Mathematical Modeling of Continuously Variable Transmission (CVT) System

DOI : 10.17577/IJERTV5IS020030

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Mathematical Modeling of Continuously Variable Transmission (CVT) System

Sarah Crosby*

Department of Engineering Mathematics Faculty of Engineering, Alexandria University Alexandria, Egypt

Hassan Elgamal

Galal Elkobrosy Department of Engineering Mathematics

Faculty of Engineering, Alexandria University Alexandria, Egypt

Department of Mechanical Engineering Faculty of Engineering, Alexandria University Alexandria, Egypt

AbstractA continuously variable transmission (CVT) provides an infinite number of gear ratios between two finite limits. This allows the engine to run at its most efficient RPM independent of the car speed, making CVT the most efficient transmission system. The CVT is also fuel efficient, because it reduces the engine speed at high vehicle speed and allows the engine to run at its optimal point.

Aiming at determining the behavior of the CVT transmission, a complete mathematical model have been constructed for the whole vehicle equipped with CVT, simulating the whole vehicle dynamics. The model takes into consideration each component of the vehicle, including: the engine dry friction disc clutch, belt type CVT, car differential, wheels and body. In this paper, the slip between the belt and the pulleys and the CVT traction curve were considered. The kinematical, geometrical and momentum equations governing the performance of the whole system were formulated and solved. The governing differential equations were numerically solved In order to predict the behavior of the different mechanical parts of the system. The results of the simulation shows that the system becomes steady after a short period of time. The analysis of the systems dynamical response demonstrates that the simulation model established represents the system efficiently.

Keywords Automatic Transmission; Continuously variable transmission; Mathematical Modeling; Transmission; ; V-belt CVT

  1. INTRODUCTION

    Over the last decades, a growing attention has been focused on the environmental question. Governments are continuously forced to set standards and to adopt actions in order to reduce the polluting emissions and the green-house gasses. In order to fulfil these requirements, car manufacturers have been obligated to dramatically reduce vehicles' gas emissions. The continuously variable transmission (CVT) represents one of the most promising solution, which is able to provide an infinite number of gear ratios between two finite limits. The CVT optimizes the engine working conditions, gets the highest efficiency, and therefore, improves fuel saving and reduces greenhouse gases emissions

    With the lack of oil and the call for reducing the environmental pollutants, the number of cars equipped with CVT has significantly increased. Up to now, more than one billon cars have been equipped with continuously variable transmission all over the world. Especially the metal belt type continuously variable transmission is applied widely. CVT has wide change range of speed ratio and it can adjust the ratio continuously and automatically according to the situation of running to maintain the engine to work in economy mode or power mode all the time. For this reason, CVT equipped cars are more economical than cars equipped with planetary gear automatic transmissions. The key advantages of a CVT that interest vehicle manufacturers and customers can be summarized as: higher engine efficiency, higher fuel economy, smooth acceleration without shift shocks and Infinite gear ratios with a small number of parts.

    A continuously variable transmission (CVT) is an automatic transmission that can change seamlessly through an infinite number of effective gear ratios between maximum and minimum values by changing the diameters of input shaft and output shaft directly, instead of going through several gears to perform gear ratio change. This contrasts with other mechanical transmissions that offer a fixed number of gear ratios. The most common type of CVT used is the pulley based CVT. The variable-diameter pulleys are main component of the pulley-based CVT. Each pulley is made of two 20-degree cones facing each other. A belt rides in the groove between them, as shown in Fig.1.

    Variable-diameter pulleys must always come in pairs, as shown in Fig.2, one of the pulleys, known as the driving pulley, is connected to the crankshaft of the engine, and the second pulley is called the driven pulley. When one pulley increases its radius, the other decreases its radius to keep the belt tight. As the two pulleys change their radii relative to one another, they create an infinite number of gear ratios.

    Aiming to achieve an optimal CVT performance and a significant reduction of fuel consumption, it is fundamental to have a very good control strategy of the transmission, which in turns needs a reliable model of the CVT mechanical behavior.

