FORMULAS FOR MOTORIZED LINEAR MOTION SYSTEMS

21
Haydon Kerk Motion Solutions Pittman Motors : 203 756 7441 : 267 933 2105 www.HaydonKerkPittman.com FORMULAS FOR MOTORIZED LINEAR MOTION SYSTEMS

Transcript of FORMULAS FOR MOTORIZED LINEAR MOTION SYSTEMS

Page 1: FORMULAS FOR MOTORIZED LINEAR MOTION SYSTEMS

Haydon Kerk Motion SolutionsPittman Motors

203 756 7441 267 933 2105

wwwHaydonKerkPittmancom

FORMULAS FOR MOTORIZED

LINEAR MOTION SYSTEMS

Symbol Description Units Symbol Description Units

a ms2linear acceleration

a rads2angular acceleration

α degCtemperature coefficient

Td Nmreverse torque required for deceleration

Tf Nmtorque required to overcome friction

Tg Nmtorque required to overcome gravity

TLR Nmlocked rotor torque

TRMS NmRMS torque required over the total duty cycle

TRMS (motor) NmRMS torque required at the motor shaft

μ nacoefficient of friction

VT Vmotor terminal voltage

Vbus VDC drive bus voltage

v mslinear velocity

vf msfinal linear velocity

msinitial linear velocity

mspeak linear velocity

W Jenergy

ω radsangular velocity

radsno load angular velocity

radspeak angular velocity

vi

vPK

ωo

ωPK

TD Nmdrag preload torque

ain rads2angular acceleration

aout rads2angular acceleration

DV Nm sradviscous damping factor

F Nforce

Fa Nlinear force required during acceleration (FJ +Ff +Fg)

Ff Nlinear force required to overcome friction

Fg Nlinear force required to overcome gravity

FJ Nlinear force required to overcome load inertia

g ms2gravitational constant (98 ms2)

I Acurrent

ILR Alocked rotor current

IO Ano-load current

IPK Apeak current

IRMS ARMS current

J Kg-m2inertia

JBKg-m2inertia of belt

JGBKg-m2inertia of the gear box

JinKg-m2reflected inertia at system input

Jout Kg-m2reflected inertia at system output

JP1Kg-m2inertia of pully 1

JP2Kg-m2inertia of pully 2

JSKg-m2lead screw inertia

K(i)

Vradsvoltage constant

K(f) NmA or V(rads)KE or KT (finalldquohotrdquo)

KE

KE or KT (initial ldquocoldrdquo)

Km Nm wmotor constant

NmA or V(rads)

KT NmAtorque constant

L mrevscrew lead

m Kgmass

N nagear ratio

RPM

ηRPMspeed

nO no load speed

nnaefficiency

RPMnPK peak speed

load orientation (horizontal = 0deg vertical = 90deg)

degCθ degrees

Θa ambient temperature

Θf degCmotor temperature (finalldquohotrdquo)

Θi degCmotor temperature (initial ldquocoldrdquo)

Θm degCmotor temperature

Θr degCmotor temperature rise

Θrated degCrated motor temperature

P Wpower

Pavg Waverage power

PCV Wpower required at constant velocity

Pin Wpower required at system input

Ploss Wdissipated power

Pmax Wmaximum power

Pmax(f) Wmaximum power (finalldquohotrdquo)

Pmax(i) Wmaximum power (initial ldquocoldrdquo)

Pout Woutput power

PPK Wpeak power

Rm RPMNmmotor regulation

Rmt Ωmotor terminal resistance

Rmt(f) Ωmotor terminal resistance (finalldquohotrdquo)

Rmt(i) Ωmotor terminal resistance (initial ldquocoldrdquo)

Rth degCWthermal resistance

r mradius

S mlinear distance

T Nmtorque

Ta Nmtorque required to overcome load inertia

Ta (motor) Nmtorque required at motor shaft during acceleration

TC Nmcontinuous rated motor torque

TCF Nmcoulomb friction torque

Tin Nmtorque required at system input

Tout Nmtorque required at system output

radsangular velocity at system inputωin

radsangular velocity at system outputωout

t stime

Symbols and Units

2 SYMBOLS AND UNITS

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1

(V)(I)Watts

F x St =

+ +MotorController

DrivePowerSupply

Mechanical Transmission

Lead Screw Support Rails

WattsLost

(V)(I)Watts

WattsLost

WattsLost

WattsLost

(T)(ω)Watts

(T)(ω)Watts

WattsLost

All analysis begins with the linear motion profile of the output and the force required

to move the load

t 1 t 2 t 3t move

Total Cycle Time

t dwell

vTime

vpk

a -a

13 13 13 Trapazoidal Motion Profile

Optimized for minimum power

t 1 = t 2= t 3

vPK =3S2t

S is the total move distancet is the total move timeArea under profile curve represents distance moved

Watts Out

Motion System Formulas

Any motion system should be broken down into its individual components

Motion Profiles

3

(V)(I)Watts

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2

MOTION SYSTEMFORMULAS

Triangular Motion Profiles

Optimized for minimum acceleration slope

t 1 = t 2

vPK =2St

S is the total move distancet is the total move time 13 13 13

Triangular

Move Profile Peak Velocity (VPK) Acceleration (a )

t 1 t 2t move

Total Cycle Time

t dwell

vTime

a -avPK

Complex Motion Profile

Linear to Rotary Formulas through a Lead Screw

(1) s = ndashndashndashndashvf - vi

2t

(2) vf = vi + a t

(3) s = vi t + ndash at 212

(4) 2 as = vf 2

- vi 2

(5) a = ( vf - vi) ( tf - ti )

= ΔvΔt

a = = rads 2 2 π a

L

ω = = rads 2 π v

L

n = = RPM 60 v

L

3 known quantities are needed to solve for the other 2 Each segment calculated individually

Linear Motion Formulas

t 1 t 2 t 3t move

Total Cycle Time

t dwell

vTime

v1

a 1

-a3

a 2

t 4 t 5

v2

Motion System Formulas4

Area under profile curve represents distance moved

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MOTION SYSTEMFORMULAS

Inertia Acceleration Velocity and Required Torque

Jin mout

ain aout ωin vout

Tin Fout

radsec 2

radsec

ain = 2 πa

L

ωin = 2 π v

L

Jin = m 2 πL 2

X η1 + JS Kg-m2

Τin= Τa+Τf +Τg+ΤD N-m

Τa= Jin ain

Τf =cosOslashmgμL

2 π ηN-m

N-m

Τg =sinOslashmgL

2 π ηN-m

ΤD = dragpreloadrefer to manufacturerrsquos data

N-m

Lead Screw System

Belt and Pulley System

ain = radsec 2

ωin = radsec Jin Jout

ain aout ωin ωout

Tin Tout

Jin = NJout

2X η

1 + JP1 Kg-m2

N-mFor a 1st approximation analysis JP1 JP2 and JB can be disregarded For higher precision pulley and belt inertias should be included as well as the reflected inertia of these components

+ JP2+ JB

(aout)(N) (ωout)(N)

Τin = NΤout X η

1

Motion System Formulas5

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MOTION SYSTEMFORMULAS

Motion System Formulas6

Gearbox System

radsec 2

ωin = radsec

Jin = NJout

2X η

1

+

Kg-m2

N-m

Drag torque can be significant (TD) depending on the viscosity of the lubricant For a first approximation analysis this can be left outGearbox inertia may be found in manufacturerrsquos data sheets or calculated using gear dimensions and materials mass

JGB

(ωout)(N) Τin = N

Τout X η1

+

ΤD

ΤRMS = Τ12t 1 + Τ2

2t2 + Τ32t 3 + Τn

2t n

t 1 + t 2 + t 3 + t n + tdwell

TωLinear Power =

FSt

= watts

Rotary Power = = watts

Mechanical Power and RMS Torque

Jin Jout

ain aout ωin ωout

Tin Tout

ain = (aout)(N)

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MOTION SYSTEMFORMULAS

Assuming IO is very small it can be ignored for a quick approximation of motor no-load speed If more accurate results are needed IO will also need to be adjuisted based on the motor Viscous Damping Factor (Dv) As the no-load speed increases with increased terminal voltage on a given motor IO will also increase based on the motorrsquos Dv value found on the manufacturerrsquos motor data sheet

Motor No Load SpeedVT ndash (IO x Rmt)no = 95493 x

KE

ndash ndash ndash ndash ndash ndash ndash ndash ndash ndash ndash ndash ndash

ndash ndash ndash

ndash ndash ndash

BreakawayTorque

Viscous Damping Torque (Dv)

Coulomb Friction

Τ

ωο

Motion System Formulas7

DC Motor Formulas

Motor Locked Rotor Current Locked Rotor Torque

ILR =VT

Rmt

TLR = ILR x KT

Motor Regulation

RmtRm = 95493 x

KE x KT

= x KT

VT

Rmt

ndash Theoretical using motor constants

nORm = TLR

ndash Regulation calculated using test data

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MOTION SYSTEMFORMULAS

Motor Power Relationships

ndash Watts lost due to winding resistance

Ploss = I2 x Rmt

ndash Output power at any point on the motor curve

Pout = ω x Tndash Motor maximum output power

Pmax = 025 x ωo x TLR

ndash Motor maximum output power (Theoretical)

Pmax = 025 x VT2

Rmt

Temperature Effects on Motor Constants

ndash Change in terminal resistance

Rmt(f) = Rmt(i) x [ 1 + αconductor (Θf ndash Θi) ]αconductor = 00040degC (copper)

K(f) = K(i) x [ 1 + αmagnet (Θf ndash Θi) ]ndash Change in torque constant and voltage constant ( KT = KE )

Magnetic Material

CeramicSmCoAlNiCoNdFeB

Θmax (degC)300degC300degC540degC150degC

The magnetic material values above represent average figures for particular material classes and estimating performance If exact values are needed consult data sheets for a specific magnet grade

Motion System Formulas8

αmagnet (degC)-00020degC-00004degC-00002degC-00012degC

In the case of a mechanically commutated motor with graphite brushes this analysis will result in a slight error because of the negative temperature coefficient of carbon For estimation purposes this can be ignored but for a more rigorous analysis the behavior of the brush material must be taken into account This is not an issue when evaluating a brushless dc motor

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MOTION SYSTEMFORMULAS

Motion System Formulas9

For PMDC motors (brush type) brush drop can be approximated as a constant resistance connected in series with the armature winding This series resistance is included in the Rmt factor found on manufacturerrsquos data sheet

Copper graphiteSilver graphiteElectrographite

02 to 04 Ω 02 to 04 Ω 08 to 10 Ω

Motor Brush Resistance

Brush Material Resistance

DC Motor Thermal Resistance and Temperature Rise

ndash Thermal resistance

ndash Motor temperature rise

Θm = Θr + Θa

ndash OR ndash

Temperature Rise degCRth =

Machine Losses W

Θr = Rth x I2 x Rmt

Θr = Rth x I

2 x Rmt

1 ndash (Rth x I2 x Rmt x α)

ndash OR ndash

Θr = Rth x(TC

Km)

2

ndash Final motor temperature

Use caution when applying these formulas Depending on the conditions used to determine motor specifications only one formula should be used Contact Appications Engineering for questions

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8

MOTION SYSTEMFORMULAS

Motion System Formulas9

For PMDC motors (brush type) brush drop can be approximated as a constant resistance connected in series with the armature winding This series resistance is included in the Rmt factor found on manufacturerrsquos data sheet

Copper graphiteSilver graphiteElectrographite

02 to 04 Ω 02 to 04 Ω 08 to 10 Ω

Motor Brush Resistance

Brush Material Resistance

DC Motor Thermal Resistance and Temperature Rise

ndash Thermal resistance

ndash Motor temperature rise

Θm = Θr + Θa

ndash OR ndash

Temperature Rise degCRth =

Machine Losses W

Θr = Rth x I2 x Rmt

Θr = Rth x I

2 x Rmt

1 ndash (Rth x I2 x Rmt x α)

ndash OR ndash

Θr = Rth x(TC

Km)

2

ndash Final motor temperature

Use caution when applying these formulas Depending on the conditions used to determine motor specifications only one formula should be used Contact Appications Engineering for questions

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MOTION SYSTEMFORMULAS

Stepper Motor Formulas

Required ldquoTime Offrdquo using an LR drive (constant voltage drive)

180Step Oslash = (phases)(rotor teeth)

ndash Motor step angle

ndash RPM (revolutionsminute) to SPS (stepssecond)

Maximum accuracy of a step motor system has nothing to do with maximum resolution of a step motor system Maximum accuracy is always based on the accuracy of 1 full step but maximum resolution is based on all system components including lead screw encoder stepper motor and step mode on the controller

ndash SPS (stepssecond) to RPM (revolutionsminute)(SPS) x (Step Oslash)

RPM = 6

(RPM) x 6 SPS = (Step Oslash)

ndash Travel per step

mstep =L

positionsrev

ndash Overdriving capability

Required ldquoTime Offrdquo using a chopper drive (constant current drive)

Time Off = (Time On) (Applied Voltage)2

(Rated Voltage)ndash Time On

Time Off = (Time On) (Applied Current)2

(Rated Current)ndash Time On

It is not unusual for a customer to drive Haydon Kerk stepper motors beyond their rated power to obtain the most force in the smallest package size Precautions must be taken to prevent the motor from exceeding its maximum temperature The ldquoonrdquo time should not exceed 2 - 3 minutes As general rule of thumb driving a motor at 25 to 3x its rated voltage or current may result in wasted energy and erratic behavior due to saturation

Special notes for can-stack motors bull more easily saturated due to less active material (steel) bull more iron losses due to unlaminated steel

Motion System Formulas10

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MOTION SYSTEMFORMULAS

Lead Screw Performance Characteristics

End Mounting Rigidity

Impact Impact

ScrewLength

Increase In Affects Increase In Affects

Critical speed

Compression load

Critical speed

Inertia

Compression load

Stiffness

Spring Rate

Load capacity

Drive torque

Angular velocity

Load cpacity

Positioning accuracy

ScrewDiameter

Lead

Load

Preload

Nut length

SI Unit Systems

LengthMassTimeElectric currentThermodynamic temperatureLuminous intensityAmount of substance

meterkilogramsecondamperekelvincandelamole

mKgsAKcdmol

Base Units

Derived Unitsareavolumefrequencymass density (density)speed velocityangular velocityaccelerationangular accelerationforcepressure (mechanical stress)kinematic viscositydynamic viscositywork energy quantity of heatpowerentropyspecific heat capacitythermal conductivity

square metercubic meterhertzkilogram per cubic metermeter per secondradian per secondmeter per second squaredradian per second squarednewtonpascalsquare meter per secondnewton-second per square meter joulewattjoule per kelvinjoule per kilogram kelvinwatt per meter kelvin

m2

m3

HzKgm3

msradsms2

rads2

NPam2sN-sm2

JWJKJKg-K)W(m-K)

Quantity Name of Unit Symbol

Motion System Reference Tables11

Critical speed

Compression load

System stiffness

Life

Positioning accuracy

System stiffness

Drag torque

Load capacity

Stiffness

sndash1

Kg-ms2

Nm2

N-mJs or N-ms

ExamplesAn INCREASE ( ) in screw length results in a DECREASE ( ) in critical speed

An INCREASE in screw diameter results in an INCREASE in critical speed

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MOTION SYSTEMREFERENCE TABLES

t 1 t 2 t 3 t dwell

vTime

vpk

a -a

MassMove distanceMove timeDwell timeMotion profileScrew speed limitLoad supportOrientationRotary to linear conversionAmbient temperature

9 Kg200 mm10 s05 s13 13 13 Trapezoidal1000 RPMμ = 001VerticalTFE lead screw 8 mm diameter 275 mm long30degC

ControlDrive Power Supply

4 quadrant with encoder32 VDC 35 Arms50 A peak

0333 s 0333 s 0333 s 05 s

vPK =3S2t =

(3)(02m)(2)(10 s) =

06m2 s =

03ms

Linear Velocity

Linear Acceleration

a = vf - vi

t = 03ms - 0 ms 0333 s 0901ms2 =

1st step is to thoroughly understand system constraints and motion profile requirements

Linear Motion ExampleData

Motion System Application Example12

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12

MOTION SYSTEMAPPLICATION EXAMPLE

Lmin =60vPK

n =(60)(03ms)1000 revmin = 0018mrev = 18mmrev

Refer to Kerk screw chart ndash closest lead in an 8mm screw diameter is

2032mmrev = 002032 mrevη = 086 free wheeling nut TFE coated screw

JS = 388 x 10ndash 7 Kg-m2 Other critical lead screw considerations bull Column loading bull Critical speed

Jm aa

ωPKvPK

TF

= ωPK = 2 π vPK

L=

(2π)(03ms)002032 mrev

9276 rads

a = =2πaL

=(2π)(0901ms2)002032 mrev

2786 rads2

Linear to Rotary Conversion ndash Velocity Acceleration Inertia Torque

Minimum screw lead to maintain lt 1000 RPM shaft speed

Linear to Rotary Conversion Velocity

Linear to Rotary Conversion Acceleration

Motion System Application Example13

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13

MOTION SYSTEMFORMULAS

Jin= m 2 πL 2

X η1 + JS= 9Kg

2 π002032 m 2

X 0861 + 388x10ndash 7Kg-m2

= 1095 X 10ndash 5 Kg-m2 + 388x10ndash 7 Kg-m2

= 1134 x 10ndash 5 Kg-m2

Τin= Τa+Τf +Τg+ΤD

Τa = Jin ain

Τg =sinOslashmgL

2 π η

= (1134 x 10ndash 5 Kg-m2)(2786 rads2)

= 00316 Nm

Τf =cosOslashmgμL

2 π η= (cos90)(9 Kg)(98ms2)(001)(002032m)

2 π (086)= 0 Nm

= (sin90)(9 Kg)(98ms2)(002032m)2 π (086)

=1792 Nm

5404 = 03316 Nm

ΤDSince a free-wheeling lead screw nut was used there is no preload force If an anti-backlash nut was used ΤD would be gt0 due to preload force Always reference manufacturerrsquos data for more information

= 0 Nm

Linear to Rotary Conversion Torque

Linear to Rotary Conversion Inertia

Motion System Application Example14

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14

MOTION SYSTEMFORMULAS

t 1 t 2 t 3 t dwell Time0333 s 0333 s 0333 s 05 s

Rotary Motion Profile

ωPK

ω

= 9276 rads a = 2786 rads2

Τ1= Τa+Τf +Τg+ΤD

= (00316 Nm) + (0 Nm) + (03316 Nm) + (0 Nm)

= 03632 Nm

Τ2= Τa+Τf +Τg+ΤD

= (0 Nm) + (0 Nm) + (03316 Nm) + (0 Nm)

= 03316 Nm

Τ3= Τa+Τf +Τg+ΤD

= (00316 Nm) + (0 Nm) + (03316 Nm) + (0 Nm)

= 03632 Nm

Τ1

Τ2

Τ3

Motion System Application Example15

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15

MOTION SYSTEMAPPLICATION EXAMPLE

Lead Screw Input Requirements

= 9276 rads= 2786 rads2

= 03632 Nm= 03316 Nm= ndash 03632 Nm= 0333 s= 0333 s= 0333 s= 05 s= 3369 W= 3076 W

ωPKaΤ1Τ2Τ3

t1t2t3tdwell

PPKPCV

PPK = (Τ1)(ωPK)

= (03632 Nm)(9276 rads)= 3369 W

PCV = (Τ2)(ωPK)

= (03316 Nm)(9276 rads)= 3076 W

RMS Torque Requirement Lead Screw Input

=

0333 s + 0333 s + 0333 s + 05 s

(03632 Nm)2(0333 s)+(03316 Nm)2(0333 s)+(- 03632 Nm)2(0333 s)

ΤRMS = Τ12t 1 + Τ2

2t2 + Τ32t 3

t 1 + t 2 + t 3 + tdwell

15

00439 + 00366 + 00439 = 00829 =

ΤRMS = 02879 Nm

Motion System Application Example16

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16

MOTION SYSTEMAPPLICATION EXAMPLE

Load Parameters at the Lead Screw Shaft

= 9276 rads= 2786 rads2

= 03632 Nm= 03316 Nm= ndash 03632 Nm= 02879 Nm= 3369 W= 3076 W

ωPKaΤ1Τ2Τ3ΤRMSPPKPCV

Motor SelectionA direct-drive motor can be used however it would be large and expensive For example a DC057B-3 brush motor

would easily meet continuous torque and continuous speed as well as a 181 load-to-rotor inertia This motor however has significantly more output power than needed at 128 watts

A smaller motor with a transmission would optimize the system Look for a motor starting at a rated power around 40 watts Pittman DC040B-6

DC040B-6Reference VoltageContinuous TorqueRated CurrentRated PowerTorque ConstantVoltage ConstantTerminal ResistanceMax Winding TemperatureRotor Inertia

240 V00812 Nm236 A37 W0042 NmA0042 Vrads185 Ω155degC847 x 10ndash6 Kg-m2

Motor TransmissionLead Screw Support Rails

ΤRMS

PPK

ωPK

= 02879 Nm= 3369 W= 9276 rads

M

Other Motor Considerationsndash Type stepper brush BLDCndash Input speed (high input speed will result in high noise levels)ndash Total motor footprintndash Ambient temperaturendash Environmental conditionsndash Load-to-rotor inertiandash Encoder ready needed

Motion System Application Example17

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17

MOTION SYSTEMAPPLICATION EXAMPLE

Gearbox Transmission Selection

When selecting a reduction ratio keep in mind that the RMS required torque at the motor shaft needs to fall below the continuous rated output torque of the motor

G30 A - Planetary GearboxMaximum output load 247 Nm

TRMS( Lead Screw)

