# Research on Transmission Characteristics of Hydromechanical Continuously Variable Transmission of Tractor

Authors: Junyan Wang; Changgao Xia; Xin Fan; Junyu Cai
Journal: Mathematical Problems in Engineering
 (2020)

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## Abstract

This paper proposes a new transmission scheme of hydromechanical continuously variable transmission (HMCVT) for tractors. The HMCVT has 4 working ranges in each of the front and rear directions. The speed characteristic and the torque characteristic of HMCVT are theoretically derived. On the basis of HMCVT power flow direction, the Крейнeс formula is used to calculate the transmission efficiency. Then, the image analysis method is used to study the influence of parameters on the transmission efficiency of HMCVT, and the main influencing factors are found. The results of theoretical derivation demonstrate that, by coordinating control of the HST displacement ratio and the engagement conditions of shifting clutches, the stepless speed regulation of HMCVT at the tractor speed of 0–50 km/h can be realized. The proposed HMCVT has the ability to continuously transmit and change torque over all working ranges. The overall transmission efficiency of HMCVT is at a high level. To verify the theoretical derivation, Amesim simulation software is used for the modeling and simulation of HMCVT. The simulation results are consistent with the theoretical analysis results. Therefore, the HMCVT proposed in this paper has the advantages of compact structure and high transmission efficiency, and it is suitable for matching tractors.


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## Body

## 1. Introduction

In addition to transportation, the main function of tractors is to drive agricultural machinery needed for agricultural production, which needs the transmission of tractors to provide constantly changing speed and torque to adapt to frequent external load changes [1].In this context, the use of a continuously variable transmission (CVT) for tractors has become a trend. The most common type of CVTs is the hydraulic static transmission (HST) [2]; it mainly includes hydraulic pumps, hydraulic motors, and control mechanism. HST has good performance at low speed and can easily switch between positive rotation and negative rotation. However, its transmission efficiency is much lower than that of gear transmission, which leads to the fact that it is rarely used in tractors.In order to overcome the above shortcomings, the HST is connected in parallel with the mechanical components to form a hydromechanical continuously variable transmission (HMCVT) [3–6]. HMCVT transmits only part of the power through the HST; the remaining power is transmitted through the mechanical components, so the efficiency is much higher than HST. HMCVT can realize stepless speed regulation through the HST, while relying on mechanical components to achieve high efficiency transmission. In addition, the HMCVT can control the engine to operate on an optimal power curve or optimal economic curve, thus achieving optimum power performance or optimum fuel economy for tractors [7–9].The research on power split continuously variable transmission technology began in the early 20th century in the world. However, limited by the level of hydraulic system manufacturing, this type of transmission was not used in armored and heavy construction vehicles until the late 1960s [10]. Literature [8] first proposes a transmission scheme for single-row planetary gears with two hydromechanical working ranges. In each of the two working ranges, the maximum speed of the tractor can reach 32 km/h and 50 km/h, respectively. Then, according to the market demand, only one hydromechanical working range transmission scheme is proposed, which can match the 75 kw engine. The forward driving speed is from 0 to 40 km/h, and the reverse driving speed is from 0 to 25 km/h [11]. The transmission developed by Steyr company has three planetary gear rows, controlled by 4 clutches and 2 brakes. This transmission can achieve speed adjustments from 0 to 50 km/h [12]. The authors of [13, 14] study a HMCVT that has four planetary gear rows and can achieve speed adjustment from 0 to 40 km/h. The author of [15] proposes a transmission scheme in which a single-row planetary gear is connected in series with a mechanical stepped transmission. This transmission has 6 ranges in the forward direction and 3 ranges in the reverse direction.In general, the more the transmission ranges, the greater the expansion of the hydraulic power [10], but it will lead to frequent switching during the ranges and reducing the transmission performance. The single-row planetary scheme helps to reduce the size of the transmission, but the hydraulic motor needs to be reversed quickly at the moment of switching, which needs high performance of the hydraulic motor.This paper will propose a transmission scheme for a new 4-range HMCVT suitable for wheeled tractors and then conduct a detailed theoretical analysis on the transmission characteristics (speed characteristic, torque characteristic, and efficiency characteristic) of HMCVT and find out the important factors that affect the efficiency of HMCVT; finally, the Amesim simulation software is used to verify the transmission scheme.


## 2. Transmission Scheme of HMCVT

The structure of HMCVT proposed in this paper is shown in Figure1. It includes the input shaft, pump-motor hydraulic system, planetary gear mechanism, gear pairs, shifting clutches, and directional clutches. The hydraulic path of HMCVT is the HST, and its mechanical path is the Ravigneaux planetary gear mechanism (PGM). The hydraulic path and the mechanical path are connected in parallel to form a closed loop and then are connected in series with the range switching mechanism. In the PGM, when the output component is the carrier c, it is an ordinary PGM; when the output component is the small sun gear s2, then it is a Ravigneaux PGM. In addition, part of the engine power is through the HST into the PGM, and the HST realizes the stepless transmission.Figure 1
The structure of HMCVT.There are 6 clutches in the HMCVT, which are 4 shifting clutches and 2 directional clutches. Different clutches can realize 4 working ranges in the front or rear directions. In order to ensure that engine power can be transferred from the input shaft to the output shaft, one shifting clutch and one directional clutch must be engaged. The engagement conditions of clutches in each range of HMCVT are shown in Table1.Table 1
Engagement conditions of clutches of HMCVT.
RangeDirectional clutchShifting clutchCFCRC1C2C3C4ForwardFirst++Second++Third++Fourth++ReverseFirst++Second++Third++Fourth++


## 3. Theoretical Analysis on Transmission Characteristics of HMCVT

### 3.1. Speed Characteristic

#### 3.1.1. Ranges HM1 and HM3

The power transmission routes of ranges HM1 and HM3 are basically the same, except for the difference between gear pairs i7 and i8. Under these two conditions, the power is input by the large sun gear s1 and the ring gear r and output by the carrier c.Ifne denotes the engine speed, the variable pump input shaft speed np is(1)np=nei1i2.Ignoring the effect of the volumetric efficiency of the volumetric speed control loop, the relationship between the motor output shaft speednm, the variable pump input shaft speed np, and the pump-motor displacement ratio ε is(2)nm=εnp=εnei1i2,where ε=Qp/Qm−1≤ε≤1, Qp is the displacement of the variable pump, and Qm is the displacement of the quantitative motor.The input speed of the large sun gear in PGMns1 is(3)ns1=nm−i3=εne−i1i2i3.The input speed of the ring gearnr equals ne. According to the PGM speed characteristic formula [16], the relationship between the input speed of large sun gear ns1, the input speed of ring gear nr, the output speed of carrier nc, and the PGM standing transmission ratio −k1 is satisfied:(4)ns1−k1nr−1−k1nc=0,where −k1=Zr/Zs1>1, Zr is the number of teeth of the ring gear, and Zs1 is the number of teeth of the large sun gear.Therefore, the output speed of the carriernc is(5)nc=−k1i1i2i3−ε1−k1i1i2i3ne.The intermediate shaft speednz is(6)nz=−k1i1i2i3−ε1−k1i1i2i3i4i8ne,HM1,−k1i1i2i3−ε1−k1i1i2i3i4i7ne,HM3.Finally, the output shaft speed of HMCVTno is(7)no=−k1i1i2i3−ε1−k1i1i2i3i4i8i11i12ne,HM1,−k1i1i2i3−ε1−k1i1i2i3i4i7i11i12ne,HM3.The transmission ratio of HMCVTig is(8)ig=neno=1−k1i1i2i3i4i8i11i12−k1i1i2i3−ε,HM1,1−k1i1i2i3i4i7i11i12−k1i1i2i3−εHM3.


