dynamic analysis of a novel geared infinitely variable

DYNAMIC ANALYSIS OF A NOVEL GEARED INFINITELY VARIABLE TRANSMISSION

Z. R. LiMechanical Engineering Department

University of Maryland, Baltimore CountyBaltimore, MD [email protected]

X. F. WangMechanical Engineering Department

University of Maryland, Baltimore CountyBaltimore, MD [email protected]

W. D. ZhuMechanical Engineering Department

University of Maryland, Baltimore CountyBaltimore, MD 21250

[email protected]

ABSTRACTA novel geared infinitely variable transmission (IVT) that

can generate a continuous output-to-input speed ratio from zero to a certain value is studied for vehicle and wind turbine applications. The principle of changing the output-to-input speed ratio is to use a crank-slider mechanism; the output-to-input speed ratio is controlled by adjusting the crank length. Since the crank-slider mechanism can lead to relatively largevariation of the output-to-input speed ratio in one rotation of the crank, the instantaneous input and output speeds and accelerations have variations and the corresponding forcesexerted on each part of the IVT can have obvious changes in one rotation of the crank. Since forces on some parts of the IVT are critical and can cause failure of the IVT, a dynamic analysis of the IVT is necessary to simulate the input and output speeds and accelerations. A method that combinesLagrangian dynamics and Newtonian dynamics is developed in this work to analyze the motion of the IVT. The dynamic analysis results can be used to evaluate the design of the IVT.

1. INTRODUCTIONTraditional transmission mainly consists of a number of

gears with multiple gear ratios and the output-to-input speedratio of the transmission is varied by switching different pairs of meshed gears [1]. In a vehicle application, one end of the transmission is connected to an engine crank shaft by a clutch,and the other end is connected to a wheel shaft. Most vehicle engines need to operate at a relatively high rotational speed,and the speed varies when a vehicle is starting, accelerating, decelerating, and stopping. The transmission reduces thehigher speed of an engine to a lower wheel speed when the vehicle starts to move, simultaneously increasing the output torque for generating enough propulsion. After starting up the vehicle, the resultant friction exerted on the wheels decreasesto a lower value and the transmission gradually increases the speed ratio to a relatively high level to yield a higher outputspeed. Generally, an engine has a maximum operating

efficiency at a particular rotational speed; traditional transmission cannot achieve high engine efficiency for many working conditions due to a discrete selection of gear ratios of the transmission.

In a wind turbine application, one end of the transmission is connected to a wind turbine shaft, and the other end is connected to a generator. The power extracted from wind by the turbine rotor is affected by the tip ratio, which is the ratio of the tip speed of the blades and the wind speed [2]. To achieve the maximum efficiency of a wind turbine, the rotor speed varies with the wind speed. Wind turbines that use the traditional transmission to transmit wind energy to generators can only operate at a constant rotor speed [3]. Consequently,wind energy at variable wind speeds cannot be extracted at the maximum efficiency. Wind turbines that use continuouslyvariable transmission (CVT) or power converters can harvestmaximum wind energy from variable wind speeds [4]-[6]. However, a power converter is expensive and has a high failure rate; CVT can overcome these disadvantages [7].

In order to achieve continuous speed variation, some types of CVT are in production. A system that consists of two variable-diameter pulleys (VDP) with one belt mounted on them is the most common type of CVT used in vehicle applications [8]. The ratio of the transmission is changed by moving two sheaves of one pulley closer together or fartherapart. The efficiency of the VDP is low due to the low efficiency of the transmission belt [9]. Another system that is made up of discs and rollers, called toroidal CVT [10], is also used in some vehicle applications. The toroidal CVT is relatively efficient, but more complex than the traditional CVT, and some toroidal CVT can achieve a zero output-to-input speed ratio. A new type of CVT that uses a magnetic principle has been developed since 2006 [11]. The magnetic transmission ratio can be adjusted by importing or exporting electrical power to change the gear ratio. While the magnetic

1 Copyright © 2014 by ASME

Proceedings of the ASME 2014 International Mechanical Engineering Congress and Exposition IMECE2014

November 14-20, 2014, Montreal, Quebec, Canada

IMECE2014-36551

CVT is highly efficient, it cannot generate a large torque due to its operating principle.

In addition to the mechanical or magnetic CVT discussed above, a hydrostatic CVT system [12] with a variable displacement pump and a hydraulic motor is developed to transmit a larger torque compared to the mechanical CVT, butit is more expensive and complex. This type of CVT is relatively easy to use and stable, and has high efficiency [13]. However, the hydrostatic CVT system is more sensitive to contamination and may cause environmental contamination.

Most of the CVT mentioned above cannot reach a zero speed ratio, which means that a vehicle with a CVT always has to be installed with a fluid coupling or torque converter [14] between the engine and the gearbox to help start the vehicle. A specific type of CVT, called infinitely variable transmission (IVT), which can continuously vary the speed ratio from zero to a specific value, has been developed. Most of IVT results from CVT combined with a planetary gear system, which is used for the speed ratio to vary from zero to the minimum speed ratio that the CVT can reach [15]. Another type of IVT that mainly consists of spur gears and two eccentric drivers, which convert an eccentric motion of cams to a concentric motion of the output shaft by using one-way clutches, is developed [16]. It can transmit a larger torque and is scalable. However, the friction between the cams cangenerate heat and the principle of the eccentric drivers causeslarge instantaneous variation of the output speed, which canresult in non-smooth motion of a vehicle. Also, small pins that are used to connect the cams and the drivers are not durable.To reduce variation of the output speed and improve the performance of the IVT, a new type of IVT is designed and analyzed [17].