02879 Nm

02879 Nm02879 Nm

Reduction

41

5161

Efficiency

090

090090

TRMS( Motor)

00798 Nm

00640 Nm00533 Nm

A 51 gearbox would allow about a 27 safety margin between what the system requires and the continuous torque output of the motor

System Summary 51η = 090

M

= 00807 Nm= 00737 Nm= ndash 00807 Nm= 00640 Nm= 4638 rads= 504 x 10 ndash 6 Kg-m2

= 3743 W

Τ1Τ2Τ3ΤRMS

PPK

ωPK

J

= 03632 Nm= 03316 Nm= ndash 03632 Nm= 02879 Nm= 9276 rads= 1134 x 10 ndash 5 Kg-m2

= 3369 W

Τ1Τ2Τ3ΤRMS

PPK

ωPK

J

Motion System Application Example18

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18

MOTION SYSTEMAPPLICATION EXAMPLE

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19

Motion Profile at the Motor

t 1 t 2 t 3 t dwell Time0333 s 0333 s 0333 s 05 s

ωΤ1

Τ2

Τ3= 4638 rads= 4429 RPM= ΤPK = 00807 Nm

ωPK

Τ1

nPK

Drive and Power Supply Requirements

Peak Current Required

IPK =TPK

KT+ IO =

00807 Nm0042 NmA

+ 0180 A = 210 A

RMS Current Required

IRMS =TRMS

KT+ IO =

00640 Nm0042 NmA

+ 0180 A = 170 A

Minimum bus Voltage Required

Vbus= IPKRmt+ ωPKKE = (210A)(185Ω) + (4638 rads)(0042 Vrads)= 389 V + 1948 V= 2337 V

0100 Nm 0200 Nm 0300 Nm 0400 Nm 0500 Nm 0600 Nm 0700 Nm

30 VDC

0100 Nm 0200 Nm 0300 Nm 0500 Nm 0600 Nm 0700 Nm

24 VDC

700060005000

RPM 4000300020001000

0400 Nm

TRMS

TPK

700060005000

RPM 4000300020001000

TRMS

TPK

Performance using 24V reference voltage

Performance using a 30V bus voltage

A 30 VDC bus will meet the application requirements as wells supply a margin of safety

MOTION SYSTEMAPPLICATION EXAMPLE

0100 Nm 0200 Nm 0300 Nm 0400 Nm 0500 Nm 0600 Nm 0700 Nm

30 VDC

0100 Nm 0200 Nm 0300 Nm 0500 Nm 0600 Nm 0700 Nm

24 VDC

700060005000

RPM 4000300020001000

0400 Nm

TRMS

TPK

700060005000

RPM 4000300020001000

TRMS

TPK

Ambient temperature Rated motor temperature Temperature rise Motor temperature

= 30degC= 155degC= 7638degC= 10638degC

Θa

Θm

ΘrΘrated

=Rth x IRMS

2 x Rmt

1 ndash (Rth x IRMS2 x Rmt x 000392degC)

Θr

=11degCω x 170A2

x 185Ω 1 ndash (11degCω x 170A2

x 185Ω x 000392degC)

= 5881 1 ndash (023)

= 5881 077

=

Θm = Θr + Θa = 7638degC + 30degC = 10638degC

7638degC

Performance using 24V reference voltage

Performance using a 30V bus voltage

Will the motor meet the temperature rise caused by the application

Motion System Application Example20

A 30 VDC bus will meet the application requirements as wells supply a margin of safety

203 756 7441 1500 Meriden Road Waterbury CT USA 06705

267 933 2105 534 Godshall DriveHarleysville PA USA 19438

wwwHaydonKerkPittmancom

copy All rights reserved AMETEK Advanced Motion Solutions No part of this doc-ument or technical information can be used reproduced or altered in any form without approval or proper authorization from AMETEK Inc and global affiliates

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MOTION SYSTEMAPPLICATION EXAMPLE

Page 2: FORMULAS FOR MOTORIZED LINEAR MOTION SYSTEMS

Symbol Description Units Symbol Description Units

a ms2linear acceleration

a rads2angular acceleration

α degCtemperature coefficient

Td Nmreverse torque required for deceleration

Tf Nmtorque required to overcome friction

Tg Nmtorque required to overcome gravity

TLR Nmlocked rotor torque

TRMS NmRMS torque required over the total duty cycle

TRMS (motor) NmRMS torque required at the motor shaft

μ nacoefficient of friction

VT Vmotor terminal voltage

Vbus VDC drive bus voltage

v mslinear velocity

vf msfinal linear velocity

msinitial linear velocity

mspeak linear velocity

W Jenergy

ω radsangular velocity

radsno load angular velocity

radspeak angular velocity

vi

vPK

ωo

ωPK

TD Nmdrag preload torque

ain rads2angular acceleration

aout rads2angular acceleration

DV Nm sradviscous damping factor

F Nforce

Fa Nlinear force required during acceleration (FJ +Ff +Fg)

Ff Nlinear force required to overcome friction

Fg Nlinear force required to overcome gravity

FJ Nlinear force required to overcome load inertia

g ms2gravitational constant (98 ms2)

I Acurrent

ILR Alocked rotor current

IO Ano-load current

IPK Apeak current

IRMS ARMS current

J Kg-m2inertia

JBKg-m2inertia of belt

JGBKg-m2inertia of the gear box

JinKg-m2reflected inertia at system input

Jout Kg-m2reflected inertia at system output

JP1Kg-m2inertia of pully 1

JP2Kg-m2inertia of pully 2

JSKg-m2lead screw inertia

K(i)

Vradsvoltage constant

K(f) NmA or V(rads)KE or KT (finalldquohotrdquo)

KE

KE or KT (initial ldquocoldrdquo)

Km Nm wmotor constant

NmA or V(rads)

KT NmAtorque constant

L mrevscrew lead

m Kgmass

N nagear ratio

RPM

ηRPMspeed

nO no load speed

nnaefficiency

RPMnPK peak speed

load orientation (horizontal = 0deg vertical = 90deg)

degCθ degrees

Θa ambient temperature

Θf degCmotor temperature (finalldquohotrdquo)

Θi degCmotor temperature (initial ldquocoldrdquo)

Θm degCmotor temperature

Θr degCmotor temperature rise

Θrated degCrated motor temperature

P Wpower

Pavg Waverage power

PCV Wpower required at constant velocity

Pin Wpower required at system input

Ploss Wdissipated power

Pmax Wmaximum power

Pmax(f) Wmaximum power (finalldquohotrdquo)

Pmax(i) Wmaximum power (initial ldquocoldrdquo)

Pout Woutput power

PPK Wpeak power

Rm RPMNmmotor regulation

Rmt Ωmotor terminal resistance

Rmt(f) Ωmotor terminal resistance (finalldquohotrdquo)

Rmt(i) Ωmotor terminal resistance (initial ldquocoldrdquo)

Rth degCWthermal resistance

r mradius

S mlinear distance

T Nmtorque

Ta Nmtorque required to overcome load inertia

Ta (motor) Nmtorque required at motor shaft during acceleration

TC Nmcontinuous rated motor torque

TCF Nmcoulomb friction torque

Tin Nmtorque required at system input

Tout Nmtorque required at system output

radsangular velocity at system inputωin

radsangular velocity at system outputωout

t stime

Symbols and Units

2 SYMBOLS AND UNITS

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1

(V)(I)Watts

F x St =

+ +MotorController

DrivePowerSupply

Mechanical Transmission

Lead Screw Support Rails

WattsLost

(V)(I)Watts

WattsLost

WattsLost

WattsLost

(T)(ω)Watts

(T)(ω)Watts

WattsLost

All analysis begins with the linear motion profile of the output and the force required

to move the load

t 1 t 2 t 3t move

Total Cycle Time

t dwell

vTime

vpk

a -a

13 13 13 Trapazoidal Motion Profile

Optimized for minimum power

t 1 = t 2= t 3

vPK =3S2t

S is the total move distancet is the total move timeArea under profile curve represents distance moved

Watts Out

Motion System Formulas

Any motion system should be broken down into its individual components

Motion Profiles

3

(V)(I)Watts

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2

MOTION SYSTEMFORMULAS

Triangular Motion Profiles

Optimized for minimum acceleration slope

t 1 = t 2

vPK =2St

S is the total move distancet is the total move time 13 13 13

Triangular

Move Profile Peak Velocity (VPK) Acceleration (a )

t 1 t 2t move

Total Cycle Time

t dwell

vTime

a -avPK

Complex Motion Profile

Linear to Rotary Formulas through a Lead Screw

(1) s = ndashndashndashndashvf - vi

2t

(2) vf = vi + a t

(3) s = vi t + ndash at 212

(4) 2 as = vf 2

- vi 2

(5) a = ( vf - vi) ( tf - ti )

= ΔvΔt

a = = rads 2 2 π a

L

ω = = rads 2 π v

L

n = = RPM 60 v

L

3 known quantities are needed to solve for the other 2 Each segment calculated individually

Linear Motion Formulas

t 1 t 2 t 3t move

Total Cycle Time

t dwell

vTime

v1

a 1

-a3

a 2

t 4 t 5

v2

Motion System Formulas4

Area under profile curve represents distance moved

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3

MOTION SYSTEMFORMULAS

Inertia Acceleration Velocity and Required Torque

Jin mout

ain aout ωin vout

Tin Fout

radsec 2

radsec

ain = 2 πa

L

ωin = 2 π v

L

Jin = m 2 πL 2

X η1 + JS Kg-m2

Τin= Τa+Τf +Τg+ΤD N-m

Τa= Jin ain

Τf =cosOslashmgμL

2 π ηN-m

N-m

Τg =sinOslashmgL

2 π ηN-m

ΤD = dragpreloadrefer to manufacturerrsquos data

N-m

Lead Screw System

Belt and Pulley System

ain = radsec 2

ωin = radsec Jin Jout

ain aout ωin ωout

Tin Tout

Jin = NJout

2X η

1 + JP1 Kg-m2

N-mFor a 1st approximation analysis JP1 JP2 and JB can be disregarded For higher precision pulley and belt inertias should be included as well as the reflected inertia of these components

+ JP2+ JB

(aout)(N) (ωout)(N)

Τin = NΤout X η

1

Motion System Formulas5

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4

MOTION SYSTEMFORMULAS

Motion System Formulas6

Gearbox System

radsec 2

ωin = radsec

Jin = NJout

2X η

1

+

Kg-m2

N-m

Drag torque can be significant (TD) depending on the viscosity of the lubricant For a first approximation analysis this can be left outGearbox inertia may be found in manufacturerrsquos data sheets or calculated using gear dimensions and materials mass

JGB

(ωout)(N) Τin = N

Τout X η1

+

ΤD

ΤRMS = Τ12t 1 + Τ2

2t2 + Τ32t 3 + Τn

2t n

t 1 + t 2 + t 3 + t n + tdwell

TωLinear Power =

FSt

= watts

Rotary Power = = watts

Mechanical Power and RMS Torque

Jin Jout

ain aout ωin ωout

Tin Tout

ain = (aout)(N)

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

5

MOTION SYSTEMFORMULAS

Assuming IO is very small it can be ignored for a quick approximation of motor no-load speed If more accurate results are needed IO will also need to be adjuisted based on the motor Viscous Damping Factor (Dv) As the no-load speed increases with increased terminal voltage on a given motor IO will also increase based on the motorrsquos Dv value found on the manufacturerrsquos motor data sheet

Motor No Load SpeedVT ndash (IO x Rmt)no = 95493 x

KE

ndash ndash ndash ndash ndash ndash ndash ndash ndash ndash ndash ndash ndash

ndash ndash ndash

ndash ndash ndash

BreakawayTorque

Viscous Damping Torque (Dv)

Coulomb Friction

Τ

ωο

Motion System Formulas7

DC Motor Formulas

Motor Locked Rotor Current Locked Rotor Torque

ILR =VT

Rmt

TLR = ILR x KT

Motor Regulation

RmtRm = 95493 x

KE x KT

= x KT

VT

Rmt

ndash Theoretical using motor constants

nORm = TLR

ndash Regulation calculated using test data

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6

MOTION SYSTEMFORMULAS

Motor Power Relationships

ndash Watts lost due to winding resistance

Ploss = I2 x Rmt

ndash Output power at any point on the motor curve

Pout = ω x Tndash Motor maximum output power

Pmax = 025 x ωo x TLR

ndash Motor maximum output power (Theoretical)

Pmax = 025 x VT2

Rmt

Temperature Effects on Motor Constants

ndash Change in terminal resistance

Rmt(f) = Rmt(i) x [ 1 + αconductor (Θf ndash Θi) ]αconductor = 00040degC (copper)

K(f) = K(i) x [ 1 + αmagnet (Θf ndash Θi) ]ndash Change in torque constant and voltage constant ( KT = KE )

Magnetic Material

CeramicSmCoAlNiCoNdFeB

Θmax (degC)300degC300degC540degC150degC

The magnetic material values above represent average figures for particular material classes and estimating performance If exact values are needed consult data sheets for a specific magnet grade

Motion System Formulas8

αmagnet (degC)-00020degC-00004degC-00002degC-00012degC

In the case of a mechanically commutated motor with graphite brushes this analysis will result in a slight error because of the negative temperature coefficient of carbon For estimation purposes this can be ignored but for a more rigorous analysis the behavior of the brush material must be taken into account This is not an issue when evaluating a brushless dc motor

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7

MOTION SYSTEMFORMULAS

Motion System Formulas9

For PMDC motors (brush type) brush drop can be approximated as a constant resistance connected in series with the armature winding This series resistance is included in the Rmt factor found on manufacturerrsquos data sheet

Copper graphiteSilver graphiteElectrographite

02 to 04 Ω 02 to 04 Ω 08 to 10 Ω

Motor Brush Resistance

Brush Material Resistance

DC Motor Thermal Resistance and Temperature Rise

ndash Thermal resistance

ndash Motor temperature rise

Θm = Θr + Θa

ndash OR ndash

Temperature Rise degCRth =

Machine Losses W

Θr = Rth x I2 x Rmt

Θr = Rth x I

2 x Rmt

1 ndash (Rth x I2 x Rmt x α)

ndash OR ndash

Θr = Rth x(TC

Km)

2

ndash Final motor temperature

Use caution when applying these formulas Depending on the conditions used to determine motor specifications only one formula should be used Contact Appications Engineering for questions

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

8

MOTION SYSTEMFORMULAS

Motion System Formulas9

For PMDC motors (brush type) brush drop can be approximated as a constant resistance connected in series with the armature winding This series resistance is included in the Rmt factor found on manufacturerrsquos data sheet

Copper graphiteSilver graphiteElectrographite

02 to 04 Ω 02 to 04 Ω 08 to 10 Ω

Motor Brush Resistance

Brush Material Resistance

DC Motor Thermal Resistance and Temperature Rise

ndash Thermal resistance

ndash Motor temperature rise

Θm = Θr + Θa

ndash OR ndash

Temperature Rise degCRth =

Machine Losses W

Θr = Rth x I2 x Rmt

Θr = Rth x I

2 x Rmt

1 ndash (Rth x I2 x Rmt x α)

ndash OR ndash

Θr = Rth x(TC

Km)

2

ndash Final motor temperature

Use caution when applying these formulas Depending on the conditions used to determine motor specifications only one formula should be used Contact Appications Engineering for questions

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

9

MOTION SYSTEMFORMULAS

Stepper Motor Formulas

Required ldquoTime Offrdquo using an LR drive (constant voltage drive)

180Step Oslash = (phases)(rotor teeth)

ndash Motor step angle

ndash RPM (revolutionsminute) to SPS (stepssecond)

Maximum accuracy of a step motor system has nothing to do with maximum resolution of a step motor system Maximum accuracy is always based on the accuracy of 1 full step but maximum resolution is based on all system components including lead screw encoder stepper motor and step mode on the controller

ndash SPS (stepssecond) to RPM (revolutionsminute)(SPS) x (Step Oslash)

RPM = 6

(RPM) x 6 SPS = (Step Oslash)

ndash Travel per step

mstep =L

positionsrev

ndash Overdriving capability

Required ldquoTime Offrdquo using a chopper drive (constant current drive)

Time Off = (Time On) (Applied Voltage)2

(Rated Voltage)ndash Time On

Time Off = (Time On) (Applied Current)2

(Rated Current)ndash Time On

It is not unusual for a customer to drive Haydon Kerk stepper motors beyond their rated power to obtain the most force in the smallest package size Precautions must be taken to prevent the motor from exceeding its maximum temperature The ldquoonrdquo time should not exceed 2 - 3 minutes As general rule of thumb driving a motor at 25 to 3x its rated voltage or current may result in wasted energy and erratic behavior due to saturation

Special notes for can-stack motors bull more easily saturated due to less active material (steel) bull more iron losses due to unlaminated steel

Motion System Formulas10

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10

MOTION SYSTEMFORMULAS

Lead Screw Performance Characteristics

End Mounting Rigidity

Impact Impact

ScrewLength

Increase In Affects Increase In Affects

Critical speed

Compression load

Critical speed

Inertia

Compression load

Stiffness

Spring Rate

Load capacity

Drive torque

Angular velocity

Load cpacity

Positioning accuracy

ScrewDiameter

Lead

Load

Preload

Nut length

SI Unit Systems

LengthMassTimeElectric currentThermodynamic temperatureLuminous intensityAmount of substance

meterkilogramsecondamperekelvincandelamole

mKgsAKcdmol

Base Units

Derived Unitsareavolumefrequencymass density (density)speed velocityangular velocityaccelerationangular accelerationforcepressure (mechanical stress)kinematic viscositydynamic viscositywork energy quantity of heatpowerentropyspecific heat capacitythermal conductivity

square metercubic meterhertzkilogram per cubic metermeter per secondradian per secondmeter per second squaredradian per second squarednewtonpascalsquare meter per secondnewton-second per square meter joulewattjoule per kelvinjoule per kilogram kelvinwatt per meter kelvin

m2

m3

HzKgm3

msradsms2

rads2

NPam2sN-sm2

JWJKJKg-K)W(m-K)

Quantity Name of Unit Symbol

Motion System Reference Tables11

Critical speed

Compression load

System stiffness

Life

Positioning accuracy

System stiffness

Drag torque

Load capacity

Stiffness

sndash1

Kg-ms2

Nm2

N-mJs or N-ms

ExamplesAn INCREASE ( ) in screw length results in a DECREASE ( ) in critical speed

An INCREASE in screw diameter results in an INCREASE in critical speed

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

11

MOTION SYSTEMREFERENCE TABLES

t 1 t 2 t 3 t dwell

vTime

vpk

a -a

MassMove distanceMove timeDwell timeMotion profileScrew speed limitLoad supportOrientationRotary to linear conversionAmbient temperature

9 Kg200 mm10 s05 s13 13 13 Trapezoidal1000 RPMμ = 001VerticalTFE lead screw 8 mm diameter 275 mm long30degC

ControlDrive Power Supply

4 quadrant with encoder32 VDC 35 Arms50 A peak

0333 s 0333 s 0333 s 05 s

vPK =3S2t =

(3)(02m)(2)(10 s) =

06m2 s =

03ms

Linear Velocity

Linear Acceleration

a = vf - vi

t = 03ms - 0 ms 0333 s 0901ms2 =

1st step is to thoroughly understand system constraints and motion profile requirements

Linear Motion ExampleData

Motion System Application Example12

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

12

MOTION SYSTEMAPPLICATION EXAMPLE

Lmin =60vPK

n =(60)(03ms)1000 revmin = 0018mrev = 18mmrev

Refer to Kerk screw chart ndash closest lead in an 8mm screw diameter is

2032mmrev = 002032 mrevη = 086 free wheeling nut TFE coated screw

JS = 388 x 10ndash 7 Kg-m2 Other critical lead screw considerations bull Column loading bull Critical speed

Jm aa

ωPKvPK

TF

= ωPK = 2 π vPK

L=

(2π)(03ms)002032 mrev

9276 rads

a = =2πaL

=(2π)(0901ms2)002032 mrev

2786 rads2

Linear to Rotary Conversion ndash Velocity Acceleration Inertia Torque

Minimum screw lead to maintain lt 1000 RPM shaft speed

Linear to Rotary Conversion Velocity

Linear to Rotary Conversion Acceleration

Motion System Application Example13

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

13

MOTION SYSTEMFORMULAS

Jin= m 2 πL 2

X η1 + JS= 9Kg

2 π002032 m 2

X 0861 + 388x10ndash 7Kg-m2

= 1095 X 10ndash 5 Kg-m2 + 388x10ndash 7 Kg-m2

= 1134 x 10ndash 5 Kg-m2

Τin= Τa+Τf +Τg+ΤD

Τa = Jin ain

Τg =sinOslashmgL

2 π η

= (1134 x 10ndash 5 Kg-m2)(2786 rads2)

= 00316 Nm

Τf =cosOslashmgμL

2 π η= (cos90)(9 Kg)(98ms2)(001)(002032m)

2 π (086)= 0 Nm

= (sin90)(9 Kg)(98ms2)(002032m)2 π (086)

=1792 Nm

5404 = 03316 Nm

ΤDSince a free-wheeling lead screw nut was used there is no preload force If an anti-backlash nut was used ΤD would be gt0 due to preload force Always reference manufacturerrsquos data for more information

= 0 Nm

Linear to Rotary Conversion Torque

Linear to Rotary Conversion Inertia

Motion System Application Example14

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

14

MOTION SYSTEMFORMULAS

t 1 t 2 t 3 t dwell Time0333 s 0333 s 0333 s 05 s

Rotary Motion Profile

ωPK

ω

= 9276 rads a = 2786 rads2

Τ1= Τa+Τf +Τg+ΤD

= (00316 Nm) + (0 Nm) + (03316 Nm) + (0 Nm)