    Fig.1 Layout of CVT V-belt

    Fig.2 Pulley based CVT structure

    Gerbert [1, 2] worked on understanding the mechanics of traction belts, especially metal pushing V-belts and rubber V- belts. He used quasi-static equilibrium analysis to develop a set of equations that capture the dynamic interactions between the belt and the pulley. Since the belt is capable of moving both radially and tangentially, variable sliding angle approach was implemented to describe friction between the belt and the pulley. Gerbert [3] also analyzed the slip behavior of a rubber belt CVT. He also discussed slip during wedging due to poor fit between the belt and the pulley.

    Kim and co-workers [4,5] investigated the metal belt behavior analytically and experimentally. They proposed a speed ratiotorque loadaxial force relationship to calculate belt slip. They obtained the equations of motion using quasi- static equilibrium conditions and reported that the gross slip points depend on the torque transmitting capacity of the driven side. Bonsen et al. [6] analyzed slip and efficiency in a metal pushing V-belt CVT. They stated that high clamping force reduces the efficiency of a CVT. However, high clamping forces are necessary to avoid excess slip between the belt and the pulley.

    Sferra et al. [7] developed a unique model of a metal V-belt CVT in order to simulate its transient behavior. The model included inertial and pulley deformation effects. Discrete and continuous shifting behaviors were simulated in order to analyze efficiency and power losses due to friction between the belt and the pulley halves. The results showed high loss of efficiency during shifting transients.

    Bullinger and Pfeiffer [8,9] developed a detailed elastic model of metal V-belt CVT system to determine its power transmission characteristics at steady state. Pulley, shaft, and belt deformations were taken into account. The frictional constraints were modeled using the theory of unilateral constraints.

    Sattler [10] analyzed the mechanics of a metal chain and V- belt considering longitudinal and transverse stiffness of the

    chain/belt, and pulley misalignment and deformations. The pulley was assumed to deform in two ways, pure axial deformation and a skew deformation. The model was primarily used to study efficiency aspects of belt and chain CVTs.

    Carbone et al. [11] proposed a model that describes both the steady-state and the shifting dynamics of the V-belt CVT. The belt was modeled as a one-dimensional continuous body with zero radial thickness and infinite axial stiffness. Later, Carbone et al. [12] investigated the influence of pulley deformation on the shifting mechanism of a metal V-belt CVT. Coulomb friction hypothesis was used to model friction betwen different surfaces. Flexural effects of the belt were neglected; however, pulley bending was considered based on Sattlers model [10].

    Although there are many researches modeling and describing the dynamics of the CVT, the majority of these researches aimed only at modeling the CVT variator without modeling the complete vehicle system.

    In this paper, a complete model for the whole vehicle equipped with CVT, including all the vehicle components; car engine, friction disc clutch, CVT, car differential, wheels and car body has been constructed.

    This paper puts forward the mathematical model of CVT system including pulley the slip between the pulley and the CVT belt. The kinematical, geometrical and momentum equations governing the performance of the CVT system were formulated and solved. The numerical solution of these equations predicts the behavior of the different mechanical parts of the system.

  2. MATHEMATICAL MODEL

    The whole vehicle is modeled including the engine, the clutch, the CVT, the car differential and the wheels. As shown in Fig. 3, the model is composed of engine, dry friction disc clutch, belt type CVT, car differential, wheels and body.

    The following dimensionless conversions will be used throughout the mathematical model:

    Dimensionless radius: Dimensionless time:

    Dimensionless angular speed: Dimensionless force: Dimensionless torque: Dimensionless moment of inertia:

    Where, is the reference angular speed, c is the center distance between the two pulleys and is the belt mass density.

    1. The Engine Model

      Engine model is developed by applying Newtons second law to the rotational dynamics of the engine:

      (1)

      160

      140

      120

      100

      Fig. 3 Simplified model of transmission system

      e

      e

      Where Te is the engine torque, Je is the rotary inertia of the engine , and . is the acceleration of the engine. The engine torque Te is assumed to be a function of the engines rotational speed e and the throttle angle opening . The

      (N.m)

      Te

      80

      60

      40

      20

      0

      -20

      -40

      relationship between Te and e is deduced from the engines

      0 100 200 300

      400 500 600 700

      e (rad/s)

      performance map shown in Fig. 4, where there is a separate curve for each position of the accelerator.

