2 backlash simulation

1

Backlash Simulation

SOLO HERMELIN

Updated: 26.11.06 25.02.11

No Backlash Backlash

http://www.solohermelin.com

http://www.solohermelin.com/

2

SOLO Backlash

Introduction

Two masses interacting with translational motion inside a moving vehicle

Backlash Mathematical Model

Spur GearMotor Moment Equation

Load Moment EquationContact MomentBacklash Error

Backlash Logic

MATLAB ProgramSimulation Results

References

Teeth Collision

3

SOLO Backlash

Gears are used to transmit torque between rotating axes using teeth. In a perfect Gear System, the tooth of one axis Gear is always in Contact with the tooth of the Gear of the other axis.

Because of production tolerances, during the rotation the teeth Contact is lost for a small angle, until is reestablished. This is the Gear Backlash.

4

SOLO Backlash

To develop the Backlash model we must deal with the following cases:

Because of production tolerances, during the rotation the teeth Contact is lost for a small angle, until is reestablished. This is the Gear Backlash.

(1) No Contact between the teeth of the two axes.

In this case the rotation dynamics of each axis is independent.

(2) Contact between the teeth of the two axes is just established.

In this case an equal impulse is transferred between the teeth in Contact that will produce a change in angular velocity of the two axes.

(3) A Continuous Contact between mating teeth exists. In this case the rotation rate of the two axes is coupled and defined by the Gear Teeth Ratio. The Moments transferred between the two axes are such that the rotation rates and rotation accelerations are coupled.

(4) Contact Lost (Disengagement), when the mating teeth lose Contact.

To get a understanding of the Backlash that involves rotation of two axes let start with a simpler example of one dimensional translation of two body that interact inside a moving vehicle, as seen in the Figure. Return to the Table of Content

5

SOLO Backlash

Consider a simple example (one dimensional translation) of a vehicle moving with a known velocity VB (t). Inside the vehicle we have two masses: 1. m1 of coordinate x1 on which an external force F1 is applied 2. m2 of coordinate x2 on which an external force F2 is applied

When the masses are in contact a internal force F12, between the two masses, applies.

We can see that we have the physical constraint: - δBL ≤ x2-x1 ≤δBL

Two Masses Interacting with Translational Motion inside a Moving Vehicle

6

SOLO Backlash

Equations of Motion

1. No contact between m1 and m2

( )

( )BLBL

BD

BD

xx

VFFm

x

VFFm

x

δδ <−<−

−−=

−−=

12

22

2

11

1

2

1

1

1

2. Contact between m1 and m2

( )

( )BL

BD

BD

xx

VFFFm

x

VFFFm

x

δ±=

−−−=

−−+=

12

1222

2

1211

1

2

1

1

1

21, FF - known applied forces on m1 and m2, respectively

21, DD FF - known disturbance forces on m1 and m2, respectively

( ) ( )21

122112

12

mm

FFmFFmxx DD −−−

=−

( ) ( )BL

DD xxFmm

mm

mm

FFmFFmxx δ±=+−

−−−=− 1212

21

21

21

121112

12


7

SOLO Backlash

Equations of Motion (continue – 1)

At collision between m1 and m2 a transfer of linear impulse ΔP occurs between the two masses

( )( ) ( )( )[ ] ( )( ) ( )( )[ ]( ) ( )[ ] ( ) ( )[ ]−−+−=−−+=

−+−++−=−+−++=∆=∆

222111

22211112

xxmxxm

xVxVmxVxVmtFP BBBB

or

( ) ( )[ ] ( ) ( )[ ] 0222111 =−−++−−+ xxmxxm

The second equation is obtained by using the elastic coefficient of collision e (e = 0, for plastic collision and e =1 for an elastic collision), that is defined as:

We obtain one equation with two unknowns ( ) ( )++ 21 , xx

( ) ( )( ) ( )

( )[ ] ( )[ ]( )[ ] ( )[ ]−+−−+

++−++−=−−−+−+−=

21

21

21

21:xVxV

xVxV

VV

VVe

BB

BB

( ) ( )( ) ( )−−−

+−+−=21

21

xx

xxe

or

We have x2 = x1 - δBL if before contact (-) ( ) ( ) 012 <−−− xx

We have x2 = x1 +δBL if before contact (-) ( ) ( ) 012 >−−− xx

2. Contact between m1 and m2 (continue – 1)


8

SOLO Backlash


Now we have two equations with two unknowns:

or

Solving for , we obtain:( ) ( )++ 21 , xx

( ) ( ) ( ) ( ) ( ) ( )−++−−=++ 22121121 1 xmexemmxmm

( ) ( ) ( ) ( ) ( ) ( )−−+−+=++ 21211221 1 xemmxmexmm

( ) ( ) ( )( ) ( ) ( )[ ]−−−

+++−=+ 12

21

211

1xx

mm

mexx

( ) ( ) ( )( ) ( ) ( )[ ]−−−

++−−=+ 12

21

122

1xx

mm

mexx


( )( ) ( ) ( )[ ]−−−

++=∆=∆ 12

21

2112

1xx

mm

mmetFP ( ) ( )[ ] ( ) ( )[ ]−−+−=−−+=∆=∆ 22211112 xxmxxmtFP or

1F

2m

1x

2x

BV

1m

BLδ

1 2P∆P∆

BLδ

0&12 >∆=− Pxx BLδ

2F

( ) ( ) ( ) ( )( ) ( ) ( ) ( )−−−=+−+

−+−=+++

1221

22112211

xexexx

xmxmxmxm


9

SOLO Backlash


( ) ( )

