficiency Expression**} & \multicolumn{2}{c}{**Efficiency**} \\ \cline{3-4}  & & & \(\eta\) = 0.985, \(\eta_{\text{out}}\) = 0.98 & \(\eta\) = 0.99, \(\eta_{\text{out}}\) = 0.98 \\  & & & \(\eta_{\text{in}}\) = 0.97 & \(\eta_{\text{in}}\) = 0.97 \\ \hline
1 & k\({}_{2}\) k\({}_{4}\) (1 + k\({}_{5}\)) & \(\eta^{4}(1+k\)\({}_{5}\)) & 0.9043 & 0.9228 \\
2 & k\({}_{2}\) k\({}_{4}\) (1 + k\({}_{5}\))/k\({}_{1}\) & \(\eta^{6}(1+k\)\({}_{5}\)) & 0.8774 & 0.9044 \\
3 & k\({}_{4}\) (1 + k\({}_{5}\)) & \(\eta^{2}(1+k\)\({}_{5}\)) & 0.9321 & 0.9416 \\
4 & k\({}_{4}\) (1 + k\({}_{5}\))/k\({}_{1}\) & \(\eta^{4}(1+k\)\({}_{5}\)) & 0.9043 & 0.9228 \\
5 & k\({}_{2}\) k\({}_{4}\) (1 + k\({}_{5}\))/k\({}_{3}\) & \(\eta^{6}(1+k\)\({}_{5}\)) & 0.8774 & 0.9044 \\ \hline \hline \end{tabular}
\end{table} 
Table 3: Efficiency of the first five gears of a two-state logic planetary transmission. Figure 5: Total efficiency of dual-state logic planetary gearbox. Compared with other studies, the efficiency loss of gear six in this paper is the largest, which is due to the large number of planetary rows involved in the working condition. Specific solutions include optimizing design parameters and changing the number of planetary rows.