The new type of IVT in Ref. [17] uses a crank-slider system and is all geared so that it can transmit a larger torque with higher efficiency than the IVT in Ref. [16]; there are no small pins and cams in the new design. The instantaneous variation of the output speed is reduced to a much lower level by using four eccentric drivers. However, each single eccentric driver of the IVT has a large speed variation in one rotation of the crank due to the eccentric motion of the driver, and the variation of the inertial force due to the speed variation can be significant. In vehicle and wind turbine applications, momentsof inertia of the output and input shafts are so large that a small speed variation of the shafts can result in large force variations in them, respectively. A dynamic analysis thatconsiders the IVT as a multi-rigid-body system is proposed in this work to provide a comprehensive understanding of the dynamic behavior of the IVT.

Dynamic analysis focuses on studying the influence of atime-varying load on the motion of a single part of a structureor the overall structure [18]. The new type of IVT is a highly complicated multi-rigid-body system with many constraints; simply using Newtonian dynamics would result in complicated analysis. To obtain the equation of motion of the IVT and calculate each force in it, a method that combinesLagrangian dynamics and Newtonian dynamics is developed.

The Lagrangian approach can express and solve the systemequation with one generalized coordinate, without considering constraint forces [19]. Time histories of the vehicle and generator speeds can be derived using the results that are obtained from the Lagrangian approach for the vehicle andwind turbine applications, respectively. With the accelerationassociated with the generalized coordinate calculated from thesystem equation, the forces exerted on each part of the IVTcan be subsequently obtained using a Newtonian approach. Acomputer program is written using MATLAB to calculate the generalized acceleration, and the velocity of and the forces on each part of the IVT, and to verify the results.

2. DESCRIPTION OF A GEARED IVTThe IVT discussed here is a fully mechanical gearbox that

can achieve a continuously varied output-to-input speed ratio from zero to a specific value. To achieve the function of theIVT, two modules are developed: a driver module thatconverts eccentric motions of drivers to a rotational motion of the output shaft, and a control module that adjusts the eccentricity of the motions of the drivers during operation of the IVT, as shown in Fig. 1.

2.1. Driver moduleThe driver module has four drivers, and each driver uses a

crank-slider mechanism shown in Fig. 2. A driver consists of two crank gears, two output gears, a joint with two side racks, a driver case, and two one-way bearings, as shown in Fig. 3.The joint installed on the crank shaft has two racks: one on the upper side that is meshed with the right crank gear, and the other on the lower side that is meshed with the left crank gear, as shown in Fig. 4. The length between the center of the crank gear, which is denoted by O , and that of the joint, which is

denoted by 1O , is the eccentricity.

When the two crank gears rotate about the center O in

Fig. 2 with the same angular velocity, the joint only rotatesabout the center O , which means that the eccentricity of the

motion of the driver remains unchanged. When the crank gears rotate with different angular velocities, the motion of the joint is a superposition of a rotation about the center O and a

translation that can change the eccentricity. The driver casethat connects the joint to the output gears converts the eccentric motion of the driver to a rotational motion of the output shaft. Since the output gears are installed on the output shaft through two one-way bearings, the output shaft rotates in a preset direction that depends on how the one-way bearings are installed. The four drivers are installed with a 90°angledifference between two adjacent joints, and there is only onedriver that drives the output shaft at each time instant. Due to this implementation, the angular velocity of the output shaft isthe maximum angular velocity of the eight output gears.

2.2. Control moduleThe control module mainly consists of two groups of

planetary gears, two control gears, and two control shafts, as


shown in Fig. 5 and Fig. 6. The carrier is installed on the input shaft by a key connection; the planetary gears that are driven by the carrier through four small shafts are meshed with the sun gears and the ring gears. The control gears that are installed on the control shaft by a key connection are used to change the angular velocity of the ring gears. When the carrier rotates about the center of the input shaft, the control shafts remain stationary, and the planetary gears drive the upper and lower sun gears to rotate about the center of the carrier withthe same angular velocity. If the ring gears have a speed difference caused by the control gears, the upper and lower sun gears correspondingly rotate with a different angular velocity. Since the sun gears are meshed with the crank gears, the eccentricity will be changed by the rotation of the control shafts.

3. LAGRANGIAN FORMULATION OF THE IVT The IVT is a one-degree-of-freedom system when the

control shaft remains stationary, and the dynamic analysis of the IVT is considered in two dimensions. To more easily calculate the velocity and acceleration of each part of the IVT,

the angle between 1OO and 2OO in Fig. 7 and Fig. 8, which

is denoted by 1 , is selected as the generalized coordinate.

Four drivers in the gearbox are alternatively engaged to work;

only one of the drivers is engaged at each time instant, and 1corresponds to the working driver.

