2 backlash simulation

Backlash Simulation

SOLO HERMELIN

Updated: 26.11.06 25.02.11

No Backlash Backlash

http://www.solohermelin.com

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

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.

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

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

SOLO Backlash

Equations of Motion

1. No contact between m1 and m2

( )BLBL

δδ <−<−

−−=

2. Contact between m1 and m2

−−−=

−−+=

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

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

( ) ( )21

122112

FFmFFmxx DD −−−

( ) ( )BL

DD xxFmm

FFmFFmxx δ±=+−

−−−=− 1212

121112

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

( ) ( )[ ] ( ) ( )[ ] 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:xVxV

( ) ( )( ) ( )−−−

+−+−=21

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)

SOLO Backlash

Now we have two equations with two unknowns:

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

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

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

( ) ( ) ( )( ) ( ) ( )[ ]−−−

+++−=+ 12

( ) ( ) ( )( ) ( ) ( )[ ]−−−

++−−=+ 12

( )( ) ( ) ( )[ ]−−−

++=∆=∆ 12

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

1 2P∆P∆

0&12 >∆=− Pxx BLδ

( ) ( ) ( ) ( )( ) ( ) ( ) ( )−−−=+−+

−+−=+++

22112211

xexexx

xmxmxmxm

SOLO Backlash

( ) ( )

FFmFFmxx

δ±≠=∆

+−−−−

122112

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

∆−∆= 0

( ) ( ) ( ) ( )−∆=−∆−−∆=−∆ 000

21 xexeaea

( ) ( ) ( ) 0011221

21 >−∆+=+=∆ txeFmm

mma δ

SOLO Backlash

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

( ) ( ) 00112

11 2 =−∆−=∆ xeTaTTx

−∆= 021

( ) ( ) ( ) ( )−∆=−∆−−∆=−∆ 000

21 xexeaea

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.

eTeTTnTTT

n −−=++=+++=

111121

12 xxx −=∆a

12 xxx −=∆

1T 2T 3T12 TeT = 13 TeT =

( ) txeatx −∆−=∆ 02

−∆= 021

12 xxx −=∆

( )−∆ 0x

( )−∆− 0xe

( ) ( )−∆=−∆ 01 xeTx

( )−∆− 1Txe

( ) ( )−∆=−∆ 12 TxeTx

( )−∆− 2Txe

( )−∆−=∆ 0xeatx

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

SOLO Backlash

12 xxx −=∆a

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

eTeTTnTTT

n −−=++=+++=

111121

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

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

−−−=

−−+=

( ) ( )( )21

1222112

FFmFFmF

FFFFxx

+−−−

−−+==

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

SOLO Backlash

( ) ( ) ( )( ) ( ) ( )[ ]−−−

++−−=+ 12

( ) ( ) ( )( ) ( ) ( )[ ]−−−

+++−=+ 12

( ) ( ) ( ) ( )1121

12 DD FF

m −+

−−+

12 xx −

0 12 xx −

BLδBLδ−1x

12F( )121 2 xxF −

1 ( )[ ]1212 xxFsign −

Stickmode

Return to the Table of Content

SOLO Backlash

Backlash Mathematical Model

Motor Moment Equation

DmrmCTBm J

TTiK −−+= ωθ

InertiaMotor

eDisturbanc

Torquection

CommandTorque

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

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 =

ππ ==

ωω==

Pitch Circles

BaseCircles

ϕR B1

PressureLine 1Pressu

Line 2

Begincontact

Endcontact

BCContact Line 1

(Involute)

B'Contact Line 2

(Involute)

rFPressure

Line 1

PressureLine 2

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.

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

θ ψA

Cθ Cr

( )θ,rB

( )CCrC θ,

( )( )βββ

βββcossin

sincos

−=+=

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.