BL

aa

DD

xxa

Fmm

mm

mm

FFmFFmxx

δ±≠=∆

+−−−−

=−

∆

12

1221

21

21

122112

0

12

2. Contact between m1 and m2 (continue – 3) Let solve at contact x2 = x1 +δBL

( ) 00 >−∆ x

Let assume, for simplicity, that during contact phase: 0>= consta

We have:

( ) ( ) 000 <−∆−=+∆ xex

( ) ( ) ( )−∆−=+∆+=∆ 00 xeatxattx

( ) ( )−∆−=∆ 02

1 2 xetattx

Next contact

( ) ( ) 00112

11 2 =−∆−=∆ xeTaTTx

( )e

a

xT

∆−∆= 0

21

( ) ( ) ( ) ( )−∆=−∆−−∆=−∆ 000

21 xexeaea

xTx

( ) ( ) ( ) 0011221

21 >−∆+=+=∆ txeFmm

mma δ


10

SOLO Backlash

Equations of Motion (continue – 4) 2. Contact between m1 and m2 (continue – 4) Second contact

( ) ( ) 00112

11 2 =−∆−=∆ xeTaTTx

( )e

a

xT

−∆= 021

( ) ( ) ( ) ( )−∆=−∆−−∆=−∆ 000

21 xexeaea

xTx

Third contact

( ) 211

22 eTea

TxT =−∆=

( ) ( ) ( )−∆=−∆=−∆ 012 2 xeTxeTx

(n+1)th contactneTTn 1=

( ) ( )−∆=−∆ 0xeTnx n

We can see that, if e < 1, the time between two successive contacts converge to zero and the two bodies stacked together.

( )e

eTeTTnTTT

nn

n −−=++=+++=

+

1

111121

1


12 xxx −=∆a

12 xxx −=∆

BLδ

1T 2T 3T12 TeT = 13 TeT =

( ) txeatx −∆−=∆ 02

1 2

( )e

a

xT

−∆= 021

12 xxx −=∆

( )−∆ 0x

( )−∆− 0xe

( ) ( )−∆=−∆ 01 xeTx

( )−∆− 1Txe

( ) ( )−∆=−∆ 12 TxeTx

( )−∆− 2Txe

( )−∆−=∆ 0xeatx

( ) ( ) ( )txea δ−∆+−=∆ 011 12 aea ∆=∆ 223 2 aeaea ∆=∆=∆

11

SOLO Backlash


12 xxx −=∆a

12 xxx −=∆

BLδ

12 xxx −=∆

( )−∆ 0x

( )−∆− 0xe The two massesstick together


We can see that, if e < 1, the time between two successive contacts converge to zero and the two bodies stacked together.

If we assume e = 0, the masses stick together instantaneously and we obtain:

( ) ( ) 000 =−∆−=+∆ xex

When we perform a numerically computationwe have for some n:

teTTn n ∆≤= 1

where Δt is the integration step, and then wewill encounter a numerical problem. Therefore we must assume that the masses stick together before this happens.

( ) ( ) 001221

21 >−∆=+=∆ txFmm

mma δ


( )e

eTeTTnTTT

nn

n −−=++=+++=

+

1

111121

1

12Serway and Jewett,“Physics for Scientists and Engineers”, 6th Ed.,

13

SOLO Backlash


Let see what are the equation of motion when the two masses stick together ( x1(t) = x2 (t) for a nonzero period of time).

In this case

( )

( )

−−−=

−−+=

BD

BD

VFFFm

x

VFFFm

x

2

1

1222

1

1211

1

1

1( )

( ) ( )( )21

1222112

21

22121

1

1

mm

FFmFFmF

Vmm

FFFFxx

DD

BDD

+−−−

=

−+

−−+==

From the Figures bellow the masses will stick together as long as ( ){ } 01212 >−⋅ xxFsign


14

SOLO Backlash


1F

1DF

12F

12F

2DF

12F

2F

( ) ( ) ( )( ) ( ) ( )[ ]−−−

++−−=+ 12

21

122

1xx

mm

mexx

2

1

m s

1

s

1

BV

2x

BL

( ) ( ) ( )( ) ( ) ( )[ ]−−−

+++−=+ 12

21

211

1xx

mm

mexx

1

1

m s

1

s

1

BV

1x

BL

( ) ( ) ( ) ( )1121

22

21

12 DD FF

mm

mFF

mm

m −+

−−+

2x

12 xx −

BL1

0 12 xx −

BLδBLδ−1x

2x

12F( )121 2 xxF −

1 ( )[ ]1212 xxFsign −

AND

BL

Stickmode


Return to the Table of Content

15

SOLO Backlash

Backlash Mathematical Model

Motor Moment Equation

m

DmrmCTBm J

TTiK −−+= ωθ

( )

InertiaMotor

Bmm

eDisturbanc

Dm

Torquection

rm

CommandTorque

put

CT JTTiK ωθ −=−−ReIm

- command currentCi- moment of inertia of the rotor along it’s rotational axis mJ- rotor gear angle and it’s first and second order derivativemmm θθθ ,,- angular rate and acceleration of the bodyBB ωω ,- reaction torque of the gimbals gear on the rotor gearrmT-disturbance torque on the rotor (friction, mass-unbalance, inertia cross-coupling) – not a function of

DmTmθ


16

Spur Gear Nomenclature

Pitch circles – Theoretical circles upon which all calculations are based with centers O1 and O2 and diameters D1 and D2.

The two pitch circles are tangent at the pitch point P.

Pressure Lines – The lines passing through P and making

an angle φ with the tangent to the pitch circles along which the tooth of one gear presses the tooth of the second gear.