3.1. Kinematic model of the driver moduleWhen both sides of the crank gear rotate with the same

angular velocity, which means that the eccentricity remains unchanged, the angular velocity of the joint, which is denoted

by J , is equal to that of the crank gear, which is denoted by

CG . The motion of the joint is the superposition of the

translation perpendicular to 1OO , which is a circular motion

of the center of the joint around that of the crank gear O , and

the rotation about the center of the joint. When the center of

the joint 1O rotates about O with the eccentricity d , that of

one side of the driver that is meshed with the joint by a needle bearing rotates about O with the eccentricity d , as shown in

Fig. 7. The other side of the driver is meshed with the output gear. The motion of the driver is the superposition of two

translations along i

, which is perpendicular to 1 2O O , and j

,

which is along 1 2O O , and the rotation about the centroid C of

the driver, as shown in Fig. 7. The angle between 1 2O O and

2OO is denoted by 2 , and the length of 1 2O O is denoted

by 2l . Hence, the translation velocity of the joint Jv , the

translation velocity of the driver jv along j

, the translation

velocity of the driver iv along i

, and the angular velocity of

the driver D are 1d , 2l , 2 2 1( )l CO , and 2 ,

respectively. Using laws of sines and cosines, the translation

velocity of the joint 1d has two components in the i

and j

directions, as shown in Fig. 8. Hence, jv , iv , and D are

1 12 1 2 2

1 1 1

sin

2 cosj

l dv l

l d l d

(1)

2 2 1

22 21 1

1 1 1 1 12 21 1 1

( )

cos( 2 cos )

2 cos

iv l CO

d l dl d l d CO

l d l d

(2)

21 1

2 1 2 21 1 1

cos

2 cosD

d l d

l d l d

(3)

respectively, where 1l is the length of 2OO . The angular

velocities of a pair of output gears, which are denoted by

1OG and 2OG , and that of the output shaft OS are

2

2 1 11 1 2 1 2 2

1 1 1

1 11 2 2

1 1 1

cos

2 cos

sin

2 cos

OGOG

OG

l d l d

R l d l d

l d

R l d l d

(4)

2

2 1 12 1 2 1 2 2

1 1 1

1 11 2 2

1 1 1

cos

2 cos

sin

2 cos

OGOG

OG

l d l d

R l d l d

l d

R l d l d

(5)

21 1 2

1 11 2 2

1 1 1

21 1

1 2 21 1 1

,

sin

2 cos

cos

2 cos

OSOG

OG

l

R

l d

R l d l d

d l d

l d l d

(6)

respectively, where OGR is the radius of the output gear.

The acceleration of each part of a driver can be obtained by differentiating the corresponding velocity with respect totime t . To calculate the acceleration of the centroid of the

driver, which is denoted by Ca , a simplified model of the

driver in a new coordinate system with the origin at 2O is

established, as shown in Fig. 9. The position of C is

2 2 1O C j l CO j (7)

Differentiating 2 1l CO j

with respect to t twice yields


2

2 12

2 2 1 2

2 2 2 1 2

22 2 1 2

2

C

da l CO j

dtd d

l j l CO idt dt

l l CO i

l l CO j

(8)

The accelerations of the centroid in the i

and j

directions,

which are denoted by ia and ja , respectively, are

2 2 2 1 22 ( )ia l l CO (9)

22 2 1 2( )ja l l CO (10)

3.2. Kinematic model of the control moduleThe angular velocity of the sun gear, which is meshed

with the crank gear and the planetary gear, is denoted by SG ,

as shown in Fig. 10, and can be expressed by

1 CGSG

SG

R

R

(11)

where CGR is the radius of the outer crank gear and SGRis the radius of the outer sun gear. The angular velocity of the planetary gear, which is meshed with the sun gear and the ring

gear, is denoted by PG . When the ring gear is fixed, the

relation between SG and PG is

2 PG PG SG SGR r (12)

which leads to

1 2

CG SGPG

SG PG

R r

R R (13)

1 2

CG SGPG

SG

R rv

R (14)

where PGv is the translation velocity of the planetary gear,

PGR is the radius of the planetary gear, and SGr is the radius

of the inner sun gear, as shown in Fig. 10. The velocity of the

planetary shaft, which is denoted by PSv , is equal to the

translation velocity of the planetary gear, as shown in Fig. 11:

1 2

CG SGPS

SG

R rv

R (15)

Since the carrier and the input shaft are connected by a keyconnection, the angular velocity of the carrier, which is

denoted by C , and that of the input shaft, which is denoted

by IS , are the same, as shown in Fig. 12:

1 2

CG SGC

SG PG SG

R r

R R r

(16)

1 2

CG SGIS

SG PG SG

R r

R R r

(17)

3.3. Lagrangian equation Based on the velocity of each part of the IVT obtained

above, the kinetic energy of each part can be obtained as shown in Table 1. Each part is considered as a rigid body and the gravitational potential energy is not considered.