Spur Gear Teeth

In the same way

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

2/Bt2/At

toothinvolute

base circle

involute

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.

tDOADOE 2

+=+∠=∠ φ

tDOBDOE 2

+=+∠=∠ φ

−+= BA

ABB invinv

tRt φφ

Width ofspace

Face width

Top land

Addendumcircle

thicknessAddendum

Dedendum

circle

Clearencecircle

Clearence

circle

Bottom la

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

Spur Gear Force Equation SOLO

If no backlash rmθm = rLθL

r θθ

ππ ==

Mating condition of the teeth on the two gears is

Circular path of gear 1 = Circular path of gear 2

r === :LLL

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

rLLawsNewton

rT ===

Pressure line 1

Base circle

Pitch circle

Pitch circleBase circle

Pinion(driver)

Gear(driven )

ϕBegin

contact

EndcontactA

Trm, TrL are the reaction moments on the two gears

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

- 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

LDLrmLBL J

TTTN −−+−= ωθ or

SOLO Backlash

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

DmrmCT

rLrm N

+−−== ωω

( )[ ] ( )2

LDLmLBLmDmCTL

TTJNNJTiKJ

TSTICK

+++++−=

DmrmCTBm J

TTiK −−=−ωθ

LDLrmLBL J

TTTN −−=+ωθ

Contact Moment

SOLO Backlash

Define the backlash error

LLmBL N θθε −=:

BLBLIF θε < 0=rmT 0== rmLrL TNT

BLBLIF θε ≥STICKrmrm TT = rmLrL TNT =

Backlash Error

SOLO Backlash

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

( ) →−== trV BmmOmM

M 1ωθρ

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

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

( ) ( )( ) →

→→

+++−=+=

trtrrV

BLLBLmLO

ωθωρρ

Teeth Collision

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

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

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

∆−=−−+=

−−−−+=∆

θθωθωθ

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

BLLBLLL

∆=−−+=

+−−++=∆

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

P θθθθ 11

Teeth collision (continue – 1)

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

P θθθθ 11

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

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

( ) ( )( ) ( )

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

( ) ( )( ) ( )

( ) ( )( ) ( )−−−

+−+−=−−−+−+−=

−−−−−−+−−+−=

−−−+−+−=

BmmmBmLL

θθθθ

ωθωθωθωθ

SOLO Backlash

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

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

−−−=+++−

−+−=+++

LLmLLm

LLmmLLLmmL

JJNJJN

θθθθθθθθ

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

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

−++−−=++

LmLLmmLGmLL

LLLmLmLmmLL

JNeJJNeJNJ

JNeJeJNJNJ

θθθ

( ) ( ) ( )( ) ( ) ( )[ ]−−−

++−−=+ LLm

Je θθθθ 2

( ) ( ) ( )( ) ( ) ( )[ ]−−−

+++−=+ LLm

mLLL N

JNe θθθθ 2

( ) ( )[ ] ( ) ( ) ( )[ ]−−−+

+=−−+−=∆=∆ LLm

Lmmmmmrrm N

JJeJrPtT

Cθθθθ

SOLO Backlash

Let compute ( ) ( )+−+ LLm N θθ

( ) ( ) ( )( ) ( ) ( )[ ]−−−

++−−=+ LLm

Je θθθθ 2

( ) ( ) ( )( ) ( ) ( )[ ]−−−

+++−=+ LLm

mLLL N

JNe θθθθ 2

( ) ( ) ( )( )

( )( ) ( ) ( )[ ]−−−

+−++−=+−+ LLm

LLLm N

JeN θθθθ

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

We can see that

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

SOLO Backlash

LLmBL N θθε −=

1T 2T 3T12 TeT = 13 TeT =

( ) teat BLBL −−∆= 02

1 2 εε

∆−= 0

( )−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

ε −= 021

( ) ( ) ( ) ( )−=−−−=− 000

21 BLBLBL

BL eeaea

T εεεε θθ

LDLrmLBL J

TTTN −−+−= ωθ

DmrmCTBm J

TTiK −−+= ωθ

( ) ( ) ( )teTJJ

mLLLLmBL

θθε

−+−+=

+−+=−=

SOLO Backlash

The kinetic energy loss due to the collision is given by

( ) ( ) ( ) ( )2

+++−

−+−=∆ LLmmLLmmk JJJJE θθθθ

( ) ( )

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

−−−+

++−−

−−−++−−−

−+−=

θθθ

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

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

LLmLLm

JNJNJNJ

+−−−+

−−−−−−+

θθθθ

( )( ) ( ) ( )[ ]2

1 −−−+

−= LLm

mL NJNJ

JJe θθ

SOLO Backlash

The kinetic energy loss due to the collision is given by

( )( ) ( ) ( )[ ]2

1 −−−+

−=∆ LLm

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

MAXθθ

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

For gear2 the lose of contact occurs every radians.2

12 Θ 13 Θ14 Θ

12 θθ

1Θ−

2Θ−

Backlash Logic

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

SOLO BacklashBacklash Logic (continue - 2)

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 →

SOLO Backlash

Dynamic computations in contact

Compute

Contact Flag = .True.

