# Mechanical Efficiency of HMCVT under Steady-State Conditions

Authors: Guangqing Zhang; Hengtong Zhang; Yanyan Ge; Wei Qiu; Maohua Xiao; Xiaomei Xu; Minghui Zhou
Journal: Shock and Vibration
 (2021)

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## Abstract

Hydromechanical continuously variable transmission (HMCVT) technology has been widely used due to its advantages of ride comfort and fuel economy. The relatively uniform efficiency expression of HMCVT is obtained by studying torque and transmission ratios to reveal steady-state characteristics and predict the output torque. Mathematical models of torque ratios are derived by analyzing the HMCVT system power flow and calculating the equivalent meshing power of epicyclic gear train and efficiency for the hydraulic system. The relationship between mechanical system transmission and hydraulic system parameters is established using the torque ratios, and a mechanical system demanding surface is proposed. Two numerical examples of the HMCVT system with single and dual variable units are demonstrated to establish an effective and convenient method. The method is validated through a physical prototype TA1-02 test.


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## Body

## 1. Introduction

An epicyclic gear train (EGT) with two DOFs is closed by a hydrostatic transmission system (HST), so that only one DOF exists overall [1]. The mechanical hydraulic components are organically combined to form a simple hydromechanical continuously variable transmission (HMCVT) [2, 3]. When the input speed of HMCVT is stable, with displacement ratio e of HST changing continuously in a certain range, the output speed of HMCVT continuously changes from the minimum speed (e.g., 0) to the maximum speed. Multistage EGTs or two EGTs with multiple segments (fixed transmission ratio) are often used in HMCVT design to obtain a large transmission ratio range. By reasonably setting the system parameters of EGTs and HST, the multistage shifting transmission ratios are almost the same to reduce the impact. In the shifting process, shift quality is improved by using the control strategy of nonspeed difference. HMCVT technology, with its characteristics of ride comfort and fuel economy, has been widely studied and applied [4–8].Wang et al. [9], in his research on the mechanical efficiency of HMCVT, performed a theoretical reasoning and calculation of the mechanical efficiency of HMCVT and made a comparative analysis with an improved scheme. Cheng et al. [10] studied, based on the improved simulated annealing algorithm, the efficiency model of the hybrid continuously variable transmission. Li et al. [11, 12] analyzed the power distribution of compound and closed planetary gear transmissions and calculated the overall transmission efficiency with and without considering the power loss. The results showed that graphical representation is a practical power analysis and efficiency calculation method that can provide a theoretical reference for the design of complex closed planetary gear transmissions. Awadallah et al. [13] established a simulation model based on the mathematical model of conventional and mild hybrid power systems and studied its transmission efficiency through simulation. Although simulation studies have high work efficiency and reliability, they lack experimental verification. In addition, transmission efficiency has consistently been the focus of research on EGT [14–17]. HMCVT has important features and advantages because of the organic combination of EGTs and HST. In the HST system [18, 19], when the working pressure Δp is improved, the mechanical efficiency ηmh of the system is gradually increased, whereas the volumetric efficiency ηv of the system is reduced. This changing regularity directly affects the working characteristics of HMCVT. Mechanical efficiency ηmh affects the output torque ratio of the hydraulic system, and volumetric efficiency ηv affects the output angular velocity ratio. Maintaining HMCVT’s continuously variable transmission ratio in the HST system involves two aspects, namely, mechanical efficiency ηmh and volumetric efficiency ηv, both of which affect the overall efficiency of HMCVT. The effects of mechanical efficiency ηmh and volumetric efficiency ηv on the overall efficiency of HMCVT are studied to obtain a relatively uniform efficiency expression. Through numerical examples, this study provides a method to analyze the overall efficiency of HMCVT.


## 2. Parameter Matching

In accordance with the role of EGTs in HMCVT, HMCVTs with three active shafts [1] are divided into input coupled planetary (summing planetary) and output coupled planetary (divider planetary). The single variable hydraulic unit of HST is selected as the foundation of this research to simplify the derivation of the relationship between system parameters. In the system efficiency analysis (Section 5.2), the double-variable hydraulic units of HST are adopted to establish a method of improving system efficiency. As shown in Figure 1, an HMCVT with three active shafts and input coupled planetary is investigated in this work. The power splitting point lies in the intersection of link 1 and shaft I. In accordance with the difference in torques, the shaft throughout transmission system is divided into three links, namely, shaft I, link 1, and link 5. The power merging point is located between links 4 and 8. Angular velocities ωL (ω6) and ωH (ω7) present a low output speed and a high output speed of HMCVT, respectively. In Figure 1, Zi (i = 1, 2, 3, 4, s, s′, p, p′, and r) is the tooth number for each meshing gear. Zc is the equivalent tooth number of planetary carriers according to the installation diameter of the planetary gears.Figure 1
Sketch of the transmission system.Links 4, 5, 6, 7, and 8 comprise two-stage EGTs (the mechanical path). Link 6 is the planet carrier of EGTs, and link 7 is the sun gear of second-stage EGT. HST (the variable path) is composed of constant displacement (qc) hydraulic unit 2 and variable displacement (qv) hydraulic unit 1, with the former being connected to link 3. Links 1, 2, 3, and 4 comprise the internal transmissions with a fixed ratio.


### 2.1. Transmission Ratio Calculation

In the transmission scheme shown in Figure1, the angular velocities of links 1 and 5 are equal to the input angular velocity.(1)ωI=ω1=ω5,where ωi denotes the actual angular velocity of the ith link. Displacement ratio e is equal to the displacement of the variable hydraulic unit (qv) divided by the displacement of the constant unit (qc), i.e., e = qv/qc. The angular velocity ratio of link 3 to link 2 is called the transmission ratio τH of the variable path. Given that the constant hydraulic unit acts as a motor or a pump, two expressions can be established as follows [1]:(2)τH=ω3ω2=qvηvqc=eηv,ΙΙΙmotor,qvqcηv=eηv,Ipump.The angular velocity ratio of link 2 to link 1 is called the fixed transmission ratioτF1, i.e., τF1 = −Z1/Z2. The angular velocity ratio of link 4 to link 3 is called the fixed transmission ratio τF2, that is, τF2 = −Z3/Z4. By using τH′ to represent the total transmission ratio of parameters τF1, τF2 and τH, we obtain(3)τH′=τHτF1τF2=eηvZ1Z3Z2Z4,III,eZ1Z3Z2Z4ηv,I.The relative angular velocity of the ith link,ωij, in a reference frame (the jth link) rotating with angular velocity ωj is defined as(4)ωij=ωi−ωj.In the EGTs,τPG and τPG′ are defined as the ratio of ring gear angular velocity to sun gear angular velocity when the planetary carrier is fixed for the first and second EGTs, respectively,(5)τPG=ω46ω56=−ZsZp′ZpZr,(6)τPG′=ω46ω76=−Zs′Zr.In the HMCVT,τHM and τHM′ are defined as the ratio of output angular velocity to input angular velocity when ωL and ωH are required, respectively,(7)τHM=ωLωI=τPG−τH′τPG−1,(8)τHM′=ωHωI=τPG′−τPGτH′+τPG1−τPG′τPG′1−τPG,where equations (3), (5), and (6) are utilized. We let k = (1 − τPG) − 1 and k′ = (τPG′- − τPG) (τPG′ − τPG′τPG) − 1; then, equations (7) and (8) can be rewritten as(9)τHM=kτH′+1−k,(10)τHM=kτH′+1−k′.