      (2)

      Tabulated torques values are used to represent the relationship between engine torque, throttle opening and engine speed. These values are entered into Matlab curve fitting tool based on the engine performance map, as depicted in Fig. 5. The torque values are obtained by polynomial curve fitting in Matlab to get torque at any speed and throttle position.

      Fig.4 Engine Performance Map [14]

      These torque values are used in Eq. (1) to get engine acceleration which is then integrated to obtain engine speed.

      To obtain the non-dimensional engine :

      (3)

      Fig.5 Curve Fitting of the Engine Performance Map

    2. The Clutch Model

      A dry friction clutch type is used in the model [13]. It consists of a housing, pressure plates, friction plates, a clutch disc with torsion dampers and a release mechanism. The clutch is normally closed as the diaphragm spring is pre-tensioned when assembled .The clutch disc is equipped with torsional dampers which are coil springs that connect various segments of the clutch disc. These coil springs aim at maximizing the comfort level for the driver, when opening and closing the clutch

      The clutch system is modelled as shown in Fig. 6. The clutch disc has inertia Jcl. The torque transmitted through the clutch is Tcl. The angular speed of the clutch disc is presented by cl.

      Fig. 6 Powertrain model

      The angular displacement of the damper springs is d. Where

      d =cl- t, where cl is the clutch angular displacement and

      t is the transmission angular displacement. The nonlinear stiffness of the damper coil springs k(d) is simplified into the form [13]:

      k(d) =

      (4)

      The differential equations governing the clutch dynamics can be expressed as:

      (5)

      (6)

      The torque through the clutch while slipping is given by:

      (7)

      In which is the friction coefficient of the clutch surface material, is the active radius of the clutch plates and the normal actuation force on the clutch plate is given .

      The dimensionless clutch torque can be obtained by:

      (8)

      Where:

      (9)

    3. CVT and drive shafts:

      1. CVT Model Assumptions

        The model that will be presented is derived by making the following assumptions and simplifications:

        1. The metal belt is considered as a one-dimensional continuous body, with locally rigid motion. This means there is no longitudinal and transversal deformation, i.e. the belt is considered to be an inextensible strip with zero radial thickness and infinite axial stiffness.

        2. The bending stiffness of the belt is neglected.

        3. The Coulomb friction coefficient , acting between the segments and the pulleys, has a constant value.

      2. CVT system Mechanics

        In Fig.7 the entire CVT system is shown, where the driving, the driven pulley and the belt are presented. The moving half pulleys are subject to active axial forces which can be changed to obtain a variation of the belts pitch radii and thus modify the speed ratio.

        Fig.7 CVT Scheme

        In Fig.8, the kinematical and geometrical quantities involved are shown, and the plane rer e is illustrated. The figure shows the sliding angle , the angular coordinate , the radial coordinate r, the radius of curvature and the slope angle . Moreover, the tangent unit vector of the belt and the corresponding normal unit vector n are represented in addition to the radial and circumferential unit vectors er and e .

      3. Geometric and Kinematic Equations

        By observing Fig.8, the following relations can be derived:

        . (10)

        (11)

        = . (12)

        r.d=.tan() (13)

        (14)

        Where d is the local sliding angular velocity of the belt, and l, are the length and the angular extension of a material element of the belt respectively. Equation (13) correlates the radial velocity of the belt with its sliding tangential velocity, while (14) relates the half-opening angle s in the sliding plane with the sliding angle .

        Fig.8 Kinematical and Geometric Quantities

      4. Continuity Equation

        According to [11], the continuity equation can be written as:

        +=0 (15)

        This equation is based on the assumption that the belt is inextensible, that the longitudinal and transverse deformations are neglected and that the elongation of the belt is also neglected.