#### 3.1.2. Ranges HM2 and HM4

The power transmission routes of ranges HM2 and HM4 are the same, except for the difference between gear pairs i6 and i9. Under these two conditions, the power is input by the large sun gear s1 and the ring gear r and output by the small sun gear s2.The same as the previous case, the input speed of the large sun gearns1 is ns1=−εne/i1i2i3; the input speed of the ring gear nr equals ne. According to the PGM speed characteristic formula (16), the relationship between the input speed of large sun gear ns1, the input speed of ring gear nr, the output speed of the small sun gear ns2, and the PGM standing transmission ratios k2 and −k3 is satisfied:(9)1−k3k2−k3nr−1−k2k2−k3ns1−ns2=0.Therefore, the output speed of the small sun gearns2 is(10)ns2=1−k3i1i2i3+1−k2εk2−k3i1i2i3ne,where k2=Zs2/Zr>0, −k3=Zs2/Zs1>0, and Zr,Zs1,Zs2 are the numbers of teeth of ring gear, large sun gear, and small sun gear, respectively.The intermediate shaft speednz is(11)nz=1−k3i1i2i3+1−k2εk2−k3i1i2i3i5i9ne,HM2,1−k3i1i2i3+1−k2εk2−k3i1i2i3i5i6ne,HM4.Finally, the output shaft speed of HMCVTno is(12)no=1−k3i1i2i3+1−k2εk2−k3i1i2i3i5i9i11i12ne,HM2,1−k3i1i2i3+1−k2εk2−k3i1i2i3i5i6i11i12ne,HM4.The transmission ratio of HMCVTig is(13)ig=neno=k2−k3i1i2i3i5i9i11i121−k3i1i2i3+1−k2ε,HM2,k2−k3i1i2i3i5i6i11i121−k3i1i2i3+1−k2ε,HM4.


### 3.2. Determination of Parameters of HMCVT

The rated speed of the engine matching with the HMCVT is 2200 r/min, the minimum speed of the tractor is 4 km/h, the maximum speed is 50 km/h, the main reducer ratio of the rear axle is 3.7, the reduction ratio of the wheel side is 5.6, the total transmission ratio of the rear axle is 20.72, and the drive wheel radius is 0.8579 m.The range of the transmission ratio in the drive train is determined by parameters such as the speed of the tractor, the rated speed of the engine, and the drive wheel radius. So the transmission ratio of each range of HMCVT is calculated as follows:(14)igj=0.377nerdv,where j = 1, 2, 3, 4; ne is the rated speed of the engine; rd is the drive wheel radius; v is the theoretical driving speed of the tractor.The HMCVT has 4 working ranges, so the common ratio of the transmission ratios is(15)φ=igmaxigmin4.The HMCVT adopts a proportional transmission, so, in each working range, the transmission ratio has the following relationship:(16)igjmaxigjmin=φ.When HMCVT is switching between adjacent ranges, the conditions must be met at the changing point: the transmission ratios are equal and the pump-motor displacement ratios are equal:(17)ig1min−1=ig2min−1,ig2min1=ig3min1,ig3min−1=ig4min−1,ig4min1=imin.The gear ratio of the connecting gears between the engine and the hydraulic variable pump needs to be met:(18)i1i2≥ne.maxnp.max.According to the previous content, ifi10, i11, and i12 are set to 1, then the relationship between the various parameters of HMCVT can be obtained:(19)k1i1i2i3=−3.2727,1−k3=k1k2−1,i5i6=0.8969k11+k1,i4i7=−1.6865k11−k1,i4i8=−5.9615k11−k1,i5i9=3.1702k11+k1.These parameters are all expressions about the PGM standing transmission ratio−k1. Once the value of −k1 is determined, the values of these parameters will also be determined, which will be described later.


### 3.3. Torque Characteristic

#### 3.3.1. Ranges HM1 and HM3

Ignoring the power losses, ifTo denotes the load torque of output shaft, the intermediate shaft torque Tz is(20)Tz=Toi11i12,and then the output torque of carrier in PGMTc is(21)Tc=Tzi4i8=Toi4i8i11i12,HM1,Tzi4i7=Toi4i7i11i12,HM3.According to the PGM torque characteristic formula (16), the relationship between the input torque of large sun gear Ts1, the input torque of ring gear Tr, the output torque of carrier Tc, and the PGM standing transmission ratio −k1 is satisfied:(22)Ts1=Tr−k1=−Tc1−k1,where −k1=Zr/Zs1>1, Zr is the number of teeth of the ring gear, and Zs1 is the number of teeth of the large sun gear.Therefore, the large sun gear torqueTs1 is(23)Ts1=−Tc1−k1=−Toi4i8i11i121−k1,HM1,−Toi4i7i11i121−k1,HM3.Then, the input torque of ring gearTr is(24)Tr=−k1Ts1=k1Toi4i8i11i121−k1,HM1,k1Toi4i7i11i121−k1,HM3.Thus, the motor shaft load torqueTm is(25)Tm=Ts1−i3=Toi3i4i8i11i121−k1,HM1,Toi3i4i7i11i121−k1,HM3.The pump shaft torqueTp is(26)Tp=εTm=εToi3i4i8i11i121−k1,HM1,εToi3i4i7i11i121−k1,HM3.The engine output torqueTe is(27)Te=Tr+Tpi1i2=k1i1i2i3+εToi1i2i3i4i8i11i121−k1,HM1,k1i1i2i3+εToi1i2i3i4i7i11i121−k1,HM3.Finally, the ratio of the output torque of HMCVT to the input torque of the engineμ is(28)μ=ToTe=1−k1i1i2i3i4i8i11i12ε+k1i1i2i3,HM1,1−k1i1i2i3i4i7i11i12ε+k1i1i2i3,HM3.


#### 3.3.2. Ranges HM2 and HM4

The same as the previous case,To denotes the load torque of output shaft, and the intermediate shaft torque Tz is Tz=To/i11i12. Then the output torque of small sun gear in PGM Ts2 is(29)Ts2=Tzi5i9=Toi5i9i11i12,HM2,Tzi5i6=Toi5i6i11i12,HM4.The external torque algebra sum of PGM is zero, soTs1+Tr+Ts2=0. According to the PGM torque characteristic formula (16), the relationship between the input torque of large sun gear Ts1, the output torque of small sun gear Ts2, and the PGM standing transmission ratios k2 and −k3 is satisfied:(30)Ts1=k2−1k3−k2Ts2=k2−1Tok3−k2i5i9i11i12,HM2,k2−1Tok3−k2i5i6i11i12,HM4.The engine output torqueTe is(31)Te=1−k2ε+1−k3i1i2i3k3−k2i1i2i3i5i9i11i12To,HM2,1−k2ε+1−k3i1i2i3k3−k2i1i2i3i5i6i11i12To,HM4.Finally, the ratio of the output torque of HMCVT to the input torque of the engineμ is(32)μ=ToTe=k3−k2i1i2i3i5i9i11i121−k2ε+1−k3i1i2i3,HM2,k3−k2i1i2i3i5i6i11i121−k2ε+1−k3i1i2i3,HM4.


### 3.4. Circulating Power Flow Calculation

#### 3.4.1. Ranges HM1 and HM3

The speed ratio of the power transmitting from the engine to the carrierice is(33)ice=ncne=−k1i1i2i3−ε1−k1i1i2i3=icer+ices1.The speed ratios of the power transmitting from the engine to the large sun gear and ring gearire/is1e are(34)ire=1,is1e=−εi1i2i3.In the PGM, the speed ratios between the input elements (large sun gears1 and ring gear r) and the output element (carrier c) are(35)ics1r=ncns1=11−k1,icrs1=ncnr=−k11−k1.Then, we can get(36)icer=ics1ris1e=−εi1i2i31−k1,ices1=icrs1ire=−k11−k1.Next two situations will be discussed as follows [17]:(1)
Whenε<0, ices1>0, icer>0, and ices1icer>0, the PGM has no circulating power PL=0.(2)
Whenε>0, ices1>0, icer<0, and ices1icer<0, the PGM has circulating power; substituting equation (19) for k1i1i2i3 in equation (33), ice is(37)ice=3.2727−ε3.2727+i1i2i3>0.Because ofPe>0,Pr=ices1/icePe>0,Ps1=icer/icePe<0, the output power P is(38)P=Pe=−Pc=Pr+Ps1=Pr−Ps1.Then,(39)Pr=P+Ps1=P+PL.The circulating powerPL is(40)PL=Ps1=icericePe.Assume that the closed power factorKL is(41)KL=PLPe=icerice=εk1i1i2i3+ε=ϕε,where ϕε=ε/k1i1i2i3+ε and the derivative of ϕε is dϕε/dε=k1i1i2i3/k1i1i2i3+ε2<0, so function ϕε is a monotonically decreasing function. Then KL0=0, KL1=0.44, and the maximum value of the circulating power is 0.44Pe.