### 2.2. Transmission Ratio Requirement

With parameterkH to show the range of variable τH′ and parameter λ to represent the asymmetry coefficient of τH′ (kH > 0, λ > 0), when τH′ increases from −kH to λkH, τHM increases gradually within [0, τHM0], where τHM0 is the maximum of τHM.WhenτH′=−kH and τHM=0, we can deduce from equation (7) that τPG=−kH. When τH′=λkH and τHM=τHM0, we can deduce from equation (7) that τHM0=1+λkH/1+kH.WhenτH′ decreases from [−kH, λkH], τHM′ increases gradually within [τHM0, τHMmax], where τHMmax is the maximum of τHM′τHM0<τHMmax. When τHM′=τHM0, we can deduce from equation (8) that(11)τHM0=1,kH=1λ.We can obtain the expression ofτHM′ from equation (8) by using equations (10) and (11). tHM′=tHMmax, and τPG′=−1/τHMmax.WhenτPG=−1/λ and τPG′=−1/τHMmax, τH′ gradually increases in the range of −1/λ,1. Correspondingly, transmission ratio τHM increases from 0 to 1, and transmission ratio τHM′ decreases from τHMmax to 1. We can deduce from equations (9) and (10) that(12)k=−1τPG−1=λλ+1,0<k<1,k′=τPG′−τPGτPG′1−τPG=−λλ+1τHMmax−1k′<0.An example is provided throughout the paper. WhenτF1τF2=1, λ=1 and τHMmax=2.7, with the above mentioned equations, we can obtain(13)τPG=−1,τPG′=−0.37,(14)τHM=0.5τH′+0.5,τHM′=−0.85τH′+1.85.The tooth numbers of the EGTs fitting equations (13) and (14) are summarized in Table 1.Table 1
Tooth numbers of the planetary gear train.
ZsZrZcZpZs′Zp′489263153429


## 2.1. Transmission Ratio Calculation

In the transmission scheme shown in Figure1, the angular velocities of links 1 and 5 are equal to the input angular velocity.(1)ωI=ω1=ω5,where ωi denotes the actual angular velocity of the ith link. Displacement ratio e is equal to the displacement of the variable hydraulic unit (qv) divided by the displacement of the constant unit (qc), i.e., e = qv/qc. The angular velocity ratio of link 3 to link 2 is called the transmission ratio τH of the variable path. Given that the constant hydraulic unit acts as a motor or a pump, two expressions can be established as follows [1]:(2)τH=ω3ω2=qvηvqc=eηv,ΙΙΙmotor,qvqcηv=eηv,Ipump.The angular velocity ratio of link 2 to link 1 is called the fixed transmission ratioτF1, i.e., τF1 = −Z1/Z2. The angular velocity ratio of link 4 to link 3 is called the fixed transmission ratio τF2, that is, τF2 = −Z3/Z4. By using τH′ to represent the total transmission ratio of parameters τF1, τF2 and τH, we obtain(3)τH′=τHτF1τF2=eηvZ1Z3Z2Z4,III,eZ1Z3Z2Z4ηv,I.The relative angular velocity of the ith link,ωij, in a reference frame (the jth link) rotating with angular velocity ωj is defined as(4)ωij=ωi−ωj.In the EGTs,τPG and τPG′ are defined as the ratio of ring gear angular velocity to sun gear angular velocity when the planetary carrier is fixed for the first and second EGTs, respectively,(5)τPG=ω46ω56=−ZsZp′ZpZr,(6)τPG′=ω46ω76=−Zs′Zr.In the HMCVT,τHM and τHM′ are defined as the ratio of output angular velocity to input angular velocity when ωL and ωH are required, respectively,(7)τHM=ωLωI=τPG−τH′τPG−1,(8)τHM′=ωHωI=τPG′−τPGτH′+τPG1−τPG′τPG′1−τPG,where equations (3), (5), and (6) are utilized. We let k = (1 − τPG) − 1 and k′ = (τPG′- − τPG) (τPG′ − τPG′τPG) − 1; then, equations (7) and (8) can be rewritten as(9)τHM=kτH′+1−k,(10)τHM=kτH′+1−k′.


## 2.2. Transmission Ratio Requirement

With parameterkH to show the range of variable τH′ and parameter λ to represent the asymmetry coefficient of τH′ (kH > 0, λ > 0), when τH′ increases from −kH to λkH, τHM increases gradually within [0, τHM0], where τHM0 is the maximum of τHM.WhenτH′=−kH and τHM=0, we can deduce from equation (7) that τPG=−kH. When τH′=λkH and τHM=τHM0, we can deduce from equation (7) that τHM0=1+λkH/1+kH.WhenτH′ decreases from [−kH, λkH], τHM′ increases gradually within [τHM0, τHMmax], where τHMmax is the maximum of τHM′τHM0<τHMmax. When τHM′=τHM0, we can deduce from equation (8) that(11)τHM0=1,kH=1λ.We can obtain the expression ofτHM′ from equation (8) by using equations (10) and (11). tHM′=tHMmax, and τPG′=−1/τHMmax.WhenτPG=−1/λ and τPG′=−1/τHMmax, τH′ gradually increases in the range of −1/λ,1. Correspondingly, transmission ratio τHM increases from 0 to 1, and transmission ratio τHM′ decreases from τHMmax to 1. We can deduce from equations (9) and (10) that(12)k=−1τPG−1=λλ+1,0<k<1,k′=τPG′−τPGτPG′1−τPG=−λλ+1τHMmax−1k′<0.An example is provided throughout the paper. WhenτF1τF2=1, λ=1 and τHMmax=2.7, with the above mentioned equations, we can obtain(13)τPG=−1,τPG′=−0.37,(14)τHM=0.5τH′+0.5,τHM′=−0.85τH′+1.85.The tooth numbers of the EGTs fitting equations (13) and (14) are summarized in Table 1.Table 1
Tooth numbers of the planetary gear train.
ZsZrZcZpZs′Zp′489263153429


## 3. Power Flows in the Absence of Losses

### 3.1. Angular Velocity Ratios in the Absence of Losses

In the example mentioned, considering the absence of losses,τH′=τH=e. By using equations (9) and (10), the relationship between output angular velocity ratio τHM of HMCVT and displacement ratio e is shown in Figure 2(a).Figure 2
Relationship between output transmission ratioτHM and displacement ratio e. (a) In the absence of losses. The working area of τHM is formed by dot-dashed and double-dot-dashed lines, and the working area of τHM′ is formed by dashed and long dashed lines. The solid line is |τH′|. HMCVT realizes continuously variable transmission in the range of [0, 2.7], shifting with no speed difference in point (1,1). (b) With the losses calculated by Sts1 in Section 5.2.
(a)(b)