      5. Momentum Equations

        The forces involved in the equilibrium of the belt involves the tension of the belt (F), the linear pressure acting on the belt sides (P), the friction force (fa), and the inertia force of the belt element due to its acceleration ( d).This is visualized in Fig.9. The friction force fa = p. And the net tension of the belt; F=T-P.

        Where T is the tension of the band and P is the compressive forces between metal segments

        According to [11], the following assumptions are made to calculate the equilibrium:

        1. It is possible to calculate the local angular acceleration of the considered belts material element with the pulleys angular velocity .

        2. From Eq. (12), it follows that r R, therefore all these three parameters are written as R.

        3. The term is neglected with respect to 2R, i.e.

          d) .

          e) The belts axial and tangential acceleration are neglected.

          With these assumptions, the two Momentum Equations are:

          (16)

          (17)

          Dimensionless linear pressure: = (23)

          Dimensionless belt radial velocity: = (24) Dimensionless belt tangential velocity: =

          (25)

          Rewriting the previously derived relations in terms of the new dimensionless terms to get simpler form; (13) can be written into:

          tan= (26)

          Equations (16) and (17) can be rephrased as follows: First Momentum Equation:

          = (27)

          Second Momentum Equation:

          = (28)

          Continuity Equation:

          Kinematic Equation: (30)

          (29)

          Fig.9 Forces acting on belt

      6. Dimensionless Equations

        A new angular coordinate is defined as the angular coordinate of the real or imaginary point at which the sliding velocitys component of the belt in rer e plane is purely radially directed. Equations (10) and (15) can be reformulated as:

        = (18)

        Vr+ =0 (19)

        Where Vr is the radial sliding velocity of the belt= r ., and V is the tangential sliding velocity of the belt= r d .All the previously derived relations can be rephrased in dimensionless form once the following dimensionless quantities can be defined:

        Dimensionless radial velocity: (20) Local sliding coefficient:

        sc=d/ (21

        Dimensionless belts tension: K= (22)

        Geometric Equation:

        tan(s)=tan().cos() (31)

      7. Graphical Representation of and

        is defined as the traction coefficient and the force ratio is defined as:

        == (32)

        The values of and [11] are entered into Matlab curve fitting tool, in order to illustrate the relation between and

        for w<0 and w>0, as shown in Fig.10. The values of the ratio S / )] =

        versus the force ratio [11], were obtained by polynomial curve fitting in Matlab to represent the relationship between

        and as shown in Fig.11.

        0.25 s= (34)

        0.2

        0.15

        0.1

        0.05

        0

        -0.05

        -0.1

        -0.15

        -0.2

        W< 0

        W> 0

        Where, is the angular speed of the secondary axle, is the angular speed of the primary axle and n is the geometrical ratio, which is defined by:

        (35)

        Using the experimental data of [6], the traction coefficient is obtained as a function of the percentage slip s, for different drive ratios.

        -0.25

        0 0.5 1 1.5 2 2.5 3

        Fig.10 Traction coefficient versus force ratio (=0.1, =18°)

        0.16

        0.14

        Overdrive (n=2.26)

        W> 0

        W< 0

        W> 0

        W< 0

        4.5

        4

        0.12

        0.1

        0.08

        Medium (n=1.1)

        Low (n=0.4)

        3.5

        3

        0.06

        0.04

        2.5

        2

        1.5

        0.02

        0

        0 1 2 3 4 5 6 7 8 9 10

        Slip s %

        1

        0.5

        0

        0 0.5 1 1.5 2 2.5 3

        Fig. 12 Traction coefficient as a function of slip measured at 300rad/s, for ratios low (0.4), Medium (1.1) and overdrive (2.26) [6]

        9) Derivation of the equation of the angular speed of the

        Fig.11 Dimensionless Ratio versus force ratio (=0.1, =18°)

      8. Slip in a Continuously Variable Transmission

        The clamping force should be large enough to prevent the slip between the belt and the pulleys. On the other hand, a large clamping force reduces the efficiency of the Continuously Variable Transmission (CVT). Hence, it is important to estimate and control the right amount of slip.