#### 3.4.2. Ranges HM2 and HM4

The speed ratio of the power transmitting from the engine to the small sun gearis2e is(42)is2e=ns2ne=1−k3i1i2i3+1−k2εk2−k3i1i2i3=is2er+is2es1.The speed ratios of the power transmitting from the engine to the large sun gear and ring gearire/is1e are(43)ire=1,is1e=−εi1i2i3.In the PGM, the speed ratios between the input elements (large sun gears1 and ring gear r) and the output element (small sun gear s2) are(44)is2s1r=k2−1k2−k3,is2rs1=1−k3k2−k3.Then, we can get(45)is2er=is2s1ris1e=−k2−1εi1i2i3k2−k3,is2es1=is2rs1ire=1−k3k2−k3.Next two situations will be discussed as follows [17]:(1)
Whenε<0, is2es1>0, is2er<0, and is2es1is2er<0, the PGM has circulating power; substituting equation (19) for k2,k3,i1i2i3 in equation (42), is2e is(46)is2e=3.2727+ε3.2727−3.2727/−k1>0.Because ofPe>0,Pr=ices1/icePe>0,Ps1=icer/icePe<0, the output power P is(47)P=Pe=−Ps2=Pr+Ps1=Pr−Ps1.Then,(48)Pr=P+Ps1=P+PL.The circulating powerPL is(49)PL=Ps1=is2eris2ePe.Assume that the closed power factorKL is(50)KL=PLPe=is2eris2e=−εk1i1i2i3−ε=ϕε,whereϕε=−ε/k1i1i2i3−ε and the derivative of ϕε is dϕε/dε=−k1i1i2i3/k1i1i2i3−ε2>0, so function ϕε is a monotonically increasing function. Then, KL0=0, KL1=0.44, and the maximum value of the circulating power is 0.44Pe.(2)
Whenε>0, is2es1>0, is2er>0, and is2es1is2er>0, the PGM has no circulating power PL=0.


### 3.5. Hydraulic Split Power Calculation

Hydraulic power-split ratioρ=PH/Pe=Ps1/Pe; hydraulic power is input to the PGM through the large sun gear s1.


#### 3.5.1. Ranges HM1 and HM3

(51)ρ=Ps1Pe=icerice=εk1i1i2i3+ε=ϕε,and as can be obtained from the above, functionϕε is a monotonically decreasing function, so(52)ρ−1=0.234>0,ρ0=0,ρ1=−0.44<0.


#### 3.5.2. Ranges HM2 and HM4

(53)ρ=Ps1Pe=is2eris2e=−εk1i1i2i3−ε=ϕε.As can be obtained from the above, functionϕε is a monotonically increasing function, so(54)ρ−1=−0.44<0,ρ0=0,ρ1=0.234>0.According to the theoretical analysis on the circulating power flow and the hydraulic split power, the power flow direction of HMCVT in each working range is shown in Figure2. This result lays the foundation for the analysis on the efficiency characteristic of HMCVT.Figure 2
Power flow direction of the PGM of HMCVT. (a) Range HM1/HM3,ε>0, power-circulation. (b) Range HM1/HM3, ε<0, power-split. (c) Range HM2/HM4, ε>0, power-split. (d) Range HM2/HM4, ε<0, power-circulation.
(a)(b)(c)(d)


### 3.6. Efficiency Characteristic

The Крейнeс formula (17) is used to calculate the transmission efficiency of HMCVT:(55)η=−μ˜ig=−To/Tene/no,where μ˜ is the ratio of output torque to input torque of HMCVT considering power losses and ig is the transmission ratio of HMCVT.The transmission efficiency calculation process of each working range of HMCVT is shown in Tables2 and 3. Among them, the calculation processes of the first range and the third range are listed in Table 2, and the calculation processes of the second range and the fourth range are listed in Table 3. In Tables 2 and 3, i1−i12 are the gear ratios of the gear pairs, η1−η12 are the transmission efficiency of the gear pairs, ηH is the hydraulic system transmission efficiency, and ηs1r, ηs1s2, and ηrs2 are the transmission efficiencies of the large sun gear to the ring gear, the large sun gear to the small sun gear, and the ring gear to the small sun gear in the PGM, respectively.Table 2
Equation of efficiency calculation for ranges HM1 and HM3.
ElementsTorque equation considering power lossesε>0ε<0HM1HM3HM1HM3EngineTe=Tr+η1η2Tp/i1i2Te=Tr+Tp/i1i2η1η2PumpTp=εηHTmTp=εTm/ηHMotorTm=η3Ts1/−i3Tm=Ts1/−η3i3PGMTs1+Tr+Tc=0PGMTr=−k1Ts1/ηs1rIntermediate shaftTc=Tz/i4i8η4η8Tc=Tz/i4i7η4η7Tc=Tz/i4i8η4η8Tc=Tz/i4i7η4η7Output shaftTz=To/i11i12η11η12Torque ratioμ˜=To/TeEfficiencyη=−μ˜/igTable 3
Equation of efficiency calculation for ranges HM2 and HM4.
ElementsTorque equation considering power lossesε>0ε<0HM2HM4HM2HM4EngineTe=Tr+Tp/i1i2η1η2Te=Tr+η1η2Tp/i1i2PumpTp=εTm/ηHTp=εηHTmMotorTm=Ts1/−η3i3Tm=η3Ts1/−i3PGMTs1+Tr+Ts2=0PGMTr=1−k3ηs1s2/k3ηs1s2−k2ηrs2Ts2Intermediate shaftTs2=Tz/i5i9η5η9Ts2=Tz/i5i6η5η6Ts2=Tz/i5i9η5η9Ts2=Tz/i5i6η5η6Output shaftTz=To/i11i12η11η12Torque ratioμ˜=To/TeEfficiencyη=−μ˜/ig


## 3.1. Speed Characteristic

### 3.1.1. Ranges HM1 and HM3

The power transmission routes of ranges HM1 and HM3 are basically the same, except for the difference between gear pairs i7 and i8. Under these two conditions, the power is input by the large sun gear s1 and the ring gear r and output by the carrier c.Ifne denotes the engine speed, the variable pump input shaft speed np is(1)np=nei1i2.Ignoring the effect of the volumetric efficiency of the volumetric speed control loop, the relationship between the motor output shaft speednm, the variable pump input shaft speed np, and the pump-motor displacement ratio ε is(2)nm=εnp=εnei1i2,where ε=Qp/Qm−1≤ε≤1, Qp is the displacement of the variable pump, and Qm is the displacement of the quantitative motor.The input speed of the large sun gear in PGMns1 is(3)ns1=nm−i3=εne−i1i2i3.The input speed of the ring gearnr equals ne. According to the PGM speed characteristic formula [16], the relationship between the input speed of large sun gear ns1, the input speed of ring gear nr, the output speed of carrier nc, and the PGM standing transmission ratio −k1 is satisfied:(4)ns1−k1nr−1−k1nc=0,where −k1=Zr/Zs1>1, Zr is the number of teeth of the ring gear, and Zs1 is the number of teeth of the large sun gear.Therefore, the output speed of the carriernc is(5)nc=−k1i1i2i3−ε1−k1i1i2i3ne.The intermediate shaft speednz is(6)nz=−k1i1i2i3−ε1−k1i1i2i3i4i8ne,HM1,−k1i1i2i3−ε1−k1i1i2i3i4i7ne,HM3.Finally, the output shaft speed of HMCVTno is(7)no=−k1i1i2i3−ε1−k1i1i2i3i4i8i11i12ne,HM1,−k1i1i2i3−ε1−k1i1i2i3i4i7i11i12ne,HM3.The transmission ratio of HMCVTig is(8)ig=neno=1−k1i1i2i3i4i8i11i12−k1i1i2i3−ε,HM1,1−k1i1i2i3i4i7i11i12−k1i1i2i3−εHM3.