### 3.2. Direction of Power Flows

According to Macmillan [20], the direction of power flow in EGT cannot be modified by meshing friction. Torque equilibrium and power conservation equations are widely used to determine the efficiency of EGT [1, 21]. In this study, the power that flows into EGT is positive, and the outflow is negative.On steady state, directions of the possible power flow [22–24] are those shown in Figure 3 as type I, type II, and type III flow.Figure 3
Power flows in HMCVT. (a) Type I power flow. (b) Type II power flow. (c) Type III power flow.
(a)(b)(c)When power conservation and torque equilibrium equations are applied to EGT, we can obtain the following from Figure3:(15)TOωO+T4ω4+T5ω5=0,(16)TO+T4+T5=0,where Ti is the torque applied on the ith link. Without loss of generality, we can assume that the output shaft of the EGTs is link 7. In this case, by using equations (1) and (10), we obtain(17)ωO=ωIτHM′=ω4k′+ω51−k′,(18)T7−1=T4k′=T51−k′.In type I power flow with output shaft 7 [22], equation (15) is rewritten as(19)T7ω7+T4ω4−T5ω5=0.By substituting equations (10), (18), and (17) in equation (19), we obtain(20)τH′=1−k′k′−1k′τHM′.As indicated in the first quadrant of Figure2(a), equation (10) is rewritten as(21)τHM′=1−k′−k′τH′k′<0τH′>0.In consideration of equations (18) and (21), equation (20) is identical. When the output shaft is link 7, the work area for generating type I power flow is shown in the first quadrant of Figure 2(a) by long dashed lines.The direction of power flow relative to every working condition can be obtained. Figure2(a) shows every type of power flow with symbols I and III.PM is the HST system (the variable path). PG is the two-stage EGTs (the mechanical path). F1 and F2 are the internal transmissions with a fixed ratio. 1, 2, 3, 4, 5, I, and O are the links of HMCVT shown in Figure1.


## 3.1. Angular Velocity Ratios in the Absence of Losses

In the example mentioned, considering the absence of losses,τH′=τH=e. By using equations (9) and (10), the relationship between output angular velocity ratio τHM of HMCVT and displacement ratio e is shown in Figure 2(a).Figure 2
Relationship between output transmission ratioτHM and displacement ratio e. (a) In the absence of losses. The working area of τHM is formed by dot-dashed and double-dot-dashed lines, and the working area of τHM′ is formed by dashed and long dashed lines. The solid line is |τH′|. HMCVT realizes continuously variable transmission in the range of [0, 2.7], shifting with no speed difference in point (1,1). (b) With the losses calculated by Sts1 in Section 5.2.
(a)(b)


## 3.2. Direction of Power Flows

According to Macmillan [20], the direction of power flow in EGT cannot be modified by meshing friction. Torque equilibrium and power conservation equations are widely used to determine the efficiency of EGT [1, 21]. In this study, the power that flows into EGT is positive, and the outflow is negative.On steady state, directions of the possible power flow [22–24] are those shown in Figure 3 as type I, type II, and type III flow.Figure 3
Power flows in HMCVT. (a) Type I power flow. (b) Type II power flow. (c) Type III power flow.
(a)(b)(c)When power conservation and torque equilibrium equations are applied to EGT, we can obtain the following from Figure3:(15)TOωO+T4ω4+T5ω5=0,(16)TO+T4+T5=0,where Ti is the torque applied on the ith link. Without loss of generality, we can assume that the output shaft of the EGTs is link 7. In this case, by using equations (1) and (10), we obtain(17)ωO=ωIτHM′=ω4k′+ω51−k′,(18)T7−1=T4k′=T51−k′.In type I power flow with output shaft 7 [22], equation (15) is rewritten as(19)T7ω7+T4ω4−T5ω5=0.By substituting equations (10), (18), and (17) in equation (19), we obtain(20)τH′=1−k′k′−1k′τHM′.As indicated in the first quadrant of Figure2(a), equation (10) is rewritten as(21)τHM′=1−k′−k′τH′k′<0τH′>0.In consideration of equations (18) and (21), equation (20) is identical. When the output shaft is link 7, the work area for generating type I power flow is shown in the first quadrant of Figure 2(a) by long dashed lines.The direction of power flow relative to every working condition can be obtained. Figure2(a) shows every type of power flow with symbols I and III.PM is the HST system (the variable path). PG is the two-stage EGTs (the mechanical path). F1 and F2 are the internal transmissions with a fixed ratio. 1, 2, 3, 4, 5, I, and O are the links of HMCVT shown in Figure1.


## 4. Efficiency of HMCVT

### 4.1. Torque Ratio of Links with Losses

The pressure efficiencyηp of the HST system is approximately expressed as ηp = pP/pM, where pP and pM are pump and motor operating pressures, respectively. In the HST system, total mechanical efficiency ηmh is expressed as ηmh = ηpηPηM, where ηM and ηP are pump and motor mechanical efficiencies, respectively. The output torque of motor TM and the input torque of pump TP are expressed as(22)TP=qPpP2πηP,TM=qMpMηM2π,where qP and qM are the displacements of the pump and motor, respectively.In type III power flow, hydraulic unit 1 acts as the variable pump with a torque ofTP, and hydraulic unit 2 acts as the motor with a torque of TM. The torque ratio ΔH of HST in type III power flow is defined as TP divided by TM.(23)ΔH=TPTM=T2T3=qPpPqMpMηMηP≈eηMηPηp=eηmh.The torque ratio Δ12 of F1 to F2 (shown in Figure 3; internal transmissions) in type III power flow is defined as(24)Δ12=τF1τF2ηF1ηF2.In type I power flow, hydraulic unit 1 acts as the variable motor with a torque ofTM, and hydraulic unit 2 acts as the pump with a torque of TP. ΔH′ is defined as TP divided by TM.(25)ΔH′=TPTM=T3T2=qPpPqMpMηMηP≈1eηMηPηp=1eηmh.The torque ratioΔ12′ of F1 to F2 in type I power flow is defined as(26)Δ12′=1ηF1ηF2τF1τF2.Given that the transmission losses and the torques applied to the links are unrelated to the observer’s motion, the method of epicyclic inversion was used by Pennestri and Freudenstein [14] and Chen and Liang [21] in different ways.This study uses the transforming mechanism model [21, 25], in which the interaction forces and the relative angular velocity are all the same as the EGTs shown in Figure 1, but the planet carrier is fixed. Thus, the losses in the two models are considered to be the same. The torque ratio of the links based on meshing power [21, 25] is applied to calculate the efficiency of the two-DOF EGTs. Meshing power pic (the ith link relative to planet carrier 6) is a virtual power that is independent of the actual power pi. Meshing power ratio ψi is defined as the ratio of meshing power pic through the ith link to actual power pi through the same link. If power pi flows into EGT, it is defined as positive (negative otherwise). The expressions of ωi6, pi6, pi and ψi are(27)ωi6=ωi−ω6,pi6=Tiωi−ω6=Tiωi6,pi=Tiωi,ψi=pi6pi=Tiωi−ω6Tiωi=ωi6ωi.In the transforming mechanism, the direction ofpic can be determined in accordance with the sign of pi and ψi. If the sign of picis positive, power pic flows into the transforming mechanism; otherwise, it flows out of the system.Without loss of generality, we assume that the output shaft of the HMCVT is link 7 in type I power flow, and the relationship amongωi, ψi and pi between links isω7>ω5>ω6>ω4>0.(28)ψ5=p56p5=T5ω5−ω6T5ω5>0,p5<0,p56>0,ψ4=p46p4=T4ω4−ω6T4ω4<0,p4<0,p46>0,ψ7=p76p7=T7ω7−ω6T7ω7>0,p7<0,p76<0.Under this condition, the meshing power is from links 5 and 4 to link 7 in the transforming mechanism. When the power conservation equation and the torque equilibrium condition are applied to the EGT, we have(29)T4ω46η486η876+T5ω56η586η876+T7ω76=0,T4+T5+T7=0,where ηij6 is the efficiency of the path, in which the meshing power flows from the ith link to the jth link with the planet carrier being fixed in the transforming mechanism.To obtain the relationship between output torque and input torque, by using equations (5) and (6), we can reform equation (29) to(30)T7=T5η486η876τPG′τPG−η586η876τPG′1−η486η876τPG′τPG,T4=−T51+η586η876τPG′τPG−η586η876τPG′1−η486η876τPG′τPG.From equation (28), torque ratios Δ75 and Δ45′ of EGT (shown in Figure 3; the mechanical path) in type I power flow are defined as(31)Δ75=η486η876τPG′τPG−η586η876τPG′1−η486η876τPG′τPG,Δ45′=−1+η486η876τPG′τPG−η586η876τPG′1−η486η876τPG′τPG.When the output shaft of the EGT is link 7 in type III power flow, in reference to the actual angular velocityωi, the relationship between links is ω7>ω5>ω6>0>ω4. The same expression of Δ75 and Δ45′ as that in equation (29) can be obtained because the sign of pic remains similar to that in type I power flow. The other case is that the output shaft of HMCVT is link 6 in types III and I power flows. In this condition, we can obtain the expressions of Δ65 and Δ45 in a similar manner.(32)Δ45=−η586η846τPG,Δ65=η586η846τPG−1.