DLmLBmDmCTLrm rF

TJNJTiKJT

STICK=

+++−= 2

DmrmCTBm J

TTiK −−=−ωθ

LDLrmLBL J

TTTN +−=+ωθ

Integrate both equations to obtain

BLBm ωθωθ +− ,

Compute and integrate to obtainLm θθ , Lm θθ ,

SOLO Backlash

Dynamic computations in contact (second way)

Compute

DLmLBmDmCTLrm rF

TJNJTiKJT

STICK=

+++−= 2

DmrmCTm J

TTiK ωθ +−−

LDLrmLL J

TTTN ωθ −+−

Lm θθ ,

and integrate to obtain Lm θθ ,

SOLO Backlash

Detection of End of Contact

Contact ends if one of the following condition occurs

1. ( ) BL

mm Nθπθθθ −≥+−=∆ 2

we must also have 0≥mθ

2. ( ) BL

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

SOLO Backlash

Dynamic Computations if No Contact

Compute

Contact Flag = .False.

0== mrrm rFT

DmrmCTBm J

TTiK −−=−ωθ

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 ≥

Contact Between Gears Teeth is Detected

( )[ ] ( )mr

LDLmLBLmDmCTLrm rF

TTJNNJTiKJT

STICK=

+++++−= 2

Compute

( ) ( ) ( )( ) ( ) ( )[ ]−−−

+++−=+ mLL

Je θθθθ 2

( ) ( ) ( )( ) ( ) ( )[ ]−−−

++−−=+ mLL

mLLL N

JNe θθθθ 2

on pressure line 1 if 210 GG →>θ

120 GG →<θ If 0>= mrrm rFT

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

Contact begins at point C

Contact begins at point D

Contact Start changes from .False. to .True.

Store ( ) mθθ =+

Reinitialize the integrators using

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

TT nBL

1 −= −

LLmBL N θθε −=θa

( )−0BLε

The two gearsstick together

LLmBL N θθε −=

( )−− 0BLe ε

LLmBL N θθε −=

1−nTnT

TT NBL

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

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

−−−+++−=+

Je θθθθ

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

−−−+

+−−=+e

mLLL N

JNe θθθθ

STICKrm

Lm TJNJ

JJa 2+

Contact Between Gears Teeth is Detected (continue - 1)

Reinitialize the integrators using

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

TT nBL

1 −= −

( )−0BLε

LLmBL N θθε −=

( )−− 0BLe ε

LLmBL N θθε −=

1−nTnT

TT NBL

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

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

−−−+++−=+

Je θθθθ

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

−−−+

+−−=+e

mLLL N

JNe θθθθ

0≠e If

STICKrm

Lm TJNJ

JJa 2+

Contact Between Gears Teeth is Detected (continue - 2)

SOLO Backlash

Equations of Motion (Summary)

SOLO Backlash

Equations of Motion (Summary 1)

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

( )−0BLε

LLmBL N θθε −=

( )−− 0BLe ε

LLmBL N θθε −=

1−nTnT

TT NBL

1 −= −

by using

Contact between gears is detected (continue - 1)

( ) ( ) ( )[ ]−−−+

+=∆=∆ LLm

Lmmrrm N

JJerPtT

( ) ( ) ( )[ ]−−−+

Lmrm N

SOLO Backlash

Equations of Motion (Summary 2)

% Backlash%

% Solo Hermelin, March 2006%

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

% 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)

% 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

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;

% 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;

%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(

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

% 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[;

%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('(,

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('(,

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('(,

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

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

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 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10

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

0.1Angular Rates of Motor and Load Versus Time

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

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

0.1Angular Rates of Motor and Load Versus Time

Backlash =0.01el =0

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

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

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