### 3.1.2. Ranges HM2 and HM4

The power transmission routes of ranges HM2 and HM4 are the same, except for the difference between gear pairs i6 and i9. Under these two conditions, the power is input by the large sun gear s1 and the ring gear r and output by the small sun gear s2.The same as the previous case, the input speed of the large sun gearns1 is ns1=−εne/i1i2i3; the input speed of the ring gear nr equals ne. According to the PGM speed characteristic formula (16), the relationship between the input speed of large sun gear ns1, the input speed of ring gear nr, the output speed of the small sun gear ns2, and the PGM standing transmission ratios k2 and −k3 is satisfied:(9)1−k3k2−k3nr−1−k2k2−k3ns1−ns2=0.Therefore, the output speed of the small sun gearns2 is(10)ns2=1−k3i1i2i3+1−k2εk2−k3i1i2i3ne,where k2=Zs2/Zr>0, −k3=Zs2/Zs1>0, and Zr,Zs1,Zs2 are the numbers of teeth of ring gear, large sun gear, and small sun gear, respectively.The intermediate shaft speednz is(11)nz=1−k3i1i2i3+1−k2εk2−k3i1i2i3i5i9ne,HM2,1−k3i1i2i3+1−k2εk2−k3i1i2i3i5i6ne,HM4.Finally, the output shaft speed of HMCVTno is(12)no=1−k3i1i2i3+1−k2εk2−k3i1i2i3i5i9i11i12ne,HM2,1−k3i1i2i3+1−k2εk2−k3i1i2i3i5i6i11i12ne,HM4.The transmission ratio of HMCVTig is(13)ig=neno=k2−k3i1i2i3i5i9i11i121−k3i1i2i3+1−k2ε,HM2,k2−k3i1i2i3i5i6i11i121−k3i1i2i3+1−k2ε,HM4.


## 3.1.1. Ranges HM1 and HM3

The power transmission routes of ranges HM1 and HM3 are basically the same, except for the difference between gear pairs i7 and i8. Under these two conditions, the power is input by the large sun gear s1 and the ring gear r and output by the carrier c.Ifne denotes the engine speed, the variable pump input shaft speed np is(1)np=nei1i2.Ignoring the effect of the volumetric efficiency of the volumetric speed control loop, the relationship between the motor output shaft speednm, the variable pump input shaft speed np, and the pump-motor displacement ratio ε is(2)nm=εnp=εnei1i2,where ε=Qp/Qm−1≤ε≤1, Qp is the displacement of the variable pump, and Qm is the displacement of the quantitative motor.The input speed of the large sun gear in PGMns1 is(3)ns1=nm−i3=εne−i1i2i3.The input speed of the ring gearnr equals ne. According to the PGM speed characteristic formula [16], the relationship between the input speed of large sun gear ns1, the input speed of ring gear nr, the output speed of carrier nc, and the PGM standing transmission ratio −k1 is satisfied:(4)ns1−k1nr−1−k1nc=0,where −k1=Zr/Zs1>1, Zr is the number of teeth of the ring gear, and Zs1 is the number of teeth of the large sun gear.Therefore, the output speed of the carriernc is(5)nc=−k1i1i2i3−ε1−k1i1i2i3ne.The intermediate shaft speednz is(6)nz=−k1i1i2i3−ε1−k1i1i2i3i4i8ne,HM1,−k1i1i2i3−ε1−k1i1i2i3i4i7ne,HM3.Finally, the output shaft speed of HMCVTno is(7)no=−k1i1i2i3−ε1−k1i1i2i3i4i8i11i12ne,HM1,−k1i1i2i3−ε1−k1i1i2i3i4i7i11i12ne,HM3.The transmission ratio of HMCVTig is(8)ig=neno=1−k1i1i2i3i4i8i11i12−k1i1i2i3−ε,HM1,1−k1i1i2i3i4i7i11i12−k1i1i2i3−εHM3.


## 3.1.2. Ranges HM2 and HM4

The power transmission routes of ranges HM2 and HM4 are the same, except for the difference between gear pairs i6 and i9. Under these two conditions, the power is input by the large sun gear s1 and the ring gear r and output by the small sun gear s2.The same as the previous case, the input speed of the large sun gearns1 is ns1=−εne/i1i2i3; the input speed of the ring gear nr equals ne. According to the PGM speed characteristic formula (16), the relationship between the input speed of large sun gear ns1, the input speed of ring gear nr, the output speed of the small sun gear ns2, and the PGM standing transmission ratios k2 and −k3 is satisfied:(9)1−k3k2−k3nr−1−k2k2−k3ns1−ns2=0.Therefore, the output speed of the small sun gearns2 is(10)ns2=1−k3i1i2i3+1−k2εk2−k3i1i2i3ne,where k2=Zs2/Zr>0, −k3=Zs2/Zs1>0, and Zr,Zs1,Zs2 are the numbers of teeth of ring gear, large sun gear, and small sun gear, respectively.The intermediate shaft speednz is(11)nz=1−k3i1i2i3+1−k2εk2−k3i1i2i3i5i9ne,HM2,1−k3i1i2i3+1−k2εk2−k3i1i2i3i5i6ne,HM4.Finally, the output shaft speed of HMCVTno is(12)no=1−k3i1i2i3+1−k2εk2−k3i1i2i3i5i9i11i12ne,HM2,1−k3i1i2i3+1−k2εk2−k3i1i2i3i5i6i11i12ne,HM4.The transmission ratio of HMCVTig is(13)ig=neno=k2−k3i1i2i3i5i9i11i121−k3i1i2i3+1−k2ε,HM2,k2−k3i1i2i3i5i6i11i121−k3i1i2i3+1−k2ε,HM4.


## 3.2. Determination of Parameters of HMCVT

The rated speed of the engine matching with the HMCVT is 2200 r/min, the minimum speed of the tractor is 4 km/h, the maximum speed is 50 km/h, the main reducer ratio of the rear axle is 3.7, the reduction ratio of the wheel side is 5.6, the total transmission ratio of the rear axle is 20.72, and the drive wheel radius is 0.8579 m.The range of the transmission ratio in the drive train is determined by parameters such as the speed of the tractor, the rated speed of the engine, and the drive wheel radius. So the transmission ratio of each range of HMCVT is calculated as follows:(14)igj=0.377nerdv,where j = 1, 2, 3, 4; ne is the rated speed of the engine; rd is the drive wheel radius; v is the theoretical driving speed of the tractor.The HMCVT has 4 working ranges, so the common ratio of the transmission ratios is(15)φ=igmaxigmin4.The HMCVT adopts a proportional transmission, so, in each working range, the transmission ratio has the following relationship:(16)igjmaxigjmin=φ.When HMCVT is switching between adjacent ranges, the conditions must be met at the changing point: the transmission ratios are equal and the pump-motor displacement ratios are equal:(17)ig1min−1=ig2min−1,ig2min1=ig3min1,ig3min−1=ig4min−1,ig4min1=imin.The gear ratio of the connecting gears between the engine and the hydraulic variable pump needs to be met:(18)i1i2≥ne.maxnp.max.According to the previous content, ifi10, i11, and i12 are set to 1, then the relationship between the various parameters of HMCVT can be obtained:(19)k1i1i2i3=−3.2727,1−k3=k1k2−1,i5i6=0.8969k11+k1,i4i7=−1.6865k11−k1,i4i8=−5.9615k11−k1,i5i9=3.1702k11+k1.These parameters are all expressions about the PGM standing transmission ratio−k1. Once the value of −k1 is determined, the values of these parameters will also be determined, which will be described later.


## 3.3. Torque Characteristic

### 3.3.1. Ranges HM1 and HM3

Ignoring the power losses, ifTo denotes the load torque of output shaft, the intermediate shaft torque Tz is(20)Tz=Toi11i12,and then the output torque of carrier in PGMTc is(21)Tc=Tzi4i8=Toi4i8i11i12,HM1,Tzi4i7=Toi4i7i11i12,HM3.According to the PGM torque characteristic formula (16), the relationship between the input torque of large sun gear Ts1, the input torque of ring gear Tr, the output torque of carrier Tc, and the PGM standing transmission ratio −k1 is satisfied:(22)Ts1=Tr−k1=−Tc1−k1,where −k1=Zr/Zs1>1, Zr is the number of teeth of the ring gear, and Zs1 is the number of teeth of the large sun gear.Therefore, the large sun gear torqueTs1 is(23)Ts1=−Tc1−k1=−Toi4i8i11i121−k1,HM1,−Toi4i7i11i121−k1,HM3.Then, the input torque of ring gearTr is(24)Tr=−k1Ts1=k1Toi4i8i11i121−k1,HM1,k1Toi4i7i11i121−k1,HM3.Thus, the motor shaft load torqueTm is(25)Tm=Ts1−i3=Toi3i4i8i11i121−k1,HM1,Toi3i4i7i11i121−k1,HM3.The pump shaft torqueTp is(26)Tp=εTm=εToi3i4i8i11i121−k1,HM1,εToi3i4i7i11i121−k1,HM3.The engine output torqueTe is(27)Te=Tr+Tpi1i2=k1i1i2i3+εToi1i2i3i4i8i11i121−k1,HM1,k1i1i2i3+εToi1i2i3i4i7i11i121−k1,HM3.Finally, the ratio of the output torque of HMCVT to the input torque of the engineμ is(28)μ=ToTe=1−k1i1i2i3i4i8i11i12ε+k1i1i2i3,HM1,1−k1i1i2i3i4i7i11i12ε+k1i1i2i3,HM3.


