Dynamic Model of a Full Toroidal Variator

Dynamic Model of a Full Toroidal Variator

• The Full Toroidal Variator (FTV) is the core element in many mechanical hybrid systems for vehicle applications like Kinetic Energy Recovery System and Infinitely Variable Transmission (IVT) because it allows to manage the power transfer with continuous variation of the speed-ratio.

• In KERS applications the FTV manages the power flows between the ve- hicle and a flywheel (the kinetic energy is stored in the flywheel during braking and then reused under acceleration of the vehicle). A schematic of the considered FTV is the following:

r2θ

ω² J2, b2 K2b

r³θ ω³

J3, b3

Kb3

τ³ ra

Jb,bb

ωb

rb

θ

r²

K12

r¹ ω1

J1, b1

τ¹

• Meaning of parameters and variables of the system:

J¹ shaft moment of inertia

b1 shaft linear friction coefficient ω¹ shaft angular velocity

τ1 torque acting on the shaft r¹ gear radius

K12 stiffness coefficient between shaft and inner disk J², J³ inner and outer disks moments of inertia

b2, b³ inner and outer disks linear friction coefficients ω², ω³ inner and outer disks angular velocities

τ3 torque acting on the outer disk r² inner disk radius

θ tilt angle of the rollers r²θ, r³θ contact radial positions K2b, K³b contact stiffness coefficients

Jb rollers moment of inertia

bb rollers linear friction coefficient ωb rollers angular velocity

rb rollers radius

ra toroid centerline radius

• The two toroidal cavities contain six rollers (three for each cavity) that transfer the power from the inner disc to the outer disc and viceversa.

The variator speed-ratio is a function of the rollers inclination: a change of the tilt angle θ of the rollers causes a speed-ratio variation.

• The contact radial positions r2θ and r3θ are related to radii ra and rb as follows:

r_2θ = ra + rbsin θ, r_3θ = ra − r_b sin θ.

• The gear ratio R(θ) relating the angular speed ω2 of the inner disk to the angular speed ω3 of the outer disk is function of the tilt angle θ as follows:

R(θ) = ω3

ω2

= r2θ

r3θ

= r_a + rb sin θ

r_a − r_bsin θ = 1 + α sin θ 1 − α sin θ where

α = r_b r_a

• Plot of the gear ratio R(θ) for different values of α ∈ [0.1 : 0.1 : 0.9]:

−100 −80 −60 −40 −20 0 20 40 60 80 100

10⁻¹ 10⁰ 10¹

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

Gear ratio R(θ) for different values of α

R(θ)

θ [degree]

• The FTV is a non-linear and time-variant system and writing its lumped parameter dynamic model is non trivial problem.

• The POG model of this physical system can be easily obtained introducing

“fictitious” elastic elements between the inertias of the system. The elastic elements are the only dynamic element that can be put between two masses or inertias respecting the integral causality of the system.

• The presence of these elastic elements simplify the writing down of a dynamic model of the system. The obtained POG model is characterized only by integral causality element suitable for simulation.

• The elastic elements can be afterwards eliminated applying a proper con- gruent transformation to the original system and so obtaining the POG dynamic model of the reduced system.

• The POG block scheme of the considered FTV system is the following:

τ1

ω1

1 Shaft

- 

?

1 s

?

J₁⁻¹

?

 -

 

b1

6

6

- -

2 Stiffness and gear radii between shaft and inner disk

- r1 ^-

 r1 ^{ -}

6

1 s

6

K12

6

f12

-   r2  - r2 ^-

3 Inner disk

- 

?

1 s

?

J₂⁻¹ ω?2

 -

 

b2

6

6

- -

4 Stiffness and gear radii between inner disk and roller

- r2θ ^-

 r2θ ^{ -}

6

1 s

6

K2b

6

- 

f2b

˙xb

· · ·

· · · f_2b

˙xb ^ r_b ^

- r_b ^-

5 Rollers

- 

?

1 s

?

J_b⁻¹ ω?_b

 -

 

b_b

6

6

- -

6 Stiffness and gear radii between rollers and outer disk

- r_b ^-

 r_b ^{ -}

6

1 s

6

K_b3

6

f_b3

-   r3θ  - r_3θ ^-

7 Outer disk

- 

?

1 s

?

J₃⁻¹

?