= 03632 Nm

Τ2= Τa+Τf +Τg+ΤD

= (0 Nm) + (0 Nm) + (03316 Nm) + (0 Nm)

= 03316 Nm

Τ3= Τa+Τf +Τg+ΤD

= (00316 Nm) + (0 Nm) + (03316 Nm) + (0 Nm)

= 03632 Nm

Τ1

Τ2

Τ3

Motion System Application Example15

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

15

MOTION SYSTEMAPPLICATION EXAMPLE

Lead Screw Input Requirements

= 9276 rads= 2786 rads2

= 03632 Nm= 03316 Nm= ndash 03632 Nm= 0333 s= 0333 s= 0333 s= 05 s= 3369 W= 3076 W

ωPKaΤ1Τ2Τ3

t1t2t3tdwell

PPKPCV

PPK = (Τ1)(ωPK)

= (03632 Nm)(9276 rads)= 3369 W

PCV = (Τ2)(ωPK)

= (03316 Nm)(9276 rads)= 3076 W

RMS Torque Requirement Lead Screw Input

=

0333 s + 0333 s + 0333 s + 05 s

(03632 Nm)2(0333 s)+(03316 Nm)2(0333 s)+(- 03632 Nm)2(0333 s)

ΤRMS = Τ12t 1 + Τ2

2t2 + Τ32t 3

t 1 + t 2 + t 3 + tdwell

15

00439 + 00366 + 00439 = 00829 =

ΤRMS = 02879 Nm

Motion System Application Example16

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

16

MOTION SYSTEMAPPLICATION EXAMPLE

Load Parameters at the Lead Screw Shaft

= 9276 rads= 2786 rads2

= 03632 Nm= 03316 Nm= ndash 03632 Nm= 02879 Nm= 3369 W= 3076 W

ωPKaΤ1Τ2Τ3ΤRMSPPKPCV

Motor SelectionA direct-drive motor can be used however it would be large and expensive For example a DC057B-3 brush motor

would easily meet continuous torque and continuous speed as well as a 181 load-to-rotor inertia This motor however has significantly more output power than needed at 128 watts

A smaller motor with a transmission would optimize the system Look for a motor starting at a rated power around 40 watts Pittman DC040B-6

DC040B-6Reference VoltageContinuous TorqueRated CurrentRated PowerTorque ConstantVoltage ConstantTerminal ResistanceMax Winding TemperatureRotor Inertia

240 V00812 Nm236 A37 W0042 NmA0042 Vrads185 Ω155degC847 x 10ndash6 Kg-m2

Motor TransmissionLead Screw Support Rails

ΤRMS

PPK

ωPK

= 02879 Nm= 3369 W= 9276 rads

M

Other Motor Considerationsndash Type stepper brush BLDCndash Input speed (high input speed will result in high noise levels)ndash Total motor footprintndash Ambient temperaturendash Environmental conditionsndash Load-to-rotor inertiandash Encoder ready needed

Motion System Application Example17

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

17

MOTION SYSTEMAPPLICATION EXAMPLE

Gearbox Transmission Selection

When selecting a reduction ratio keep in mind that the RMS required torque at the motor shaft needs to fall below the continuous rated output torque of the motor

G30 A - Planetary GearboxMaximum output load 247 Nm

TRMS( Lead Screw)

02879 Nm

02879 Nm02879 Nm

Reduction

41

5161

Efficiency

090

090090

TRMS( Motor)

00798 Nm

00640 Nm00533 Nm

A 51 gearbox would allow about a 27 safety margin between what the system requires and the continuous torque output of the motor

System Summary 51η = 090

M

= 00807 Nm= 00737 Nm= ndash 00807 Nm= 00640 Nm= 4638 rads= 504 x 10 ndash 6 Kg-m2

= 3743 W

Τ1Τ2Τ3ΤRMS

PPK

ωPK

J

= 03632 Nm= 03316 Nm= ndash 03632 Nm= 02879 Nm= 9276 rads= 1134 x 10 ndash 5 Kg-m2

= 3369 W

Τ1Τ2Τ3ΤRMS

PPK

ωPK

J

Motion System Application Example18

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

18

MOTION SYSTEMAPPLICATION EXAMPLE

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19

Motion Profile at the Motor

t 1 t 2 t 3 t dwell Time0333 s 0333 s 0333 s 05 s

ωΤ1

Τ2

Τ3= 4638 rads= 4429 RPM= ΤPK = 00807 Nm

ωPK

Τ1

nPK

Drive and Power Supply Requirements

Peak Current Required

IPK =TPK

KT+ IO =

00807 Nm0042 NmA

+ 0180 A = 210 A

RMS Current Required

IRMS =TRMS

KT+ IO =

00640 Nm0042 NmA

+ 0180 A = 170 A

Minimum bus Voltage Required

Vbus= IPKRmt+ ωPKKE = (210A)(185Ω) + (4638 rads)(0042 Vrads)= 389 V + 1948 V= 2337 V

0100 Nm 0200 Nm 0300 Nm 0400 Nm 0500 Nm 0600 Nm 0700 Nm

30 VDC

0100 Nm 0200 Nm 0300 Nm 0500 Nm 0600 Nm 0700 Nm

24 VDC

700060005000

RPM 4000300020001000

0400 Nm

TRMS

TPK

700060005000

RPM 4000300020001000

TRMS

TPK

Performance using 24V reference voltage

Performance using a 30V bus voltage

A 30 VDC bus will meet the application requirements as wells supply a margin of safety

MOTION SYSTEMAPPLICATION EXAMPLE

0100 Nm 0200 Nm 0300 Nm 0400 Nm 0500 Nm 0600 Nm 0700 Nm

30 VDC

0100 Nm 0200 Nm 0300 Nm 0500 Nm 0600 Nm 0700 Nm

24 VDC

700060005000

RPM 4000300020001000

0400 Nm

TRMS

TPK

700060005000

RPM 4000300020001000

TRMS

TPK

Ambient temperature Rated motor temperature Temperature rise Motor temperature

= 30degC= 155degC= 7638degC= 10638degC

Θa

Θm

ΘrΘrated

=Rth x IRMS

2 x Rmt

1 ndash (Rth x IRMS2 x Rmt x 000392degC)

Θr

=11degCω x 170A2

x 185Ω 1 ndash (11degCω x 170A2

x 185Ω x 000392degC)

= 5881 1 ndash (023)

= 5881 077

=

Θm = Θr + Θa = 7638degC + 30degC = 10638degC

7638degC

Performance using 24V reference voltage

Performance using a 30V bus voltage

Will the motor meet the temperature rise caused by the application

Motion System Application Example20

A 30 VDC bus will meet the application requirements as wells supply a margin of safety

203 756 7441 1500 Meriden Road Waterbury CT USA 06705

267 933 2105 534 Godshall DriveHarleysville PA USA 19438

wwwHaydonKerkPittmancom

copy All rights reserved AMETEK Advanced Motion Solutions No part of this doc-ument or technical information can be used reproduced or altered in any form without approval or proper authorization from AMETEK Inc and global affiliates

NOW ASK us the tough motion control questions ONLINE

ASK THE EXPERTS

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MOTION SYSTEMAPPLICATION EXAMPLE

Page 3: FORMULAS FOR MOTORIZED LINEAR MOTION SYSTEMS

(V)(I)Watts

F x St =

+ +MotorController

DrivePowerSupply

Mechanical Transmission

Lead Screw Support Rails

WattsLost

(V)(I)Watts

WattsLost

WattsLost

WattsLost

(T)(ω)Watts

(T)(ω)Watts

WattsLost

All analysis begins with the linear motion profile of the output and the force required

to move the load

t 1 t 2 t 3t move

Total Cycle Time

t dwell

vTime

vpk

a -a

13 13 13 Trapazoidal Motion Profile

Optimized for minimum power

t 1 = t 2= t 3

vPK =3S2t

S is the total move distancet is the total move timeArea under profile curve represents distance moved

Watts Out

Motion System Formulas

Any motion system should be broken down into its individual components

Motion Profiles

3

(V)(I)Watts

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

2

MOTION SYSTEMFORMULAS

Triangular Motion Profiles

Optimized for minimum acceleration slope

t 1 = t 2

vPK =2St

S is the total move distancet is the total move time 13 13 13

Triangular

Move Profile Peak Velocity (VPK) Acceleration (a )

t 1 t 2t move

Total Cycle Time

t dwell

vTime

a -avPK

Complex Motion Profile

Linear to Rotary Formulas through a Lead Screw

(1) s = ndashndashndashndashvf - vi

2t

(2) vf = vi + a t

(3) s = vi t + ndash at 212

(4) 2 as = vf 2

- vi 2

(5) a = ( vf - vi) ( tf - ti )

= ΔvΔt

a = = rads 2 2 π a

L

ω = = rads 2 π v

L

n = = RPM 60 v

L

3 known quantities are needed to solve for the other 2 Each segment calculated individually

Linear Motion Formulas

t 1 t 2 t 3t move

Total Cycle Time

t dwell

vTime

v1

a 1

-a3

a 2

t 4 t 5

v2

Motion System Formulas4

Area under profile curve represents distance moved

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3

MOTION SYSTEMFORMULAS

Inertia Acceleration Velocity and Required Torque

Jin mout

ain aout ωin vout

Tin Fout

radsec 2

radsec

ain = 2 πa

L

ωin = 2 π v

L

Jin = m 2 πL 2

X η1 + JS Kg-m2

Τin= Τa+Τf +Τg+ΤD N-m

Τa= Jin ain

Τf =cosOslashmgμL

2 π ηN-m

N-m

Τg =sinOslashmgL

2 π ηN-m

ΤD = dragpreloadrefer to manufacturerrsquos data

N-m

Lead Screw System

Belt and Pulley System

ain = radsec 2

ωin = radsec Jin Jout

ain aout ωin ωout

Tin Tout

Jin = NJout

2X η

1 + JP1 Kg-m2

N-mFor a 1st approximation analysis JP1 JP2 and JB can be disregarded For higher precision pulley and belt inertias should be included as well as the reflected inertia of these components

+ JP2+ JB

(aout)(N) (ωout)(N)

Τin = NΤout X η

1

Motion System Formulas5

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4

MOTION SYSTEMFORMULAS

Motion System Formulas6

Gearbox System

radsec 2

ωin = radsec

Jin = NJout

2X η

1

+

Kg-m2

N-m

Drag torque can be significant (TD) depending on the viscosity of the lubricant For a first approximation analysis this can be left outGearbox inertia may be found in manufacturerrsquos data sheets or calculated using gear dimensions and materials mass

JGB

(ωout)(N) Τin = N

Τout X η1

+

ΤD

ΤRMS = Τ12t 1 + Τ2

2t2 + Τ32t 3 + Τn

2t n

t 1 + t 2 + t 3 + t n + tdwell

TωLinear Power =

FSt

= watts

Rotary Power = = watts

Mechanical Power and RMS Torque

Jin Jout

ain aout ωin ωout

Tin Tout

ain = (aout)(N)

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5

MOTION SYSTEMFORMULAS

Assuming IO is very small it can be ignored for a quick approximation of motor no-load speed If more accurate results are needed IO will also need to be adjuisted based on the motor Viscous Damping Factor (Dv) As the no-load speed increases with increased terminal voltage on a given motor IO will also increase based on the motorrsquos Dv value found on the manufacturerrsquos motor data sheet

Motor No Load SpeedVT ndash (IO x Rmt)no = 95493 x

KE

ndash ndash ndash ndash ndash ndash ndash ndash ndash ndash ndash ndash ndash

ndash ndash ndash

ndash ndash ndash

BreakawayTorque

Viscous Damping Torque (Dv)

Coulomb Friction

Τ

ωο

Motion System Formulas7

DC Motor Formulas

Motor Locked Rotor Current Locked Rotor Torque

ILR =VT

Rmt

TLR = ILR x KT

Motor Regulation

RmtRm = 95493 x

KE x KT

= x KT

VT

Rmt

ndash Theoretical using motor constants

nORm = TLR

ndash Regulation calculated using test data

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6

MOTION SYSTEMFORMULAS

Motor Power Relationships

ndash Watts lost due to winding resistance

Ploss = I2 x Rmt

ndash Output power at any point on the motor curve

Pout = ω x Tndash Motor maximum output power

Pmax = 025 x ωo x TLR

ndash Motor maximum output power (Theoretical)

Pmax = 025 x VT2

Rmt

Temperature Effects on Motor Constants

ndash Change in terminal resistance

Rmt(f) = Rmt(i) x [ 1 + αconductor (Θf ndash Θi) ]αconductor = 00040degC (copper)

K(f) = K(i) x [ 1 + αmagnet (Θf ndash Θi) ]ndash Change in torque constant and voltage constant ( KT = KE )

Magnetic Material

CeramicSmCoAlNiCoNdFeB

Θmax (degC)300degC300degC540degC150degC

The magnetic material values above represent average figures for particular material classes and estimating performance If exact values are needed consult data sheets for a specific magnet grade

Motion System Formulas8

αmagnet (degC)-00020degC-00004degC-00002degC-00012degC

In the case of a mechanically commutated motor with graphite brushes this analysis will result in a slight error because of the negative temperature coefficient of carbon For estimation purposes this can be ignored but for a more rigorous analysis the behavior of the brush material must be taken into account This is not an issue when evaluating a brushless dc motor

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7

MOTION SYSTEMFORMULAS

Motion System Formulas9

For PMDC motors (brush type) brush drop can be approximated as a constant resistance connected in series with the armature winding This series resistance is included in the Rmt factor found on manufacturerrsquos data sheet

Copper graphiteSilver graphiteElectrographite

02 to 04 Ω 02 to 04 Ω 08 to 10 Ω

Motor Brush Resistance

Brush Material Resistance

DC Motor Thermal Resistance and Temperature Rise

ndash Thermal resistance

ndash Motor temperature rise

Θm = Θr + Θa

ndash OR ndash

Temperature Rise degCRth =

Machine Losses W

Θr = Rth x I2 x Rmt

Θr = Rth x I

2 x Rmt

1 ndash (Rth x I2 x Rmt x α)

ndash OR ndash

Θr = Rth x(TC

Km)

2

ndash Final motor temperature

Use caution when applying these formulas Depending on the conditions used to determine motor specifications only one formula should be used Contact Appications Engineering for questions

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

8

MOTION SYSTEMFORMULAS

Motion System Formulas9

For PMDC motors (brush type) brush drop can be approximated as a constant resistance connected in series with the armature winding This series resistance is included in the Rmt factor found on manufacturerrsquos data sheet

Copper graphiteSilver graphiteElectrographite

02 to 04 Ω 02 to 04 Ω 08 to 10 Ω

Motor Brush Resistance

Brush Material Resistance

DC Motor Thermal Resistance and Temperature Rise

ndash Thermal resistance

ndash Motor temperature rise

Θm = Θr + Θa

ndash OR ndash

Temperature Rise degCRth =

Machine Losses W

Θr = Rth x I2 x Rmt

Θr = Rth x I

2 x Rmt

1 ndash (Rth x I2 x Rmt x α)

ndash OR ndash

Θr = Rth x(TC

Km)

2

ndash Final motor temperature

Use caution when applying these formulas Depending on the conditions used to determine motor specifications only one formula should be used Contact Appications Engineering for questions

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

9

MOTION SYSTEMFORMULAS

Stepper Motor Formulas

Required ldquoTime Offrdquo using an LR drive (constant voltage drive)

180Step Oslash = (phases)(rotor teeth)

ndash Motor step angle

ndash RPM (revolutionsminute) to SPS (stepssecond)

Maximum accuracy of a step motor system has nothing to do with maximum resolution of a step motor system Maximum accuracy is always based on the accuracy of 1 full step but maximum resolution is based on all system components including lead screw encoder stepper motor and step mode on the controller

ndash SPS (stepssecond) to RPM (revolutionsminute)(SPS) x (Step Oslash)

RPM = 6

(RPM) x 6 SPS = (Step Oslash)

ndash Travel per step

mstep =L

positionsrev

ndash Overdriving capability

Required ldquoTime Offrdquo using a chopper drive (constant current drive)

Time Off = (Time On) (Applied Voltage)2

(Rated Voltage)ndash Time On

Time Off = (Time On) (Applied Current)2

(Rated Current)ndash Time On

It is not unusual for a customer to drive Haydon Kerk stepper motors beyond their rated power to obtain the most force in the smallest package size Precautions must be taken to prevent the motor from exceeding its maximum temperature The ldquoonrdquo time should not exceed 2 - 3 minutes As general rule of thumb driving a motor at 25 to 3x its rated voltage or current may result in wasted energy and erratic behavior due to saturation

Special notes for can-stack motors bull more easily saturated due to less active material (steel) bull more iron losses due to unlaminated steel

Motion System Formulas10

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10

MOTION SYSTEMFORMULAS

Lead Screw Performance Characteristics

End Mounting Rigidity

Impact Impact

ScrewLength

Increase In Affects Increase In Affects

Critical speed

Compression load

Critical speed

Inertia

Compression load

Stiffness

Spring Rate

Load capacity

Drive torque

Angular velocity

Load cpacity

Positioning accuracy

ScrewDiameter

Lead

Load

Preload

Nut length

SI Unit Systems

LengthMassTimeElectric currentThermodynamic temperatureLuminous intensityAmount of substance

meterkilogramsecondamperekelvincandelamole

mKgsAKcdmol

Base Units

Derived Unitsareavolumefrequencymass density (density)speed velocityangular velocityaccelerationangular accelerationforcepressure (mechanical stress)kinematic viscositydynamic viscositywork energy quantity of heatpowerentropyspecific heat capacitythermal conductivity

square metercubic meterhertzkilogram per cubic metermeter per secondradian per secondmeter per second squaredradian per second squarednewtonpascalsquare meter per secondnewton-second per square meter joulewattjoule per kelvinjoule per kilogram kelvinwatt per meter kelvin

m2

m3

HzKgm3

msradsms2

rads2

NPam2sN-sm2

JWJKJKg-K)W(m-K)

Quantity Name of Unit Symbol

Motion System Reference Tables11

Critical speed

Compression load

System stiffness

Life

Positioning accuracy

System stiffness

Drag torque

Load capacity

Stiffness

sndash1

Kg-ms2

Nm2

N-mJs or N-ms

ExamplesAn INCREASE ( ) in screw length results in a DECREASE ( ) in critical speed

An INCREASE in screw diameter results in an INCREASE in critical speed

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11

MOTION SYSTEMREFERENCE TABLES

t 1 t 2 t 3 t dwell

vTime

vpk

a -a

MassMove distanceMove timeDwell timeMotion profileScrew speed limitLoad supportOrientationRotary to linear conversionAmbient temperature

9 Kg200 mm10 s05 s13 13 13 Trapezoidal1000 RPMμ = 001VerticalTFE lead screw 8 mm diameter 275 mm long30degC

ControlDrive Power Supply

4 quadrant with encoder32 VDC 35 Arms50 A peak

0333 s 0333 s 0333 s 05 s

vPK =3S2t =

(3)(02m)(2)(10 s) =

06m2 s =

03ms

Linear Velocity

Linear Acceleration

a = vf - vi

t = 03ms - 0 ms 0333 s 0901ms2 =

1st step is to thoroughly understand system constraints and motion profile requirements

Linear Motion ExampleData

Motion System Application Example12

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12

MOTION SYSTEMAPPLICATION EXAMPLE

Lmin =60vPK

n =(60)(03ms)1000 revmin = 0018mrev = 18mmrev

Refer to Kerk screw chart ndash closest lead in an 8mm screw diameter is

2032mmrev = 002032 mrevη = 086 free wheeling nut TFE coated screw

JS = 388 x 10ndash 7 Kg-m2 Other critical lead screw considerations bull Column loading bull Critical speed

Jm aa

ωPKvPK

TF

= ωPK = 2 π vPK

L=

(2π)(03ms)002032 mrev

9276 rads

a = =2πaL

=(2π)(0901ms2)002032 mrev

2786 rads2

Linear to Rotary Conversion ndash Velocity Acceleration Inertia Torque

Minimum screw lead to maintain lt 1000 RPM shaft speed

Linear to Rotary Conversion Velocity

Linear to Rotary Conversion Acceleration

Motion System Application Example13

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13

MOTION SYSTEMFORMULAS

Jin= m 2 πL 2

X η1 + JS= 9Kg

2 π002032 m 2

X 0861 + 388x10ndash 7Kg-m2

= 1095 X 10ndash 5 Kg-m2 + 388x10ndash 7 Kg-m2

= 1134 x 10ndash 5 Kg-m2

Τin= Τa+Τf +Τg+ΤD

Τa = Jin ain

Τg =sinOslashmgL

2 π η

= (1134 x 10ndash 5 Kg-m2)(2786 rads2)

= 00316 Nm

Τf =cosOslashmgμL

2 π η= (cos90)(9 Kg)(98ms2)(001)(002032m)

2 π (086)= 0 Nm

= (sin90)(9 Kg)(98ms2)(002032m)2 π (086)

=1792 Nm

5404 = 03316 Nm

ΤDSince a free-wheeling lead screw nut was used there is no preload force If an anti-backlash nut was used ΤD would be gt0 due to preload force Always reference manufacturerrsquos data for more information

= 0 Nm

Linear to Rotary Conversion Torque

Linear to Rotary Conversion Inertia

Motion System Application Example14

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14

MOTION SYSTEMFORMULAS

t 1 t 2 t 3 t dwell Time0333 s 0333 s 0333 s 05 s

Rotary Motion Profile

ωPK

ω

= 9276 rads a = 2786 rads2

Τ1= Τa+Τf +Τg+ΤD

= (00316 Nm) + (0 Nm) + (03316 Nm) + (0 Nm)