### 3.3.2. Ranges HM2 and HM4

The same as the previous case,To denotes the load torque of output shaft, and the intermediate shaft torque Tz is Tz=To/i11i12. Then the output torque of small sun gear in PGM Ts2 is(29)Ts2=Tzi5i9=Toi5i9i11i12,HM2,Tzi5i6=Toi5i6i11i12,HM4.The external torque algebra sum of PGM is zero, soTs1+Tr+Ts2=0. According to the PGM torque characteristic formula (16), the relationship between the input torque of large sun gear Ts1, the output torque of small sun gear Ts2, and the PGM standing transmission ratios k2 and −k3 is satisfied:(30)Ts1=k2−1k3−k2Ts2=k2−1Tok3−k2i5i9i11i12,HM2,k2−1Tok3−k2i5i6i11i12,HM4.The engine output torqueTe is(31)Te=1−k2ε+1−k3i1i2i3k3−k2i1i2i3i5i9i11i12To,HM2,1−k2ε+1−k3i1i2i3k3−k2i1i2i3i5i6i11i12To,HM4.Finally, the ratio of the output torque of HMCVT to the input torque of the engineμ is(32)μ=ToTe=k3−k2i1i2i3i5i9i11i121−k2ε+1−k3i1i2i3,HM2,k3−k2i1i2i3i5i6i11i121−k2ε+1−k3i1i2i3,HM4.


## 3.3.1. Ranges HM1 and HM3

Ignoring the power losses, ifTo denotes the load torque of output shaft, the intermediate shaft torque Tz is(20)Tz=Toi11i12,and then the output torque of carrier in PGMTc is(21)Tc=Tzi4i8=Toi4i8i11i12,HM1,Tzi4i7=Toi4i7i11i12,HM3.According to the PGM torque characteristic formula (16), the relationship between the input torque of large sun gear Ts1, the input torque of ring gear Tr, the output torque of carrier Tc, and the PGM standing transmission ratio −k1 is satisfied:(22)Ts1=Tr−k1=−Tc1−k1,where −k1=Zr/Zs1>1, Zr is the number of teeth of the ring gear, and Zs1 is the number of teeth of the large sun gear.Therefore, the large sun gear torqueTs1 is(23)Ts1=−Tc1−k1=−Toi4i8i11i121−k1,HM1,−Toi4i7i11i121−k1,HM3.Then, the input torque of ring gearTr is(24)Tr=−k1Ts1=k1Toi4i8i11i121−k1,HM1,k1Toi4i7i11i121−k1,HM3.Thus, the motor shaft load torqueTm is(25)Tm=Ts1−i3=Toi3i4i8i11i121−k1,HM1,Toi3i4i7i11i121−k1,HM3.The pump shaft torqueTp is(26)Tp=εTm=εToi3i4i8i11i121−k1,HM1,εToi3i4i7i11i121−k1,HM3.The engine output torqueTe is(27)Te=Tr+Tpi1i2=k1i1i2i3+εToi1i2i3i4i8i11i121−k1,HM1,k1i1i2i3+εToi1i2i3i4i7i11i121−k1,HM3.Finally, the ratio of the output torque of HMCVT to the input torque of the engineμ is(28)μ=ToTe=1−k1i1i2i3i4i8i11i12ε+k1i1i2i3,HM1,1−k1i1i2i3i4i7i11i12ε+k1i1i2i3,HM3.


## 3.3.2. Ranges HM2 and HM4

The same as the previous case,To denotes the load torque of output shaft, and the intermediate shaft torque Tz is Tz=To/i11i12. Then the output torque of small sun gear in PGM Ts2 is(29)Ts2=Tzi5i9=Toi5i9i11i12,HM2,Tzi5i6=Toi5i6i11i12,HM4.The external torque algebra sum of PGM is zero, soTs1+Tr+Ts2=0. According to the PGM torque characteristic formula (16), the relationship between the input torque of large sun gear Ts1, the output torque of small sun gear Ts2, and the PGM standing transmission ratios k2 and −k3 is satisfied:(30)Ts1=k2−1k3−k2Ts2=k2−1Tok3−k2i5i9i11i12,HM2,k2−1Tok3−k2i5i6i11i12,HM4.The engine output torqueTe is(31)Te=1−k2ε+1−k3i1i2i3k3−k2i1i2i3i5i9i11i12To,HM2,1−k2ε+1−k3i1i2i3k3−k2i1i2i3i5i6i11i12To,HM4.Finally, the ratio of the output torque of HMCVT to the input torque of the engineμ is(32)μ=ToTe=k3−k2i1i2i3i5i9i11i121−k2ε+1−k3i1i2i3,HM2,k3−k2i1i2i3i5i6i11i121−k2ε+1−k3i1i2i3,HM4.


## 3.4. Circulating Power Flow Calculation

### 3.4.1. Ranges HM1 and HM3

The speed ratio of the power transmitting from the engine to the carrierice is(33)ice=ncne=−k1i1i2i3−ε1−k1i1i2i3=icer+ices1.The speed ratios of the power transmitting from the engine to the large sun gear and ring gearire/is1e are(34)ire=1,is1e=−εi1i2i3.In the PGM, the speed ratios between the input elements (large sun gears1 and ring gear r) and the output element (carrier c) are(35)ics1r=ncns1=11−k1,icrs1=ncnr=−k11−k1.Then, we can get(36)icer=ics1ris1e=−εi1i2i31−k1,ices1=icrs1ire=−k11−k1.Next two situations will be discussed as follows [17]:(1)
Whenε<0, ices1>0, icer>0, and ices1icer>0, the PGM has no circulating power PL=0.(2)
Whenε>0, ices1>0, icer<0, and ices1icer<0, the PGM has circulating power; substituting equation (19) for k1i1i2i3 in equation (33), ice is(37)ice=3.2727−ε3.2727+i1i2i3>0.Because ofPe>0,Pr=ices1/icePe>0,Ps1=icer/icePe<0, the output power P is(38)P=Pe=−Pc=Pr+Ps1=Pr−Ps1.Then,(39)Pr=P+Ps1=P+PL.The circulating powerPL is(40)PL=Ps1=icericePe.Assume that the closed power factorKL is(41)KL=PLPe=icerice=εk1i1i2i3+ε=ϕε,where ϕε=ε/k1i1i2i3+ε and the derivative of ϕε is dϕε/dε=k1i1i2i3/k1i1i2i3+ε2<0, so function ϕε is a monotonically decreasing function. Then KL0=0, KL1=0.44, and the maximum value of the circulating power is 0.44Pe.