The total energy totalT of the IVT is

3

10

3

10

1 2

5 0.5

0.5 2 8

4 5 4

total OS IS CG Dn

OG SG PGn

C PS TS TS TG J

T T T T T n

T n T T

T T T T T T

(18)

where n corresponds to the driver number, and

1 0.5 with 0,1,2,3D DT T n n (19)

1 0.5 with 0,1,2,3OG OGT T n n (20)

The system equation of the IVT is

1 1

1 1 1

1 1

( ) ( , )

total total

IS OSinput output

T TdQ

dt

(21)

where Q is the generalized force, and input and output are

the input and output torques, respectively. The rotational angle

of the joint in the first driver module is denoted by 1 . Since

the rotational angles of two adjacent joints has a 90°angle

difference, the relation between 1 and 1 can be expressed

by

1 1

1

1 1

1 3, if ,

2 4 2 4 2

5 3, if ,

4 2 4 2 2 2

for 0,1,2,3

n n

n n

n

(22)

as shown in Fig. 13.To derive the system equation, define a function

d A AS A

dt

(23)

Substituting totalT for A in Eq. (23) and using Eq. (18 ) yield


3

10

3

10

1 2

5

0.5

0.5

2 8

4

5 4

total OS IS CG

Dn

OGn

SG PG C

PS TS TS

TG J

S T S T S T S T

S T n

S T n

S T S T S T

S T S T S T

S T S T

(24)

Split totalS T into three parts, which are denoted by 1P , 2P ,

and 3P . The first part is

1

2 21 1 2

2 2

2

2 2

2

5 2

8 4

5

4

( ) ( )

5 ( ) (5 5 )

5 2 ( )

[8 ( ) 8 ( ) ]2 2

4 ( ) [2

IS CG SG

PG C PS

ST LT TG

J

CG CGTS TS

TG TG

CGTG J J

TG

CGCG SG

SG

CG SG CG SGPG PG

SG PG SG

CG SGPS C

SG

P S T S T S T

S T S T S T

S T S T S T

S T

R RJ J

R R

RJ J m d

R

RJ J

R

R r R rJ m

R R R

R r Rm J

R

2

2

1 1

]2

[ ]2

CG SG

SG PG SG

CG SGIS

SG PG SG

r

R R r

R rJ

R R r

L

(25)

where differentiation has been carried out and 1L is a

constant. The second part is

3

2 10

3

10

0.5

0.5

Dn

OGn

P S T n

S T n

(26)

Since differentiation in 2P and 3P is complicated, two

functions 1f and 2f are defined to simplify the expressions

of 2P and 3P :

21 1

1 1 2 21 1 1

1 12 1 2 2

1 1 1

cos

2 cos

sin

2 cos

d l df

l d l d

l df

l d l d

(27)

Differentiating 1 1f and 2 1f with respect to 1 yields

21 1 1 1' 1 1

1 1 22 2 2 21 1 1 1 1 1

2

1 1' 1 12 1 3 2 2

2 2 2 1 1 11 1 1

2 sin cossin

2 cos 2 cos

sin cos

2 cos2 cos

l d d l dl df

l d l d l d l d

l d l df

l d l dl d l d

(28)

Substituting DT and OGT in Table 1 for A in Eq. (23) yields

2 21 1 2 1

21 12 2 21 1 1 1 1 1

'1 1 1 1 2

1' '1 1 1 1 2 1 2 1

2 cos

D D

D

D

D

D

J f m fS T

m f l d l d CO

J f f

m f f f f

(29)

22 1 2

1 1 1 12

'2 1 2 1 ' 2

1 1 1 1 12

22

2

OGOG

OG

OGOG

J fS T f

R

f fJ f f

R

(30)

where differentiation has been carried out. Let

22 1 1 1

22 2 2

1 1 1 1 1

22 12 2

2 1 1 12

'3 1 1 1 1 1

' '1 1 1 1 2 1 2 1

'2 1 2 1 '

1 1 1 12

2 cos

22

2

D

D

OGD

OG

D

D

OGOG

L J f

m f l d l d CO

J fm f f

R

L J f f

m f f f f

f fJ f f

R

(31)

Substituting Eq. (31) into Eqs. (29) and (30) and adding the two equations yield

21 1 2 1 1 3 1 1D OGS T S T L L (32)

Hence,

3

22 2 1 1 3 1 1

0

0.5 0.5n

P L n L n

(33)

The third part is


3

22 1 1 1 2 12

1 1 12 2

'2 1 2 1

2

' 21 1 1 1 1

' '1 1 2 1 2 1 1 1

2

OS

OSOG OG

OG

OS

P S T

f f fJ f

R R

f f

R

J f f

f f f f

(34)

where differentiation has been carried out. Let

22 1 1 1 2 12

4 1 1 12 2

'2 1 2 1

2

'5 1 1 1 1 1

' '1 1 2 1 2 1 1 1

2OS

OG OG

OG

OS

f f fL J f

R R

f f

R

L J f f

f f f f

(35)

Where 10 , and

22 1 1 1 2 12

6 1 1 12 2

'2 1 2 1

2

'7 1 1 1 1 1

' '1 1 2 1 2 1 1 1

2OS

OG OG

OG

OS

f f fL J f

R R

f f

R

L J f f

f f f f

(36)

Where 1 2 . Substituting Eqs. (35) and (36) into Eq.