Base circles – Circles tangent to pressure line with centers O1 and O2 and diameters

DB1 and DB2.

ϕϕ cos&cos 2211 DDDD BB ==

Circular pitch p – The distance measured along the pitch circle from a point on one tooth to the

corresponding point on the adjacent point. Since we assume that the two

gears always maintain contact during rotation the circular pitches are equal.

N1, N2 – Number of teeth on the two gears.

ω1, ω2 – Angular rates of the two gears.

2211 ωω NN =

2

2

1

1

N

D

N

Dp

ππ ==

1

2

2

1

2

1

ωω==

N

N

D

D

SOLO

Pitch Circles

P

O 1

O 2

R 1

R 2

BaseCircles

ϕR B1

R B2

ϕ

PressureLine 1Pressu

re

Line 2

17

A

Begincontact

Endcontact

BCContact Line 1

(Involute)

A'

B'Contact Line 2

(Involute)

rFPressure

Line 1

PressureLine 2

ReactionForce

rF

ReactionForce

Spur Gear Teeth

Spur Gear Tooth shape must be such that contact between the tooth of one gear to thecorresponding tooth of the second gear iscontinuously maintained until the contact occurs with the adjacent tooth.

If the shapes of the teeth on the two gearshave this property they are called conjugates.

From the figure we can see that during therotation of the driver the contact begins atpoint A and continues until it ends at point B.

The segment AB is on the pressure line.

Among infinite possibilities of conjugateshapes the one that is almost exclusively usedin the gear design is the involute.

SOLO

18

Spur Gear Teeth

Among infinite possibilities of conjugateshapes the one that is almost exclusively usedin the gear design is the involute.

The involute of a circle (base circle) is obtainedby an imaginary string IA (see Figures) wound on

the circle and then unwounded (IC, IB(, while holding it taut.

Base circleO

β

I

θ ψA

β1BR

β1BR

r

x

y

Cβ

Cθ Cr

( )θ,rB

( )CCrC θ,

( )( )βββ

βββcossin

sincos

1

1

−=+=

B

B

Ry

Rx

The involute equation

The involute has the following properties:

1. All lines normal to the involute are tangent to the base circle

2. The radius of curvature of the involute at a point P (r,θ) is given by ρ=RB1β. The base circle is the locus of the center of curvature

of the involute.

SOLO

19

Spur Gear Teeth

In the same way

Point A on the involute has the radius RA and thethickness tA along the circle RA.

SOLO

AB

AR

BR

bRAφ

Bφ

2/Bt2/At

O

E

D

FG

toothinvolute

base circle

B

involute

OG

BG

OG

DGDOG φtan===∠

∩

BBBB invDOGDOB φφφφ =−=−∠=∠ :tan

AAA invDOA φφφ =−=∠ :tan

Point B on the involute has the radius RB and thethickness tB along the circle RB.

A

A

AA

A

R

tinv

R

tDOADOE 2

121

+=+∠=∠ φ

B

B

BB

B

R

tinv

R

tDOBDOE 2

121

+=+∠=∠ φ

−+= BA

B

ABB invinv

R

tRt φφ

22

soloh, 07/18/2005

Hamilton H. Mabie and Fred W. Ocvirk"Mechanics and Dynamics of Machinary" SI version3th edition, pp.98-99

20

Width ofspace

Face width

Top land

Addendumcircle

Tooth

thicknessAddendum

Dedendum

Dedendum

circle

Clearencecircle

Clearence

Pitch

circle

Face

Flank

Bottom la

nd

Spur Gear Teeth SOLO

Addendum – the radial distance between the top land of the tooth and the pitch circle

Dedendum – the radial distance between the bottom land of the tooth and the pitch circle

Addendum circle – the circle passing through the top land of the tooth

Dedendum circle – the circle passing through the bottom land of the tooth

Clearance circle – the circle tangent to the addendum of the mating gear

21

Spur Gear Force Equation SOLO

If no backlash rmθm = rLθL

Lm

Lm r

r θθ

=

load

L

motor

m

N

D

N

Dp

ππ ==

Mating condition of the teeth on the two gears is

Circular path of gear 1 = Circular path of gear 2

L

motor

load

m

L

m

L NN

N

D

D

r

r === :LLL

motor

loadL

m

Lm N

N

N

r

r θθθθ =

=

=

The force applied by gear 1 on gear 2 is Fr . Since the tooth surface shape is an involute and the force at the point of contact is normal to the surface, it will be tangent to the base circle that defines the involute. Therefore Fr will always be on the pressure line and

tangent to both base circles.

ϕϕ cos

3'

cos L

th

m r

bL

rLLawsNewton

r

r

bm

rm

r

TF

r

T ==

L

rLrL

L

mrL

L

mrm N

TT

N

NT

r

rT ===

Pressure line 1

Base circle

Pitch circle

Pitch circleBase circle

P B

Pinion(driver)

Gear(driven )

C

O1

ϕBegin

contact

EndcontactA

rF

rF

bmrmr

bLr

Lr

ϕ

ϕ

Trm, TrL are the reaction moments on the two gears


22

SOLO Backlash

Load Moment Equation

( ) LDLBmL

LLDLBLLrmLLm TTN

JTTJTNT ++

+=+++== ωθωθ 1

- moment of inertia of the load (including all the parts attached to it) along it’s rotational axis

LJ

- load gear angle and it’s first and second order derivativeLLL θθθ ,,

- torque applied by the motor gear on the load gearrLTwhere

-disturbance torque on the load (friction, mass-unbalance, inertia cross-coupling) – not a function of