### 3.4.2. Ranges HM2 and HM4

The speed ratio of the power transmitting from the engine to the small sun gearis2e is(42)is2e=ns2ne=1−k3i1i2i3+1−k2εk2−k3i1i2i3=is2er+is2es1.The speed ratios of the power transmitting from the engine to the large sun gear and ring gearire/is1e are(43)ire=1,is1e=−εi1i2i3.In the PGM, the speed ratios between the input elements (large sun gears1 and ring gear r) and the output element (small sun gear s2) are(44)is2s1r=k2−1k2−k3,is2rs1=1−k3k2−k3.Then, we can get(45)is2er=is2s1ris1e=−k2−1εi1i2i3k2−k3,is2es1=is2rs1ire=1−k3k2−k3.Next two situations will be discussed as follows [17]:(1)
Whenε<0, is2es1>0, is2er<0, and is2es1is2er<0, the PGM has circulating power; substituting equation (19) for k2,k3,i1i2i3 in equation (42), is2e is(46)is2e=3.2727+ε3.2727−3.2727/−k1>0.Because ofPe>0,Pr=ices1/icePe>0,Ps1=icer/icePe<0, the output power P is(47)P=Pe=−Ps2=Pr+Ps1=Pr−Ps1.Then,(48)Pr=P+Ps1=P+PL.The circulating powerPL is(49)PL=Ps1=is2eris2ePe.Assume that the closed power factorKL is(50)KL=PLPe=is2eris2e=−εk1i1i2i3−ε=ϕε,whereϕε=−ε/k1i1i2i3−ε and the derivative of ϕε is dϕε/dε=−k1i1i2i3/k1i1i2i3−ε2>0, so function ϕε is a monotonically increasing function. Then, KL0=0, KL1=0.44, and the maximum value of the circulating power is 0.44Pe.(2)
Whenε>0, is2es1>0, is2er>0, and is2es1is2er>0, the PGM has no circulating power PL=0.


## 3.4.1. Ranges HM1 and HM3

The speed ratio of the power transmitting from the engine to the carrierice is(33)ice=ncne=−k1i1i2i3−ε1−k1i1i2i3=icer+ices1.The speed ratios of the power transmitting from the engine to the large sun gear and ring gearire/is1e are(34)ire=1,is1e=−εi1i2i3.In the PGM, the speed ratios between the input elements (large sun gears1 and ring gear r) and the output element (carrier c) are(35)ics1r=ncns1=11−k1,icrs1=ncnr=−k11−k1.Then, we can get(36)icer=ics1ris1e=−εi1i2i31−k1,ices1=icrs1ire=−k11−k1.Next two situations will be discussed as follows [17]:(1)
Whenε<0, ices1>0, icer>0, and ices1icer>0, the PGM has no circulating power PL=0.(2)
Whenε>0, ices1>0, icer<0, and ices1icer<0, the PGM has circulating power; substituting equation (19) for k1i1i2i3 in equation (33), ice is(37)ice=3.2727−ε3.2727+i1i2i3>0.Because ofPe>0,Pr=ices1/icePe>0,Ps1=icer/icePe<0, the output power P is(38)P=Pe=−Pc=Pr+Ps1=Pr−Ps1.Then,(39)Pr=P+Ps1=P+PL.The circulating powerPL is(40)PL=Ps1=icericePe.Assume that the closed power factorKL is(41)KL=PLPe=icerice=εk1i1i2i3+ε=ϕε,where ϕε=ε/k1i1i2i3+ε and the derivative of ϕε is dϕε/dε=k1i1i2i3/k1i1i2i3+ε2<0, so function ϕε is a monotonically decreasing function. Then KL0=0, KL1=0.44, and the maximum value of the circulating power is 0.44Pe.


## 3.4.2. Ranges HM2 and HM4

The speed ratio of the power transmitting from the engine to the small sun gearis2e is(42)is2e=ns2ne=1−k3i1i2i3+1−k2εk2−k3i1i2i3=is2er+is2es1.The speed ratios of the power transmitting from the engine to the large sun gear and ring gearire/is1e are(43)ire=1,is1e=−εi1i2i3.In the PGM, the speed ratios between the input elements (large sun gears1 and ring gear r) and the output element (small sun gear s2) are(44)is2s1r=k2−1k2−k3,is2rs1=1−k3k2−k3.Then, we can get(45)is2er=is2s1ris1e=−k2−1εi1i2i3k2−k3,is2es1=is2rs1ire=1−k3k2−k3.Next two situations will be discussed as follows [17]:(1)
Whenε<0, is2es1>0, is2er<0, and is2es1is2er<0, the PGM has circulating power; substituting equation (19) for k2,k3,i1i2i3 in equation (42), is2e is(46)is2e=3.2727+ε3.2727−3.2727/−k1>0.Because ofPe>0,Pr=ices1/icePe>0,Ps1=icer/icePe<0, the output power P is(47)P=Pe=−Ps2=Pr+Ps1=Pr−Ps1.Then,(48)Pr=P+Ps1=P+PL.The circulating powerPL is(49)PL=Ps1=is2eris2ePe.Assume that the closed power factorKL is(50)KL=PLPe=is2eris2e=−εk1i1i2i3−ε=ϕε,whereϕε=−ε/k1i1i2i3−ε and the derivative of ϕε is dϕε/dε=−k1i1i2i3/k1i1i2i3−ε2>0, so function ϕε is a monotonically increasing function. Then, KL0=0, KL1=0.44, and the maximum value of the circulating power is 0.44Pe.(2)
Whenε>0, is2es1>0, is2er>0, and is2es1is2er>0, the PGM has no circulating power PL=0.


## 3.5. Hydraulic Split Power Calculation

Hydraulic power-split ratioρ=PH/Pe=Ps1/Pe; hydraulic power is input to the PGM through the large sun gear s1.


### 3.5.1. Ranges HM1 and HM3

(51)ρ=Ps1Pe=icerice=εk1i1i2i3+ε=ϕε,and as can be obtained from the above, functionϕε is a monotonically decreasing function, so(52)ρ−1=0.234>0,ρ0=0,ρ1=−0.44<0.


### 3.5.2. Ranges HM2 and HM4

(53)ρ=Ps1Pe=is2eris2e=−εk1i1i2i3−ε=ϕε.As can be obtained from the above, functionϕε is a monotonically increasing function, so(54)ρ−1=−0.44<0,ρ0=0,ρ1=0.234>0.According to the theoretical analysis on the circulating power flow and the hydraulic split power, the power flow direction of HMCVT in each working range is shown in Figure2. This result lays the foundation for the analysis on the efficiency characteristic of HMCVT.Figure 2
Power flow direction of the PGM of HMCVT. (a) Range HM1/HM3,ε>0, power-circulation. (b) Range HM1/HM3, ε<0, power-split. (c) Range HM2/HM4, ε>0, power-split. (d) Range HM2/HM4, ε<0, power-circulation.
(a)(b)(c)(d)


## 3.5.1. Ranges HM1 and HM3

(51)ρ=Ps1Pe=icerice=εk1i1i2i3+ε=ϕε,and as can be obtained from the above, functionϕε is a monotonically decreasing function, so(52)ρ−1=0.234>0,ρ0=0,ρ1=−0.44<0.


## 3.5.2. Ranges HM2 and HM4

(53)ρ=Ps1Pe=is2eris2e=−εk1i1i2i3−ε=ϕε.As can be obtained from the above, functionϕε is a monotonically increasing function, so(54)ρ−1=−0.44<0,ρ0=0,ρ1=0.234>0.According to the theoretical analysis on the circulating power flow and the hydraulic split power, the power flow direction of HMCVT in each working range is shown in Figure2. This result lays the foundation for the analysis on the efficiency characteristic of HMCVT.Figure 2
Power flow direction of the PGM of HMCVT. (a) Range HM1/HM3,ε>0, power-circulation. (b) Range HM1/HM3, ε<0, power-split. (c) Range HM2/HM4, ε>0, power-split. (d) Range HM2/HM4, ε<0, power-circulation.
(a)(b)(c)(d)


