## 4 The Efficiency Analysis of the Matching Pattern A  
### Theoretical Calculation  
Through the above analysis, the best matching pattern is established. In this section, the efficiency of pattern a is analyzed concretely. Two schemes of pattern a have two power transmitting modes, as shown in the Figure 14. The nodal power method can be used to establish their efficiency formulas. In the Mode I in Figure 14a, the power transmitting has no recirculation. Both the hydrostatic CVT power, \(P_{CVT}\), and the mechanical transmission power, \(P_{MT}\), are forward transferring, i.e., \(-1<\rho<0\). The efficiency  
Figure 14: Power transmitting Mode: (**a**) Mode \(I\); (**b**) Mode _II_. \[\eta_{I}=\left|\frac{P_{CVT}\cdot\eta_{CVT}\eta_{p}\eta_{1}\eta_{10}^{2}+P_{MT} \cdot\eta_{2}\eta_{20}^{1}}{P_{e}}\right| \tag{30}\]  
The following formula can be concluded by using Equations (15) and (30):  
\[\eta_{I}=\left|\frac{i_{CVT}i_{PC}i_{2}\cdot\eta_{CVT}\eta_{p}\eta_{1}\eta_{1 0}^{2}+i_{1}i_{p}(1-i_{PC})\cdot\eta_{2}\eta_{20}^{1}}{i_{CVT}i_{PC}i_{2}-i_{1 }i_{p}i_{PC}+i_{1}i_{p}}\right| \tag{31}\]  
In the Mode II in Figure 14b, the power transmitting has recirculation of hydrostatic CVT power. The mechanical transmission power, \(P_{MT}\), is forward transferring, i.e., \(\rho>0\). The efficiency formula of Mode II is:  
\[\eta_{II}=\frac{|P_{o}|}{P_{e}}=\left|\frac{\eta_{2}\eta_{20}^{1}}{i_{CVT}i_{ PC}i_{2}-i_{1}i_{p}i_{PC}+i_{1}i_{p}}\right[i_{1}i_{p}(1-i_{PC})+\frac{i_{CVT}i_{ PC}i_{2}}{\eta_{CVT}\eta_{p}\eta_{1}\eta_{20}^{o}}\left|\right] \tag{32}\]  
where \(\eta_{I}\) is the efficiency of Mode I, \(\eta_{II}\) is the efficiency of Mode II, \(\eta_{CVT}\) is hydrostatic CVT efficiency, \(\eta_{1}\) is the efficiency of gear set 1, \(\eta_{2}\) is the efficiency of gear set 2, \(\eta_{p}\) is the efficiency of gear set p, \(\eta_{21}^{o}\) is the efficiency between shaft 2 and shaft 1 of PG, \(\eta_{10}^{2}\) is the efficiency between shaft 1 and output shaft of PG, and \(\eta_{20}^{1}\) is the efficiency between shaft 2 and output shaft of PG. As shown in Figure 14, there are efficiency problems in Variable Displacement Hydraulic Pump (VP), Fixed Displacement Hydraulic Motor (FM), the control valve group, and the connecting pipes, which affect the efficiency of hydrostatic CVT. Therefore, the volumetric efficiency of hydrostatic CVT is a dynamic changing process. Without regard to the efficiency loss of connecting pipes and the control valve group, the efficiency calculation is simplified as follows:  
\[\eta_{CVT}^{p}=\eta_{p}^{p}\cdot\eta_{M}^{p};\ \eta_{CVT}^{m}=\eta_{p}^{m} \cdot\eta_{M}^{m};\ \eta_{CVT}=\eta_{p}^{l}\cdot\eta_{M}^{l}=\eta_{p}^{p}\eta_{I}^{m} \cdot\eta_{M}^{m}\eta_{M}^{p} \tag{33}\]  
where \(\eta_{CVT}^{p}\) is volumetric efficiency of hydrostatic CVT, \(\eta_{CVT}^{m}\) is mechanical efficiency of hydrostatic CVT, \(\eta_{P}^{p}\) is volumetric efficiency of VP, \(\eta_{P}^{m}\) is mechanical efficiency of VP, \(\eta_{P}^{t}\) is total efficiency of VP, \(\eta_{M}^{p}\) is volumetric efficiency of FM, \(\eta_{M}^{m}\) is mechanical efficiency of FM, and \(\eta_{M}^{l}\) is total efficiency of FM. The HMCVT efficiency of Schemes 1 and 3 is as follows:  
As seen Schemes 1 and 3 in Table 1, \(i_{p}=1\). Let \(i_{CVTM}=1\), \(i_{CVTm}=-1\), then \(i_{1}=i_{2}=i\), \(\eta_{p}=1\), \(\eta_{1}=\eta_{2}=0.97\). In Scheme 1, when \(0\leq i_{CVT}\leq 1\), \(-1<\rho<0\), power transmitting mode is Mode I. When \(-1\leq i_{CVT}\leq 0\), \(\rho>0\), power transmitting mode is Mode II. The change rule of Scheme 3 is contrary to Scheme 1. The efficiency of the mechanical gear transmission is usually calculated by meshing power method. The efficiency of outer gearing is 0.97, and the efficiency of inner gearing is 0.98. Then, the HMCVT efficiency of Scheme 1, \(\eta_{S1}\), and Scheme 3, \(\eta_{S3}\), in Table 1 are:  
\[\eta_{S1}=\left\{