= 03632 Nm

Τ2= Τa+Τf +Τg+ΤD

= (0 Nm) + (0 Nm) + (03316 Nm) + (0 Nm)

= 03316 Nm

Τ3= Τa+Τf +Τg+ΤD

= (00316 Nm) + (0 Nm) + (03316 Nm) + (0 Nm)

= 03632 Nm

Τ1

Τ2

Τ3

Motion System Application Example15

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15

MOTION SYSTEMAPPLICATION EXAMPLE

Lead Screw Input Requirements

= 9276 rads= 2786 rads2

= 03632 Nm= 03316 Nm= ndash 03632 Nm= 0333 s= 0333 s= 0333 s= 05 s= 3369 W= 3076 W

ωPKaΤ1Τ2Τ3

t1t2t3tdwell

PPKPCV

PPK = (Τ1)(ωPK)

= (03632 Nm)(9276 rads)= 3369 W

PCV = (Τ2)(ωPK)

= (03316 Nm)(9276 rads)= 3076 W

RMS Torque Requirement Lead Screw Input

=

0333 s + 0333 s + 0333 s + 05 s

(03632 Nm)2(0333 s)+(03316 Nm)2(0333 s)+(- 03632 Nm)2(0333 s)

ΤRMS = Τ12t 1 + Τ2

2t2 + Τ32t 3

t 1 + t 2 + t 3 + tdwell

15

00439 + 00366 + 00439 = 00829 =

ΤRMS = 02879 Nm

Motion System Application Example16

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16

MOTION SYSTEMAPPLICATION EXAMPLE

Load Parameters at the Lead Screw Shaft

= 9276 rads= 2786 rads2

= 03632 Nm= 03316 Nm= ndash 03632 Nm= 02879 Nm= 3369 W= 3076 W

ωPKaΤ1Τ2Τ3ΤRMSPPKPCV

Motor SelectionA direct-drive motor can be used however it would be large and expensive For example a DC057B-3 brush motor

would easily meet continuous torque and continuous speed as well as a 181 load-to-rotor inertia This motor however has significantly more output power than needed at 128 watts

A smaller motor with a transmission would optimize the system Look for a motor starting at a rated power around 40 watts Pittman DC040B-6

DC040B-6Reference VoltageContinuous TorqueRated CurrentRated PowerTorque ConstantVoltage ConstantTerminal ResistanceMax Winding TemperatureRotor Inertia

240 V00812 Nm236 A37 W0042 NmA0042 Vrads185 Ω155degC847 x 10ndash6 Kg-m2

Motor TransmissionLead Screw Support Rails

ΤRMS

PPK

ωPK

= 02879 Nm= 3369 W= 9276 rads

M

Other Motor Considerationsndash Type stepper brush BLDCndash Input speed (high input speed will result in high noise levels)ndash Total motor footprintndash Ambient temperaturendash Environmental conditionsndash Load-to-rotor inertiandash Encoder ready needed

Motion System Application Example17

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17

MOTION SYSTEMAPPLICATION EXAMPLE

Gearbox Transmission Selection

When selecting a reduction ratio keep in mind that the RMS required torque at the motor shaft needs to fall below the continuous rated output torque of the motor

G30 A - Planetary GearboxMaximum output load 247 Nm

TRMS( Lead Screw)

02879 Nm

02879 Nm02879 Nm

Reduction

41

5161

Efficiency

090

090090

TRMS( Motor)

00798 Nm

00640 Nm00533 Nm

A 51 gearbox would allow about a 27 safety margin between what the system requires and the continuous torque output of the motor

System Summary 51η = 090

M

= 00807 Nm= 00737 Nm= ndash 00807 Nm= 00640 Nm= 4638 rads= 504 x 10 ndash 6 Kg-m2

= 3743 W

Τ1Τ2Τ3ΤRMS

PPK

ωPK

J

= 03632 Nm= 03316 Nm= ndash 03632 Nm= 02879 Nm= 9276 rads= 1134 x 10 ndash 5 Kg-m2

= 3369 W

Τ1Τ2Τ3ΤRMS

PPK

ωPK

J

Motion System Application Example18

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18

MOTION SYSTEMAPPLICATION EXAMPLE

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19

Motion Profile at the Motor

t 1 t 2 t 3 t dwell Time0333 s 0333 s 0333 s 05 s

ωΤ1

Τ2

Τ3= 4638 rads= 4429 RPM= ΤPK = 00807 Nm

ωPK

Τ1

nPK

Drive and Power Supply Requirements

Peak Current Required

IPK =TPK

KT+ IO =

00807 Nm0042 NmA

+ 0180 A = 210 A

RMS Current Required

IRMS =TRMS

KT+ IO =

00640 Nm0042 NmA

+ 0180 A = 170 A

Minimum bus Voltage Required

Vbus= IPKRmt+ ωPKKE = (210A)(185Ω) + (4638 rads)(0042 Vrads)= 389 V + 1948 V= 2337 V

0100 Nm 0200 Nm 0300 Nm 0400 Nm 0500 Nm 0600 Nm 0700 Nm

30 VDC

0100 Nm 0200 Nm 0300 Nm 0500 Nm 0600 Nm 0700 Nm

24 VDC

700060005000

RPM 4000300020001000

0400 Nm

TRMS

TPK

700060005000

RPM 4000300020001000

TRMS

TPK

Performance using 24V reference voltage

Performance using a 30V bus voltage

A 30 VDC bus will meet the application requirements as wells supply a margin of safety

MOTION SYSTEMAPPLICATION EXAMPLE

0100 Nm 0200 Nm 0300 Nm 0400 Nm 0500 Nm 0600 Nm 0700 Nm

30 VDC

0100 Nm 0200 Nm 0300 Nm 0500 Nm 0600 Nm 0700 Nm

24 VDC

700060005000

RPM 4000300020001000

0400 Nm

TRMS

TPK

700060005000

RPM 4000300020001000

TRMS

TPK

Ambient temperature Rated motor temperature Temperature rise Motor temperature

= 30degC= 155degC= 7638degC= 10638degC

Θa

Θm

ΘrΘrated

=Rth x IRMS

2 x Rmt

1 ndash (Rth x IRMS2 x Rmt x 000392degC)

Θr

=11degCω x 170A2

x 185Ω 1 ndash (11degCω x 170A2

x 185Ω x 000392degC)

= 5881 1 ndash (023)

= 5881 077

=

Θm = Θr + Θa = 7638degC + 30degC = 10638degC

7638degC

Performance using 24V reference voltage

Performance using a 30V bus voltage

Will the motor meet the temperature rise caused by the application

Motion System Application Example20

A 30 VDC bus will meet the application requirements as wells supply a margin of safety

203 756 7441 1500 Meriden Road Waterbury CT USA 06705

267 933 2105 534 Godshall DriveHarleysville PA USA 19438

wwwHaydonKerkPittmancom

copy All rights reserved AMETEK Advanced Motion Solutions No part of this doc-ument or technical information can be used reproduced or altered in any form without approval or proper authorization from AMETEK Inc and global affiliates

NOW ASK us the tough motion control questions ONLINE

ASK THE EXPERTS

HaydonKerkPittmancomwww

MOTION SYSTEMAPPLICATION EXAMPLE

Page 4: FORMULAS FOR MOTORIZED LINEAR MOTION SYSTEMS

Triangular Motion Profiles

Optimized for minimum acceleration slope

t 1 = t 2

vPK =2St

S is the total move distancet is the total move time 13 13 13

Triangular

Move Profile Peak Velocity (VPK) Acceleration (a )

t 1 t 2t move

Total Cycle Time

t dwell

vTime

a -avPK

Complex Motion Profile

Linear to Rotary Formulas through a Lead Screw

(1) s = ndashndashndashndashvf - vi

2t

(2) vf = vi + a t

(3) s = vi t + ndash at 212

(4) 2 as = vf 2

- vi 2

(5) a = ( vf - vi) ( tf - ti )

= ΔvΔt

a = = rads 2 2 π a

L

ω = = rads 2 π v

L

n = = RPM 60 v

L

3 known quantities are needed to solve for the other 2 Each segment calculated individually

Linear Motion Formulas

t 1 t 2 t 3t move

Total Cycle Time

t dwell

vTime

v1

a 1

-a3

a 2

t 4 t 5

v2

Motion System Formulas4

Area under profile curve represents distance moved

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

3

MOTION SYSTEMFORMULAS

Inertia Acceleration Velocity and Required Torque

Jin mout

ain aout ωin vout

Tin Fout

radsec 2

radsec

ain = 2 πa

L

ωin = 2 π v

L

Jin = m 2 πL 2

X η1 + JS Kg-m2

Τin= Τa+Τf +Τg+ΤD N-m

Τa= Jin ain

Τf =cosOslashmgμL

2 π ηN-m

N-m

Τg =sinOslashmgL

2 π ηN-m

ΤD = dragpreloadrefer to manufacturerrsquos data

N-m

Lead Screw System

Belt and Pulley System

ain = radsec 2

ωin = radsec Jin Jout

ain aout ωin ωout

Tin Tout

Jin = NJout

2X η

1 + JP1 Kg-m2

N-mFor a 1st approximation analysis JP1 JP2 and JB can be disregarded For higher precision pulley and belt inertias should be included as well as the reflected inertia of these components

+ JP2+ JB

(aout)(N) (ωout)(N)

Τin = NΤout X η

1

Motion System Formulas5

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

4

MOTION SYSTEMFORMULAS

Motion System Formulas6

Gearbox System

radsec 2

ωin = radsec

Jin = NJout

2X η

1

+

Kg-m2

N-m

Drag torque can be significant (TD) depending on the viscosity of the lubricant For a first approximation analysis this can be left outGearbox inertia may be found in manufacturerrsquos data sheets or calculated using gear dimensions and materials mass

JGB

(ωout)(N) Τin = N

Τout X η1

+

ΤD

ΤRMS = Τ12t 1 + Τ2

2t2 + Τ32t 3 + Τn

2t n

t 1 + t 2 + t 3 + t n + tdwell

TωLinear Power =

FSt

= watts

Rotary Power = = watts

Mechanical Power and RMS Torque

Jin Jout

ain aout ωin ωout

Tin Tout

ain = (aout)(N)

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

5

MOTION SYSTEMFORMULAS

Assuming IO is very small it can be ignored for a quick approximation of motor no-load speed If more accurate results are needed IO will also need to be adjuisted based on the motor Viscous Damping Factor (Dv) As the no-load speed increases with increased terminal voltage on a given motor IO will also increase based on the motorrsquos Dv value found on the manufacturerrsquos motor data sheet

Motor No Load SpeedVT ndash (IO x Rmt)no = 95493 x

KE

ndash ndash ndash ndash ndash ndash ndash ndash ndash ndash ndash ndash ndash

ndash ndash ndash

ndash ndash ndash

BreakawayTorque

Viscous Damping Torque (Dv)

Coulomb Friction

Τ

ωο

Motion System Formulas7

DC Motor Formulas

Motor Locked Rotor Current Locked Rotor Torque

ILR =VT

Rmt

TLR = ILR x KT

Motor Regulation

RmtRm = 95493 x

KE x KT

= x KT

VT

Rmt

ndash Theoretical using motor constants

nORm = TLR

ndash Regulation calculated using test data

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

6

MOTION SYSTEMFORMULAS

Motor Power Relationships

ndash Watts lost due to winding resistance

Ploss = I2 x Rmt

ndash Output power at any point on the motor curve

Pout = ω x Tndash Motor maximum output power

Pmax = 025 x ωo x TLR

ndash Motor maximum output power (Theoretical)

Pmax = 025 x VT2

Rmt

Temperature Effects on Motor Constants

ndash Change in terminal resistance

Rmt(f) = Rmt(i) x [ 1 + αconductor (Θf ndash Θi) ]αconductor = 00040degC (copper)

K(f) = K(i) x [ 1 + αmagnet (Θf ndash Θi) ]ndash Change in torque constant and voltage constant ( KT = KE )

Magnetic Material

CeramicSmCoAlNiCoNdFeB

Θmax (degC)300degC300degC540degC150degC

The magnetic material values above represent average figures for particular material classes and estimating performance If exact values are needed consult data sheets for a specific magnet grade

Motion System Formulas8

αmagnet (degC)-00020degC-00004degC-00002degC-00012degC

In the case of a mechanically commutated motor with graphite brushes this analysis will result in a slight error because of the negative temperature coefficient of carbon For estimation purposes this can be ignored but for a more rigorous analysis the behavior of the brush material must be taken into account This is not an issue when evaluating a brushless dc motor

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

7

MOTION SYSTEMFORMULAS

Motion System Formulas9

For PMDC motors (brush type) brush drop can be approximated as a constant resistance connected in series with the armature winding This series resistance is included in the Rmt factor found on manufacturerrsquos data sheet

Copper graphiteSilver graphiteElectrographite

02 to 04 Ω 02 to 04 Ω 08 to 10 Ω

Motor Brush Resistance

Brush Material Resistance

DC Motor Thermal Resistance and Temperature Rise

ndash Thermal resistance

ndash Motor temperature rise

Θm = Θr + Θa

ndash OR ndash

Temperature Rise degCRth =

Machine Losses W

Θr = Rth x I2 x Rmt

Θr = Rth x I

2 x Rmt

1 ndash (Rth x I2 x Rmt x α)

ndash OR ndash

Θr = Rth x(TC

Km)

2

ndash Final motor temperature

Use caution when applying these formulas Depending on the conditions used to determine motor specifications only one formula should be used Contact Appications Engineering for questions

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

8

MOTION SYSTEMFORMULAS

Motion System Formulas9

For PMDC motors (brush type) brush drop can be approximated as a constant resistance connected in series with the armature winding This series resistance is included in the Rmt factor found on manufacturerrsquos data sheet

Copper graphiteSilver graphiteElectrographite

02 to 04 Ω 02 to 04 Ω 08 to 10 Ω

Motor Brush Resistance

Brush Material Resistance

DC Motor Thermal Resistance and Temperature Rise

ndash Thermal resistance

ndash Motor temperature rise

Θm = Θr + Θa

ndash OR ndash

Temperature Rise degCRth =

Machine Losses W

Θr = Rth x I2 x Rmt

Θr = Rth x I

2 x Rmt

1 ndash (Rth x I2 x Rmt x α)

ndash OR ndash

Θr = Rth x(TC

Km)

2

ndash Final motor temperature

Use caution when applying these formulas Depending on the conditions used to determine motor specifications only one formula should be used Contact Appications Engineering for questions

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

9

MOTION SYSTEMFORMULAS

Stepper Motor Formulas

Required ldquoTime Offrdquo using an LR drive (constant voltage drive)

180Step Oslash = (phases)(rotor teeth)

ndash Motor step angle

ndash RPM (revolutionsminute) to SPS (stepssecond)

Maximum accuracy of a step motor system has nothing to do with maximum resolution of a step motor system Maximum accuracy is always based on the accuracy of 1 full step but maximum resolution is based on all system components including lead screw encoder stepper motor and step mode on the controller

ndash SPS (stepssecond) to RPM (revolutionsminute)(SPS) x (Step Oslash)

RPM = 6

(RPM) x 6 SPS = (Step Oslash)

ndash Travel per step

mstep =L

positionsrev

ndash Overdriving capability

Required ldquoTime Offrdquo using a chopper drive (constant current drive)

Time Off = (Time On) (Applied Voltage)2

(Rated Voltage)ndash Time On

Time Off = (Time On) (Applied Current)2

(Rated Current)ndash Time On

It is not unusual for a customer to drive Haydon Kerk stepper motors beyond their rated power to obtain the most force in the smallest package size Precautions must be taken to prevent the motor from exceeding its maximum temperature The ldquoonrdquo time should not exceed 2 - 3 minutes As general rule of thumb driving a motor at 25 to 3x its rated voltage or current may result in wasted energy and erratic behavior due to saturation

Special notes for can-stack motors bull more easily saturated due to less active material (steel) bull more iron losses due to unlaminated steel

Motion System Formulas10

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

10

MOTION SYSTEMFORMULAS

Lead Screw Performance Characteristics

End Mounting Rigidity

Impact Impact

ScrewLength

Increase In Affects Increase In Affects

Critical speed

Compression load

Critical speed

Inertia

Compression load

Stiffness

Spring Rate

Load capacity

Drive torque

Angular velocity

Load cpacity

Positioning accuracy

ScrewDiameter

Lead

Load

Preload

Nut length

SI Unit Systems

LengthMassTimeElectric currentThermodynamic temperatureLuminous intensityAmount of substance

meterkilogramsecondamperekelvincandelamole

mKgsAKcdmol

Base Units

Derived Unitsareavolumefrequencymass density (density)speed velocityangular velocityaccelerationangular accelerationforcepressure (mechanical stress)kinematic viscositydynamic viscositywork energy quantity of heatpowerentropyspecific heat capacitythermal conductivity

square metercubic meterhertzkilogram per cubic metermeter per secondradian per secondmeter per second squaredradian per second squarednewtonpascalsquare meter per secondnewton-second per square meter joulewattjoule per kelvinjoule per kilogram kelvinwatt per meter kelvin

m2

m3

HzKgm3

msradsms2

rads2

NPam2sN-sm2

JWJKJKg-K)W(m-K)

Quantity Name of Unit Symbol

Motion System Reference Tables11

Critical speed

Compression load

System stiffness

Life

Positioning accuracy

System stiffness

Drag torque

Load capacity

Stiffness

sndash1

Kg-ms2

Nm2

N-mJs or N-ms

ExamplesAn INCREASE ( ) in screw length results in a DECREASE ( ) in critical speed

An INCREASE in screw diameter results in an INCREASE in critical speed

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

11

MOTION SYSTEMREFERENCE TABLES

t 1 t 2 t 3 t dwell

vTime

vpk

a -a

MassMove distanceMove timeDwell timeMotion profileScrew speed limitLoad supportOrientationRotary to linear conversionAmbient temperature

9 Kg200 mm10 s05 s13 13 13 Trapezoidal1000 RPMμ = 001VerticalTFE lead screw 8 mm diameter 275 mm long30degC

ControlDrive Power Supply

4 quadrant with encoder32 VDC 35 Arms50 A peak

0333 s 0333 s 0333 s 05 s

vPK =3S2t =

(3)(02m)(2)(10 s) =

06m2 s =

03ms

Linear Velocity

Linear Acceleration

a = vf - vi

t = 03ms - 0 ms 0333 s 0901ms2 =

1st step is to thoroughly understand system constraints and motion profile requirements

Linear Motion ExampleData

Motion System Application Example12

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

12

MOTION SYSTEMAPPLICATION EXAMPLE

Lmin =60vPK

n =(60)(03ms)1000 revmin = 0018mrev = 18mmrev

Refer to Kerk screw chart ndash closest lead in an 8mm screw diameter is

2032mmrev = 002032 mrevη = 086 free wheeling nut TFE coated screw

JS = 388 x 10ndash 7 Kg-m2 Other critical lead screw considerations bull Column loading bull Critical speed

Jm aa

ωPKvPK

TF

= ωPK = 2 π vPK

L=

(2π)(03ms)002032 mrev

9276 rads

a = =2πaL

=(2π)(0901ms2)002032 mrev

2786 rads2

Linear to Rotary Conversion ndash Velocity Acceleration Inertia Torque

Minimum screw lead to maintain lt 1000 RPM shaft speed

Linear to Rotary Conversion Velocity

Linear to Rotary Conversion Acceleration

Motion System Application Example13

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

13

MOTION SYSTEMFORMULAS

Jin= m 2 πL 2

X η1 + JS= 9Kg

2 π002032 m 2

X 0861 + 388x10ndash 7Kg-m2

= 1095 X 10ndash 5 Kg-m2 + 388x10ndash 7 Kg-m2

= 1134 x 10ndash 5 Kg-m2

Τin= Τa+Τf +Τg+ΤD

Τa = Jin ain

Τg =sinOslashmgL

2 π η

= (1134 x 10ndash 5 Kg-m2)(2786 rads2)

= 00316 Nm

Τf =cosOslashmgμL

2 π η= (cos90)(9 Kg)(98ms2)(001)(002032m)

2 π (086)= 0 Nm

= (sin90)(9 Kg)(98ms2)(002032m)2 π (086)

=1792 Nm

5404 = 03316 Nm

ΤDSince a free-wheeling lead screw nut was used there is no preload force If an anti-backlash nut was used ΤD would be gt0 due to preload force Always reference manufacturerrsquos data for more information

= 0 Nm

Linear to Rotary Conversion Torque

Linear to Rotary Conversion Inertia

Motion System Application Example14

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

14

MOTION SYSTEMFORMULAS

t 1 t 2 t 3 t dwell Time0333 s 0333 s 0333 s 05 s

Rotary Motion Profile

ωPK

ω

= 9276 rads a = 2786 rads2

Τ1= Τa+Τf +Τg+ΤD

= (00316 Nm) + (0 Nm) + (03316 Nm) + (0 Nm)

= 03632 Nm

Τ2= Τa+Τf +Τg+ΤD

= (0 Nm) + (0 Nm) + (03316 Nm) + (0 Nm)

= 03316 Nm

Τ3= Τa+Τf +Τg+ΤD

= (00316 Nm) + (0 Nm) + (03316 Nm) + (0 Nm)

= 03632 Nm

Τ1

Τ2

Τ3

Motion System Application Example15

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

15

MOTION SYSTEMAPPLICATION EXAMPLE

Lead Screw Input Requirements

= 9276 rads= 2786 rads2

= 03632 Nm= 03316 Nm= ndash 03632 Nm= 0333 s= 0333 s= 0333 s= 05 s= 3369 W= 3076 W