 -

 

b3

6

6

- -

8

τ3

ω3

• The state space model L ˙x = A x + Bu of the POG system is:















J1 ˙ω1 1 K12

f˙12

J2 ˙ω2 1 K2b

f˙2b

J_b ˙ωb 1 K_b3 f˙_b3

J3 ˙ω3















| {z }

L ˙x

=













−b1−r1

r1 0 −r2

r2 −b2 −r_2θ r_2θ 0 −rb

r_b −b_b −r_b

r_b 0 −r3θ

r3θ −b3













| {z }

A











 ω1

f12

ω2

f_2b ω_b f_b3

ω3













| {z }

x

+













1 0 0 0 0 0 0 0 0 0 0 0 0 −1













| {z }

B

 τ1

τ3



| {z }

u

• The matrices and vectors L, A, B, x, u can be directly read from the POG block scheme.

• When K12 → ∞, K2b → ∞ and Kb3 → ∞ the state variables are linked by the following constraint:











0 = r1ω1 −r2ω2

0 = r2θω2 − r_bω_b 0 = rbω_b − r3θ ω3.

⇒















ω2 = r1

r2

ω1

ω_b = r_2θ

r_b ω2 = r_2θ r1

r_b r2

ω1

ω3 = r_b

r_3θ ω_b = r2θ r1

r_3θ r2

ω1.

• The following state space congruent transformation can be used:

x = T(θ) ω1 ⇔











 ω1

f12

ω2

f2b

ω_b f_b3

ω3













| {z } x

=











 1 0 R2

0 R_b(θ)

0 R3(θ)













| {z } T(θ)

ω1

where gear ratios R2, Rb(θ) and R3(θ) have the following structure:

R2 = r1

r2

, R_b(θ) = r1r_2θ

r2r_b = r1(ra + rbsin θ) r2r_b , R3(θ) = r1r2θ

r2r_3θ = r1(ra + rbsin θ) r2(ra − r_bsin θ).

• Since the transformation matrix T(θ) is a column matrix, the transformed and reduced model is a first order dynamic model that can be written in the following explicit form:

J(θ) ˙ω1 + N (θ, ˙θ) ω1 = −b(θ) ω1 +

1 −R3(θ)

| {z }

B¯

τ1

τ3



(1)

where ¯B = T^TB, coefficients J(θ) and b(θ) are

J(θ) = T^TLT = J1 + R²₂J2 + R²_b(θ) Jb + R²₃(θ) J3

b(θ) = −T^TAT = b1 + R₂²b2 + R²_b(θ) bb + R²₃(θ) b3

and

N(θ, ˙θ) = T^TL ˙T = r2θ r_b ˙θ cos θ R2²

J_b

r_b² + 2 raJ3

r₃³_θ



= J˙(θ) 2 .

• The transformed and reduced system (1) can be graphically represented using the following POG block scheme:

τ1

ω1

1

- 

J⁻¹(θ)

1 s

?

?

?

 -

 

N(θ, ˙θ)

6

6

- -

 

b(θ)

6

6

- - -R3(θ) ^-

 R3(θ)

8

τ3

ω3

where

J(θ) = J1 + R²₂J2 + R²_b(θ) Jb + R₃²(θ) J3

b(θ) = b1 + R²2b2 + R²_b(θ) bb + R²3(θ) b3

N(θ, ˙θ) = r2θ r_b ˙θ cos θ R²2

J_b

r_b² + ^2r_r^a3J3

3θ

 and

R2 = r1

r2

R_b(θ) = r1(ra + rb sin θ) r2r_b

R3(θ) = r1(ra + rbsin θ) r2(ra − r_bsin θ).

• The system can be graphically represented by the following POG scheme:

...

...