### 4.2. Efficiency of HMCVT

In the steady state of HMCVT, we assume that output torqueTo of HMCVT is the result of the calculation and not of the actual load, such that the input torque and angular velocity remain similar to those for diesel working under rated conditions. When the power pi contributed to Ti flows into the Split point in Figure 1, Ti is defined as positive; otherwise, it is negative. Here, T5 is negative in type I and III power flows. To maintain consistency with the signs of T5 and T1 for EGTs, when the torque equilibrium condition is applied to the Split point, we have(33)TI−T5−T1=0.Depending on the type of power flow with output shaft 6, two kinds of relationship exist betweenT1 and T5.(34)T1=ΔHΔ45Δ12T,III,−Δ45ΔH′Δ12′T,I.The relationship betweenT5 and TI is(35)T5=TI1+ΔHΔ45Δ12,III,TI1−Δ45/ΔH′Δ12′,I.We obtain the transmission efficiency of the HMCVT with output shaft 6 by using the relationship expressions ofTI with T5 and equation (7) as follows:(36)ηHM=−T6ω6TIωI=−Δ65TIω6/1+ΔHΔ45Δ12TIωI=−Δ65τHM1+ΔHΔ45Δ12,III,−Δ65TIω6/1−Δ45/ΔH′Δ12′TIωI=−Δ65τHM1−Δ45/ΔH′ΔH′,I.The working pressure of the hydraulic system is represented by the working pressure of the fixed displacement hydraulic unit 2.(37)Δp=2πT3qc=2πτF2Δ45T5qcηF2=2πτF2Δ45TIqcηF21+Δ45Δ12ΔH,III,2πτF2Δ45T5ηF2qc=2πτF2Δ45ηF2TIqc1−Δ45/Δ12′ΔH′,I.When the output shaft is link 7, equations (32), (34), and (35) remain the same, but Δ45, Δ65 and τHM are replaced by Δ45′, Δ75, and τHM′, respectively:(38)T1=ΔHΔ45′Δ12T5,III,−Δ45′ΔH′Δ12′T5,I,(39)ηHM′=−T7ω7TIωI=−Δ75τHM′1+ΔHΔ45′Δ12,III,−Δ75τHM′1−Δ45′/ΔH′Δ12′,I,(40)Δ′p=2πT3qc=2πΔ45′τF2TIqcηF21+Δ45′Δ12ΔH,III,2πΔ45′τF2ηF2TIqc1−Δ45′/Δ12′ΔH′I.When the losses from the churning of the lubricating oil and shaft bearing friction are not considered, once the characteristics of the HMCVT have been determined, torque ratios Δ12, Δ12′, Δ75, Δ45′, Δ65, and Δ45 become constant. However, firstly, torque ratios ΔH and ΔH′ vary with the change in mechanical efficiency ηmh and displacement ratio e for the hydraulic system. Secondly, τHM and τHM′ vary with the change in volumetric efficiency ηv and displacement ratio e. Thirdly, volumetric efficiency ηv and mechanical efficiency ηmh change with displacement ratio e and system working pressure Δp. According to equation (37) or (40), the curved surface of ηmh relative to e and Δp is called the mechanical system demanding surface.


## 4.1. Torque Ratio of Links with Losses

The pressure efficiencyηp of the HST system is approximately expressed as ηp = pP/pM, where pP and pM are pump and motor operating pressures, respectively. In the HST system, total mechanical efficiency ηmh is expressed as ηmh = ηpηPηM, where ηM and ηP are pump and motor mechanical efficiencies, respectively. The output torque of motor TM and the input torque of pump TP are expressed as(22)TP=qPpP2πηP,TM=qMpMηM2π,where qP and qM are the displacements of the pump and motor, respectively.In type III power flow, hydraulic unit 1 acts as the variable pump with a torque ofTP, and hydraulic unit 2 acts as the motor with a torque of TM. The torque ratio ΔH of HST in type III power flow is defined as TP divided by TM.(23)ΔH=TPTM=T2T3=qPpPqMpMηMηP≈eηMηPηp=eηmh.The torque ratio Δ12 of F1 to F2 (shown in Figure 3; internal transmissions) in type III power flow is defined as(24)Δ12=τF1τF2ηF1ηF2.In type I power flow, hydraulic unit 1 acts as the variable motor with a torque ofTM, and hydraulic unit 2 acts as the pump with a torque of TP. ΔH′ is defined as TP divided by TM.(25)ΔH′=TPTM=T3T2=qPpPqMpMηMηP≈1eηMηPηp=1eηmh.The torque ratioΔ12′ of F1 to F2 in type I power flow is defined as(26)Δ12′=1ηF1ηF2τF1τF2.Given that the transmission losses and the torques applied to the links are unrelated to the observer’s motion, the method of epicyclic inversion was used by Pennestri and Freudenstein [14] and Chen and Liang [21] in different ways.This study uses the transforming mechanism model [21, 25], in which the interaction forces and the relative angular velocity are all the same as the EGTs shown in Figure 1, but the planet carrier is fixed. Thus, the losses in the two models are considered to be the same. The torque ratio of the links based on meshing power [21, 25] is applied to calculate the efficiency of the two-DOF EGTs. Meshing power pic (the ith link relative to planet carrier 6) is a virtual power that is independent of the actual power pi. Meshing power ratio ψi is defined as the ratio of meshing power pic through the ith link to actual power pi through the same link. If power pi flows into EGT, it is defined as positive (negative otherwise). The expressions of ωi6, pi6, pi and ψi are(27)ωi6=ωi−ω6,pi6=Tiωi−ω6=Tiωi6,pi=Tiωi,ψi=pi6pi=Tiωi−ω6Tiωi=ωi6ωi.In the transforming mechanism, the direction ofpic can be determined in accordance with the sign of pi and ψi. If the sign of picis positive, power pic flows into the transforming mechanism; otherwise, it flows out of the system.Without loss of generality, we assume that the output shaft of the HMCVT is link 7 in type I power flow, and the relationship amongωi, ψi and pi between links isω7>ω5>ω6>ω4>0.(28)ψ5=p56p5=T5ω5−ω6T5ω5>0,p5<0,p56>0,ψ4=p46p4=T4ω4−ω6T4ω4<0,p4<0,p46>0,ψ7=p76p7=T7ω7−ω6T7ω7>0,p7<0,p76<0.Under this condition, the meshing power is from links 5 and 4 to link 7 in the transforming mechanism. When the power conservation equation and the torque equilibrium condition are applied to the EGT, we have(29)T4ω46η486η876+T5ω56η586η876+T7ω76=0,T4+T5+T7=0,where ηij6 is the efficiency of the path, in which the meshing power flows from the ith link to the jth link with the planet carrier being fixed in the transforming mechanism.To obtain the relationship between output torque and input torque, by using equations (5) and (6), we can reform equation (29) to(30)T7=T5η486η876τPG′τPG−η586η876τPG′1−η486η876τPG′τPG,T4=−T51+η586η876τPG′τPG−η586η876τPG′1−η486η876τPG′τPG.From equation (28), torque ratios Δ75 and Δ45′ of EGT (shown in Figure 3; the mechanical path) in type I power flow are defined as(31)Δ75=η486η876τPG′τPG−η586η876τPG′1−η486η876τPG′τPG,Δ45′=−1+η486η876τPG′τPG−η586η876τPG′1−η486η876τPG′τPG.When the output shaft of the EGT is link 7 in type III power flow, in reference to the actual angular velocityωi, the relationship between links is ω7>ω5>ω6>0>ω4. The same expression of Δ75 and Δ45′ as that in equation (29) can be obtained because the sign of pic remains similar to that in type I power flow. The other case is that the output shaft of HMCVT is link 6 in types III and I power flows. In this condition, we can obtain the expressions of Δ65 and Δ45 in a similar manner.(32)Δ45=−η586η846τPG,Δ65=η586η846τPG−1.