DLTLθ

- torque applied on the load gearLT

L

LDLrmLBL J

TTTN −−+−= ωθ or


23

SOLO Backlash

From this equation we can find the moment Trm necessary to stick the motor gear to the gimbals gear is:

L

LDLBB

m

DmrmCT

LL

L

L

rLrm N

TT

J

TTiK

NN

J

N

TT

++

+

+−−== ωω

1

( )[ ] ( )2

2

1

11

1

LmL

LDLmLBLmDmCTL

Lm

L

L

LDLBB

m

DmCT

LL

L

rm

NJJ

TTJNNJTiKJ

NJJ

N

TT

J

TiK

NN

J

TSTICK

+++++−=

+

++

+

+−

=

ω

ωω

m

DmrmCTBm J

TTiK −−=−ωθ

L

LDLrmLBL J

TTTN −−=+ωθ

Contact Moment


24

SOLO Backlash

Define the backlash error

LLmBL N θθε −=:

Then

BLBLIF θε < 0=rmT 0== rmLrL TNT

BLBLIF θε ≥STICKrmrm TT = rmLrL TNT =

Backlash Error


25

SOLO Backlash

Let find what happens at the collision of the teeth of the gears.

( ) →−== trV BmmOmM

I

M 1ωθρ

The velocity at the point M on the motor gear is:

The velocity at the point L on the load gear is:

( ) ( )( ) →

→→

−=

+++−=+=

trr

trtrrV

BmLL

BLLBLmLO

I

OO

I

L LLm

1

11

ωθ

ωθωρρ

Teeth Collision

(1) No Contact between the teeth of the two axes.

In this case the rotation dynamics of each axis is independent.

26

SOLO Backlash

The collision between toot M and L will produce an equal and opposite impulse Pr on the gear tooth, that will produce reaction impulses on the gear axes, and a change in angular impulse on both gears (ΔHm, ΔHL)

( )[ ] ( )[ ]( ) ( )[ ] mrmmm

BmmBmmm

rPJ

JJH

∆−=−−+=

−−−−+=∆

θθωθωθ

( )[ ] ( )[ ]( ) ( )[ ] LrLLL

BLLBLLL

rPJ

JJH

∆=−−+=

+−−++=∆

θθωθωθ

( ) ( )−− Lm θθ , and are the angular rates of the motor and gimbals gears, respectively, before (-) and after(+) the collision.

( ) ( )++ Lm θθ ,

We obtain the following equations

( ) ( )[ ] ( ) ( )[ ]−−+−=−−+=∆ mmmm

LLLL

r Jr

Jr

P θθθθ 11

Teeth collision (continue – 1)

27

SOLO Backlash

The second equation is obtained by using the elastic coefficient of collision e (e = 0, for plastic collision and e =1 for an elastic collision), that is defined as:

( ) ( )[ ] ( ) ( )[ ]−−+−=−−+=∆ mmmm

LLLL

r Jr

Jr

P θθθθ 11

or

( ) ( )[ ] ( ) ( )[ ] 0=−−++−−+ mmmLLLL JNJ θθθθ

We obtain one equation with two unknowns ( ) ( )++ Lm θθ ,

( ) ( )( ) ( )

( )[ ] ( )[ ]( )[ ] ( )[ ]

( ) ( )( ) ( )

( ) ( )( ) ( )−−−

+−+−=−−−+−+−=

−−−−−−+−−+−=

−−−+−+−=

mLL

mLL

mmLL

mmLL

BmmmBmLL

BmmmBmLL

ML

ML

N

N

rr

rr

rrrr

rrrr

VV

VVe

θθθθ

θθθθ

ωθωθωθωθ

:


28

SOLO Backlash

or:

Now we have two equations with two unknowns ( ) ( )++ Lm θθ ,

( ) ( ) ( ) ( )( ) ( ) ( ) ( )

−−−=+++−

−+−=+++

LLmLLm

LLmmLLLmmL

eNeN

JJNJJN

θθθθθθθθ

Solving for we obtain( ) ( )++ Lm θθ ,

( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( )−−+−+=++

−++−−=++

LmLLmmLGmLL

LLLmLmLmmLL

JNeJJNeJNJ

JNeJeJNJNJ

θθθ

θθθ

22

22

1

1

( ) ( ) ( )( ) ( ) ( )[ ]−−−

++−−=+ LLm

mLL

Lmm N

JNJ

Je θθθθ 2

1

( ) ( ) ( )( ) ( ) ( )[ ]−−−

+++−=+ LLm

mLL

mLLL N

JNJ

JNe θθθθ 2

1


Also

( ) ( )[ ] ( ) ( ) ( )[ ]−−−+

+=−−+−=∆=∆ LLm

mLL

Lmmmmmrrm N

JNJ

JJeJrPtT

Cθθθθ

2

1

29

SOLO Backlash

Let compute ( ) ( )+−+ LLm N θθ

( ) ( ) ( )( ) ( ) ( )[ ]−−−

++−−=+ LLm

mLL

Lmm N

JNJ

Je θθθθ 2

1

( ) ( ) ( )( ) ( ) ( )[ ]−−−

+++−=+ LLm

mLL

mLLL N

JNJ

JNe θθθθ 2

1


( ) ( ) ( )( )