T1 J¹

w¹ b¹

T³ T4

v⁵

T = r¹

K¹² F⁶

v⁷

T⁸

T = r²

J²

w² b²

T¹⁰ T11

v¹²

T = r2th

K2b

F¹³

v14

T15 Jb

wb bb

T¹⁷ T18

v¹⁹

T = rb

T = rb

Kb3

F²⁰

v21

T²²

T = r³th

J³

w³ b³

T²⁴ 1

a

3

c

4 5

e 7

7

g

g

8 9

i

11

m

12 13

o 2

b d

6

f

h

10

l n

• This scheme has been obtained using the following source code:

**, Gr, Si, Sn, Si, As, Si, POG, Si, EQN, Si, SLX, Si

**, pSplit, 42, Sr, No, pSr, No, pEsu, No, Split, 7 rT, 1, a, An,-90, Ln, 1.2

rJ, 1, a, Sh, 0.5, Kn, J 1, En, w 1, Kn0, 13, Qn0, 2 - -, 1, 2

- -, a, b

rB, 2, b, Kn, b 1

CB, [2;3],[b;c], Kn, E2=r 1*E1, Sh, [0.6;-0.2]

mK, 3, 4, Ln, 0.8, Kn, K 1 2 - -, c, d, Ln, 0.8

CB, [4;5],[d;e], Kn, E1=r 2*E2, Sh, [0.2;-0.6]

rJ, 5, e, Kn, J 2, En, w 2 - -, 5, 6, Ln, 0.5

- -, e, f, Ln, 0.5 rB, 6, f, Kn, b 2

CB, [6;7],[f;g], Kn, E2=r 2 t h*E1, Sh, [0.6;-0.2]

mK, 7, 8, Ln, 0.8, Kn, K 2 b - -, g, h, Ln, 0.8

CB, [8;9],[h;i], Kn, F2=r b*F1, Sh, [0.2;-0.6]

rJ, 9, i, Kn, J b, En, w b - -, 9, 10, Ln, 0.5

- -, i, l, Ln, 0.5

rB, 10, l, Kn, b b

CB, [10;11],[l;m], Kn, E2=r b*E1, Sh, [0.6;-0.2]

mK, 11, 12, Ln, 0.8, Kn, K b 3 - -, m, n, Ln, 0.8

CB, [12;13],[n;o], Kn, F2=r 3 t h*F1, Sh, [0.2;-0.6]

rJ, 13, o, Kn, J 3, En, w 3 rB, 13, o, Sh, 0.5, Kn, b 3

• The POG Modeler provides the following POG block scheme:

T¹

1 J1

w¹ w1

b¹ T³

r¹

r¹

K¹² F⁶

r² r²

1 J2

w²

b² T¹⁰

r²th

r²th

...

...

K²b

F¹³

1 Jb

wb

bb

T¹⁷

rb

rb

rb

rb

Kb3

F²⁰

r³th

r³th

1 s

1 s 1

s

1 s

1 s

1 s 1

s

1 J3

w³

b³ T²⁴

• The POG state space equations L ˙x = Ax + Bu and y = Cx + Du are:

L ˙x =

















−b1 −r1 0 0 0 0 0

r¹ 0 −r² 0 0 0 0

0 r² −b² −r_2th 0 0 0

0 0 r_2th 0 −rb 0 0

0 0 0 rb −bb −rb 0

0 0 0 0 rb 0 −r_3th

0 0 0 0 0 r_3th −b³































 w1

F⁶ w² F13

wb

F20

w³















 +















 1 0 0 0 0 0 0















 T¹

where

L =

















J¹ 0 0 0 0 0 0

0 ¹

K1 2

0 0 0 0 0

0 0 J² 0 0 0 0

0 0 0 ¹

K2 b

0 0 0

0 0 0 0 Jb 0 0

0 0 0 0 0 ¹

Kb 3

0

0 0 0 0 0 0 J3

















• System parameters:

-- Matlab commands --- (POG_KERS_Par_SLX_m) ---

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

syms J_1 b_1 r_a r_b r_1 r_2 J_2 b_2 r_b J_b b_b J_3 b_3 th R2=r_1/r_2;

r2th=r_a+r_b*sin(th); r3th=r_a-r_b*sin(th);

Rbth=r_1*r2th/(r_2*r_b);

R3th=r_1*r2th/(r_2*r3th);

Jth=J_1+J_2*R2^2+J_b*Rbth^2+J_3*R3th^2;

bth=b_1+b_2*R2^2+b_b*Rbth^2+b_3*R3th^2;

dJ_dth=diff(Jth,’th’)/2;

%%%%%%%%%%%% SYSTEM PARAMETERS %%%%%%%%%%%%%%%%%%%%%%%%%%

J_1=0.026*kg*meters^2; % 2. Inertia. Internal parameter.

b_1=1.06*Nm/(4000*rpm); % 3. Angular friction. Internal parameter.

r_1=17.2*mm; % 5. Parameter. Transformer/Gyrator.