ωPKaΤ1Τ2Τ3

t1t2t3tdwell

PPKPCV

PPK = (Τ1)(ωPK)

= (03632 Nm)(9276 rads)= 3369 W

PCV = (Τ2)(ωPK)

= (03316 Nm)(9276 rads)= 3076 W

RMS Torque Requirement Lead Screw Input

=

0333 s + 0333 s + 0333 s + 05 s

(03632 Nm)2(0333 s)+(03316 Nm)2(0333 s)+(- 03632 Nm)2(0333 s)

ΤRMS = Τ12t 1 + Τ2

2t2 + Τ32t 3

t 1 + t 2 + t 3 + tdwell

15

00439 + 00366 + 00439 = 00829 =

ΤRMS = 02879 Nm

Motion System Application Example16

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

16

MOTION SYSTEMAPPLICATION EXAMPLE

Load Parameters at the Lead Screw Shaft

= 9276 rads= 2786 rads2

= 03632 Nm= 03316 Nm= ndash 03632 Nm= 02879 Nm= 3369 W= 3076 W

ωPKaΤ1Τ2Τ3ΤRMSPPKPCV

Motor SelectionA direct-drive motor can be used however it would be large and expensive For example a DC057B-3 brush motor

would easily meet continuous torque and continuous speed as well as a 181 load-to-rotor inertia This motor however has significantly more output power than needed at 128 watts

A smaller motor with a transmission would optimize the system Look for a motor starting at a rated power around 40 watts Pittman DC040B-6

DC040B-6Reference VoltageContinuous TorqueRated CurrentRated PowerTorque ConstantVoltage ConstantTerminal ResistanceMax Winding TemperatureRotor Inertia

240 V00812 Nm236 A37 W0042 NmA0042 Vrads185 Ω155degC847 x 10ndash6 Kg-m2

Motor TransmissionLead Screw Support Rails

ΤRMS

PPK

ωPK

= 02879 Nm= 3369 W= 9276 rads

M

Other Motor Considerationsndash Type stepper brush BLDCndash Input speed (high input speed will result in high noise levels)ndash Total motor footprintndash Ambient temperaturendash Environmental conditionsndash Load-to-rotor inertiandash Encoder ready needed

Motion System Application Example17

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

17

MOTION SYSTEMAPPLICATION EXAMPLE

Gearbox Transmission Selection

When selecting a reduction ratio keep in mind that the RMS required torque at the motor shaft needs to fall below the continuous rated output torque of the motor

G30 A - Planetary GearboxMaximum output load 247 Nm

TRMS( Lead Screw)

02879 Nm

02879 Nm02879 Nm

Reduction

41

5161

Efficiency

090

090090

TRMS( Motor)

00798 Nm

00640 Nm00533 Nm

A 51 gearbox would allow about a 27 safety margin between what the system requires and the continuous torque output of the motor

System Summary 51η = 090

M

= 00807 Nm= 00737 Nm= ndash 00807 Nm= 00640 Nm= 4638 rads= 504 x 10 ndash 6 Kg-m2

= 3743 W

Τ1Τ2Τ3ΤRMS

PPK

ωPK

J

= 03632 Nm= 03316 Nm= ndash 03632 Nm= 02879 Nm= 9276 rads= 1134 x 10 ndash 5 Kg-m2

= 3369 W

Τ1Τ2Τ3ΤRMS

PPK

ωPK

J

Motion System Application Example18

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

18

MOTION SYSTEMAPPLICATION EXAMPLE

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

19

Motion Profile at the Motor

t 1 t 2 t 3 t dwell Time0333 s 0333 s 0333 s 05 s

ωΤ1

Τ2

Τ3= 4638 rads= 4429 RPM= ΤPK = 00807 Nm

ωPK

Τ1

nPK

Drive and Power Supply Requirements

Peak Current Required

IPK =TPK

KT+ IO =

00807 Nm0042 NmA

+ 0180 A = 210 A

RMS Current Required

IRMS =TRMS

KT+ IO =

00640 Nm0042 NmA

+ 0180 A = 170 A

Minimum bus Voltage Required

Vbus= IPKRmt+ ωPKKE = (210A)(185Ω) + (4638 rads)(0042 Vrads)= 389 V + 1948 V= 2337 V

0100 Nm 0200 Nm 0300 Nm 0400 Nm 0500 Nm 0600 Nm 0700 Nm

30 VDC

0100 Nm 0200 Nm 0300 Nm 0500 Nm 0600 Nm 0700 Nm

24 VDC

700060005000

RPM 4000300020001000

0400 Nm

TRMS

TPK

700060005000

RPM 4000300020001000

TRMS

TPK

Performance using 24V reference voltage

Performance using a 30V bus voltage

A 30 VDC bus will meet the application requirements as wells supply a margin of safety

MOTION SYSTEMAPPLICATION EXAMPLE

0100 Nm 0200 Nm 0300 Nm 0400 Nm 0500 Nm 0600 Nm 0700 Nm

30 VDC

0100 Nm 0200 Nm 0300 Nm 0500 Nm 0600 Nm 0700 Nm

24 VDC

700060005000

RPM 4000300020001000

0400 Nm

TRMS

TPK

700060005000

RPM 4000300020001000

TRMS

TPK

Ambient temperature Rated motor temperature Temperature rise Motor temperature

= 30degC= 155degC= 7638degC= 10638degC

Θa

Θm

ΘrΘrated

=Rth x IRMS

2 x Rmt

1 ndash (Rth x IRMS2 x Rmt x 000392degC)

Θr

=11degCω x 170A2

x 185Ω 1 ndash (11degCω x 170A2

x 185Ω x 000392degC)

= 5881 1 ndash (023)

= 5881 077

=

Θm = Θr + Θa = 7638degC + 30degC = 10638degC

7638degC

Performance using 24V reference voltage

Performance using a 30V bus voltage

Will the motor meet the temperature rise caused by the application

Motion System Application Example20

A 30 VDC bus will meet the application requirements as wells supply a margin of safety

203 756 7441 1500 Meriden Road Waterbury CT USA 06705

267 933 2105 534 Godshall DriveHarleysville PA USA 19438

wwwHaydonKerkPittmancom

copy All rights reserved AMETEK Advanced Motion Solutions No part of this doc-ument or technical information can be used reproduced or altered in any form without approval or proper authorization from AMETEK Inc and global affiliates

NOW ASK us the tough motion control questions ONLINE

ASK THE EXPERTS

HaydonKerkPittmancomwww

MOTION SYSTEMAPPLICATION EXAMPLE

Page 5: FORMULAS FOR MOTORIZED LINEAR MOTION SYSTEMS

Inertia Acceleration Velocity and Required Torque

Jin mout

ain aout ωin vout

Tin Fout

radsec 2

radsec

ain = 2 πa

L

ωin = 2 π v

L

Jin = m 2 πL 2

X η1 + JS Kg-m2

Τin= Τa+Τf +Τg+ΤD N-m

Τa= Jin ain

Τf =cosOslashmgμL

2 π ηN-m

N-m

Τg =sinOslashmgL

2 π ηN-m

ΤD = dragpreloadrefer to manufacturerrsquos data

N-m

Lead Screw System

Belt and Pulley System

ain = radsec 2

ωin = radsec Jin Jout

ain aout ωin ωout

Tin Tout

Jin = NJout

2X η

1 + JP1 Kg-m2

N-mFor a 1st approximation analysis JP1 JP2 and JB can be disregarded For higher precision pulley and belt inertias should be included as well as the reflected inertia of these components

+ JP2+ JB

(aout)(N) (ωout)(N)

Τin = NΤout X η

1

Motion System Formulas5

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

4

MOTION SYSTEMFORMULAS

Motion System Formulas6

Gearbox System

radsec 2

ωin = radsec

Jin = NJout

2X η

1

+

Kg-m2

N-m

Drag torque can be significant (TD) depending on the viscosity of the lubricant For a first approximation analysis this can be left outGearbox inertia may be found in manufacturerrsquos data sheets or calculated using gear dimensions and materials mass

JGB

(ωout)(N) Τin = N

Τout X η1

+

ΤD

ΤRMS = Τ12t 1 + Τ2

2t2 + Τ32t 3 + Τn

2t n

t 1 + t 2 + t 3 + t n + tdwell

TωLinear Power =

FSt

= watts

Rotary Power = = watts

Mechanical Power and RMS Torque

Jin Jout

ain aout ωin ωout

Tin Tout

ain = (aout)(N)

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

5

MOTION SYSTEMFORMULAS

Assuming IO is very small it can be ignored for a quick approximation of motor no-load speed If more accurate results are needed IO will also need to be adjuisted based on the motor Viscous Damping Factor (Dv) As the no-load speed increases with increased terminal voltage on a given motor IO will also increase based on the motorrsquos Dv value found on the manufacturerrsquos motor data sheet

Motor No Load SpeedVT ndash (IO x Rmt)no = 95493 x

KE

ndash ndash ndash ndash ndash ndash ndash ndash ndash ndash ndash ndash ndash

ndash ndash ndash

ndash ndash ndash

BreakawayTorque

Viscous Damping Torque (Dv)

Coulomb Friction

Τ

ωο

Motion System Formulas7

DC Motor Formulas

Motor Locked Rotor Current Locked Rotor Torque

ILR =VT

Rmt

TLR = ILR x KT

Motor Regulation

RmtRm = 95493 x

KE x KT

= x KT

VT

Rmt

ndash Theoretical using motor constants

nORm = TLR

ndash Regulation calculated using test data

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

6

MOTION SYSTEMFORMULAS

Motor Power Relationships

ndash Watts lost due to winding resistance

Ploss = I2 x Rmt

ndash Output power at any point on the motor curve

Pout = ω x Tndash Motor maximum output power

Pmax = 025 x ωo x TLR

ndash Motor maximum output power (Theoretical)

Pmax = 025 x VT2

Rmt

Temperature Effects on Motor Constants

ndash Change in terminal resistance

Rmt(f) = Rmt(i) x [ 1 + αconductor (Θf ndash Θi) ]αconductor = 00040degC (copper)

K(f) = K(i) x [ 1 + αmagnet (Θf ndash Θi) ]ndash Change in torque constant and voltage constant ( KT = KE )

Magnetic Material

CeramicSmCoAlNiCoNdFeB

Θmax (degC)300degC300degC540degC150degC

The magnetic material values above represent average figures for particular material classes and estimating performance If exact values are needed consult data sheets for a specific magnet grade

Motion System Formulas8

αmagnet (degC)-00020degC-00004degC-00002degC-00012degC

In the case of a mechanically commutated motor with graphite brushes this analysis will result in a slight error because of the negative temperature coefficient of carbon For estimation purposes this can be ignored but for a more rigorous analysis the behavior of the brush material must be taken into account This is not an issue when evaluating a brushless dc motor

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

7

MOTION SYSTEMFORMULAS

Motion System Formulas9

For PMDC motors (brush type) brush drop can be approximated as a constant resistance connected in series with the armature winding This series resistance is included in the Rmt factor found on manufacturerrsquos data sheet

Copper graphiteSilver graphiteElectrographite

02 to 04 Ω 02 to 04 Ω 08 to 10 Ω

Motor Brush Resistance

Brush Material Resistance

DC Motor Thermal Resistance and Temperature Rise

ndash Thermal resistance

ndash Motor temperature rise

Θm = Θr + Θa

ndash OR ndash

Temperature Rise degCRth =

Machine Losses W

Θr = Rth x I2 x Rmt

Θr = Rth x I

2 x Rmt

1 ndash (Rth x I2 x Rmt x α)

ndash OR ndash

Θr = Rth x(TC

Km)

2

ndash Final motor temperature

Use caution when applying these formulas Depending on the conditions used to determine motor specifications only one formula should be used Contact Appications Engineering for questions

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

8

MOTION SYSTEMFORMULAS

Motion System Formulas9

For PMDC motors (brush type) brush drop can be approximated as a constant resistance connected in series with the armature winding This series resistance is included in the Rmt factor found on manufacturerrsquos data sheet

Copper graphiteSilver graphiteElectrographite

02 to 04 Ω 02 to 04 Ω 08 to 10 Ω

Motor Brush Resistance

Brush Material Resistance

DC Motor Thermal Resistance and Temperature Rise

ndash Thermal resistance

ndash Motor temperature rise

Θm = Θr + Θa

ndash OR ndash

Temperature Rise degCRth =

Machine Losses W

Θr = Rth x I2 x Rmt

Θr = Rth x I

2 x Rmt

1 ndash (Rth x I2 x Rmt x α)

ndash OR ndash

Θr = Rth x(TC

Km)

2

ndash Final motor temperature

Use caution when applying these formulas Depending on the conditions used to determine motor specifications only one formula should be used Contact Appications Engineering for questions

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

9

MOTION SYSTEMFORMULAS

Stepper Motor Formulas

Required ldquoTime Offrdquo using an LR drive (constant voltage drive)

180Step Oslash = (phases)(rotor teeth)

ndash Motor step angle

ndash RPM (revolutionsminute) to SPS (stepssecond)

Maximum accuracy of a step motor system has nothing to do with maximum resolution of a step motor system Maximum accuracy is always based on the accuracy of 1 full step but maximum resolution is based on all system components including lead screw encoder stepper motor and step mode on the controller

ndash SPS (stepssecond) to RPM (revolutionsminute)(SPS) x (Step Oslash)

RPM = 6

(RPM) x 6 SPS = (Step Oslash)

ndash Travel per step

mstep =L

positionsrev

ndash Overdriving capability

Required ldquoTime Offrdquo using a chopper drive (constant current drive)

Time Off = (Time On) (Applied Voltage)2

(Rated Voltage)ndash Time On

Time Off = (Time On) (Applied Current)2

(Rated Current)ndash Time On

It is not unusual for a customer to drive Haydon Kerk stepper motors beyond their rated power to obtain the most force in the smallest package size Precautions must be taken to prevent the motor from exceeding its maximum temperature The ldquoonrdquo time should not exceed 2 - 3 minutes As general rule of thumb driving a motor at 25 to 3x its rated voltage or current may result in wasted energy and erratic behavior due to saturation

Special notes for can-stack motors bull more easily saturated due to less active material (steel) bull more iron losses due to unlaminated steel

Motion System Formulas10

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

10

MOTION SYSTEMFORMULAS

Lead Screw Performance Characteristics

End Mounting Rigidity

Impact Impact

ScrewLength

Increase In Affects Increase In Affects

Critical speed

Compression load

Critical speed

Inertia

Compression load

Stiffness

Spring Rate

Load capacity

Drive torque

Angular velocity

Load cpacity

Positioning accuracy

ScrewDiameter

Lead

Load

Preload

Nut length

SI Unit Systems

LengthMassTimeElectric currentThermodynamic temperatureLuminous intensityAmount of substance

meterkilogramsecondamperekelvincandelamole

mKgsAKcdmol

Base Units

Derived Unitsareavolumefrequencymass density (density)speed velocityangular velocityaccelerationangular accelerationforcepressure (mechanical stress)kinematic viscositydynamic viscositywork energy quantity of heatpowerentropyspecific heat capacitythermal conductivity

square metercubic meterhertzkilogram per cubic metermeter per secondradian per secondmeter per second squaredradian per second squarednewtonpascalsquare meter per secondnewton-second per square meter joulewattjoule per kelvinjoule per kilogram kelvinwatt per meter kelvin

m2

m3

HzKgm3

msradsms2

rads2

NPam2sN-sm2

JWJKJKg-K)W(m-K)

Quantity Name of Unit Symbol

Motion System Reference Tables11

Critical speed

Compression load

System stiffness

Life

Positioning accuracy

System stiffness

Drag torque

Load capacity

Stiffness

sndash1

Kg-ms2

Nm2

N-mJs or N-ms

ExamplesAn INCREASE ( ) in screw length results in a DECREASE ( ) in critical speed

An INCREASE in screw diameter results in an INCREASE in critical speed

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

11

MOTION SYSTEMREFERENCE TABLES

t 1 t 2 t 3 t dwell

vTime

vpk

a -a

MassMove distanceMove timeDwell timeMotion profileScrew speed limitLoad supportOrientationRotary to linear conversionAmbient temperature

9 Kg200 mm10 s05 s13 13 13 Trapezoidal1000 RPMμ = 001VerticalTFE lead screw 8 mm diameter 275 mm long30degC

ControlDrive Power Supply

4 quadrant with encoder32 VDC 35 Arms50 A peak

0333 s 0333 s 0333 s 05 s

vPK =3S2t =

(3)(02m)(2)(10 s) =

06m2 s =

03ms

Linear Velocity

Linear Acceleration

a = vf - vi

t = 03ms - 0 ms 0333 s 0901ms2 =

1st step is to thoroughly understand system constraints and motion profile requirements

Linear Motion ExampleData

Motion System Application Example12

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12

MOTION SYSTEMAPPLICATION EXAMPLE

Lmin =60vPK

n =(60)(03ms)1000 revmin = 0018mrev = 18mmrev

Refer to Kerk screw chart ndash closest lead in an 8mm screw diameter is

2032mmrev = 002032 mrevη = 086 free wheeling nut TFE coated screw

JS = 388 x 10ndash 7 Kg-m2 Other critical lead screw considerations bull Column loading bull Critical speed

Jm aa

ωPKvPK

TF

= ωPK = 2 π vPK

L=

(2π)(03ms)002032 mrev

9276 rads

a = =2πaL

=(2π)(0901ms2)002032 mrev

2786 rads2

Linear to Rotary Conversion ndash Velocity Acceleration Inertia Torque

Minimum screw lead to maintain lt 1000 RPM shaft speed

Linear to Rotary Conversion Velocity

Linear to Rotary Conversion Acceleration

Motion System Application Example13

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13

MOTION SYSTEMFORMULAS

Jin= m 2 πL 2

X η1 + JS= 9Kg

2 π002032 m 2

X 0861 + 388x10ndash 7Kg-m2

= 1095 X 10ndash 5 Kg-m2 + 388x10ndash 7 Kg-m2

= 1134 x 10ndash 5 Kg-m2

Τin= Τa+Τf +Τg+ΤD

Τa = Jin ain

Τg =sinOslashmgL

2 π η

= (1134 x 10ndash 5 Kg-m2)(2786 rads2)

= 00316 Nm

Τf =cosOslashmgμL

2 π η= (cos90)(9 Kg)(98ms2)(001)(002032m)

2 π (086)= 0 Nm

= (sin90)(9 Kg)(98ms2)(002032m)2 π (086)

=1792 Nm

5404 = 03316 Nm

ΤDSince a free-wheeling lead screw nut was used there is no preload force If an anti-backlash nut was used ΤD would be gt0 due to preload force Always reference manufacturerrsquos data for more information

= 0 Nm

Linear to Rotary Conversion Torque

Linear to Rotary Conversion Inertia

Motion System Application Example14

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14

MOTION SYSTEMFORMULAS

t 1 t 2 t 3 t dwell Time0333 s 0333 s 0333 s 05 s

Rotary Motion Profile

ωPK

ω

= 9276 rads a = 2786 rads2

Τ1= Τa+Τf +Τg+ΤD

= (00316 Nm) + (0 Nm) + (03316 Nm) + (0 Nm)

= 03632 Nm

Τ2= Τa+Τf +Τg+ΤD

= (0 Nm) + (0 Nm) + (03316 Nm) + (0 Nm)

= 03316 Nm

Τ3= Τa+Τf +Τg+ΤD

= (00316 Nm) + (0 Nm) + (03316 Nm) + (0 Nm)

= 03632 Nm

Τ1

Τ2

Τ3

Motion System Application Example15

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15

MOTION SYSTEMAPPLICATION EXAMPLE

Lead Screw Input Requirements

= 9276 rads= 2786 rads2

= 03632 Nm= 03316 Nm= ndash 03632 Nm= 0333 s= 0333 s= 0333 s= 05 s= 3369 W= 3076 W

ωPKaΤ1Τ2Τ3

t1t2t3tdwell

PPKPCV

PPK = (Τ1)(ωPK)

= (03632 Nm)(9276 rads)= 3369 W

PCV = (Τ2)(ωPK)

= (03316 Nm)(9276 rads)= 3076 W

RMS Torque Requirement Lead Screw Input

=

0333 s + 0333 s + 0333 s + 05 s

(03632 Nm)2(0333 s)+(03316 Nm)2(0333 s)+(- 03632 Nm)2(0333 s)

ΤRMS = Τ12t 1 + Τ2

2t2 + Τ32t 3

t 1 + t 2 + t 3 + tdwell

15

00439 + 00366 + 00439 = 00829 =

ΤRMS = 02879 Nm

Motion System Application Example16

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16

MOTION SYSTEMAPPLICATION EXAMPLE

Load Parameters at the Lead Screw Shaft

= 9276 rads= 2786 rads2

= 03632 Nm= 03316 Nm= ndash 03632 Nm= 02879 Nm= 3369 W= 3076 W

ωPKaΤ1Τ2Τ3ΤRMSPPKPCV

Motor SelectionA direct-drive motor can be used however it would be large and expensive For example a DC057B-3 brush motor

would easily meet continuous torque and continuous speed as well as a 181 load-to-rotor inertia This motor however has significantly more output power than needed at 128 watts

A smaller motor with a transmission would optimize the system Look for a motor starting at a rated power around 40 watts Pittman DC040B-6

DC040B-6Reference VoltageContinuous TorqueRated CurrentRated PowerTorque ConstantVoltage ConstantTerminal ResistanceMax Winding TemperatureRotor Inertia