( )( ) ( ) ( )[ ]−−−

+

+−++−=+−+ LLm

mLL

mL

mLL

LLLm N

JNJ

JNe

JNJ

JeN θθθθ

2

2

2

111

or

( ) ( ) ( ) ( )[ ]−−−−=+−+ LLmLLm NeN θθθθ

We can see that

( ) ( )[ ] ( ) ( )[ ]−−−−=+−+ LLmLLm NsignNsign θθθθ

30

SOLO Backlash

LLmBL N θθε −=

θa

BLθ

1T 2T 3T12 TeT = 13 TeT =

( ) teat BLBL −−∆= 02

1 2 εε

( )e

aT BL

∆−= 0

21ε

( )−0BLε

( ) ( )−+−=∆ 011 BLea εθ

12 θθ aea ∆=∆23 θθ aea ∆=∆

LLmBL N θθε −=

( )−− 0BLe ε( )−− 1Te BLε ( )−− 2Te BLε

( ) ( )−=− 12 TeT BLBL εε

( ) ( )−=− 01 BLBL eT εε

LLmBL N θθε −=

( )−−= 0BLBL eat εε θ

Let solve at collision

( ) 00 >−BLε

Let assume, for simplicity, that during contact phase: 0>= constaθ

We have:

( ) ( ) 000 <−−=+ BLBL e εε

( ) ( ) ( )−−=++= 00 BLBLBL eatatt εεε θθ

( ) ( )−−= 02

1 2BLBL etatt εε θ

Next contact

( ) ( ) 00112

11 2 =−−= BLBL eTaTT εε θ

( )e

aT BL

θ

ε −= 021

( ) ( ) ( ) ( )−=−−−=− 000

21 BLBLBL

BL eeaea

T εεεε θθ


L

LDLrmLBL J

TTTN −−+−= ωθ

m

DmrmCTBm J

TTiK −−+= ωθ

( ) ( ) ( )teTJJ

JNJ

TJJ

JNJT

JJ

JNJN

BL

a

rmmL

mLL

rmmL

mLLrm

mL

mLLLLmBL

STICK

STICK

δε

θθε

θ

−+−+=

+−+=−=

012

22

31

SOLO Backlash

The kinetic energy loss due to the collision is given by

( ) ( ) ( ) ( )2

2222

2

1

2

1

2

1

2

1

+++−

−+−=∆ LLmmLLmmk JJJJE θθθθ

( ) ( )

( ) ( )( ) ( ) ( )[ ]

( ) ( )( ) ( ) ( )[ ]

2

2

2

2

22

1

2

1

1

2

1

2

1

2

1

−−−+

++−−

−−−++−−−

−+−=

LLm

mLL

mLLL

LLm

mLL

Lmm

LLmm

NJNJ

JNeJ

NJNJ

JeJ

JJ

θθθ

θθθ

θθ

( )( ) ( ) ( )[ ] ( ) ( )[ ]

( )( ) ( ) ( )[ ] ( )mLLLLm

mLL

mL

LLmLLm

mLL

mL

JNJNJNJ

JJe

NNJNJ

JJe

22

22

2

2

2

1

1

+−−−+

+−

−−−−−−+

+−=

θθ

θθθθ

( )( ) ( ) ( )[ ]2

2

2

2

1 −−−+

−= LLm

mLL

mL NJNJ

JJe θθ


32

SOLO Backlash

The kinetic energy loss due to the collision is given by

( )( ) ( ) ( )[ ]2

2

2

2

1 −−−+

−=∆ LLm

mLL

mLk N

JNJ

JJeE θθ

We can see that

1. If (soft touch) then( ) ( )−=− mLLN θθ

( ) ( )−=+ mm θθ

( ) ( )−=+ LL θθ

0=∆ kE

2. If e = 1 (elastic collision)

0=∆ kE 3. The maximum kinetic energy loss is obtained when e = 0 (plastic collision)

( ) ( ) ( )[ ]2

22−−−

+=∆ LLm

mLL

mLk N

JNJ

JJE

MAXθθ



33

SOLO Backlash

The backlash of two gears occurs because of thermal expansion and tooth error at predefined angles. When backlash occurs the two gears loose contact for a small angle.

For gear1 the lose of contact occurs every radians.1

1

2

N

π=Θ

For gear2 the lose of contact occurs every radians.2

2

2

N

π=Θ

11

2

N

π=Θ

22

2

N

π=Θ

12 Θ 13 Θ14 Θ

24 Θ

23 Θ

22 Θ

1θ

2θ2

2

12 θθ

N

N=

1Θ−

2Θ−

BLθ2

Backlash Logic

34

SOLO Backlash

Backlash Logic (continue - 1)

IF Backlash = .True. & Gears dynamically independent BLk θθ ≤Θ− 11

IF Backlash = .False. & Gears stick togetherBLk θθ >Θ− 11

No Backlash Backlash

35

SOLO BacklashBacklash Logic (continue - 2)

A

C C’

Contact Line1

(Involute)

Contact Line2

(Involute)A’0>rF 0<rF

Each tooth has two contact lines:

Contact line 1 between points A and CContact line 2 between points A’ and C’

Contact line 1 of gear 1 (G1) touchescontact line 1 of gear 2 (G2) along thepressure line 1. To keep the contact wemust have Fr > 0 (the reaction force between G1 and G2).

Contact line 2 of gear 1 (G1) touchescontact line 2 of gear 2 (G2) along thepressure line 2. To keep the contact wemust have Fr < 0 (the reaction force between G1 and G2).

On pressure line 1 if 210 GG →>θ

120 GG →<θ

On pressure line 2 if 120 GG →>θ

210 GG →<θ

Gear Gi pushes gear Gj GjGi →

36

SOLO Backlash

Dynamic computations in contact

Compute

Contact Flag = .True.