        The traction curve shown in Fig.12 [6], gives the relation between the transmitted torque and the slip. The traction coefficient is defined as [16]:

        p= , s (33)

        In which Tp is the primary pulley torque, Ts is the secondary pulley torque, Rs is the secondary running radius of the belt on the pulley, Rp is the primary running radius of the belt on the pulley, Fs is the secondary clamping force, and Fp is the primary clamping force. p and s are assumed to have the same value, i.e. (p= s= The second variable in the traction curve is the slip in the variator; slip is defined as:

        CVT primary and secondary pulley

        From Fig.3 we can obtain the dynamic equation of the angular speed of the CVT secondary pulley as follows:

        (36)

        To obtain the governing equation of the primary pulley angular speed:

        First, by differentiating the slip equation (34), we have:

        (37)

        Then, by substituting from (36) and substituting from the slip equation (34); (37) becomes:

        (38)

        Therefore,

        – (39)

        And in dimensionless form:

        – (

        (40)

        Dividing the previous equation by the center distance, we have:

        In order to :

        By differentiation the traction coefficient equation (33) w.r.t time we get:

        The belt length in dimensionless form is given by:

        (49)

        (41)

        From the clutch model equation, (6) and from (41), we have:

        (42)

        But, . Let

        Therefore,

        (43)

        (50)

        Assuming the value of and getting the roots of the non- linear equation of y using Matlab, the values of and can be calculated.

        1. Evaluation of the values of and

          The non-dimensional ratio is obtained by using the

          versus slip relation, in Fig.12, knowing the value of speed ratio n and the allowed value of slip percentage. And is obtained from the relation between and in Fig.10, according to the sign of dimensionless radial .

          Where:

          From the traction coefficient equation (33):

          Substituting from (44) into (42) we get,

          Substituting from (45) into (34) we get,

          (44)

          (45)

          (51)

          The ratio is obtained from the relation between and in Fig.11, according to the sign of the dimensionless radial .

        2. Evaluation of the primary and secondary torques

        Since,

        = (46)

        In dimensionless form:

        (47)

        10) Calculation of dimensionless primary and secondary radius

        The values of the radii of the primary and secondary pulleys are functions of the ratio between the belt length and the center distance of the two pulleys. The length of the belt can be calculated from the following relation:

        L = + sin-1 +

        Hence, Since,

        Hence,

        From (33) we have,

        Substituting from (54) into (52):

        (52)

        (53)

        (54)

        (48)

        Substituting From (53) into (55)

        (55)

        From (56):

        From Eq. (56) into Eq. (53)

        (56)

        (57)

        1. Gradient resistance

          The gradient resistance depends on the angle of the road inclination and the weight of the car. The gradient resistance is given by:

          Gradient resistance= ± G sin (64)

          Where:

          ±

          Hence,

          And,

    4. Car Differential, wheels and Resistive Load

      (58)

      (59)

      (60)

      (61)

      1. Inertia Resistance

      When the car changes its velocity (accelerate or decelerate), it needs a force, this force is represented by the car esistance to change its speed (inertia force). This force depends on the mass of the car and the value of the car acceleration.

      Inertia Resistance= ± (m+meq). a (65) Where:

      a = car acceleration [m/s2]

      meq =equivalent mass of rotating parts [kg]

      The car differential splits the engine torque on the left and right transmission branches while allowing for different revolution speeds on the two shafts. The two powered wheels (usually the front wheels) are described with a single inertia Jw. A resistance load Torque is acting on the vehicle and impeding its motion. This resistive torque consists of aerodynamic drag resistance, rolling resistance, gradient resistance and inertia resistance.