## 4.2. Efficiency of HMCVT

In the steady state of HMCVT, we assume that output torqueTo of HMCVT is the result of the calculation and not of the actual load, such that the input torque and angular velocity remain similar to those for diesel working under rated conditions. When the power pi contributed to Ti flows into the Split point in Figure 1, Ti is defined as positive; otherwise, it is negative. Here, T5 is negative in type I and III power flows. To maintain consistency with the signs of T5 and T1 for EGTs, when the torque equilibrium condition is applied to the Split point, we have(33)TI−T5−T1=0.Depending on the type of power flow with output shaft 6, two kinds of relationship exist betweenT1 and T5.(34)T1=ΔHΔ45Δ12T,III,−Δ45ΔH′Δ12′T,I.The relationship betweenT5 and TI is(35)T5=TI1+ΔHΔ45Δ12,III,TI1−Δ45/ΔH′Δ12′,I.We obtain the transmission efficiency of the HMCVT with output shaft 6 by using the relationship expressions ofTI with T5 and equation (7) as follows:(36)ηHM=−T6ω6TIωI=−Δ65TIω6/1+ΔHΔ45Δ12TIωI=−Δ65τHM1+ΔHΔ45Δ12,III,−Δ65TIω6/1−Δ45/ΔH′Δ12′TIωI=−Δ65τHM1−Δ45/ΔH′ΔH′,I.The working pressure of the hydraulic system is represented by the working pressure of the fixed displacement hydraulic unit 2.(37)Δp=2πT3qc=2πτF2Δ45T5qcηF2=2πτF2Δ45TIqcηF21+Δ45Δ12ΔH,III,2πτF2Δ45T5ηF2qc=2πτF2Δ45ηF2TIqc1−Δ45/Δ12′ΔH′,I.When the output shaft is link 7, equations (32), (34), and (35) remain the same, but Δ45, Δ65 and τHM are replaced by Δ45′, Δ75, and τHM′, respectively:(38)T1=ΔHΔ45′Δ12T5,III,−Δ45′ΔH′Δ12′T5,I,(39)ηHM′=−T7ω7TIωI=−Δ75τHM′1+ΔHΔ45′Δ12,III,−Δ75τHM′1−Δ45′/ΔH′Δ12′,I,(40)Δ′p=2πT3qc=2πΔ45′τF2TIqcηF21+Δ45′Δ12ΔH,III,2πΔ45′τF2ηF2TIqc1−Δ45′/Δ12′ΔH′I.When the losses from the churning of the lubricating oil and shaft bearing friction are not considered, once the characteristics of the HMCVT have been determined, torque ratios Δ12, Δ12′, Δ75, Δ45′, Δ65, and Δ45 become constant. However, firstly, torque ratios ΔH and ΔH′ vary with the change in mechanical efficiency ηmh and displacement ratio e for the hydraulic system. Secondly, τHM and τHM′ vary with the change in volumetric efficiency ηv and displacement ratio e. Thirdly, volumetric efficiency ηv and mechanical efficiency ηmh change with displacement ratio e and system working pressure Δp. According to equation (37) or (40), the curved surface of ηmh relative to e and Δp is called the mechanical system demanding surface.


## 5. Efficiency Analysis

### 5.1. Numerical Example

The mechanical parameters of the example are displayed in equations (13) and (14) and Table 1. The loss of a conventional gear pair is estimated via the following formula [13]:(41)ψmnC=0.231Z1±1Z2,where ψmnC is the meshing loss of the power flowing through the mth and nth links when the gear carrier is fixed and Z1 and Z2 are the teeth numbers of the small and large gears of this pair, respectively.By using the equations above, we can obtain constant torque ratios. The parameters of the example are listed in Table2.Table 2
Parameters of HMCVT.
Δ12Δ12′Δ75Δ45′Δ651.021.02−0.528−0.472−1.975Δ45τF2TI (N·m)qc (cm3·r−1)Δ450.9750.729702.51070.975By using the hydraulic parametersηH, ηv and ηmh relative to Δp and e [26] listed in Tables 1 and 2, the expressions of the fitting surfaces are obtained, with the displacement of hydraulic unit 2 being fixed to the maximum value.(42)ηH=Δp2Δp1⋅−0.0002030.000244−0.0004020.010453−0.0112010.022677−0.3764060.6189120.267755e2e1T,ηv=Δp2Δp1⋅−0.0001450.000263−0.000143−0.0005940.001655−0.0023490.001936−0.0083380.998824e2e1T,ηmh=Δp2Δp1⋅−0.000039−0.000055−0.0002510.008926−0.0087870.023435−0.3771700.6213470.263924e2e1T.The analysis steps (Sts1) for obtaining the HMCVT efficiency with the single variable unit are shown below, and the results are listed in Figure4 and Table 3.(1)
Two curved surfaces in the same coordinate system are established. One is the curved surface of the hydraulic system’s mechanical efficiency, and the other is the mechanical system demanding surface. To obtain the relationships between hydraulic system mechanical efficiencyηmh and system pressure Δp with displacement ratio e, the data on the intersection of surfaces are fitted by a polynomial.(2)
Torque ratios ΔH and ΔH′ are obtained from the equations above. By substituting the polynomial of Δp relative to |e| in the fitting formula for ηv, the relationship between ηv and τH with |e| is obtained.(3)
The overall angular velocity ratiosτHM and τHM′ of the HMCVT are obtained.(4)
The overall efficienciesηHM and ηHM′ of the HMCVT are obtained.Figure 4
Results of the efficiency analysis. (a) Intersection of surfaces in type III power flow with output shaft 6. (b) Intersection of surfaces in type I power flow with output shaft 6. (c) Intersection of surfaces in type III power flow with output shaft 7. (d) Intersection of surfaces in type I power flow with output shaft 7. (e) Overall efficiency of HMCVT. (f) Overall efficiency of HMCVTwith output shaft 7.
(a)(b)(c)(d)(e)(f)Table 3
Polynomial coefficients of ratios.
TypeΔH or 1/ΔH′τH6 III ΔH[0.3515, −0.4770, 1.4554, −0.0042][−0.1093, 0.2033, 0.8792, −0.0027]6 I 1/ΔH′[−0.0908, 0.1886, 0.8034, −0.0028][−0.4537, 0.6241, −0.0107, −1.2757, 0.0027]7 III ΔH[0.0112, 0.6150, −0.9826, 1.8224, 0.0002][0.0324, −0.0628, −0.9496, 0.0005]7 I 1/ΔH′[−0.1881, 0.5067, 0.5532, −0.0005][0.0497, −0.0399, −0.0171, 1.0616, 0.0002]By observing the overall efficiency of output shaft 6 in Figure4(e), we find that the least efficiency appears at the start of the vehicle, and −1 < e < −0.9. The actual numerical result is negative. As shown in Figure 5(b), when −1 < e < −0.9, τH < −1. In this condition, the direction of output shaft 6 is reversed. Therefore, we can assume that the vehicle should be started at an e of about −0.9, and |e| should be gradually reduced to improve the speed of the vehicle.Figure 5
Curvesηv and τH of the numerical results for type I power flow with output shaft 6.
(a)(b)(1) The dotted line indicates that the curve is out of the fitting data source. (2) Denoted by a line of long dashes, this working area without the full input power is called traction limiting, indicating that the efficiency curve is highly inaccurate at the start of the entire machine. It is calculated with a peak pressure of 40 MPa of the hydraulic system.When the vehicle speed is lower than 6 km/h, the working condition without the full input power is called traction limiting. When Δp > 40 MPa in type I power flow with output shaft 6, the efficiency is calculated using the peak pressure (40 MPa) of the hydraulic system, as shown in Figures 4(b) and 5(a). This kind of calculation reflects the carrying capacity of HMCVT, which is independent of the weight of the whole vehicle and the ground adhesion performance.In the steady state, with the vehicle speed being larger than the maximum speed in type I power flow with output shaft 6 (about 7.4 km/h in the example), the maximum working pressure of the hydraulic system is 29.57 MPa when the system output shaft is link 6 ande = 0. Considering that the coefficient of torque reservation for the diesel engine is about 40%, when the system is working under rated conditions, the suitable hydraulic system pressure is about 30 MPa.By substituting the polynomials forτH (in Table 3) in equation (14), the relationship between angular velocity ratio τHM of HMCVT and displacement ratio e is obtained, as shown in Figure 2(b). When the system output shaft is link 7, the hydraulic system pressure is generally lower than that for output link 6 because torque ratio Δ45′ is smaller than Δ45. The output angular velocity of link 7 is greater than that of link 6, as shown in Figure 2(b).