[ ]mr

LmL

DLmLBmDmCTLrm rF

NJJ

TJNJTiKJT

STICK=

+++−= 2

ω

m

DmrmCTBm J

TTiK −−=−ωθ

L

LDLrmLBL J

TTTN +−=+ωθ

Integrate both equations to obtain

BLBm ωθωθ +− ,

Compute and integrate to obtainLm θθ , Lm θθ ,


37

SOLO Backlash

Dynamic computations in contact (second way)

Compute


[ ]mr

LmL

DLmLBmDmCTLrm rF

NJJ

TJNJTiKJT

STICK=

+++−= 2

ω

B

m

DmrmCTm J

TTiK ωθ +−−

=

B

L

LDLrmLL J

TTTN ωθ −+−

=


Lm θθ ,

and integrate to obtain Lm θθ ,


38

SOLO Backlash

Detection of End of Contact

Contact ends if one of the following condition occurs


1. ( ) BL

m

mm Nθπθθθ −≥+−=∆ 2

we must also have 0≥mθ

2. ( ) BL

m

mm Nθπθθθ +−≤+−=∆ 2

we must also have 0≤mθ

When one of those events occurs Contact Flag changes from .True. to .False.

4. changes signSTICKrmT

0=BLεInitialize

3. ( )+−=∆ mm θθθ

changes sign

it is enough to compute ( )+−=∆ mm θθθ Since during contact we have LLm N θθ =

and to check if change sign.mrmSTICKT θ, Here is the motor gear position when Start Contact.( )+mθ

39

SOLO Backlash

Dynamic Computations if No Contact

Compute

Contact Flag = .False.

0== mrrm rFT

m

DmrmCTBm J

TTiK −−=−ωθ

L

LDLrmLBL J

TTTN +−=+ωθ


BLBm ωθωθ +− , Compute and integrate to obtainLm θθ , Lm θθ ,


Compute LLmBL N θθε −=:

Integrate to obtainBLε BLε

If Start Contact Flag changes from .False. to .True. for one computation cycle and in the next one back to .False.

BLBL θε ≥

40


Contact Between Gears Teeth is Detected

( )[ ] ( )mr

LmL

LDLmLBLmDmCTLrm rF

NJJ

TTJNNJTiKJT

STICK=

+++++−= 2

1 ω

Compute

( ) ( ) ( )( ) ( ) ( )[ ]−−−

+++−=+ mLL

mLL

Lmm N

JNJ

Je θθθθ 2

1

( ) ( ) ( )( ) ( ) ( )[ ]−−−

++−−=+ mLL

mLL

mLLL N

JNJ

JNe θθθθ 2

1

on pressure line 1 if 210 GG →>θ

120 GG →<θ If 0>= mrrm rFT

STICK

Contact begins at point A

Contact begins at point B

on pressure line 2 if 120 GG →>θ

210 GG →<θ If 0<= mrrm rFT

STICK

Contact begins at point C

Contact begins at point D

Contact Start changes from .False. to .True.

Store ( ) mθθ =+

Reinitialize the integrators using

41


When Contact Start = .True. the elastic collisions between gears teeth will occur and the time between two successive collisions will be

( )e

a

TT nBL

nθ

ε2

1 −= −

LLmBL N θθε −=θa

BLθ

( )−0BLε

The two gearsstick together

LLmBL N θθε −=

( )−− 0BLe ε

LLmBL N θθε −=

1−nTnT

( )e

a

TT NBL

nθ

ε2

1 −= −

We can see that if e=0 (plastic collision), we have T1 = 0 and the teeth remain in contact. In this case


when Contact Start = .True.

0=e If

( ) ( ) ( )( ) ( ) ( )[ ]

0

2

1

=

−−−+++−=+

e

mLL

mLL

Lmm N

JNJ

Je θθθθ

( ) ( ) ( )( ) ( ) ( )[ ]

0

2

1

=

−−−+

+−−=+e

mLL

mLL

mLLL N

JNJ

JNe θθθθ

STICKrm

mLL

Lm TJNJ

JJa 2+

=θ

Contact Between Gears Teeth is Detected (continue - 1)

Reinitialize the integrators using

42

SOLO Backlash

To prevent numerical problems, when



When Contact Start = .True. the elastic collisions between gears teeth will occur and the time between two successive collisions will be

( )e

a

TT nBL

nθ

ε2

1 −= −


BLθ

( )−0BLε


LLmBL N θθε −=

( )−− 0BLe ε

LLmBL N θθε −=

1−nTnT

( )e

a

TT NBL

nθ

ε2

1 −= −

tTn ∆≤ (where Δt is related to the integration time), we say that the two gears teeth are in continuous contact and we declare

To assure that the teeth gears are in continuous contact, we will reinitialize using e=0, for this computation cycle.

( ) ( )++ 0,0 Lm θθ

( ) ( ) ( )( ) ( ) ( )[ ]

0

2

1

=

−−−+++−=+

e

mLL

mLL

Lmm N

JNJ

Je θθθθ

( ) ( ) ( )( ) ( ) ( )[ ]

0

2

1

=

−−−+

+−−=+e

mLL

mLL

mLLL N

JNJ

JNe θθθθ

0≠e If

STICKrm

mLL

Lm TJNJ

JJa 2+

=θ


Contact Between Gears Teeth is Detected (continue - 2)

43

SOLO Backlash

Equations of Motion (Summary)

44

SOLO Backlash

Equations of Motion (Summary 1)


45


When Contact Start = .True. instead of performing reinitialization of integrals we can, for one computation cycle, simulate the impulseat collision.