240 V00812 Nm236 A37 W0042 NmA0042 Vrads185 Ω155degC847 x 10ndash6 Kg-m2

Motor TransmissionLead Screw Support Rails

ΤRMS

PPK

ωPK

= 02879 Nm= 3369 W= 9276 rads

M

Other Motor Considerationsndash Type stepper brush BLDCndash Input speed (high input speed will result in high noise levels)ndash Total motor footprintndash Ambient temperaturendash Environmental conditionsndash Load-to-rotor inertiandash Encoder ready needed

Motion System Application Example17

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17

MOTION SYSTEMAPPLICATION EXAMPLE

Gearbox Transmission Selection

When selecting a reduction ratio keep in mind that the RMS required torque at the motor shaft needs to fall below the continuous rated output torque of the motor

G30 A - Planetary GearboxMaximum output load 247 Nm

TRMS( Lead Screw)

02879 Nm

02879 Nm02879 Nm

Reduction

41

5161

Efficiency

090

090090

TRMS( Motor)

00798 Nm

00640 Nm00533 Nm

A 51 gearbox would allow about a 27 safety margin between what the system requires and the continuous torque output of the motor

System Summary 51η = 090

M

= 00807 Nm= 00737 Nm= ndash 00807 Nm= 00640 Nm= 4638 rads= 504 x 10 ndash 6 Kg-m2

= 3743 W

Τ1Τ2Τ3ΤRMS

PPK

ωPK

J

= 03632 Nm= 03316 Nm= ndash 03632 Nm= 02879 Nm= 9276 rads= 1134 x 10 ndash 5 Kg-m2

= 3369 W

Τ1Τ2Τ3ΤRMS

PPK

ωPK

J

Motion System Application Example18

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18

MOTION SYSTEMAPPLICATION EXAMPLE

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19

Motion Profile at the Motor

t 1 t 2 t 3 t dwell Time0333 s 0333 s 0333 s 05 s

ωΤ1

Τ2

Τ3= 4638 rads= 4429 RPM= ΤPK = 00807 Nm

ωPK

Τ1

nPK

Drive and Power Supply Requirements

Peak Current Required

IPK =TPK

KT+ IO =

00807 Nm0042 NmA

+ 0180 A = 210 A

RMS Current Required

IRMS =TRMS

KT+ IO =

00640 Nm0042 NmA

+ 0180 A = 170 A

Minimum bus Voltage Required

Vbus= IPKRmt+ ωPKKE = (210A)(185Ω) + (4638 rads)(0042 Vrads)= 389 V + 1948 V= 2337 V

0100 Nm 0200 Nm 0300 Nm 0400 Nm 0500 Nm 0600 Nm 0700 Nm

30 VDC

0100 Nm 0200 Nm 0300 Nm 0500 Nm 0600 Nm 0700 Nm

24 VDC

700060005000

RPM 4000300020001000

0400 Nm

TRMS

TPK

700060005000

RPM 4000300020001000

TRMS

TPK

Performance using 24V reference voltage

Performance using a 30V bus voltage

A 30 VDC bus will meet the application requirements as wells supply a margin of safety

MOTION SYSTEMAPPLICATION EXAMPLE

0100 Nm 0200 Nm 0300 Nm 0400 Nm 0500 Nm 0600 Nm 0700 Nm

30 VDC

0100 Nm 0200 Nm 0300 Nm 0500 Nm 0600 Nm 0700 Nm

24 VDC

700060005000

RPM 4000300020001000

0400 Nm

TRMS

TPK

700060005000

RPM 4000300020001000

TRMS

TPK

Ambient temperature Rated motor temperature Temperature rise Motor temperature

= 30degC= 155degC= 7638degC= 10638degC

Θa

Θm

ΘrΘrated

=Rth x IRMS

2 x Rmt

1 ndash (Rth x IRMS2 x Rmt x 000392degC)

Θr

=11degCω x 170A2

x 185Ω 1 ndash (11degCω x 170A2

x 185Ω x 000392degC)

= 5881 1 ndash (023)

= 5881 077

=

Θm = Θr + Θa = 7638degC + 30degC = 10638degC

7638degC

Performance using 24V reference voltage

Performance using a 30V bus voltage

Will the motor meet the temperature rise caused by the application

Motion System Application Example20

A 30 VDC bus will meet the application requirements as wells supply a margin of safety

203 756 7441 1500 Meriden Road Waterbury CT USA 06705

267 933 2105 534 Godshall DriveHarleysville PA USA 19438

wwwHaydonKerkPittmancom

copy All rights reserved AMETEK Advanced Motion Solutions No part of this doc-ument or technical information can be used reproduced or altered in any form without approval or proper authorization from AMETEK Inc and global affiliates

NOW ASK us the tough motion control questions ONLINE

ASK THE EXPERTS

HaydonKerkPittmancomwww

MOTION SYSTEMAPPLICATION EXAMPLE

Page 6: FORMULAS FOR MOTORIZED LINEAR MOTION SYSTEMS

Motion System Formulas6

Gearbox System

radsec 2

ωin = radsec

Jin = NJout

2X η

1

+

Kg-m2

N-m

Drag torque can be significant (TD) depending on the viscosity of the lubricant For a first approximation analysis this can be left outGearbox inertia may be found in manufacturerrsquos data sheets or calculated using gear dimensions and materials mass

JGB

(ωout)(N) Τin = N

Τout X η1

+

ΤD

ΤRMS = Τ12t 1 + Τ2

2t2 + Τ32t 3 + Τn

2t n

t 1 + t 2 + t 3 + t n + tdwell

TωLinear Power =

FSt

= watts

Rotary Power = = watts

Mechanical Power and RMS Torque

Jin Jout

ain aout ωin ωout

Tin Tout

ain = (aout)(N)

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5

MOTION SYSTEMFORMULAS

Assuming IO is very small it can be ignored for a quick approximation of motor no-load speed If more accurate results are needed IO will also need to be adjuisted based on the motor Viscous Damping Factor (Dv) As the no-load speed increases with increased terminal voltage on a given motor IO will also increase based on the motorrsquos Dv value found on the manufacturerrsquos motor data sheet

Motor No Load SpeedVT ndash (IO x Rmt)no = 95493 x

KE

ndash ndash ndash ndash ndash ndash ndash ndash ndash ndash ndash ndash ndash

ndash ndash ndash

ndash ndash ndash

BreakawayTorque

Viscous Damping Torque (Dv)

Coulomb Friction

Τ

ωο

Motion System Formulas7

DC Motor Formulas

Motor Locked Rotor Current Locked Rotor Torque

ILR =VT

Rmt

TLR = ILR x KT

Motor Regulation

RmtRm = 95493 x

KE x KT

= x KT

VT

Rmt

ndash Theoretical using motor constants

nORm = TLR

ndash Regulation calculated using test data

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6

MOTION SYSTEMFORMULAS

Motor Power Relationships

ndash Watts lost due to winding resistance

Ploss = I2 x Rmt

ndash Output power at any point on the motor curve

Pout = ω x Tndash Motor maximum output power

Pmax = 025 x ωo x TLR

ndash Motor maximum output power (Theoretical)

Pmax = 025 x VT2

Rmt

Temperature Effects on Motor Constants

ndash Change in terminal resistance

Rmt(f) = Rmt(i) x [ 1 + αconductor (Θf ndash Θi) ]αconductor = 00040degC (copper)

K(f) = K(i) x [ 1 + αmagnet (Θf ndash Θi) ]ndash Change in torque constant and voltage constant ( KT = KE )

Magnetic Material

CeramicSmCoAlNiCoNdFeB

Θmax (degC)300degC300degC540degC150degC

The magnetic material values above represent average figures for particular material classes and estimating performance If exact values are needed consult data sheets for a specific magnet grade

Motion System Formulas8

αmagnet (degC)-00020degC-00004degC-00002degC-00012degC

In the case of a mechanically commutated motor with graphite brushes this analysis will result in a slight error because of the negative temperature coefficient of carbon For estimation purposes this can be ignored but for a more rigorous analysis the behavior of the brush material must be taken into account This is not an issue when evaluating a brushless dc motor

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7

MOTION SYSTEMFORMULAS

Motion System Formulas9

For PMDC motors (brush type) brush drop can be approximated as a constant resistance connected in series with the armature winding This series resistance is included in the Rmt factor found on manufacturerrsquos data sheet

Copper graphiteSilver graphiteElectrographite

02 to 04 Ω 02 to 04 Ω 08 to 10 Ω

Motor Brush Resistance

Brush Material Resistance

DC Motor Thermal Resistance and Temperature Rise

ndash Thermal resistance

ndash Motor temperature rise

Θm = Θr + Θa

ndash OR ndash

Temperature Rise degCRth =

Machine Losses W

Θr = Rth x I2 x Rmt

Θr = Rth x I

2 x Rmt

1 ndash (Rth x I2 x Rmt x α)

ndash OR ndash

Θr = Rth x(TC

Km)

2

ndash Final motor temperature

Use caution when applying these formulas Depending on the conditions used to determine motor specifications only one formula should be used Contact Appications Engineering for questions

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8

MOTION SYSTEMFORMULAS

Motion System Formulas9

For PMDC motors (brush type) brush drop can be approximated as a constant resistance connected in series with the armature winding This series resistance is included in the Rmt factor found on manufacturerrsquos data sheet

Copper graphiteSilver graphiteElectrographite

02 to 04 Ω 02 to 04 Ω 08 to 10 Ω

Motor Brush Resistance

Brush Material Resistance

DC Motor Thermal Resistance and Temperature Rise

ndash Thermal resistance

ndash Motor temperature rise

Θm = Θr + Θa

ndash OR ndash

Temperature Rise degCRth =

Machine Losses W

Θr = Rth x I2 x Rmt

Θr = Rth x I

2 x Rmt

1 ndash (Rth x I2 x Rmt x α)

ndash OR ndash

Θr = Rth x(TC

Km)

2

ndash Final motor temperature

Use caution when applying these formulas Depending on the conditions used to determine motor specifications only one formula should be used Contact Appications Engineering for questions

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9

MOTION SYSTEMFORMULAS

Stepper Motor Formulas

Required ldquoTime Offrdquo using an LR drive (constant voltage drive)

180Step Oslash = (phases)(rotor teeth)

ndash Motor step angle

ndash RPM (revolutionsminute) to SPS (stepssecond)

Maximum accuracy of a step motor system has nothing to do with maximum resolution of a step motor system Maximum accuracy is always based on the accuracy of 1 full step but maximum resolution is based on all system components including lead screw encoder stepper motor and step mode on the controller

ndash SPS (stepssecond) to RPM (revolutionsminute)(SPS) x (Step Oslash)

RPM = 6

(RPM) x 6 SPS = (Step Oslash)

ndash Travel per step

mstep =L

positionsrev

ndash Overdriving capability

Required ldquoTime Offrdquo using a chopper drive (constant current drive)

Time Off = (Time On) (Applied Voltage)2

(Rated Voltage)ndash Time On

Time Off = (Time On) (Applied Current)2

(Rated Current)ndash Time On

It is not unusual for a customer to drive Haydon Kerk stepper motors beyond their rated power to obtain the most force in the smallest package size Precautions must be taken to prevent the motor from exceeding its maximum temperature The ldquoonrdquo time should not exceed 2 - 3 minutes As general rule of thumb driving a motor at 25 to 3x its rated voltage or current may result in wasted energy and erratic behavior due to saturation

Special notes for can-stack motors bull more easily saturated due to less active material (steel) bull more iron losses due to unlaminated steel

Motion System Formulas10

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10

MOTION SYSTEMFORMULAS

Lead Screw Performance Characteristics

End Mounting Rigidity

Impact Impact

ScrewLength

Increase In Affects Increase In Affects

Critical speed

Compression load

Critical speed

Inertia

Compression load

Stiffness

Spring Rate

Load capacity

Drive torque

Angular velocity

Load cpacity

Positioning accuracy

ScrewDiameter

Lead

Load

Preload

Nut length

SI Unit Systems

LengthMassTimeElectric currentThermodynamic temperatureLuminous intensityAmount of substance

meterkilogramsecondamperekelvincandelamole

mKgsAKcdmol

Base Units

Derived Unitsareavolumefrequencymass density (density)speed velocityangular velocityaccelerationangular accelerationforcepressure (mechanical stress)kinematic viscositydynamic viscositywork energy quantity of heatpowerentropyspecific heat capacitythermal conductivity

square metercubic meterhertzkilogram per cubic metermeter per secondradian per secondmeter per second squaredradian per second squarednewtonpascalsquare meter per secondnewton-second per square meter joulewattjoule per kelvinjoule per kilogram kelvinwatt per meter kelvin

m2

m3

HzKgm3

msradsms2

rads2

NPam2sN-sm2

JWJKJKg-K)W(m-K)

Quantity Name of Unit Symbol

Motion System Reference Tables11

Critical speed

Compression load

System stiffness

Life

Positioning accuracy

System stiffness

Drag torque

Load capacity

Stiffness

sndash1

Kg-ms2

Nm2

N-mJs or N-ms

ExamplesAn INCREASE ( ) in screw length results in a DECREASE ( ) in critical speed

An INCREASE in screw diameter results in an INCREASE in critical speed

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

11

MOTION SYSTEMREFERENCE TABLES

t 1 t 2 t 3 t dwell

vTime

vpk

a -a

MassMove distanceMove timeDwell timeMotion profileScrew speed limitLoad supportOrientationRotary to linear conversionAmbient temperature

9 Kg200 mm10 s05 s13 13 13 Trapezoidal1000 RPMμ = 001VerticalTFE lead screw 8 mm diameter 275 mm long30degC

ControlDrive Power Supply

4 quadrant with encoder32 VDC 35 Arms50 A peak

0333 s 0333 s 0333 s 05 s

vPK =3S2t =

(3)(02m)(2)(10 s) =

06m2 s =

03ms

Linear Velocity

Linear Acceleration

a = vf - vi

t = 03ms - 0 ms 0333 s 0901ms2 =

1st step is to thoroughly understand system constraints and motion profile requirements

Linear Motion ExampleData

Motion System Application Example12

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

12

MOTION SYSTEMAPPLICATION EXAMPLE

Lmin =60vPK

n =(60)(03ms)1000 revmin = 0018mrev = 18mmrev

Refer to Kerk screw chart ndash closest lead in an 8mm screw diameter is

2032mmrev = 002032 mrevη = 086 free wheeling nut TFE coated screw

JS = 388 x 10ndash 7 Kg-m2 Other critical lead screw considerations bull Column loading bull Critical speed

Jm aa

ωPKvPK

TF

= ωPK = 2 π vPK

L=

(2π)(03ms)002032 mrev

9276 rads

a = =2πaL

=(2π)(0901ms2)002032 mrev

2786 rads2

Linear to Rotary Conversion ndash Velocity Acceleration Inertia Torque

Minimum screw lead to maintain lt 1000 RPM shaft speed

Linear to Rotary Conversion Velocity

Linear to Rotary Conversion Acceleration

Motion System Application Example13

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13

MOTION SYSTEMFORMULAS

Jin= m 2 πL 2

X η1 + JS= 9Kg

2 π002032 m 2

X 0861 + 388x10ndash 7Kg-m2

= 1095 X 10ndash 5 Kg-m2 + 388x10ndash 7 Kg-m2

= 1134 x 10ndash 5 Kg-m2

Τin= Τa+Τf +Τg+ΤD

Τa = Jin ain

Τg =sinOslashmgL

2 π η

= (1134 x 10ndash 5 Kg-m2)(2786 rads2)

= 00316 Nm

Τf =cosOslashmgμL

2 π η= (cos90)(9 Kg)(98ms2)(001)(002032m)

2 π (086)= 0 Nm

= (sin90)(9 Kg)(98ms2)(002032m)2 π (086)

=1792 Nm

5404 = 03316 Nm

ΤDSince a free-wheeling lead screw nut was used there is no preload force If an anti-backlash nut was used ΤD would be gt0 due to preload force Always reference manufacturerrsquos data for more information

= 0 Nm

Linear to Rotary Conversion Torque

Linear to Rotary Conversion Inertia

Motion System Application Example14

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14

MOTION SYSTEMFORMULAS

t 1 t 2 t 3 t dwell Time0333 s 0333 s 0333 s 05 s

Rotary Motion Profile

ωPK

ω

= 9276 rads a = 2786 rads2

Τ1= Τa+Τf +Τg+ΤD

= (00316 Nm) + (0 Nm) + (03316 Nm) + (0 Nm)

= 03632 Nm

Τ2= Τa+Τf +Τg+ΤD

= (0 Nm) + (0 Nm) + (03316 Nm) + (0 Nm)

= 03316 Nm

Τ3= Τa+Τf +Τg+ΤD

= (00316 Nm) + (0 Nm) + (03316 Nm) + (0 Nm)

= 03632 Nm

Τ1

Τ2

Τ3

Motion System Application Example15

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

15

MOTION SYSTEMAPPLICATION EXAMPLE

Lead Screw Input Requirements

= 9276 rads= 2786 rads2

= 03632 Nm= 03316 Nm= ndash 03632 Nm= 0333 s= 0333 s= 0333 s= 05 s= 3369 W= 3076 W

ωPKaΤ1Τ2Τ3

t1t2t3tdwell

PPKPCV

PPK = (Τ1)(ωPK)

= (03632 Nm)(9276 rads)= 3369 W

PCV = (Τ2)(ωPK)

= (03316 Nm)(9276 rads)= 3076 W

RMS Torque Requirement Lead Screw Input

=

0333 s + 0333 s + 0333 s + 05 s

(03632 Nm)2(0333 s)+(03316 Nm)2(0333 s)+(- 03632 Nm)2(0333 s)

ΤRMS = Τ12t 1 + Τ2

2t2 + Τ32t 3

t 1 + t 2 + t 3 + tdwell

15

00439 + 00366 + 00439 = 00829 =

ΤRMS = 02879 Nm

Motion System Application Example16

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

16

MOTION SYSTEMAPPLICATION EXAMPLE

Load Parameters at the Lead Screw Shaft

= 9276 rads= 2786 rads2

= 03632 Nm= 03316 Nm= ndash 03632 Nm= 02879 Nm= 3369 W= 3076 W

ωPKaΤ1Τ2Τ3ΤRMSPPKPCV

Motor SelectionA direct-drive motor can be used however it would be large and expensive For example a DC057B-3 brush motor

would easily meet continuous torque and continuous speed as well as a 181 load-to-rotor inertia This motor however has significantly more output power than needed at 128 watts

A smaller motor with a transmission would optimize the system Look for a motor starting at a rated power around 40 watts Pittman DC040B-6

DC040B-6Reference VoltageContinuous TorqueRated CurrentRated PowerTorque ConstantVoltage ConstantTerminal ResistanceMax Winding TemperatureRotor Inertia

240 V00812 Nm236 A37 W0042 NmA0042 Vrads185 Ω155degC847 x 10ndash6 Kg-m2

Motor TransmissionLead Screw Support Rails

ΤRMS

PPK

ωPK

= 02879 Nm= 3369 W= 9276 rads

M

Other Motor Considerationsndash Type stepper brush BLDCndash Input speed (high input speed will result in high noise levels)ndash Total motor footprintndash Ambient temperaturendash Environmental conditionsndash Load-to-rotor inertiandash Encoder ready needed

Motion System Application Example17

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

17

MOTION SYSTEMAPPLICATION EXAMPLE

Gearbox Transmission Selection

When selecting a reduction ratio keep in mind that the RMS required torque at the motor shaft needs to fall below the continuous rated output torque of the motor

G30 A - Planetary GearboxMaximum output load 247 Nm

TRMS( Lead Screw)

02879 Nm

02879 Nm02879 Nm

Reduction

41

5161

Efficiency

090

090090

TRMS( Motor)

00798 Nm

00640 Nm00533 Nm

A 51 gearbox would allow about a 27 safety margin between what the system requires and the continuous torque output of the motor

System Summary 51η = 090

M

= 00807 Nm= 00737 Nm= ndash 00807 Nm= 00640 Nm= 4638 rads= 504 x 10 ndash 6 Kg-m2

= 3743 W

Τ1Τ2Τ3ΤRMS

PPK

ωPK

J

= 03632 Nm= 03316 Nm= ndash 03632 Nm= 02879 Nm= 9276 rads= 1134 x 10 ndash 5 Kg-m2

= 3369 W

Τ1Τ2Τ3ΤRMS

PPK

ωPK

J

Motion System Application Example18

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

18

MOTION SYSTEMAPPLICATION EXAMPLE

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

19

Motion Profile at the Motor

t 1 t 2 t 3 t dwell Time0333 s 0333 s 0333 s 05 s

ωΤ1

Τ2

Τ3= 4638 rads= 4429 RPM= ΤPK = 00807 Nm

ωPK

Τ1

nPK

Drive and Power Supply Requirements

Peak Current Required

IPK =TPK

KT+ IO =

00807 Nm0042 NmA

+ 0180 A = 210 A

RMS Current Required

IRMS =TRMS

KT+ IO =

00640 Nm0042 NmA

+ 0180 A = 170 A

Minimum bus Voltage Required

Vbus= IPKRmt+ ωPKKE = (210A)(185Ω) + (4638 rads)(0042 Vrads)= 389 V + 1948 V= 2337 V

0100 Nm 0200 Nm 0300 Nm 0400 Nm 0500 Nm 0600 Nm 0700 Nm

30 VDC

0100 Nm 0200 Nm 0300 Nm 0500 Nm 0600 Nm 0700 Nm

24 VDC

700060005000

RPM 4000300020001000

0400 Nm

TRMS

TPK

700060005000

RPM 4000300020001000

TRMS

TPK

Performance using 24V reference voltage

Performance using a 30V bus voltage

A 30 VDC bus will meet the application requirements as wells supply a margin of safety