(34) yields

24 1 1 5 1 1 1

32

6 1 1 7 1 1 1

1 3, ,

2 4

5 3, ,

4 2

L L

P

L L

(37)

Substituting Eqs. (37)-(39) into Eq. (21) yields

1 1 1

1 2 31 1

32 1 1

1 1 20 3 1 1

24 1 1 5 1 1 1

32 1 1

1 1 20 3 1 1

26 1 1 7 1 1 1

,

0.5

0.5

1 3, ,

2 4

0.5

0.5

,

IS OS

input output

n

n

P P P

L nL

L n

L L

L nL

L n

L L

5 3,

4 2

(38)

Solving Eq. (38) for 1 yields

1

1

1 1

1

32

3 1 1 5 1 1 10

1 3

1 2 1 4 10

1

,

0.5 , ,

,0.5

1 3,

2 4

IS

input

OS

output

n

n

L n L

L L n L

(39)

1

1

1 1

1

32

3 1 1 7 1 1 10

1 3

1 2 1 6 10

1

,

0.5 , ,

,0.5

5 3,

4 2

IS

input

OS

output

n

n

L n L

L L n L

(40)

4. FORCE ANALYSISGiven the input and output torques, the acceleration of

each part of the IVT can be calculated from Eqs. (39) and (40). Forces exerted on each part for an unchanged eccentricity can be calculated using a Newtonian approach. The pressure angle of all the spur gears in the IVT is 20 deg and gravity is not considered.


4.1. Driver moduleThe torque exerted by the output gear on the output shaft

is denoted by OS , and the forces exerted by the driver on the

output gear are denoted by 0F and 0f , respectively, as shown

in Fig. 14. By Newton’s second law, the equations of motion of the output shaft and the working output gear are

OS output OS OSI (41)

0 1cos 20 OG OS OG OGF r J (42)

respectively, which lead to

OS OS OS outputJ (43)

10 cos 20

OG OG OS

OG

JF

r

(44)

The force exerted by the output gear on the driver case in

the i

direction is denoted by 0N , and that exerted by the

joint on the driver case is denoted by F ; the angle between

the direction of F and 1 2O O is denoted by , as shown in

Fig. 15. The equations of motion of the driver are

0 0

0 0 0

0 0 0 2 12

1 0 0

( cos cos 20 cos 20 )

( sin sin 20 sin 20 )

( sin 20 sin 20 )( )

sin ( ) cos 20

D j

D i

D

OG

F F f m a

F F N f m a

F f N l OOJ

FOO F f R

(45)

which lead to

0 0 0 2 1 2

0 0

cos 20

sin 20

OG D D iN F f R J m a OO l

F f

(46)

0 0 0

0 0

sin 20arctan

cos 20D i

D j

F f N m a

F f m a

(47)

0 0 cos 20

cosD jF f m a

F

(48)

The forces exerted by the crank gear on the joint are denoted

by 1F , 2F , and 1N , as shown in Fig. 15. The equations of

motion of the joint are

1 1 2 1 2 1

21 2 1 2 1

1 2 1 2 1 1

( ) cos 20 ( )sin 20

cos( ) ( ) cos 20

sin( ) ( )sin 20

CG J

J

J

N d F F r F F d J

F F F m d

F F F N m d

(49)

Substituting Eqs. (47) and (48) into Eq. (49) yields

21 1 1

1 2

1 2

[ cos 20 sin 20

sin 20 cos

cos 20 sin ] / cos 20

j jN m d m d

F

F

(50)

1 1

22 1

1 2

2cos 20cos 20cos 20

cos sin 20

j

CGj CG

J N d

F rm d r

F d

(51)

21

1 1 2

2

cos cos 20

cos 20

jm d

F F

F

(52)

Three transmission gears are installed on the transmission shaft 1 to keep the crank gear 1, crank gear 3, and crank gear 5 working synchronously. Two transmission gears are installed on the transmission shaft 2 to keep the crank gear 2 and thecrank gear 4 working synchronously, as shown in Fig. 17. The crank shaft is used to support the bending moments caused by the crank gears.

The force exerted by the transmission gear 5 on the crank

gear 5 is denoted by 5TF , as shown in Fig. 18. The equation

of motion of the crank gear 5 is

5 1 1cos 20 1.5 cos 20T CG CG CG CGF R F r J (53)

which leads to

51 1

cos 201.5 cos20

CG CG

T CGCG

JF R

F r

(54)




4

1 1 2 1

cos 20

1.5 cos 20

T CG

CG CG

CG

F RJ

F F r

(55)

which leads to

1 14

2 1

cos 20cos 20

1.5

CG CG

T CGCG

J

FF Rr

F

(56)




3

1 1 2 1

cos20

0.5 cos20

T CG

CG CG

CG

F RJ

F F r

(57)

which leads to

1 13

2 1

0.5 cos 20cos 20

CG CG

T CGCG

J

FF Rr

F

(58)

The force exerted by the sun gear 2 on the crank gear 2 is

denoted by 3F , and that exerted by the transmission gear 2 on

the crank gear 2 is denoted by 2TF , as shown in Fig. 21. The

equations of motion of the transmission shaft 2 and the crank gear 2 are


2 4 2cos 20 2T T TG S TG TGF F R J J a (59)

3 2

1 1 2 1

cos 20

0.5 cos 20

T CG

CG CG

CG

F F RJ

F F r

(60)


2 2 42 cos 20T S TG TG TG TF J J a R F (61)

1 13 2

2 1

cos 20cos 20

0.5

CG CG

CG TCG

J

FF R Fr

F

(62)

The force exerted by the sun gear 1 on the crank gear 1 is

denoted by 4F , and that exerted by the transmission gear 1 on

the crank gear 1 is denoted by 1TF , as shown in Fig. 22. The

equations of motion of the transmission shaft 1 and the crank gear 1 are

1 3 5 2cos 20 3T T T TG S TG TGF F F R J J a (63)

4 1 2 1cos 20 cos 20T CG CG CG CGF F R F r J (64)


1 2 3 53 cos 20T S TG TG TG T TF J J a R F F (65)

4 2 1 1cos20 cos20CG CG CG CG TF J F r R F (66)

4.2. Control moduleThe force exerted by the planetary gear on the upper sun

gear is denoted by 6F , and that exerted by the planetary gear

on the lower sun gear is denoted by 5F , as shown in Fig. 23.