BLθ

( )−0BLε


LLmBL N θθε −=

( )−− 0BLe ε

LLmBL N θθε −=

1−nTnT

( )e

a

TT NBL

nθ

ε2

1 −= −

by using

Contact between gears is detected (continue - 1)

( ) ( ) ( )[ ]−−−+

+=∆=∆ LLm

mLL

Lmmrrm N

JNJ

JJerPtT

Cθθ

2

1

( ) ( ) ( )[ ]−−−+

+∆

= LLm

mLL

Lmrm N

JNJ

JJe

tT

Cθθ

2

11

46

SOLO Backlash

Equations of Motion (Summary 2)


47

% Backlash%

% Solo Hermelin, March 2006%

% Simulates Backlash Effects of a Gear Transmission Driven by an Electric % Motor to Move a Predefined Load

clear

% Default Parameters dt=0.001;

DegtoRad = pi/180; %Transform fro Degrees to Radians %Motor Parameters

Jmotor = 1.14e-6; %Motor Inertia [Kg*m^2[ %Jmotor = 1.; %Motor Inertia [Kg*m^2[

Nmotor = 36; % Number of theeth on motor gear pmotor = 360/Nmotor; % Angle between two consecutive theeth on motor Gear (deg(

pmotorR = pmotor*DegtoRad; % Angle between two consecutive theeth on motor Gear (deg( Backlash = .25; % Percent of Motor Gear Angle for which theeth lose contact

%Load Parameters Jload = 8.5e-5; % Load Inertia [Kg*m^2[

%Jload = 1; % Load Inertia [Kg*m^2[ Nload = 72; % Number of theeth on load gear

NL = Nload/Nmotor; % Theeth Ratio NLJm = NL*Jmotor; NL2Jm = NL*NLJm;

JlNLJ2m=Jload+NL2Jm; TfM = 0.; % Friction Moment on Motor

TfL = 0.; % Friction Moment on Load

BacklashMATLAB Program (page 1)

48

% Initialization of State-Variables ThetM = 0.;

ThetM1Dot = 0.; PhiM2Dot = 0.;

ThetL = 0.; ThetL1Dot = 0.;

PhiL2Dot = 0.; epsBL = ThetM - NL*ThetL;

StartContact = 0; % Has value 1 only for one integration cycle, when gear theeth make contact Contact = 0.; % Change to 1 when gear theeth are in contact

k=0;

dt=input('Enter Integration Time Step (s(: '(; tend=input('Enter Final Simultion Time (s(: '(;

Backlash=input('Enter Backlash (0-1(: '(; el = input('Elasticity of Gear Theeth (0-1(: '(;

InMoment = input('Amplitude of Input Sinusoidal Moment (N*m(: '(; freqin = input('Frequency of Sinusoidal Input (Hz(: '(;

Aomegab = input('Amplitude of Sinusoidal Body Rate (rad/sec(:'(; fomegab = input('Frequency of Sinusoidal Body Rate (rad/sec(:'(;

omgi = 2*pi*freqin; omgob =2*pi*fomegab;

labelBacklash=['Backlash =' num2str(Backlash( [; labelel=['el =' num2str(el( [;

ThetBL = Backlash*pmotor; % Backlash angle at motor gear [deg[ ThetBLR = ThetBL*DegtoRad;

t=0;


49

% Loop on time while t<=tend k = k+1;%plot index % Inputs KtIc = InMoment*cos(omgi*t(; %Sinusoidal Input [N*m[ %KtIc = InMoment; Tload = 0.; % Load Moment [N*m[ OmegaB = Aomegab*sin(omgob*t(; % Body rate [rad/s[ OmegaDotB = Aomegab*omgob*cos(omgob*t(; % Body angular acceleration [rad/s^2[

epsDotBL = ThetM1Dot -NL*ThetL1Dot; %Computation of Stiction Moment TrmStick = ( Jload*(KtIc-TfM+Jmotor*(1+NL(*OmegaDotB(+NLJm*(Tload+TfL( (/JlNLJ2m;


50

%Backlash Logic if (Contact==0( & (abs(epsBL(>= ThetBLR( StartContact = 1; % Integration cycle where theeth contact occurs TMP =0.5* Jmotor*Jload*abs(epsDotBL(/(abs(TrmStick(*JlNLJ2m(; if TMP<=dt Contact = 1; EndCntct1 = 0; EndCntct2 = 0; EndCntct3 = 0; ThetM1Dot =ThetM1Dot -Jload*epsDotBL/JlNLJ2m; %Rate change due to Plastic Impact ThetL1Dot =ThetL1Dot +NLJm*epsDotBL/JlNLJ2m; %Rate change due to Plastic Impact Thetcntct = ThetM; sgnTstck = sign (TrmStick(; epsBL = 0.; p = 0; % index else ThetM1Dot =ThetM1Dot -(1+el(*Jload*epsDotBL/JlNLJ2m; %Rate change due to Elastic Impact ThetL1Dot =ThetL1Dot +(1+el(*NLJm*epsDotBL/JlNLJ2m; %Rate change due to Elastic Impact Contact = 0; end %TMP<=dt end %(Contact==0( & (abs(epsBL(>= ThetBLR(


51

if (Contact==1( & (StartContact==0( %Check end of Contact

p = p + 1; DelThetM = ThetM - Thetcntct;

if p==1 sgnDelThM = sign (DelThetM(;

end if p>1

if (sign (DelThetM(~=sgnDelThM ( Contact = 0;

EndCntct1 = 1 elseif abs(DelThetM(>=(pmotorR-ThetBLR(

Contact = 0; EndCntct2 = 1

end end % p>1

if sign(TrmStick(~=sgnTstck Contact = 0;

EndCntct3 = 1 end

end % (Contact==1( & (StartContact==0(

if (Contact==1( Trm = TrmStick; else % Contact=0

Trm = 0; end % Contact==1

%End Backlash Logic


52

% Motor DynamicsThetM2Dot = (KtIc-TfM-Trm(/Jmotor+OmegaDotB ;