### 5.2. HMCVT with Double-Variable Hydraulic Units

In the HMCVT system with double-variable hydraulic units, hydraulic unit 2 is the variable displacement (qv′q'v) hydraulic unit connected to link 3 in Figure 1. We can obtain a relatively stable e through different combinations of variable qv and variable qv′, which allow hydraulic units 1 and 2 to work in a relatively high-efficiency area. Without consideration of fuel consumption and vehicle dynamic characteristics, the problem of HMCVT efficiency changes. How variables qv and qv′ should be varied to obtain the maximum system efficiency in the steady state under the rated conditions of the diesel engine is the new problem to be solved.The analysis steps (Sts2) for obtaining the curve of variablesqv′ (qv = qv′e) are shown as follows, and the results obtained with the example above are listed in Figure 6:(1)
When variablee = constant, e′ = qv′/qvmax′, where qvmax′ is the maximum value of variable qv′. The fitting formulas (curved surfaces) of ηv and ηmh relative to Δp and e′ are obtained with the changed displacement of hydraulic unit 2.(2)
To obtain the relationship between overall efficiencyη and variable e′, the steps of Sts1 are completed with qv′ replacing qc. Attention is given to the difference between e and e′.(3)
In the range of variablee, processes (1) and (2) are repeated. In the three-dimensional coordinate system, the curved surface that represents η relative to e and e′ is obtained, and the points (e, e′, η) that have the highest efficiency for each e in step (b) are connected.Figure 6
Results of Sts2 with double-variable hydraulic units. (a) In type III with output link 6. (b) In type I with output link 6. (c) In type III with output link 7. (d) In type I with output link 7. (e) Comparison of efficiency with 6. (f) Comparison of efficiency with 7. (g) Curves of Δp relative to e. (h) Comparison of To relative to v.
(a)(b)(c)(d)(e)(f)(g)(h)By substituting the polynomialΔH′ or ΔH, equations (30) and (31) in equation (34), we obtain the relationship between output torque To of HMCVT and the foreword speed of the vehicle, whilst assuming that the minimum speed is 0, and the maximum speed is 39.96 km/h, as shown in Figure 6(h).When the machine is started with type I power flow with output shaft 6, most of the time, it works one′ = 1 or Δ′p = 40 MPa, and the efficiency curve is almost the same as that of the single variable unit system. In addition, from the comparisons of the efficiency curves with outputs 6 and 7 in Figures 6(e) and 6(f), we find that the overall efficiency of HMCVT with double-variable hydraulic units is improved. The maximum value of qv′ is about 91 cm3·r−1, and the system variable qv′ is effectively reduced.


#### 5.2.1. The Solid Curves Belong to the HST with a Single Variable

As shown by the curves of Δp relative to e, the HST of HMCVT always works at the medium-pressure stage, in which the relative maximum efficiency of the hydraulic system can be obtained. However, the working pressure is unstable, mainly because the maximum values of mechanical efficiency ηmh and volumetric efficiency ηv cannot be obtained at the same time, and because the positions of ηv and ηmh in the efficiency expression are different. As shown by the curves of Δp relative to e, when e = 0, the mutations of working pressure are mainly due to the mutations of e′.At the start of the operation of the entire machine in type I power flow with output shaft 6, an effective way to improve system efficiency is to increase the q'vmax of hydraulic unit 2, as shown in Figure7, by using the same curved surfaces of ηv and ηmh without consideration of the influence of maximum displacement qvmax′ on system efficiency ηv and ηmh.Figure 7
Relationship ofηHM with e and qvmax′ in type I power flow with output shaft 6.To increase the maximum speed of the vehicle with dual variable units, a maximum speed greater than 60 kph can be obtained by setting variablee < −2.6 in the type III power flow with output shaft 7. Let e″ = qv/qvmax, where qvmax is the maximum value of variable qv. To obtain the curved surface that represents η relative to e and e″, the processes of Sts2 are repeated, with e″ (=ee′) replacing e′. The points (e, e″, η) that have the highest efficiency for each e are connected.Whene < −1.8, e″ = 1, as shown in Figure 8. This condition indicates that the transmission efficiency of HMCVT can be improved when the value of qvmax is increased and e < −1.8, but the improvement is not significant.Figure 8
Curved surfaces ofηHM′ relative to e and e″ in type III power flow with output shaft 7 when e < −1.