The equations of motion of the sun gear 1 and the sun gear 2 are

6 44 cos 20 cos 20SG SG SG SGF r F R J (67)

5 34 cos 20 cos 20SG SG SG SGF r F R J (68)


6 4 cos 20 4cos 20SG SG SG SGF J F R r (69)

5 3 cos 20 4cos 20SG SG SG SGF J F R r (70)

The force exerted by the planetary shaft on the upper

planetary gear is denoted by 8XF , that exerted by the

planetary shaft on the lower planetary gear is denoted by

7 XF , those exerted by the upper ring gear on the upper sun

gear are denoted by rF and 8YF , respectively, and those

exerted by the lower ring gear on the lower sun gear are

denoted by RF and 7YF , respectively, as shown in Fig. 24.

The equations of motion of the upper planetary gear and thelower planetary gear are

8 6

6

28 6

cos20 cos20

cos 20 cos 20

sin 20 sin 20

X r PG PG

R PG PG PG

Y r PG PG PG SG

F F F m v

F F R J

F F F m R r

(71)

7 5

5

27 5

cos 20 cos20

cos20 cos 20

sin 20 sin 20

X R PG PG

R PG PG PG

Y R PG PG PG SG

F F F m v

F F R J

F F F m R r

(72)


5cos 20

PG PGR

PG

JF F

R

(73)

7 5 cos 20X PG PG RF m v F F (74)

27 5 sin 20Y PG PG PG SG RF m R r F F (75)

6cos 20

PG PGr

PG

JF F

R

(76)

8 6 cos20X PG PG RF m v F F (77)

28 6 sin 20Y PG PG PG SG RF m R r F F (78)

The force exerted by the carrier on the planetary shaft is

denoted by 9F , as shown in Fig. 25. The equation of motion

of the planetary shaft is

9 7 8X X PS PSF F F J (79)

which leads to

9 7 8PS PS X XF J F F (80)

The force exerted by the input shaft on the carrier is

denoted by 10F , as shown in Fig. 26. The equation of motion

of the carrier is 10 94OS PG SG C CF R F R r J (81)

which leads to

10 94 PG SG C C ISF F R r J R (82)

The equation of motion of the input shaft is

10input IS IS ISF R J (83)

Hence, the force 10F can also be expressed by

10 input IS IS ISF J R (84)

5. RESULTS AND DISCUSSION

5.1. Verification of analytical expressions in Secs. 3 and 4Equations (82) and (84) are two different forms of

expressing 10F , which can be used to verify the correctness of

the analytical expressions in Secs. 3 and 4. Subtracting Eq. (82) from Eq. (81) yields the force difference, which should bezero. Based on the kinematic, dynamic, and force analyses in Secs. 3 and 4, a computer program for the mathematical model of the IVT is written in MATLAB to simulation the

performance of the IVT. With 200input Nm , 0output ,

10d mm , and the other parameters shown in Table 2, the

calculated force difference mentioned above is almost zero, as shown in Fig. 27; it is not exactly zero due to the calculation accuracy of MATLAB. The parameters here with different


operating conditions are used in Secs. 5.2 and 5.3 for vehicle and wind turbine applications to understand the performance of the IVT.

5.2. Kinematic analysis resultsThe output-to-input speed ratio of a one-driver system is

1 1,output OS

input IS

(85)

and that of a two-driver system with a 90 deg angle differencebetween the two joints is

1 1 1 1max , , ,OS OSoutput

input IS

(86)The output-to-input speed ratio of the four-driver system is

1 1 1 1

1 1 1 1

, , 0.5 , ,max

, , 1.5 ,

OS OS

OS OSoutput

input IS

(87)

The simulation results of Eqs. (85)-(87)are shown in Fig. 28through Fig. 30, respectively. It is seen that the output-to-input speed ratio of the four-driver system is smoother than those ofthe one-driver and two-driver systems. The peak-to-peak instantaneous variation of the output-to-input speed ratio of the two-driver system is 25.67% and that the four-driver system is 9.3%.

5.3. Dynamic analysis resultsFor a vehicle application with the initial angular velocity

of the input shaft 33.3 /IS rad s , and the initial value of

the generalized coordinate 1 0 , the dynamic input and

output speeds are calculated from Eqs. (39) and (40) with 23inputI Nms and 224outputI Nms , and shown in Fig.

31. The peak-to-peak instantaneous variation of the input speed is 9.57% relative to the average input speed and that of the output speed is 1.24% relative to the average output speed.Since the moment inertia of the output load outputI is eight

times that of the input load inputI , the variation of the input

speed is almost eight times that of the output speed. The variation of the output speed from the dynamic analysis is much smaller than that from the kinematic analysis, which indicates that a realistic performance of the output speed of the IVT can be much better than the performance predicted by the kinematic analysis.