      1. Aerodynamic Drag Resistance

        The force due to aerodynamic drag depends mainly on the shape of the vehicle, the density of the surrounding air, and the velocity of the vehicle. The equation for the aerodynamic drag force is:

        Drag Force = (62)

        Where:

        CD = Coefficient of aerodynamics resistance (drag coefficient)

        A= Car frontal area [m2 air = Air density [kg/m3] u = Car velocity [m/s]

      2. Rolling resistance

      Rolling resistance= G. f. cos (63)

      Where:

      f = coefficient of rolling resistance G= car weight [N]

      m = car mass [kg]

      g = Acceleration due to gravity

      =the angle of road inclination

      Where:

      = The equivalent rotary inertia of wheels

      = The equivalent rotary inertia of car differential

      = Transmission efficiency

      = Differential gear ratio

      = tire radius [m]

      ± :

      5) The Total Resistive Force

      The total resistance = Drag Force + Rolling resistance + Gradient resistance+ Inertia Resistance

      Hence,

      The total resistance = + G. f. cos ± G sin ± (m+ meq). a

      Resistive Torque= Tv = . Resistive Force

      Tv = * ( + G. f. cos ± G sin ± (m + meq). a)

      (66)

      In dimensionless form:

      (67)

      Where:

      A1= , A2= , A3= +

      From Fig.3 we can obtain the dynamic equation of the vehicle angular speed as follows:

      (68)

    5. Evaluation of the load torque on CVT

    can be obtained from the following relation :

    1.2

    1.1

    1

    Dimensionless Angular Speeds

    Dimensionless Angular Speeds

    0.9

    0.8

    0.7

    0.6

    0.5

    0.4

    0.3

    0.2

    cl

    e

    p

    s

    v

    0 500 1000 1500 2000 2500 3000 3500 4000 4500 5000

    Time (t*)

  3. NUMERICAL SOLUTION, RESULTS AND DISCUSSION

    From (1), (5), (36), (40) and (68), the governing ordinary differential equations of the vehicle mathematical model can be written as follows:

    (

    In which, Je, is the rotary inertia of engine; Jcl, is the rotary inertia of clutch; Js is the rotary inertia of the secondary pulley of CVT; Jr is the rotary inertia of the car differential; and Jw is the rotary inertia of the wheels. And Be, Bcl, Bs and Br represent the equivalent damping coefficient of each axis respectively

    1. The initial Conditions

      At the beginning of the motion when t*=0, the initial conditions are:

    2. Numerical Solution

      The ordinary differential equations were numerically solved using ode23 MATLAB solver. The ode23 solver uses second and third order Runge-Kutta-Fehlberg integration with variable step size. The angular speeds of each component of the model were integrated with time. The results are plotted in Fig.13

      From Fig.13, the following results are obtained:

      1. The dimensionless angular speed of engine , initially starts at (equivalent to an angular speed of 210 rad/s). It then decreases until it becomes steady at a dimensionless angular speed of 0.82, after a time of approximately 5 seconds (equivalent to a non- dimensional time = 1050).

      2. The dimensionless angular speed of clutch , starts . It then decreases until it becomes steady at a dimensionless angular speed of 0.78, after a time of approximately 1 second.

      3. The dimensionless angular speed of CVT primary pulley , starts . It then decreases until it becomes steady at a dimensionless angular speed of about 0.75, after an actual time of approximately 1 second.

      4. The dimensionless angular speed of CVT secondary pulley , starts . It then decreases until it becomes steady at a dimensionless angular speed of about 0.7, after a time of approximately 14 seconds.

      5. The dimensionless angular speed of vehicle , starts . It then

    decreases until it becomes steady at a dimensionless angular speed of about 0.3, after a time of approximately 0.7 seconds.

  4. CONCLUSIONS

In this paper, a mathematical model for a complete vehicle equipped with CVT is constructed. The model simulates the whole vehicle dynamics. A separate model for every component of the vehicle; the engine, clutch, CVT, car differentials, wheels and car body, was constructed.

The slip between the belt and the pulley along with the CVT traction curve were taken into consideration in the mathematical model. The kinematic and geometric equations were formulated for the whole system.

The differential equations for the angular velocities for every component in the system were formulated. These equations were numerically solved in order to predict the behavior of the system. Plotting the angular speeds with time, the numerically simulated results show that the system becomes steady after a short period of time. The graphical representation of the systems dynamical response demonstrates that the simulation model established represent the system efficiently.

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