## 5.1. Numerical Example

The mechanical parameters of the example are displayed in equations (13) and (14) and Table 1. The loss of a conventional gear pair is estimated via the following formula [13]:(41)ψmnC=0.231Z1±1Z2,where ψmnC is the meshing loss of the power flowing through the mth and nth links when the gear carrier is fixed and Z1 and Z2 are the teeth numbers of the small and large gears of this pair, respectively.By using the equations above, we can obtain constant torque ratios. The parameters of the example are listed in Table2.Table 2
Parameters of HMCVT.
Δ12Δ12′Δ75Δ45′Δ651.021.02−0.528−0.472−1.975Δ45τF2TI (N·m)qc (cm3·r−1)Δ450.9750.729702.51070.975By using the hydraulic parametersηH, ηv and ηmh relative to Δp and e [26] listed in Tables 1 and 2, the expressions of the fitting surfaces are obtained, with the displacement of hydraulic unit 2 being fixed to the maximum value.(42)ηH=Δp2Δp1⋅−0.0002030.000244−0.0004020.010453−0.0112010.022677−0.3764060.6189120.267755e2e1T,ηv=Δp2Δp1⋅−0.0001450.000263−0.000143−0.0005940.001655−0.0023490.001936−0.0083380.998824e2e1T,ηmh=Δp2Δp1⋅−0.000039−0.000055−0.0002510.008926−0.0087870.023435−0.3771700.6213470.263924e2e1T.The analysis steps (Sts1) for obtaining the HMCVT efficiency with the single variable unit are shown below, and the results are listed in Figure4 and Table 3.(1)
Two curved surfaces in the same coordinate system are established. One is the curved surface of the hydraulic system’s mechanical efficiency, and the other is the mechanical system demanding surface. To obtain the relationships between hydraulic system mechanical efficiencyηmh and system pressure Δp with displacement ratio e, the data on the intersection of surfaces are fitted by a polynomial.(2)
Torque ratios ΔH and ΔH′ are obtained from the equations above. By substituting the polynomial of Δp relative to |e| in the fitting formula for ηv, the relationship between ηv and τH with |e| is obtained.(3)
The overall angular velocity ratiosτHM and τHM′ of the HMCVT are obtained.(4)
The overall efficienciesηHM and ηHM′ of the HMCVT are obtained.Figure 4
Results of the efficiency analysis. (a) Intersection of surfaces in type III power flow with output shaft 6. (b) Intersection of surfaces in type I power flow with output shaft 6. (c) Intersection of surfaces in type III power flow with output shaft 7. (d) Intersection of surfaces in type I power flow with output shaft 7. (e) Overall efficiency of HMCVT. (f) Overall efficiency of HMCVTwith output shaft 7.
(a)(b)(c)(d)(e)(f)Table 3
Polynomial coefficients of ratios.
TypeΔH or 1/ΔH′τH6 III ΔH[0.3515, −0.4770, 1.4554, −0.0042][−0.1093, 0.2033, 0.8792, −0.0027]6 I 1/ΔH′[−0.0908, 0.1886, 0.8034, −0.0028][−0.4537, 0.6241, −0.0107, −1.2757, 0.0027]7 III ΔH[0.0112, 0.6150, −0.9826, 1.8224, 0.0002][0.0324, −0.0628, −0.9496, 0.0005]7 I 1/ΔH′[−0.1881, 0.5067, 0.5532, −0.0005][0.0497, −0.0399, −0.0171, 1.0616, 0.0002]By observing the overall efficiency of output shaft 6 in Figure4(e), we find that the least efficiency appears at the start of the vehicle, and −1 < e < −0.9. The actual numerical result is negative. As shown in Figure 5(b), when −1 < e < −0.9, τH < −1. In this condition, the direction of output shaft 6 is reversed. Therefore, we can assume that the vehicle should be started at an e of about −0.9, and |e| should be gradually reduced to improve the speed of the vehicle.Figure 5
Curvesηv and τH of the numerical results for type I power flow with output shaft 6.
(a)(b)(1) The dotted line indicates that the curve is out of the fitting data source. (2) Denoted by a line of long dashes, this working area without the full input power is called traction limiting, indicating that the efficiency curve is highly inaccurate at the start of the entire machine. It is calculated with a peak pressure of 40 MPa of the hydraulic system.When the vehicle speed is lower than 6 km/h, the working condition without the full input power is called traction limiting. When Δp > 40 MPa in type I power flow with output shaft 6, the efficiency is calculated using the peak pressure (40 MPa) of the hydraulic system, as shown in Figures 4(b) and 5(a). This kind of calculation reflects the carrying capacity of HMCVT, which is independent of the weight of the whole vehicle and the ground adhesion performance.In the steady state, with the vehicle speed being larger than the maximum speed in type I power flow with output shaft 6 (about 7.4 km/h in the example), the maximum working pressure of the hydraulic system is 29.57 MPa when the system output shaft is link 6 ande = 0. Considering that the coefficient of torque reservation for the diesel engine is about 40%, when the system is working under rated conditions, the suitable hydraulic system pressure is about 30 MPa.By substituting the polynomials forτH (in Table 3) in equation (14), the relationship between angular velocity ratio τHM of HMCVT and displacement ratio e is obtained, as shown in Figure 2(b). When the system output shaft is link 7, the hydraulic system pressure is generally lower than that for output link 6 because torque ratio Δ45′ is smaller than Δ45. The output angular velocity of link 7 is greater than that of link 6, as shown in Figure 2(b).


## 5.2. HMCVT with Double-Variable Hydraulic Units

In the HMCVT system with double-variable hydraulic units, hydraulic unit 2 is the variable displacement (qv′q'v) hydraulic unit connected to link 3 in Figure 1. We can obtain a relatively stable e through different combinations of variable qv and variable qv′, which allow hydraulic units 1 and 2 to work in a relatively high-efficiency area. Without consideration of fuel consumption and vehicle dynamic characteristics, the problem of HMCVT efficiency changes. How variables qv and qv′ should be varied to obtain the maximum system efficiency in the steady state under the rated conditions of the diesel engine is the new problem to be solved.The analysis steps (Sts2) for obtaining the curve of variablesqv′ (qv = qv′e) are shown as follows, and the results obtained with the example above are listed in Figure 6:(1)
When variablee = constant, e′ = qv′/qvmax′, where qvmax′ is the maximum value of variable qv′. The fitting formulas (curved surfaces) of ηv and ηmh relative to Δp and e′ are obtained with the changed displacement of hydraulic unit 2.(2)
To obtain the relationship between overall efficiencyη and variable e′, the steps of Sts1 are completed with qv′ replacing qc. Attention is given to the difference between e and e′.(3)
In the range of variablee, processes (1) and (2) are repeated. In the three-dimensional coordinate system, the curved surface that represents η relative to e and e′ is obtained, and the points (e, e′, η) that have the highest efficiency for each e in step (b) are connected.Figure 6
Results of Sts2 with double-variable hydraulic units. (a) In type III with output link 6. (b) In type I with output link 6. (c) In type III with output link 7. (d) In type I with output link 7. (e) Comparison of efficiency with 6. (f) Comparison of efficiency with 7. (g) Curves of Δp relative to e. (h) Comparison of To relative to v.
(a)(b)(c)(d)(e)(f)(g)(h)By substituting the polynomialΔH′ or ΔH, equations (30) and (31) in equation (34), we obtain the relationship between output torque To of HMCVT and the foreword speed of the vehicle, whilst assuming that the minimum speed is 0, and the maximum speed is 39.96 km/h, as shown in Figure 6(h).When the machine is started with type I power flow with output shaft 6, most of the time, it works one′ = 1 or Δ′p = 40 MPa, and the efficiency curve is almost the same as that of the single variable unit system. In addition, from the comparisons of the efficiency curves with outputs 6 and 7 in Figures 6(e) and 6(f), we find that the overall efficiency of HMCVT with double-variable hydraulic units is improved. The maximum value of qv′ is about 91 cm3·r−1, and the system variable qv′ is effectively reduced.