For a wind turbine application with 212inputI Nms ,

21outputI Nms , the initial angular velocity of the input

shaft 33.3 /IS rad s , and the initial value of the

generalized coordinate 1 0 , the dynamic input and output

speeds are shown in Fig. 32. The peak-to-peak instantaneous variation of the input speed is 1.75% relative to the average

input speed and that of the output speed is 9.26% relative to the average output speed. Since the moment of inertia of the output load is one-twelfth of that of the input load, the variation of the output speed is much larger than that of the input speed. The variation of the output speed from thedynamic analysis is close to that from the kinematic analysis, and it can generate a varied electric voltage of the generator. The varied electric voltage can be reduced by a filter following the generator.

To generate smoother input and output speeds for vehicle and wind turbine applications, respectively, a pair of non-circular gears that are installed between a prime mover and the input shaft of the IVT are proposed [17]. The non-circular gears that have a changing gear ratio can generate a variedoutput speed with a constant input speed, which means that thevariations of the input and output speeds can be compensatedby the non-circular gears, as illustrated in Fig. 33 and Fig. 34.

The eccentric motion of the driver module introduces a speed variation in the IVT, and the forces exerted on each part

of the IVT can be large. The maximum acceleration of 1 is

calculated using Eqs. (39) and (40) for the vehicle application when the input speed varies from 0 to 100 RPM, and itincreases with the input speed, as shown in Fig. 35. Since the moment of inertia of the output shaft is very large, the maximum torque occurs at the output shaft of the IVT. The

torque OS on the output shaft is calculated using Eq. (44);

the maximum torque on the output shaft is shown in Fig. 36, and it increases with the input speed. Strong parts of the IVTare required to withstand high torques at high input speeds.

6. CONCLUSIONSBy combining the Lagrangian and Newtonian approaches,

dynamic and force analyses can be easily carried out for the IVT. The results calculated from the dynamic analysis aremore realistic than those from the kinematic analysis. In the vehicle application, the variation of the input speed is larger than that of the output speed, and the variation of the output speed from the dynamic analysis is much smaller than that from the kinematic analysis. In the wind turbine application, the variation of the output speed is larger than that of the input speed, and the variation of the output speed from the dynamic analysis is close to that from the kinematic analysis. The variations of the input and output speeds can be further reduced by adding a pair of non-circular gears. The force analysis shows that the forces exerted on each part of the IVT increase with the input speed and strong parts are required to withstand higher torques at high input speeds.

7. ACKNOWLEDGEMENTSThe authors are grateful to the financial support from the

National Science Foundation through Award CMMI-1335397. They would also like to thank Eric Meyer for his assistance onthe project.


8. REFERENCES[1] Lechner, G., Naunheimer, H., and Ryborz, J., 1999, Automotive Transmissions: Fundamentals, Selection, Design and Application, Springer, pp. 111-159. [2] Manwell, J.F., Mcgowan, J.G., and Rogers, A.L., 2010, Wind Energy Explained: Theory, Design and Application, 2nd Edition, Wiley, pp. 91-157.[3] Hansen, A.D., Iov, F., Blaabjerg, F., and Hansen, L.H., 2004, "Review of Contemporary Wind Turbine Concepts and their Market Penetration", Wind Engineering, 28(3), pp. 247-263.[4] Muljadi, E., Pierce, K., and Migliore, P., 1998, "Control Strategy for Variable-speed, Stall-regulated Wind Turbines", Proceedings of the 1998 American Control Conference, 3, pp. 1710-1714.[5] Ko, H., Yoon, G., Kyung, N., and Hong, W., 2008, "Modeling and Control of DFIG-based Variable-speed Wind-turbine", Electric Power Systems Research, 78(11), pp. 1841-1849.[6] Blaabjerg, F., Chen, Z., Teodorescu, R., and Iov, F., 2006, "Power Electronics in Wind Turbine Systems", 2006 CES/IEEE 5th International Power Electronics and Motion Control Conference, 23(20-21), pp. 1-11.[7] Miltenovic, V., Velimirovic, M., Banic, M., and Miltenovic, A., 2011, "Design of Wind Turbines Drive Train Based on CVT", Balkan Journal of Mechanical Transmissions, 1(1), pp. 46-56.[8] Fischetti, M., 2006, "No More Gears", Scientific American 294, pp. 92-93.[9] Chen, T. F., Lee, D. W., and Sung, C. K., 1998, "An experimental study on transmission efficiency of a rubber V-belt CVT", Mechanism and machine theory, 33(4), pp. 351-363.[10] Machida, H., and Murakami, Y., 2000, "Development of the PowerToros unit half toroidal CVT", Motion and Control, 9, 15-26.[11] Wang, J., Atallah, K., and Carvley, S. D., 2011, "A Magnetic Continuously Variable Transmission Device", IEEE Transactions on Magnetics, 47(10), 2815-2818.[12] Rydberg, K., 1998, "Hydrostatic Drives in Heavy Mobile Machinery–New Concepts and Development Trends," SAE Technical Paper 981989.[13] Gorla, C., and Cesana, Paolo, 2011, "Efficiency Models of Wind Turbines Gearboxes with Hydrostatic CVT", Balkan Journal of Mechanical Transmissions, Vol. 1, No. 2, pp. 17-24.[14] Kotwicki, A. J., 1982, "Dynamic models for torque converter equipped vehicles", Society of Automotive Engineers.[15] Lahr, D. F., and Hong, D. W., 2006, "The operation and kinematic analysis of a novel cam-based infinitely variable transmission", Proceedings of the 30th ASME Mechanisms and Robotics Conference, September 10-13, Philadelphia, Pennsylvania, USA.[16] Zhu, W.D., and Wang, X.F., 2014, "Modeling and Control of an Infinitely Variable Speed Converter with