K_1_2=150*Newton/(0.2*mm); % 6. Stiffness. Internal parameter.

r_2=51.5*mm; % 12. Parameter. Transformer/Gyrator.

J_2=0.675*kg*r_2^2; % 9. Inertia. Internal parameter.

b_2=2.2*Nm/(4000*rpm); % 10. Angular friction. Internal parameter.

K_2_b=80*Newton/(0.2*mm); % 13. Stiffness. Internal parameter.

r_a=34.3*mm; r_b=27.5*mm;

J_b=1.0*kg*r_b^2; % 16. Inertia. Internal parameter.

b_b=1.9*Nm/(4000*rpm); % 17. Angular friction. Internal parameter.

K_b_3=80*Newton/(0.2*mm); % 20. Stiffness. Internal parameter.

J_3=100*J_2; % 23. Inertia. Internal parameter.

b_3=1.3*Nm/(4000*rpm); % 24. Angular friction. Internal parameter.

%%%%%%%%%%%% INPUT VALUES %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

th_A=70*degrees; % Maximum amplitude of theta

th_w=3; % Frequency of theta

tt=(0:0.001:10)’; % Time vector th=th_A*sin(th_w*tt); % theta(tt) dot_th=th_w*th_A*cos(th_w*tt); % dot_theta(tt)

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

th_0=th(1); R2=r_1/r_2;

r_2_t_h=r_a+r_b*sin(th_0);

r_3_t_h=r_a-r_b*sin(th_0);

%%%%%%%%%%%% INITIAL CONDITIONS %%%%%%%%%%%%%%%%%%%%%%%%%

T_1_0=0*Nm; % 1. Torque. Input value.

w_1_0=2000*rpm; % 2. Angular momentum. Initial condition.

w_2_0=w_1_0*r_1/r_2; % 9. Angular momentum. Initial condition.

w_b_0=w_2_0*r_2_t_h/r_b; % 16. Angular momentum. Initial condition.

w_3_0=w_b_0*r_b/r_3_t_h; % 23. Angular momentum. Initial condition.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

---

• Angular velocities ω1, ω2, ωb, ω3 when b1 = b2 = bb = b3 = 0:

0 1 2 3

0 500 1000 1500 2000 2500 3000

w_1 [rpm]

Angular Velocity w 1

0 1 2 3

0 200 400 600 800 1000

w_2 [rpm]

Angular Velocity w 2

0 1 2 3

0 200 400 600 800 1000

w_b [rpm]

Angular Velocity w b

Time [s]

0 1 2 3

0 200 400 600 800 1000 1200

w_3 [rpm]

Angular Velocity w 3

Time [s]

• Energy stored in the system when b1 = b2 = bb = b3 = 0:

0 0.5 1 1.5 2 2.5 3

0 0.2 0.4 0.6 0.8 1 1.2

Total Energy En, Energy En1 stored in in J_1, Energy En3 stored in in J_3

En, En1, En3 [kJ]

0 0.5 1 1.5 2 2.5 3

0 0.2 0.4 0.6 0.8 1 1.2 1.4

Total Inertia J_th

J_th [kg*m2 ]

Time [s]

Time [s]

• Angular velocities ω1, ω2, ωb, ω3 of the system with dissipations:

0 1 2 3

0 500 1000 1500 2000 2500

w_1 [rpm]

Angular Velocity w 1

0 1 2 3

0 200 400 600 800 1000

w_2 [rpm]

Angular Velocity w 2

0 1 2 3

0 200 400 600 800 1000

w_b [rpm]

Angular Velocity w b

Time [s]

0 1 2 3

0 200 400 600 800 1000

w_3 [rpm]

Angular Velocity w 3

Time [s]

• Stored energy of the system with dissipations:

0 0.5 1 1.5 2 2.5 3

0 0.2 0.4 0.6 0.8 1

Total Energy En, Energy En1 stored in in J_1, Energy En3 stored in in J_3

En, En1, En3 [kJ]

0 0.5 1 1.5 2 2.5 3

0 0.2 0.4 0.6 0.8 1 1.2 1.4

Total Inertia J_th

J_th [kg*m2 ]

Time [s]

Time [s]