MOTION SYSTEMAPPLICATION EXAMPLE

0100 Nm 0200 Nm 0300 Nm 0400 Nm 0500 Nm 0600 Nm 0700 Nm

30 VDC

0100 Nm 0200 Nm 0300 Nm 0500 Nm 0600 Nm 0700 Nm

24 VDC

700060005000

RPM 4000300020001000

0400 Nm

TRMS

TPK

700060005000

RPM 4000300020001000

TRMS

TPK

Ambient temperature Rated motor temperature Temperature rise Motor temperature

= 30degC= 155degC= 7638degC= 10638degC

Θa

Θm

ΘrΘrated

=Rth x IRMS

2 x Rmt

1 ndash (Rth x IRMS2 x Rmt x 000392degC)

Θr

=11degCω x 170A2

x 185Ω 1 ndash (11degCω x 170A2

x 185Ω x 000392degC)

= 5881 1 ndash (023)

= 5881 077

=

Θm = Θr + Θa = 7638degC + 30degC = 10638degC

7638degC

Performance using 24V reference voltage

Performance using a 30V bus voltage

Will the motor meet the temperature rise caused by the application

Motion System Application Example20

A 30 VDC bus will meet the application requirements as wells supply a margin of safety

203 756 7441 1500 Meriden Road Waterbury CT USA 06705

267 933 2105 534 Godshall DriveHarleysville PA USA 19438

wwwHaydonKerkPittmancom

copy All rights reserved AMETEK Advanced Motion Solutions No part of this doc-ument or technical information can be used reproduced or altered in any form without approval or proper authorization from AMETEK Inc and global affiliates

NOW ASK us the tough motion control questions ONLINE

ASK THE EXPERTS

HaydonKerkPittmancomwww

MOTION SYSTEMAPPLICATION EXAMPLE

Page 7: FORMULAS FOR MOTORIZED LINEAR MOTION SYSTEMS

Assuming IO is very small it can be ignored for a quick approximation of motor no-load speed If more accurate results are needed IO will also need to be adjuisted based on the motor Viscous Damping Factor (Dv) As the no-load speed increases with increased terminal voltage on a given motor IO will also increase based on the motorrsquos Dv value found on the manufacturerrsquos motor data sheet

Motor No Load SpeedVT ndash (IO x Rmt)no = 95493 x

KE

ndash ndash ndash ndash ndash ndash ndash ndash ndash ndash ndash ndash ndash

ndash ndash ndash

ndash ndash ndash

BreakawayTorque

Viscous Damping Torque (Dv)

Coulomb Friction

Τ

ωο

Motion System Formulas7

DC Motor Formulas

Motor Locked Rotor Current Locked Rotor Torque

ILR =VT

Rmt

TLR = ILR x KT

Motor Regulation

RmtRm = 95493 x

KE x KT

= x KT

VT

Rmt

ndash Theoretical using motor constants

nORm = TLR

ndash Regulation calculated using test data

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

6

MOTION SYSTEMFORMULAS

Motor Power Relationships

ndash Watts lost due to winding resistance

Ploss = I2 x Rmt

ndash Output power at any point on the motor curve

Pout = ω x Tndash Motor maximum output power

Pmax = 025 x ωo x TLR

ndash Motor maximum output power (Theoretical)

Pmax = 025 x VT2

Rmt

Temperature Effects on Motor Constants

ndash Change in terminal resistance

Rmt(f) = Rmt(i) x [ 1 + αconductor (Θf ndash Θi) ]αconductor = 00040degC (copper)

K(f) = K(i) x [ 1 + αmagnet (Θf ndash Θi) ]ndash Change in torque constant and voltage constant ( KT = KE )

Magnetic Material

CeramicSmCoAlNiCoNdFeB

Θmax (degC)300degC300degC540degC150degC

The magnetic material values above represent average figures for particular material classes and estimating performance If exact values are needed consult data sheets for a specific magnet grade

Motion System Formulas8

αmagnet (degC)-00020degC-00004degC-00002degC-00012degC

In the case of a mechanically commutated motor with graphite brushes this analysis will result in a slight error because of the negative temperature coefficient of carbon For estimation purposes this can be ignored but for a more rigorous analysis the behavior of the brush material must be taken into account This is not an issue when evaluating a brushless dc motor

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

7

MOTION SYSTEMFORMULAS

Motion System Formulas9

For PMDC motors (brush type) brush drop can be approximated as a constant resistance connected in series with the armature winding This series resistance is included in the Rmt factor found on manufacturerrsquos data sheet

Copper graphiteSilver graphiteElectrographite

02 to 04 Ω 02 to 04 Ω 08 to 10 Ω

Motor Brush Resistance

Brush Material Resistance

DC Motor Thermal Resistance and Temperature Rise

ndash Thermal resistance

ndash Motor temperature rise

Θm = Θr + Θa

ndash OR ndash

Temperature Rise degCRth =

Machine Losses W

Θr = Rth x I2 x Rmt

Θr = Rth x I

2 x Rmt

1 ndash (Rth x I2 x Rmt x α)

ndash OR ndash

Θr = Rth x(TC

Km)

2

ndash Final motor temperature

Use caution when applying these formulas Depending on the conditions used to determine motor specifications only one formula should be used Contact Appications Engineering for questions

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

8

MOTION SYSTEMFORMULAS

Motion System Formulas9

For PMDC motors (brush type) brush drop can be approximated as a constant resistance connected in series with the armature winding This series resistance is included in the Rmt factor found on manufacturerrsquos data sheet

Copper graphiteSilver graphiteElectrographite

02 to 04 Ω 02 to 04 Ω 08 to 10 Ω

Motor Brush Resistance

Brush Material Resistance

DC Motor Thermal Resistance and Temperature Rise

ndash Thermal resistance

ndash Motor temperature rise

Θm = Θr + Θa

ndash OR ndash

Temperature Rise degCRth =

Machine Losses W

Θr = Rth x I2 x Rmt

Θr = Rth x I

2 x Rmt

1 ndash (Rth x I2 x Rmt x α)

ndash OR ndash

Θr = Rth x(TC

Km)

2

ndash Final motor temperature

Use caution when applying these formulas Depending on the conditions used to determine motor specifications only one formula should be used Contact Appications Engineering for questions

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

9

MOTION SYSTEMFORMULAS

Stepper Motor Formulas

Required ldquoTime Offrdquo using an LR drive (constant voltage drive)

180Step Oslash = (phases)(rotor teeth)

ndash Motor step angle

ndash RPM (revolutionsminute) to SPS (stepssecond)

Maximum accuracy of a step motor system has nothing to do with maximum resolution of a step motor system Maximum accuracy is always based on the accuracy of 1 full step but maximum resolution is based on all system components including lead screw encoder stepper motor and step mode on the controller

ndash SPS (stepssecond) to RPM (revolutionsminute)(SPS) x (Step Oslash)

RPM = 6

(RPM) x 6 SPS = (Step Oslash)

ndash Travel per step

mstep =L

positionsrev

ndash Overdriving capability

Required ldquoTime Offrdquo using a chopper drive (constant current drive)

Time Off = (Time On) (Applied Voltage)2

(Rated Voltage)ndash Time On

Time Off = (Time On) (Applied Current)2

(Rated Current)ndash Time On

It is not unusual for a customer to drive Haydon Kerk stepper motors beyond their rated power to obtain the most force in the smallest package size Precautions must be taken to prevent the motor from exceeding its maximum temperature The ldquoonrdquo time should not exceed 2 - 3 minutes As general rule of thumb driving a motor at 25 to 3x its rated voltage or current may result in wasted energy and erratic behavior due to saturation

Special notes for can-stack motors bull more easily saturated due to less active material (steel) bull more iron losses due to unlaminated steel

Motion System Formulas10

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

10

MOTION SYSTEMFORMULAS

Lead Screw Performance Characteristics

End Mounting Rigidity

Impact Impact

ScrewLength

Increase In Affects Increase In Affects

Critical speed

Compression load

Critical speed

Inertia

Compression load

Stiffness

Spring Rate

Load capacity

Drive torque

Angular velocity

Load cpacity

Positioning accuracy

ScrewDiameter

Lead

Load

Preload

Nut length

SI Unit Systems

LengthMassTimeElectric currentThermodynamic temperatureLuminous intensityAmount of substance

meterkilogramsecondamperekelvincandelamole

mKgsAKcdmol

Base Units

Derived Unitsareavolumefrequencymass density (density)speed velocityangular velocityaccelerationangular accelerationforcepressure (mechanical stress)kinematic viscositydynamic viscositywork energy quantity of heatpowerentropyspecific heat capacitythermal conductivity

square metercubic meterhertzkilogram per cubic metermeter per secondradian per secondmeter per second squaredradian per second squarednewtonpascalsquare meter per secondnewton-second per square meter joulewattjoule per kelvinjoule per kilogram kelvinwatt per meter kelvin

m2

m3

HzKgm3

msradsms2

rads2

NPam2sN-sm2

JWJKJKg-K)W(m-K)

Quantity Name of Unit Symbol

Motion System Reference Tables11

Critical speed

Compression load

System stiffness

Life

Positioning accuracy

System stiffness

Drag torque

Load capacity

Stiffness

sndash1

Kg-ms2

Nm2

N-mJs or N-ms

ExamplesAn INCREASE ( ) in screw length results in a DECREASE ( ) in critical speed

An INCREASE in screw diameter results in an INCREASE in critical speed

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

11

MOTION SYSTEMREFERENCE TABLES

t 1 t 2 t 3 t dwell

vTime

vpk

a -a

MassMove distanceMove timeDwell timeMotion profileScrew speed limitLoad supportOrientationRotary to linear conversionAmbient temperature

9 Kg200 mm10 s05 s13 13 13 Trapezoidal1000 RPMμ = 001VerticalTFE lead screw 8 mm diameter 275 mm long30degC

ControlDrive Power Supply

4 quadrant with encoder32 VDC 35 Arms50 A peak

0333 s 0333 s 0333 s 05 s

vPK =3S2t =

(3)(02m)(2)(10 s) =

06m2 s =

03ms

Linear Velocity

Linear Acceleration

a = vf - vi

t = 03ms - 0 ms 0333 s 0901ms2 =

1st step is to thoroughly understand system constraints and motion profile requirements

Linear Motion ExampleData

Motion System Application Example12

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

12

MOTION SYSTEMAPPLICATION EXAMPLE

Lmin =60vPK

n =(60)(03ms)1000 revmin = 0018mrev = 18mmrev

Refer to Kerk screw chart ndash closest lead in an 8mm screw diameter is

2032mmrev = 002032 mrevη = 086 free wheeling nut TFE coated screw

JS = 388 x 10ndash 7 Kg-m2 Other critical lead screw considerations bull Column loading bull Critical speed

Jm aa

ωPKvPK

TF

= ωPK = 2 π vPK

L=

(2π)(03ms)002032 mrev

9276 rads

a = =2πaL

=(2π)(0901ms2)002032 mrev

2786 rads2

Linear to Rotary Conversion ndash Velocity Acceleration Inertia Torque

Minimum screw lead to maintain lt 1000 RPM shaft speed

Linear to Rotary Conversion Velocity

Linear to Rotary Conversion Acceleration

Motion System Application Example13

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

13

MOTION SYSTEMFORMULAS

Jin= m 2 πL 2

X η1 + JS= 9Kg

2 π002032 m 2

X 0861 + 388x10ndash 7Kg-m2

= 1095 X 10ndash 5 Kg-m2 + 388x10ndash 7 Kg-m2

= 1134 x 10ndash 5 Kg-m2

Τin= Τa+Τf +Τg+ΤD

Τa = Jin ain

Τg =sinOslashmgL

2 π η

= (1134 x 10ndash 5 Kg-m2)(2786 rads2)

= 00316 Nm

Τf =cosOslashmgμL

2 π η= (cos90)(9 Kg)(98ms2)(001)(002032m)

2 π (086)= 0 Nm

= (sin90)(9 Kg)(98ms2)(002032m)2 π (086)

=1792 Nm

5404 = 03316 Nm

ΤDSince a free-wheeling lead screw nut was used there is no preload force If an anti-backlash nut was used ΤD would be gt0 due to preload force Always reference manufacturerrsquos data for more information

= 0 Nm

Linear to Rotary Conversion Torque

Linear to Rotary Conversion Inertia

Motion System Application Example14

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

14

MOTION SYSTEMFORMULAS

t 1 t 2 t 3 t dwell Time0333 s 0333 s 0333 s 05 s

Rotary Motion Profile

ωPK

ω

= 9276 rads a = 2786 rads2

Τ1= Τa+Τf +Τg+ΤD

= (00316 Nm) + (0 Nm) + (03316 Nm) + (0 Nm)

= 03632 Nm

Τ2= Τa+Τf +Τg+ΤD

= (0 Nm) + (0 Nm) + (03316 Nm) + (0 Nm)

= 03316 Nm

Τ3= Τa+Τf +Τg+ΤD

= (00316 Nm) + (0 Nm) + (03316 Nm) + (0 Nm)

= 03632 Nm

Τ1

Τ2

Τ3

Motion System Application Example15

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

15

MOTION SYSTEMAPPLICATION EXAMPLE

Lead Screw Input Requirements

= 9276 rads= 2786 rads2

= 03632 Nm= 03316 Nm= ndash 03632 Nm= 0333 s= 0333 s= 0333 s= 05 s= 3369 W= 3076 W

ωPKaΤ1Τ2Τ3

t1t2t3tdwell

PPKPCV

PPK = (Τ1)(ωPK)

= (03632 Nm)(9276 rads)= 3369 W

PCV = (Τ2)(ωPK)

= (03316 Nm)(9276 rads)= 3076 W

RMS Torque Requirement Lead Screw Input

=

0333 s + 0333 s + 0333 s + 05 s

(03632 Nm)2(0333 s)+(03316 Nm)2(0333 s)+(- 03632 Nm)2(0333 s)

ΤRMS = Τ12t 1 + Τ2

2t2 + Τ32t 3

t 1 + t 2 + t 3 + tdwell

15

00439 + 00366 + 00439 = 00829 =

ΤRMS = 02879 Nm

Motion System Application Example16

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

16

MOTION SYSTEMAPPLICATION EXAMPLE

Load Parameters at the Lead Screw Shaft

= 9276 rads= 2786 rads2

= 03632 Nm= 03316 Nm= ndash 03632 Nm= 02879 Nm= 3369 W= 3076 W

ωPKaΤ1Τ2Τ3ΤRMSPPKPCV

Motor SelectionA direct-drive motor can be used however it would be large and expensive For example a DC057B-3 brush motor

would easily meet continuous torque and continuous speed as well as a 181 load-to-rotor inertia This motor however has significantly more output power than needed at 128 watts

A smaller motor with a transmission would optimize the system Look for a motor starting at a rated power around 40 watts Pittman DC040B-6

DC040B-6Reference VoltageContinuous TorqueRated CurrentRated PowerTorque ConstantVoltage ConstantTerminal ResistanceMax Winding TemperatureRotor Inertia

240 V00812 Nm236 A37 W0042 NmA0042 Vrads185 Ω155degC847 x 10ndash6 Kg-m2

Motor TransmissionLead Screw Support Rails

ΤRMS

PPK

ωPK

= 02879 Nm= 3369 W= 9276 rads

M

Other Motor Considerationsndash Type stepper brush BLDCndash Input speed (high input speed will result in high noise levels)ndash Total motor footprintndash Ambient temperaturendash Environmental conditionsndash Load-to-rotor inertiandash Encoder ready needed

Motion System Application Example17

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

17

MOTION SYSTEMAPPLICATION EXAMPLE

Gearbox Transmission Selection

When selecting a reduction ratio keep in mind that the RMS required torque at the motor shaft needs to fall below the continuous rated output torque of the motor

G30 A - Planetary GearboxMaximum output load 247 Nm

TRMS( Lead Screw)

02879 Nm

02879 Nm02879 Nm

Reduction

41

5161

Efficiency

090

090090

TRMS( Motor)

00798 Nm

00640 Nm00533 Nm

A 51 gearbox would allow about a 27 safety margin between what the system requires and the continuous torque output of the motor

System Summary 51η = 090

M

= 00807 Nm= 00737 Nm= ndash 00807 Nm= 00640 Nm= 4638 rads= 504 x 10 ndash 6 Kg-m2

= 3743 W

Τ1Τ2Τ3ΤRMS

PPK

ωPK

J

= 03632 Nm= 03316 Nm= ndash 03632 Nm= 02879 Nm= 9276 rads= 1134 x 10 ndash 5 Kg-m2

= 3369 W

Τ1Τ2Τ3ΤRMS

PPK

ωPK

J

Motion System Application Example18

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

18

MOTION SYSTEMAPPLICATION EXAMPLE

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

19

Motion Profile at the Motor

t 1 t 2 t 3 t dwell Time0333 s 0333 s 0333 s 05 s

ωΤ1

Τ2

Τ3= 4638 rads= 4429 RPM= ΤPK = 00807 Nm

ωPK

Τ1

nPK

Drive and Power Supply Requirements

Peak Current Required

IPK =TPK

KT+ IO =

00807 Nm0042 NmA

+ 0180 A = 210 A

RMS Current Required

IRMS =TRMS

KT+ IO =

00640 Nm0042 NmA

+ 0180 A = 170 A

Minimum bus Voltage Required

Vbus= IPKRmt+ ωPKKE = (210A)(185Ω) + (4638 rads)(0042 Vrads)= 389 V + 1948 V= 2337 V

0100 Nm 0200 Nm 0300 Nm 0400 Nm 0500 Nm 0600 Nm 0700 Nm

30 VDC

0100 Nm 0200 Nm 0300 Nm 0500 Nm 0600 Nm 0700 Nm

24 VDC

700060005000

RPM 4000300020001000

0400 Nm

TRMS

TPK

700060005000

RPM 4000300020001000

TRMS

TPK

Performance using 24V reference voltage

Performance using a 30V bus voltage

A 30 VDC bus will meet the application requirements as wells supply a margin of safety

MOTION SYSTEMAPPLICATION EXAMPLE

0100 Nm 0200 Nm 0300 Nm 0400 Nm 0500 Nm 0600 Nm 0700 Nm

30 VDC

0100 Nm 0200 Nm 0300 Nm 0500 Nm 0600 Nm 0700 Nm

24 VDC

700060005000

RPM 4000300020001000

0400 Nm

TRMS

TPK

700060005000

RPM 4000300020001000

TRMS

TPK

Ambient temperature Rated motor temperature Temperature rise Motor temperature

= 30degC= 155degC= 7638degC= 10638degC

Θa

Θm

ΘrΘrated

=Rth x IRMS

2 x Rmt

1 ndash (Rth x IRMS2 x Rmt x 000392degC)

Θr

=11degCω x 170A2

x 185Ω 1 ndash (11degCω x 170A2

x 185Ω x 000392degC)

= 5881 1 ndash (023)

= 5881 077

=

Θm = Θr + Θa = 7638degC + 30degC = 10638degC

7638degC

Performance using 24V reference voltage

Performance using a 30V bus voltage

Will the motor meet the temperature rise caused by the application

Motion System Application Example20

A 30 VDC bus will meet the application requirements as wells supply a margin of safety

203 756 7441 1500 Meriden Road Waterbury CT USA 06705

267 933 2105 534 Godshall DriveHarleysville PA USA 19438

wwwHaydonKerkPittmancom

copy All rights reserved AMETEK Advanced Motion Solutions No part of this doc-ument or technical information can be used reproduced or altered in any form without approval or proper authorization from AMETEK Inc and global affiliates

NOW ASK us the tough motion control questions ONLINE

ASK THE EXPERTS

HaydonKerkPittmancomwww

MOTION SYSTEMAPPLICATION EXAMPLE

Page 8: FORMULAS FOR MOTORIZED LINEAR MOTION SYSTEMS

Motor Power Relationships

ndash Watts lost due to winding resistance

Ploss = I2 x Rmt

ndash Output power at any point on the motor curve

Pout = ω x Tndash Motor maximum output power

Pmax = 025 x ωo x TLR

ndash Motor maximum output power (Theoretical)

Pmax = 025 x VT2

Rmt

Temperature Effects on Motor Constants

ndash Change in terminal resistance

Rmt(f) = Rmt(i) x [ 1 + αconductor (Θf ndash Θi) ]αconductor = 00040degC (copper)

K(f) = K(i) x [ 1 + αmagnet (Θf ndash Θi) ]ndash Change in torque constant and voltage constant ( KT = KE )

Magnetic Material

CeramicSmCoAlNiCoNdFeB

Θmax (degC)300degC300degC540degC150degC

The magnetic material values above represent average figures for particular material classes and estimating performance If exact values are needed consult data sheets for a specific magnet grade

Motion System Formulas8

αmagnet (degC)-00020degC-00004degC-00002degC-00012degC

In the case of a mechanically commutated motor with graphite brushes this analysis will result in a slight error because of the negative temperature coefficient of carbon For estimation purposes this can be ignored but for a more rigorous analysis the behavior of the brush material must be taken into account This is not an issue when evaluating a brushless dc motor

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

7

MOTION SYSTEMFORMULAS

Motion System Formulas9

For PMDC motors (brush type) brush drop can be approximated as a constant resistance connected in series with the armature winding This series resistance is included in the Rmt factor found on manufacturerrsquos data sheet

Copper graphiteSilver graphiteElectrographite

02 to 04 Ω 02 to 04 Ω 08 to 10 Ω

Motor Brush Resistance

Brush Material Resistance

DC Motor Thermal Resistance and Temperature Rise

ndash Thermal resistance

ndash Motor temperature rise

Θm = Θr + Θa

ndash OR ndash

Temperature Rise degCRth =

Machine Losses W

Θr = Rth x I2 x Rmt

Θr = Rth x I

2 x Rmt

1 ndash (Rth x I2 x Rmt x α)

ndash OR ndash

Θr = Rth x(TC

Km)