## 3.6. Efficiency Characteristic

The Крейнeс formula (17) is used to calculate the transmission efficiency of HMCVT:(55)η=−μ˜ig=−To/Tene/no,where μ˜ is the ratio of output torque to input torque of HMCVT considering power losses and ig is the transmission ratio of HMCVT.The transmission efficiency calculation process of each working range of HMCVT is shown in Tables2 and 3. Among them, the calculation processes of the first range and the third range are listed in Table 2, and the calculation processes of the second range and the fourth range are listed in Table 3. In Tables 2 and 3, i1−i12 are the gear ratios of the gear pairs, η1−η12 are the transmission efficiency of the gear pairs, ηH is the hydraulic system transmission efficiency, and ηs1r, ηs1s2, and ηrs2 are the transmission efficiencies of the large sun gear to the ring gear, the large sun gear to the small sun gear, and the ring gear to the small sun gear in the PGM, respectively.Table 2
Equation of efficiency calculation for ranges HM1 and HM3.
ElementsTorque equation considering power lossesε>0ε<0HM1HM3HM1HM3EngineTe=Tr+η1η2Tp/i1i2Te=Tr+Tp/i1i2η1η2PumpTp=εηHTmTp=εTm/ηHMotorTm=η3Ts1/−i3Tm=Ts1/−η3i3PGMTs1+Tr+Tc=0PGMTr=−k1Ts1/ηs1rIntermediate shaftTc=Tz/i4i8η4η8Tc=Tz/i4i7η4η7Tc=Tz/i4i8η4η8Tc=Tz/i4i7η4η7Output shaftTz=To/i11i12η11η12Torque ratioμ˜=To/TeEfficiencyη=−μ˜/igTable 3
Equation of efficiency calculation for ranges HM2 and HM4.
ElementsTorque equation considering power lossesε>0ε<0HM2HM4HM2HM4EngineTe=Tr+Tp/i1i2η1η2Te=Tr+η1η2Tp/i1i2PumpTp=εTm/ηHTp=εηHTmMotorTm=Ts1/−η3i3Tm=η3Ts1/−i3PGMTs1+Tr+Ts2=0PGMTr=1−k3ηs1s2/k3ηs1s2−k2ηrs2Ts2Intermediate shaftTs2=Tz/i5i9η5η9Ts2=Tz/i5i6η5η6Ts2=Tz/i5i9η5η9Ts2=Tz/i5i6η5η6Output shaftTz=To/i11i12η11η12Torque ratioμ˜=To/TeEfficiencyη=−μ˜/ig


## 4. Analysis on the Influence of Parameters on the Transmission Efficiency of HMCVT

The results of the transmission efficiency in each working range of HMCVT are as follows:(56)ηHM1=ηs1r−k1η4η8η11η12ε+k1i1i2i31−k1η1η2η3ηHηs1rε+k1i1i2i3,ε>0,ηs1r−k1η1η2η3ηHη4η8η11η12ε+k1i1i2i31−k1i1i2i3η1η2η3ηHk1+εηs1r,ε≤0,ηHM2=k3ηs1s2−k2ηrs2η5η9η11η121−k3i1i2i3+1−k2εk3−k21−k3ηs1s2i1i2i3−η1η2η3ηHεk2ηrs2−1,ε<0,k3ηs1s2−k2ηrs2η1η2η3η5η9η11η12ηH1−k3i1i2i3+1−k2εk3−k21−k3ηs1s2i1i2i3η1η2η3ηH−εk2ηrs2−1,ε≥0,ηHM3=ηs1r−k1η4η7η11η12ε+k1i1i2i31−k1η1η2η3ηHηs1rε+k1i1i2i3,ε>0,ηs1r−k1η1η2η3ηHη4η7η11η12ε+k1i1i2i31−k1i1i2i3η1η2η3ηHk1+εηs1r,ε≤0,ηHM4=k3ηs1s2−k2ηrs2η5η6η11η121−k3i1i2i3+1−k2εk3−k21−k3ηs1s2i1i2i3−η1η2η3ηHεk2ηrs2−1,ε<0,k3ηs1s2−k2ηrs2η1η2η3η5η6η11η12ηH1−k3i1i2i3+1−k2εk3−k21−k3ηs1s2i1i2i3η1η2η3ηH−εk2ηrs2−1,ε≥0.When HMCVT is working in HM1 or HM3 range, if the transmission efficiencies of the gear pairs i9 and i8 are equal η7=η8, the transmission efficiencies of HMCVT in HM1 and HM3 ranges are equal.When HMCVT is working in HM2 or HM4 range, if the transmission efficiencies of the gear pairs i9 and i6 are equal η9=η6, the transmission efficiencies of HMCVT in HM2 and HM4 ranges are also equal.Due to the complexity of the above results, it is necessary to evaluate the influence of displacement ratioε, standing transmission ratio of PGM −k1, gear ratio i1i2i3, and hydraulic system efficiency ηH on the transmission efficiency of HMCVT by means of image analysis method. We assume that η1−η12 are 0.98, ηs1r is 0.98, ηs1s2 and ηrs2 are 0.97, and k1i1i2i3 is −3.2727. The evaluation results are shown in Figure 3.Figure 3
The influence of parameters on the transmission efficiency of HMCVT. (a) The influence of−k1 on HM1 and HM3 ranges. (b) The influence of −k1 on HM2 and HM4 ranges. (c) The influence of ηH on HM1 and HM3 ranges. (d) The influence of ηH on HM2 and HM4 ranges. (e) The influence of i1i2i3 on HM1 and HM3 ranges. (f) The influence of i1i2i3 on HM2 and HM4 ranges.
(a)(b)(c)(d)(e)(f)According to the analysis on Figure3, the following rules are obtained.In each range of HMCVT, when the displacement ratioε is 0, the transmission efficiency reaches the maximum value. Now the hydraulic path (HST) does not transmit power, and the power is completely transmitted by the mechanical path. When the displacement ratio ε is −1 or 1, the transmission efficiency reaches the minimum value, and the hydraulic path transmits more power at this time, resulting in a lower overall efficiency of HMCVT.In each range of HMCVT, the transmission efficiency value in the power circulation phase is lower than the transmission efficiency value in the power split phase, which indicates that the power circulation will cause the transmission efficiency of HMCVT to decrease.The larger the value of the PGM standing transmission ratio−k1, the higher the transmission efficiency of HMCVT, but the influence on improving the transmission efficiency is not obvious. The PGM standing transmission ratio −k1 ranges from 1.5 to 3. The gear ratio i1i2i3 and the PGM standing transmission ratio −k1 are in an inversely proportional relationship; the smaller the value of the gear ratio i1i2i3, the higher the transmission efficiency of HMCVT, and the influence on improving the transmission efficiency is also not obvious. The gear ratio i1i2i3 ranges from 1.1 to 2.2.The greater the value of HST efficiencyηH, the higher the overall transmission efficiency of HMCVT. The influence of ηH on improving transmission efficiency of HMCVT is significant.


## 5. Analysis on the Transmission Characteristics Results of HMCVT

According to the previous content, the value of the PGM standing transmission ratio−k1 is set to 2.1875, and the HST efficiency value ηH is set to 0.81; then the other parameters of HMCVT are k2=0.2714, −k3=0.5938, i1=i2=1, i3=1.5, i4=i5=1, i6=1.66, i7=1.16, i8=4.10, i9=5.84, and i10=i11=i12=1.The relationship between the transmission ratio and displacement ratio is shown in Figure4. The curve of the tractor speed is shown in Figure 5. In the first range, the displacement ratio changes from 1 to −1, the transmission ratio decreases from 8.62 to 4.59, and the tractor speed increases from 0 to 7.50 km/h. In the second range, the displacement ratio changes from −1 to 1, the transmission ratio decreases from 4.59 to 2.43, and the tractor speed increases from 7.50 km/h to 14.12 km/h. In the third range, the displacement ratio changes from 1 to −1, the transmission ratio decreases from 2.43 to 1.30, and the tractor speed increases from 14.12 km/h to 26.44 km/h. In the fourth range, the displacement ratio changes from −1 to 1, the transmission ratio decreases from 1.30 to 0.69, and the tractor speed increases from 26.44 km/h to 50 km/h. The results show that the stepless speed regulation of HMCVT in the tractor speed range of 0–50 km/h can be realized by the coordinated control of the HST displacement ratio and the clutches.Figure 4
Relationship between transmission ratio and displacement ratio.Figure 5
The curve of the tractor speed.The ratios of the output shaft torque to the input shaft torque in each range of HMCVT are shown in Figure6(a). The continuous change of torque can be realized in the entire working ranges. The ratios of the motor shaft torque to the output shaft torque in each range of HMCVT are shown in Figure 6(b). The torque ratio remains constant in each range and varies between different ranges. The values of torque ratios in the four ranges are 5.14%, −9.72%, 18.49%, and −35.18%, respectively.Figure 6
The curve of torque characteristic. (a) Ratios of output shaft torque to input shaft torque. (b) Ratios of motor shaft torque to output shaft torque.
(a)(b)The relationship between the tractor speed and the overall transmission efficiency of HMCVT is shown in Figure7. Each of the four ranges has a transmission efficiency maximum value of 0.917, 0.902, 0.917, and 0.902, respectively. The efficiency value always increases first with the increase of the tractor speed and then decreases after reaching the maximum value. The maximum transmission efficiency in each range is obtained when the HST displacement ratio is 0. Now, the hydraulic path does not transmit power and all power is transmitted by the mechanical path. The transmission efficiency of HMCVT is higher than 83%, so the overall transmission efficiency is at a high level.Figure 7
Overall transmission efficiency of HMCVT.