### 5.2.1. The Solid Curves Belong to the HST with a Single Variable

As shown by the curves of Δp relative to e, the HST of HMCVT always works at the medium-pressure stage, in which the relative maximum efficiency of the hydraulic system can be obtained. However, the working pressure is unstable, mainly because the maximum values of mechanical efficiency ηmh and volumetric efficiency ηv cannot be obtained at the same time, and because the positions of ηv and ηmh in the efficiency expression are different. As shown by the curves of Δp relative to e, when e = 0, the mutations of working pressure are mainly due to the mutations of e′.At the start of the operation of the entire machine in type I power flow with output shaft 6, an effective way to improve system efficiency is to increase the q'vmax of hydraulic unit 2, as shown in Figure7, by using the same curved surfaces of ηv and ηmh without consideration of the influence of maximum displacement qvmax′ on system efficiency ηv and ηmh.Figure 7
Relationship ofηHM with e and qvmax′ in type I power flow with output shaft 6.To increase the maximum speed of the vehicle with dual variable units, a maximum speed greater than 60 kph can be obtained by setting variablee < −2.6 in the type III power flow with output shaft 7. Let e″ = qv/qvmax, where qvmax is the maximum value of variable qv. To obtain the curved surface that represents η relative to e and e″, the processes of Sts2 are repeated, with e″ (=ee′) replacing e′. The points (e, e″, η) that have the highest efficiency for each e are connected.Whene < −1.8, e″ = 1, as shown in Figure 8. This condition indicates that the transmission efficiency of HMCVT can be improved when the value of qvmax is increased and e < −1.8, but the improvement is not significant.Figure 8
Curved surfaces ofηHM′ relative to e and e″ in type III power flow with output shaft 7 when e < −1.


## 5.2.1. The Solid Curves Belong to the HST with a Single Variable

As shown by the curves of Δp relative to e, the HST of HMCVT always works at the medium-pressure stage, in which the relative maximum efficiency of the hydraulic system can be obtained. However, the working pressure is unstable, mainly because the maximum values of mechanical efficiency ηmh and volumetric efficiency ηv cannot be obtained at the same time, and because the positions of ηv and ηmh in the efficiency expression are different. As shown by the curves of Δp relative to e, when e = 0, the mutations of working pressure are mainly due to the mutations of e′.At the start of the operation of the entire machine in type I power flow with output shaft 6, an effective way to improve system efficiency is to increase the q'vmax of hydraulic unit 2, as shown in Figure7, by using the same curved surfaces of ηv and ηmh without consideration of the influence of maximum displacement qvmax′ on system efficiency ηv and ηmh.Figure 7
Relationship ofηHM with e and qvmax′ in type I power flow with output shaft 6.To increase the maximum speed of the vehicle with dual variable units, a maximum speed greater than 60 kph can be obtained by setting variablee < −2.6 in the type III power flow with output shaft 7. Let e″ = qv/qvmax, where qvmax is the maximum value of variable qv. To obtain the curved surface that represents η relative to e and e″, the processes of Sts2 are repeated, with e″ (=ee′) replacing e′. The points (e, e″, η) that have the highest efficiency for each e are connected.Whene < −1.8, e″ = 1, as shown in Figure 8. This condition indicates that the transmission efficiency of HMCVT can be improved when the value of qvmax is increased and e < −1.8, but the improvement is not significant.Figure 8
Curved surfaces ofηHM′ relative to e and e″ in type III power flow with output shaft 7 when e < −1.


## 6. Test Verification

The effectiveness of the HMCVT transmission efficiency calculation method proposed in this study is verified, and an HMCVT gearbox with a multisegment univariate unit similar to that in Figure1 is selected as the test object. The test site of the physical prototype TA1-02 is shown in Figure 9. Torque and speed sensors HBM3000 and HBM5000 with ranges of 3000 and 5000 N m, respectively, are installed on the output shaft of the DC variable frequency motor, the output shaft of the HMCVT, and the hydraulic motor of the energy recovery device. System pressure test data from the transmission TCU are read via CAN bus. The test arrangement is applied in accordance with the gear shift strategy of the tractor’s traction conditions. The input torque is stable at 1030 N m, and the input speed is stable at 1700 r·min−1. These values are gradually increased according to the transmission ratio, and A, B, C, D, E, and F are selected for a total of six working conditions for the bench test.Figure 9
Graph of systematic experiments.Some of the data collected on site are shown in Figure10. The input torque is represented by a blue line, the pressure of the B channel between the hydraulic units is represented by the red line, and the high-pressure-side system pressure is represented by a yellow line.Figure 10
Measurement test program.
(a)(b)The transmission efficiency comparison is shown in Figure11, and the system pressure comparison is shown in Figure 12. In both figures, the star point denotes the test data of the bench under six working conditions, and the solid line is the numerical calculation result of the 22 point positions. The maximum point of the simulation efficiency error is at point C of the minimum displacement ratio of the type I power flow, and the error is 4.7%. The error for the other operating conditions is less than 2%. Working conditions C and D are at the minimum displacement ratio, and the torque of hydraulic units 1 and 2 is small. The efficiency simulation error is large due to the relatively large simulation torque control error and the relatively large influence of the nonlinear fitting effect of the HST hydraulic system. The efficiency simulation values of the operating conditions, except for operating conditions A, C, and D, are greater than the actual measured values, because the simulation does not consider transmission churning loss and bearing loss. In Figure 12, the simulated pressure curve effectively reflects the working characteristics of the hydraulic system. As the transmission ratio increases, the HST system pressure gradually decreases with a hyperbolic curve, and the system works in type I and III power flow states. The high-pressure side of the system shows no changes. The simulated pressure curve is lower than the actual measured value, and the transmission power of the HST system is lower than that in the actual working conditions, which is the main reason for the high efficiency of the simulation.Figure 11
Data on system efficiency.Figure 12
Data on system pressure.


## 7. Conclusions

Two parameters, namely, hydraulic transmission asymmetryλ and maximum angular velocity ratio τHMmax, are introduced to express the structural characteristics of hydromechanical continuously variable transmission (HMCVT) with a variable speed range of 0 to the maximum. This approach provides a means to solve the classification problem of HMCVT.The relationship between mechanical system transmission and the parameters of the hydraulic system is established by the mechanical system demanding surface. The theory of meshing power is applied to determine the torque ratio coefficient Δi of an epicyclic gear train (EGT) with two DOFs. The expression of HMCVT overall efficiency is simplified and unified by torque ratio coefficient Δi in the steady state under the rated conditions of the diesel engine.The control theory of parameterse′ and e″ is revealed to obtain the maximum transmission efficiency of HMCVT. The application method of this theory is introduced through numerical examples, so that hydraulic system components can be selected properly, and the output torque can be predicted.The physical prototype bench test conducted using TA1-02 multisection univariate-unit HMCVT and the simulation show that the maximum efficiency simulation error is at point C of the minimum displacement ratio of the type I power flow, and the error is 4.7%. The error for the other operating conditions is less than 2%. The simulated pressure curve effectively reflects the working characteristics of the hydraulic system. As the transmission ratio increases, the HST system pressure gradually decreases in a hyperbolic curve, which verifies the effectiveness of the HMCVT efficiency analysis steps.For the study of transmission efficiency of future transmissions, the influence of diesel engine working characteristics, overall slip rate, and changes in traction force on transmission efficiency will be comprehensively considered.



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*Source: 4275922-2021-09-07.xml*