Application to Wind Turbines", ASME Journal of Dynamic Systems, Measurement, and Control, in press.[17] Wang, X.F., and Zhu, W.D., 2013, "Design, Modeling and Simulation of a Geared Infinitely Variable Transmision", ASME Journal of Mechanical Design, in press.[18] Amirouche, and Farid M. L., 2006, Fundamentals of Multibody Dynamics: Theory and Applications, Springer, pp. 287-303. [19] Brizard A. J., 2008, An Introduction to Lagrangian Mechanics, Singapore: World Scientific.


ANNEX A

FIGURES AND TABLES

Fig. 1 Layout of the IVT

Fig. 2 Schematic of a driver that uses a crank-slider mechanism


Fig. 3 Three-dimensional view of a driver

Fig. 4 Three-dimensional view of the joint and the crank gears


Fig. 5 Exploded view of two groups of planetary gears

Fig. 6 Assembly of the control module and two groups of planetary gears


Fig. 7 Kinematic relation of a driver

Fig. 8 Schematic of the kinematic model of a driver


Fig. 9 Simplified model of a driver in a new coordinate system

Fig. 10 Kinematic relation between the crank gear and the sun gear


Fig. 11 Velocity relation of a planetary gear system

Fig. 12 Velocity relations of the input shaft, the planetary gear, and the carrier


Fig. 13 Relation between 1 and 1

Fig. 14 Force analysis of the output gear


Fig. 15 Force analysis of the driver

Fig. 16 Force analysis of the joint


Fig. 17 Layout of the crank gears and the transmission gears

Fig. 18 Force analysis of the transmission gear 5 and the crank gear 5


Fig. 21 Force analysis of the transmission gear 2, the crank gear 2, and the sun gear 2

Fig. 22 Force analysis of the transmission gear 1, crank gear 1, and sun gear 1


Fig. 23 Force analysis of a sun gear

Fig. 24 Force analysis of a planetary gear

Fig. 25 Force analysis of the planetary shaft


Fig. 26 Force analysis of the carrier

Fig. 27 Force difference versus the input angle

Fig. 28 Output-to-input speed ratio of the one-driver system


Fig. 29 Output-to-input speed ratio of the two-driver system

Fig. 30 Output-to-input speed ratio of the four-driver system


Fig. 31 Dynamic input and output speeds for the vehicle application

Fig. 32 Dynamic input and output speeds for the wind turbine application


Fig. 33 Relation between the input and output speeds of the IVT without use of the non-circular gears between the prime mover and the input shaft

Fig. 34 Relation between the input and output speeds with use of the non-circular gears between the prime mover and the input shaft

Fig. 35 Relation between the input speed and the maximum acceleration of 1


Fig. 36 Relation between the input speed and the maximum torque on the output shaft

Table 1. Kinetic energy of each part of the IVT and its quantity

No. Part Kinetic energy Quantity

1Input

shaft21

2IS IS IST J 1

2 Carrier 21

2C C CT J 1

3Planetary gear

2 21 1

2 2PG PG PG PG PGT J m v 8

4 Sun gear 21

2SG SG SGT J 2

5Crank

gear21

2CG CG CGT J 5

6 Joint 2 21 1

2 2J J J J JT J m v 4

7 Driver 2 2 21 1

2 2D D i j D DT m v v J 4

8Output

gear 2 21 2

1

2OG OG OG OGT J 4

9Output

shaft21

2OS OS OST J 1

10Transmis

sion gear21

2TG TG TGT J 5

11TransmissionShaft 2

22 2 2

1

2TS TS TST J 1


12TransmissionShaft 1

21 1 1

1

2TS TS TST J 1

13Planetary

shaft21

2PS PS PST m v 4

Note: T, J, and m with subscripts represent the kinetic energy, the moment of inertia, and the mass of each part, respectively. ISJ is

the combination of the moments of inertia of the input shaft and a prime mover that is connected to the input shaft, which is denoted

by inputI . OSJ is the combination of the moments of inertia of the output shaft and an output load that is connected to the output

shaft, which is denoted by outputI .

Table 2. The moment of inertia and mass of each part of the IVT

Part name Joint DriverPlanetary shaft

Planetary gear

Moment of

Inertia (Nms²)41.24 10 427.1 10 72.3 10

6 1.18 10

Mass (kg)11.47 10 1.15 21.88 10

21.88 10

Part nameInput

shaftCarrier Sun gear Crank gear

Moment of

Inertia (Nms²)40.25 10 44.91 10 48.03 10

415.87 10

Part nameOutput

gearOutput shaft

Transmission shaft 1

Transmission shaft 2

Moment of

Inertia (Nms²)40.69 10 40.60 10 67.79 10

67.79 10

Part nameTransmis

sion gearMoment of

Inertia (Nms²)65.96 10

Note: The above values are calculated by Solidworks using dimensions of a prototype of the IVT and the mass density of steel; the dimensions of the prototype are not listed here due to space limitation.


dynamic analysis of a novel geared infinitely variable

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