## 6. Verification of HMCVT Simulation Model

The modeling and simulation of HMCVT are accomplished by using Amesim simulation software. According to Figure1, a simulation model of HMCVT is established, as shown in Figure 8.Figure 8
HMCVT simulation model.


### 6.1. Hydraulic System

The hydraulic system is mainly composed of a variable displacement hydraulic pump, a fixed displacement hydraulic motor, two relief valves, two check valves, and a charge hydraulic pump with a relief valve. The aim of the charge pump is to compensate for the leakage loss in the pipeline and to establish the lowest pressure in the pump’s suction pipeline. Two relief valves are used to limit the maximum pressure in the system piping, and its adjustment pressure should be above the normal working pressure of the system. Volumetric losses of the pump and motor are neglected, and their hydromechanical efficiencies are set at 0.9.


### 6.2. Mechanical Components

Mechanical components are assumed to be constant efficiency components throughout the transmission. The power loss of each gear pair is 2%, so the efficiency is set to 0.98. In the Ravigneaux planetary gear mechanism, the efficiencies of ring gear-long planetary gear couple, long planetary gear-large sun gear couple, long planetary gear-short planetary gear couple, and short planetary gear-small sun gear couple are set at 0.99. The main reducer, differential, and wheel reducer are combined and modeled as a gear reducer with an efficiency of 0.965. To simplify the model, the friction losses of the shaft, gears, and bearings were ignored during the simulation modeling process.


### 6.3. Model Control and Simulation Results

The engine speed is assumed to have a constant value of 2200 r/min; the load torque of the output shaft of HMCVT is set to 100 Nm; the model simulation time is 32 s, including range HM1: 0–8 s, range HM2: 8–16 s, range HM3: 16–24 s, and range HM4: 24–32 s.The simulation results are shown in Figure9; through the variable pump displacement and coordinated control of the clutches [18, 19], the transmission gradually changes from range HM1 to range HM4, and the tractor achieves stepless speed change from 0 to 50 km/h. The simulation results are consistent with the previous theoretical analysis results, which show that the 4-range HMCVT transmission scheme proposed in this paper is suitable for matching wheeled tractors.Figure 9
Simulation results. (a) Simulation result of the tractor speed. (b) Simulation result of the transmission efficiency of HMCVT.
(a)(b)


### 6.4. Comparison

Xu LY’s transmission scheme [15] has been applied to mass production tractors. This transmission scheme has 6 working ranges in the forward direction and 3 working ranges in the reverse direction, which is controlled by 8 clutches. The speed range in the forward direction is 0∼30 km/h, while the reverse direction is only 0∼10 km/h. The transmission efficiency of the hydraulic range is low (less than 80%); the transmission efficiency of the hydromechanical ranges is high (higher than 90%) but is only in the forward direction.The scheme proposed in this paper has 4 working ranges and a speed range of 0∼50 km/h in the forward and reverse directions, which is controlled by 6 clutches. The fewer the transmission ranges, the less switching during the working ranges, so the transmission performance is improved. The smaller number of clutches ensures that the system is easier to control. The transmission efficiencies in the forward and reverse directions are close, the overall forward transmission efficiency is higher than 83%, and the highest transmission efficiency is higher than 91%.Therefore, the application of HMCVT proposed in this paper is wider; it can not only match conventional tractors but also match two-way operation tractors and other engineering vehicles.


## 6.1. Hydraulic System

The hydraulic system is mainly composed of a variable displacement hydraulic pump, a fixed displacement hydraulic motor, two relief valves, two check valves, and a charge hydraulic pump with a relief valve. The aim of the charge pump is to compensate for the leakage loss in the pipeline and to establish the lowest pressure in the pump’s suction pipeline. Two relief valves are used to limit the maximum pressure in the system piping, and its adjustment pressure should be above the normal working pressure of the system. Volumetric losses of the pump and motor are neglected, and their hydromechanical efficiencies are set at 0.9.


## 6.2. Mechanical Components

Mechanical components are assumed to be constant efficiency components throughout the transmission. The power loss of each gear pair is 2%, so the efficiency is set to 0.98. In the Ravigneaux planetary gear mechanism, the efficiencies of ring gear-long planetary gear couple, long planetary gear-large sun gear couple, long planetary gear-short planetary gear couple, and short planetary gear-small sun gear couple are set at 0.99. The main reducer, differential, and wheel reducer are combined and modeled as a gear reducer with an efficiency of 0.965. To simplify the model, the friction losses of the shaft, gears, and bearings were ignored during the simulation modeling process.


## 6.3. Model Control and Simulation Results

The engine speed is assumed to have a constant value of 2200 r/min; the load torque of the output shaft of HMCVT is set to 100 Nm; the model simulation time is 32 s, including range HM1: 0–8 s, range HM2: 8–16 s, range HM3: 16–24 s, and range HM4: 24–32 s.The simulation results are shown in Figure9; through the variable pump displacement and coordinated control of the clutches [18, 19], the transmission gradually changes from range HM1 to range HM4, and the tractor achieves stepless speed change from 0 to 50 km/h. The simulation results are consistent with the previous theoretical analysis results, which show that the 4-range HMCVT transmission scheme proposed in this paper is suitable for matching wheeled tractors.Figure 9
Simulation results. (a) Simulation result of the tractor speed. (b) Simulation result of the transmission efficiency of HMCVT.
(a)(b)


## 6.4. Comparison

Xu LY’s transmission scheme [15] has been applied to mass production tractors. This transmission scheme has 6 working ranges in the forward direction and 3 working ranges in the reverse direction, which is controlled by 8 clutches. The speed range in the forward direction is 0∼30 km/h, while the reverse direction is only 0∼10 km/h. The transmission efficiency of the hydraulic range is low (less than 80%); the transmission efficiency of the hydromechanical ranges is high (higher than 90%) but is only in the forward direction.The scheme proposed in this paper has 4 working ranges and a speed range of 0∼50 km/h in the forward and reverse directions, which is controlled by 6 clutches. The fewer the transmission ranges, the less switching during the working ranges, so the transmission performance is improved. The smaller number of clutches ensures that the system is easier to control. The transmission efficiencies in the forward and reverse directions are close, the overall forward transmission efficiency is higher than 83%, and the highest transmission efficiency is higher than 91%.Therefore, the application of HMCVT proposed in this paper is wider; it can not only match conventional tractors but also match two-way operation tractors and other engineering vehicles.


## 7. Conclusions

This paper proposes a new type of transmission scheme of HMCVT. The Ravigneaux PGM and the HST are connected in parallel to form a closed loop and then connected in series with the range switching mechanism. This new HMCVT features a double-row PGM with 4 hydromechanical working ranges. By coordinating control of the HST displacement ratio and the engagement conditions of shifting clutches, the stepless speed regulation of HMCVT at the tractor speed of 0–50 km/h is realized.According to the theoretical analysis on the new HMCVT, the speed characteristic curve and the torque characteristic curve are obtained. Based on the power flow direction of HMCVT, the transmission efficiency curve is obtained by using the Крейнeс formula.Using image analysis method to study the influence of four parameters (displacement ratioε, PGM standing transmission ratio −k1, gear ratio i1i2i3, and hydraulic system efficiency ηH) on the transmission efficiency of HMCVT, the results show that the standing transmission ratio −k1 and gear ratio i1i2i3 have little effect on the transmission efficiency, whereas the displacement ratio ε and hydraulic system efficiency ηH have a greater impact on the transmission efficiency.The parameters of the new HMCVT are finally determined. The results of the transmission characteristics show that the new HMCVT proposed in this paper has the stepless speed regulation capability for tractors, which can also continuously transmit and change the torque. The transmission efficiency of HMCVT is at a high level.In order to verify the theoretical derivation, Amesim simulation software is used for the modeling and simulation of HMCVT. The simulation results are consistent with the theoretical analysis results. Therefore, we believe that the transmission scheme is reasonable and effective for tractors.



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*Source: 6978329-2020-08-12.xml*