%Load DynamicsThetL2Dot = (NL*Trm-Tload-TfL(/Jload-OmegaDotB ;

%Plot Data PreparationT (k( = [t[;ktic (k( = [KtIc[;thetm (k( = [ThetM/DegtoRad[;thethm1d (k( = [ThetM1Dot[;thetm2d (k( = [ThetM2Dot[;thetl (k( = [NL*ThetL/DegtoRad[;thethl1d (k( = [NL*ThetL1Dot[;thetl2d (k( = [NL*ThetL2Dot[;epsBLD (k(= [epsBL/DegtoRad[;epsD (k( = [(ThetM-NL*ThetL(/DegtoRad[;epsdotBL (k( = [epsDotBL[;strtcontct (k( = [StartContact[;contct (k( = [Contact[;


53

%Integration of State Variables if StartContact == 1

StartContact =0; else

ThetM1Dot = ThetM1Dot+ThetM2Dot*dt; ThetL1Dot = ThetL1Dot+ThetL2Dot*dt;

end if Contact==0

epsBL =epsBL + epsDotBL*dt; end

ThetM = ThetM + ThetM1Dot*dt; ThetL = ThetL + ThetL1Dot*dt;

t=t+dt;end % of Time Loop t<=tend

figure(1(plot(T,thetm2d ,'g-'(;

legend(labelBacklash,labelel(;hold on;

plot(T,thetl2d ,'r-.'(;hold off;title('Angular Acceleration of Motor and Load Versus Time'(,

%axis([0 5 -1 1[(,grid, xlabel('Time (s('(, ylabel('ThetM2dot, NL*ThetL2dot (rad/s^2('(,


54

figure(2(plot(T,thethm1d,'b-'(;


plot(T,thethl1d,'r-.'(;hold off;

title('Angular Rates of Motor and Load Versus Time '(,%axis([0 1 -0.1 0.1[(,

grid, xlabel('Time (s('(, ylabel('ThetM1dot, NL*ThetL1dot (rad/s('(,

figure(3(plot(T,thetm ,'b-'(;


plot(T,thetl,'r-.'(;hold off;title('Angles of Motor and Load Versus Time'(,

%axis([0 5 -180 180[(,grid, xlabel('Time (s('(, ylabel('Theta_Motor, NL*Theta_Load (deg('(,

figure(4(plot(T,epsBLD ,'b-'(;

legend(labelBacklash,labelel(;hold off;title('Angle Errors EpsBL (deg( of Motor and Load Versus Time'(,

%axis([0 5 -180 180[(,grid, xlabel('Time (s('(, ylabel('EpsBL (deg('(,


55

figure(5(plot(T,epsBLD ,'b-'(;


plot(T,epsD,'r-.'(;hold off;title('Angle Errors EpsBL & Eps (deg( of Motor and Load Versus Time'(,

%axis([0 5 -180 180[(,grid, xlabel('Time (s('(, ylabel('EpsBL, Eps (deg('(,

figure(6(plot(T,epsdotBL,'b--'(;

legend(labelBacklash,labelel(;hold off;title('Rate of Angles of Errors (rad/s( Versus Time'(,

%axis([0 1 -0.1 0.1[(,grid, xlabel('Time (s('(, ylabel('EpsDotBL (rad/s('(,

pause



56

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1-0.06

-0.04

-0.02

0

0.02

0.04

0.06Rate of Angles of Errors (rad/s) Versus Time

Time

Eps

Dot

BL

Backlash =0.01el =0.9

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1-0.03

-0.02

-0.01

0

0.01

0.02

0.03

0.04

0.05


Time

Eps

Dot

BL


0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10

0.01

0.02

0.03

0.04

0.05


Time

Eps

Dot

BL

Backlash =0.01el =0

Simulation Results

570 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

-0.06

-0.04

-0.02

0

0.02

0.04


Time

Eps

Dot

BL


0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-0.1

-0.08

-0.06

-0.04

-0.02

0

0.02

0.04

0.06

0.08

0.1Angular Rates of Motor and Load Versus Time

Time

The

tM1d

ot,

The

tL1d

otBacklash =0.01el =0.9

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-0.06

-0.04

-0.02

0

0.02

0.04


Time

Eps

Dot

BL

Backlash =0.01el =0

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-0.1

-0.08

-0.06

-0.04

-0.02

0

0.02

0.04

0.06

0.08

0.1Angular Rates of Motor and Load Versus Time

Time

The

tM1d

ot,

The

tL1d

ot

Backlash =0.01el =0


58

SOLO Backlash

References

1. Donald T. Greenwood, “Principles of Dynamics”, Prentice Hall, 1965

2. Wolfram Stadler, “Analytical Robotics and Mechatronics”, McGraw-Hill, 1995

3. Hamilton H. Mabie and Fred W. Ocvirk, “Mechanism and Dynamics of Machinery”, SI Version, Third Edition, John Wiley & Sons, 1978

4. Joseph E. Shigley, Charles R. Mischke, Richard G. Budynas, “Mechanical Engineering Design”, Seventh Edition, McGraw-Hill, 2004


5. Solo Hermelin, “Gears” Presentation, http://www.solohermelin.com

January 4, 2015 59

SOLO

TechnionIsraeli Institute of Technology

1964 – 1968 BSc EE1968 – 1971 MSc EE

Israeli Air Force1970 – 1974

RAFAELIsraeli Armament Development Authority

1974 – 2013

Stanford University1983 – 1986 PhD AA

2 backlash simulation

Science