2

ndash Final motor temperature

Use caution when applying these formulas Depending on the conditions used to determine motor specifications only one formula should be used Contact Appications Engineering for questions

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

8

MOTION SYSTEMFORMULAS

Motion System Formulas9

For PMDC motors (brush type) brush drop can be approximated as a constant resistance connected in series with the armature winding This series resistance is included in the Rmt factor found on manufacturerrsquos data sheet

Copper graphiteSilver graphiteElectrographite

02 to 04 Ω 02 to 04 Ω 08 to 10 Ω

Motor Brush Resistance

Brush Material Resistance

DC Motor Thermal Resistance and Temperature Rise

ndash Thermal resistance

ndash Motor temperature rise

Θm = Θr + Θa

ndash OR ndash

Temperature Rise degCRth =

Machine Losses W

Θr = Rth x I2 x Rmt

Θr = Rth x I

2 x Rmt

1 ndash (Rth x I2 x Rmt x α)

ndash OR ndash

Θr = Rth x(TC

Km)

2

ndash Final motor temperature

Use caution when applying these formulas Depending on the conditions used to determine motor specifications only one formula should be used Contact Appications Engineering for questions

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

9

MOTION SYSTEMFORMULAS

Stepper Motor Formulas

Required ldquoTime Offrdquo using an LR drive (constant voltage drive)

180Step Oslash = (phases)(rotor teeth)

ndash Motor step angle

ndash RPM (revolutionsminute) to SPS (stepssecond)

Maximum accuracy of a step motor system has nothing to do with maximum resolution of a step motor system Maximum accuracy is always based on the accuracy of 1 full step but maximum resolution is based on all system components including lead screw encoder stepper motor and step mode on the controller

ndash SPS (stepssecond) to RPM (revolutionsminute)(SPS) x (Step Oslash)

RPM = 6

(RPM) x 6 SPS = (Step Oslash)

ndash Travel per step

mstep =L

positionsrev

ndash Overdriving capability

Required ldquoTime Offrdquo using a chopper drive (constant current drive)

Time Off = (Time On) (Applied Voltage)2

(Rated Voltage)ndash Time On

Time Off = (Time On) (Applied Current)2

(Rated Current)ndash Time On

It is not unusual for a customer to drive Haydon Kerk stepper motors beyond their rated power to obtain the most force in the smallest package size Precautions must be taken to prevent the motor from exceeding its maximum temperature The ldquoonrdquo time should not exceed 2 - 3 minutes As general rule of thumb driving a motor at 25 to 3x its rated voltage or current may result in wasted energy and erratic behavior due to saturation

Special notes for can-stack motors bull more easily saturated due to less active material (steel) bull more iron losses due to unlaminated steel

Motion System Formulas10

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

10

MOTION SYSTEMFORMULAS

Lead Screw Performance Characteristics

End Mounting Rigidity

Impact Impact

ScrewLength

Increase In Affects Increase In Affects

Critical speed

Compression load

Critical speed

Inertia

Compression load

Stiffness

Spring Rate

Load capacity

Drive torque

Angular velocity

Load cpacity

Positioning accuracy

ScrewDiameter

Lead

Load

Preload

Nut length

SI Unit Systems

LengthMassTimeElectric currentThermodynamic temperatureLuminous intensityAmount of substance

meterkilogramsecondamperekelvincandelamole

mKgsAKcdmol

Base Units

Derived Unitsareavolumefrequencymass density (density)speed velocityangular velocityaccelerationangular accelerationforcepressure (mechanical stress)kinematic viscositydynamic viscositywork energy quantity of heatpowerentropyspecific heat capacitythermal conductivity

square metercubic meterhertzkilogram per cubic metermeter per secondradian per secondmeter per second squaredradian per second squarednewtonpascalsquare meter per secondnewton-second per square meter joulewattjoule per kelvinjoule per kilogram kelvinwatt per meter kelvin

m2

m3

HzKgm3

msradsms2

rads2

NPam2sN-sm2

JWJKJKg-K)W(m-K)

Quantity Name of Unit Symbol

Motion System Reference Tables11

Critical speed

Compression load

System stiffness

Life

Positioning accuracy

System stiffness

Drag torque

Load capacity

Stiffness

sndash1

Kg-ms2

Nm2

N-mJs or N-ms

ExamplesAn INCREASE ( ) in screw length results in a DECREASE ( ) in critical speed

An INCREASE in screw diameter results in an INCREASE in critical speed

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11

MOTION SYSTEMREFERENCE TABLES

t 1 t 2 t 3 t dwell

vTime

vpk

a -a

MassMove distanceMove timeDwell timeMotion profileScrew speed limitLoad supportOrientationRotary to linear conversionAmbient temperature

9 Kg200 mm10 s05 s13 13 13 Trapezoidal1000 RPMμ = 001VerticalTFE lead screw 8 mm diameter 275 mm long30degC

ControlDrive Power Supply

4 quadrant with encoder32 VDC 35 Arms50 A peak

0333 s 0333 s 0333 s 05 s

vPK =3S2t =

(3)(02m)(2)(10 s) =

06m2 s =

03ms

Linear Velocity

Linear Acceleration

a = vf - vi

t = 03ms - 0 ms 0333 s 0901ms2 =

1st step is to thoroughly understand system constraints and motion profile requirements

Linear Motion ExampleData

Motion System Application Example12

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12

MOTION SYSTEMAPPLICATION EXAMPLE

Lmin =60vPK

n =(60)(03ms)1000 revmin = 0018mrev = 18mmrev

Refer to Kerk screw chart ndash closest lead in an 8mm screw diameter is

2032mmrev = 002032 mrevη = 086 free wheeling nut TFE coated screw

JS = 388 x 10ndash 7 Kg-m2 Other critical lead screw considerations bull Column loading bull Critical speed

Jm aa

ωPKvPK

TF

= ωPK = 2 π vPK

L=

(2π)(03ms)002032 mrev

9276 rads

a = =2πaL

=(2π)(0901ms2)002032 mrev

2786 rads2

Linear to Rotary Conversion ndash Velocity Acceleration Inertia Torque

Minimum screw lead to maintain lt 1000 RPM shaft speed

Linear to Rotary Conversion Velocity

Linear to Rotary Conversion Acceleration

Motion System Application Example13

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13

MOTION SYSTEMFORMULAS

Jin= m 2 πL 2

X η1 + JS= 9Kg

2 π002032 m 2

X 0861 + 388x10ndash 7Kg-m2

= 1095 X 10ndash 5 Kg-m2 + 388x10ndash 7 Kg-m2

= 1134 x 10ndash 5 Kg-m2

Τin= Τa+Τf +Τg+ΤD

Τa = Jin ain

Τg =sinOslashmgL

2 π η

= (1134 x 10ndash 5 Kg-m2)(2786 rads2)

= 00316 Nm

Τf =cosOslashmgμL

2 π η= (cos90)(9 Kg)(98ms2)(001)(002032m)

2 π (086)= 0 Nm

= (sin90)(9 Kg)(98ms2)(002032m)2 π (086)

=1792 Nm

5404 = 03316 Nm

ΤDSince a free-wheeling lead screw nut was used there is no preload force If an anti-backlash nut was used ΤD would be gt0 due to preload force Always reference manufacturerrsquos data for more information

= 0 Nm

Linear to Rotary Conversion Torque

Linear to Rotary Conversion Inertia

Motion System Application Example14

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14

MOTION SYSTEMFORMULAS

t 1 t 2 t 3 t dwell Time0333 s 0333 s 0333 s 05 s

Rotary Motion Profile

ωPK

ω

= 9276 rads a = 2786 rads2

Τ1= Τa+Τf +Τg+ΤD

= (00316 Nm) + (0 Nm) + (03316 Nm) + (0 Nm)

= 03632 Nm

Τ2= Τa+Τf +Τg+ΤD

= (0 Nm) + (0 Nm) + (03316 Nm) + (0 Nm)

= 03316 Nm

Τ3= Τa+Τf +Τg+ΤD

= (00316 Nm) + (0 Nm) + (03316 Nm) + (0 Nm)

= 03632 Nm

Τ1

Τ2

Τ3

Motion System Application Example15

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15

MOTION SYSTEMAPPLICATION EXAMPLE

Lead Screw Input Requirements

= 9276 rads= 2786 rads2

= 03632 Nm= 03316 Nm= ndash 03632 Nm= 0333 s= 0333 s= 0333 s= 05 s= 3369 W= 3076 W

ωPKaΤ1Τ2Τ3

t1t2t3tdwell

PPKPCV

PPK = (Τ1)(ωPK)

= (03632 Nm)(9276 rads)= 3369 W

PCV = (Τ2)(ωPK)

= (03316 Nm)(9276 rads)= 3076 W

RMS Torque Requirement Lead Screw Input

=

0333 s + 0333 s + 0333 s + 05 s

(03632 Nm)2(0333 s)+(03316 Nm)2(0333 s)+(- 03632 Nm)2(0333 s)

ΤRMS = Τ12t 1 + Τ2

2t2 + Τ32t 3

t 1 + t 2 + t 3 + tdwell

15

00439 + 00366 + 00439 = 00829 =

ΤRMS = 02879 Nm

Motion System Application Example16

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16

MOTION SYSTEMAPPLICATION EXAMPLE

Load Parameters at the Lead Screw Shaft

= 9276 rads= 2786 rads2

= 03632 Nm= 03316 Nm= ndash 03632 Nm= 02879 Nm= 3369 W= 3076 W

ωPKaΤ1Τ2Τ3ΤRMSPPKPCV

Motor SelectionA direct-drive motor can be used however it would be large and expensive For example a DC057B-3 brush motor

would easily meet continuous torque and continuous speed as well as a 181 load-to-rotor inertia This motor however has significantly more output power than needed at 128 watts

A smaller motor with a transmission would optimize the system Look for a motor starting at a rated power around 40 watts Pittman DC040B-6

DC040B-6Reference VoltageContinuous TorqueRated CurrentRated PowerTorque ConstantVoltage ConstantTerminal ResistanceMax Winding TemperatureRotor Inertia

240 V00812 Nm236 A37 W0042 NmA0042 Vrads185 Ω155degC847 x 10ndash6 Kg-m2

Motor TransmissionLead Screw Support Rails

ΤRMS

PPK

ωPK

= 02879 Nm= 3369 W= 9276 rads

M

Other Motor Considerationsndash Type stepper brush BLDCndash Input speed (high input speed will result in high noise levels)ndash Total motor footprintndash Ambient temperaturendash Environmental conditionsndash Load-to-rotor inertiandash Encoder ready needed

Motion System Application Example17

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17

MOTION SYSTEMAPPLICATION EXAMPLE

Gearbox Transmission Selection

When selecting a reduction ratio keep in mind that the RMS required torque at the motor shaft needs to fall below the continuous rated output torque of the motor

G30 A - Planetary GearboxMaximum output load 247 Nm

TRMS( Lead Screw)

02879 Nm

02879 Nm02879 Nm

Reduction

41

5161

Efficiency

090

090090

TRMS( Motor)

00798 Nm

00640 Nm00533 Nm

A 51 gearbox would allow about a 27 safety margin between what the system requires and the continuous torque output of the motor

System Summary 51η = 090

M

= 00807 Nm= 00737 Nm= ndash 00807 Nm= 00640 Nm= 4638 rads= 504 x 10 ndash 6 Kg-m2

= 3743 W

Τ1Τ2Τ3ΤRMS

PPK

ωPK

J

= 03632 Nm= 03316 Nm= ndash 03632 Nm= 02879 Nm= 9276 rads= 1134 x 10 ndash 5 Kg-m2

= 3369 W

Τ1Τ2Τ3ΤRMS

PPK

ωPK

J

Motion System Application Example18

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18

MOTION SYSTEMAPPLICATION EXAMPLE

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19

Motion Profile at the Motor

t 1 t 2 t 3 t dwell Time0333 s 0333 s 0333 s 05 s

ωΤ1

Τ2

Τ3= 4638 rads= 4429 RPM= ΤPK = 00807 Nm

ωPK

Τ1

nPK

Drive and Power Supply Requirements

Peak Current Required

IPK =TPK

KT+ IO =

00807 Nm0042 NmA

+ 0180 A = 210 A

RMS Current Required

IRMS =TRMS

KT+ IO =

00640 Nm0042 NmA

+ 0180 A = 170 A

Minimum bus Voltage Required

Vbus= IPKRmt+ ωPKKE = (210A)(185Ω) + (4638 rads)(0042 Vrads)= 389 V + 1948 V= 2337 V

0100 Nm 0200 Nm 0300 Nm 0400 Nm 0500 Nm 0600 Nm 0700 Nm

30 VDC

0100 Nm 0200 Nm 0300 Nm 0500 Nm 0600 Nm 0700 Nm

24 VDC

700060005000

RPM 4000300020001000

0400 Nm

TRMS

TPK

700060005000

RPM 4000300020001000

TRMS

TPK

Performance using 24V reference voltage

Performance using a 30V bus voltage

A 30 VDC bus will meet the application requirements as wells supply a margin of safety

MOTION SYSTEMAPPLICATION EXAMPLE

0100 Nm 0200 Nm 0300 Nm 0400 Nm 0500 Nm 0600 Nm 0700 Nm

30 VDC

0100 Nm 0200 Nm 0300 Nm 0500 Nm 0600 Nm 0700 Nm

24 VDC

700060005000

RPM 4000300020001000

0400 Nm

TRMS

TPK

700060005000

RPM 4000300020001000

TRMS

TPK

Ambient temperature Rated motor temperature Temperature rise Motor temperature

= 30degC= 155degC= 7638degC= 10638degC

Θa

Θm

ΘrΘrated

=Rth x IRMS

2 x Rmt

1 ndash (Rth x IRMS2 x Rmt x 000392degC)

Θr

=11degCω x 170A2

x 185Ω 1 ndash (11degCω x 170A2

x 185Ω x 000392degC)

= 5881 1 ndash (023)

= 5881 077

=

Θm = Θr + Θa = 7638degC + 30degC = 10638degC

7638degC

Performance using 24V reference voltage

Performance using a 30V bus voltage

Will the motor meet the temperature rise caused by the application

Motion System Application Example20

A 30 VDC bus will meet the application requirements as wells supply a margin of safety

203 756 7441 1500 Meriden Road Waterbury CT USA 06705

267 933 2105 534 Godshall DriveHarleysville PA USA 19438

wwwHaydonKerkPittmancom

copy All rights reserved AMETEK Advanced Motion Solutions No part of this doc-ument or technical information can be used reproduced or altered in any form without approval or proper authorization from AMETEK Inc and global affiliates

NOW ASK us the tough motion control questions ONLINE

ASK THE EXPERTS

HaydonKerkPittmancomwww

MOTION SYSTEMAPPLICATION EXAMPLE

Page 9: FORMULAS FOR MOTORIZED LINEAR MOTION SYSTEMS

Motion System Formulas9

For PMDC motors (brush type) brush drop can be approximated as a constant resistance connected in series with the armature winding This series resistance is included in the Rmt factor found on manufacturerrsquos data sheet

Copper graphiteSilver graphiteElectrographite

02 to 04 Ω 02 to 04 Ω 08 to 10 Ω

Motor Brush Resistance

Brush Material Resistance

DC Motor Thermal Resistance and Temperature Rise

ndash Thermal resistance

ndash Motor temperature rise

Θm = Θr + Θa

ndash OR ndash

Temperature Rise degCRth =

Machine Losses W

Θr = Rth x I2 x Rmt

Θr = Rth x I

2 x Rmt

1 ndash (Rth x I2 x Rmt x α)

ndash OR ndash

Θr = Rth x(TC

Km)

2

ndash Final motor temperature

Use caution when applying these formulas Depending on the conditions used to determine motor specifications only one formula should be used Contact Appications Engineering for questions

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8

MOTION SYSTEMFORMULAS

Motion System Formulas9

For PMDC motors (brush type) brush drop can be approximated as a constant resistance connected in series with the armature winding This series resistance is included in the Rmt factor found on manufacturerrsquos data sheet

Copper graphiteSilver graphiteElectrographite

02 to 04 Ω 02 to 04 Ω 08 to 10 Ω

Motor Brush Resistance

Brush Material Resistance

DC Motor Thermal Resistance and Temperature Rise

ndash Thermal resistance

ndash Motor temperature rise

Θm = Θr + Θa

ndash OR ndash

Temperature Rise degCRth =

Machine Losses W

Θr = Rth x I2 x Rmt

Θr = Rth x I

2 x Rmt

1 ndash (Rth x I2 x Rmt x α)

ndash OR ndash

Θr = Rth x(TC

Km)

2

ndash Final motor temperature

Use caution when applying these formulas Depending on the conditions used to determine motor specifications only one formula should be used Contact Appications Engineering for questions

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9

MOTION SYSTEMFORMULAS

Stepper Motor Formulas

Required ldquoTime Offrdquo using an LR drive (constant voltage drive)

180Step Oslash = (phases)(rotor teeth)

ndash Motor step angle

ndash RPM (revolutionsminute) to SPS (stepssecond)

Maximum accuracy of a step motor system has nothing to do with maximum resolution of a step motor system Maximum accuracy is always based on the accuracy of 1 full step but maximum resolution is based on all system components including lead screw encoder stepper motor and step mode on the controller

ndash SPS (stepssecond) to RPM (revolutionsminute)(SPS) x (Step Oslash)

RPM = 6

(RPM) x 6 SPS = (Step Oslash)

ndash Travel per step

mstep =L

positionsrev

ndash Overdriving capability

Required ldquoTime Offrdquo using a chopper drive (constant current drive)

Time Off = (Time On) (Applied Voltage)2

(Rated Voltage)ndash Time On

Time Off = (Time On) (Applied Current)2

(Rated Current)ndash Time On

It is not unusual for a customer to drive Haydon Kerk stepper motors beyond their rated power to obtain the most force in the smallest package size Precautions must be taken to prevent the motor from exceeding its maximum temperature The ldquoonrdquo time should not exceed 2 - 3 minutes As general rule of thumb driving a motor at 25 to 3x its rated voltage or current may result in wasted energy and erratic behavior due to saturation

Special notes for can-stack motors bull more easily saturated due to less active material (steel) bull more iron losses due to unlaminated steel

Motion System Formulas10

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10

MOTION SYSTEMFORMULAS

Lead Screw Performance Characteristics

End Mounting Rigidity

Impact Impact

ScrewLength

Increase In Affects Increase In Affects

Critical speed

Compression load

Critical speed

Inertia

Compression load

Stiffness

Spring Rate

Load capacity

Drive torque

Angular velocity

Load cpacity

Positioning accuracy

ScrewDiameter

Lead

Load

Preload

Nut length

SI Unit Systems

LengthMassTimeElectric currentThermodynamic temperatureLuminous intensityAmount of substance

meterkilogramsecondamperekelvincandelamole

mKgsAKcdmol

Base Units

Derived Unitsareavolumefrequencymass density (density)speed velocityangular velocityaccelerationangular accelerationforcepressure (mechanical stress)kinematic viscositydynamic viscositywork energy quantity of heatpowerentropyspecific heat capacitythermal conductivity

square metercubic meterhertzkilogram per cubic metermeter per secondradian per secondmeter per second squaredradian per second squarednewtonpascalsquare meter per secondnewton-second per square meter joulewattjoule per kelvinjoule per kilogram kelvinwatt per meter kelvin

m2

m3

HzKgm3

msradsms2

rads2

NPam2sN-sm2

JWJKJKg-K)W(m-K)

Quantity Name of Unit Symbol

Motion System Reference Tables11

Critical speed

Compression load

System stiffness

Life

Positioning accuracy

System stiffness

Drag torque

Load capacity

Stiffness

sndash1

Kg-ms2

Nm2

N-mJs or N-ms

ExamplesAn INCREASE ( ) in screw length results in a DECREASE ( ) in critical speed

An INCREASE in screw diameter results in an INCREASE in critical speed

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

11

MOTION SYSTEMREFERENCE TABLES

t 1 t 2 t 3 t dwell

vTime

vpk

a -a

MassMove distanceMove timeDwell timeMotion profileScrew speed limitLoad supportOrientationRotary to linear conversionAmbient temperature

9 Kg200 mm10 s05 s13 13 13 Trapezoidal1000 RPMμ = 001VerticalTFE lead screw 8 mm diameter 275 mm long30degC

ControlDrive Power Supply

4 quadrant with encoder32 VDC 35 Arms50 A peak

0333 s 0333 s 0333 s 05 s

vPK =3S2t =

(3)(02m)(2)(10 s) =

06m2 s =

03ms

Linear Velocity

Linear Acceleration

a = vf - vi

t = 03ms - 0 ms 0333 s 0901ms2 =

1st step is to thoroughly understand system constraints and motion profile requirements

Linear Motion ExampleData

Motion System Application Example12

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

12

MOTION SYSTEMAPPLICATION EXAMPLE

Lmin =60vPK

n =(60)(03ms)1000 revmin = 0018mrev = 18mmrev

Refer to Kerk screw chart ndash closest lead in an 8mm screw diameter is

2032mmrev = 002032 mrevη = 086 free wheeling nut TFE coated screw

JS = 388 x 10ndash 7 Kg-m2 Other critical lead screw considerations bull Column loading bull Critical speed

Jm aa

ωPKvPK

TF

= ωPK = 2 π vPK

L=

(2π)(03ms)002032 mrev

9276 rads

a = =2πaL

=(2π)(0901ms2)002032 mrev

2786 rads2

Linear to Rotary Conversion ndash Velocity Acceleration Inertia Torque

Minimum screw lead to maintain lt 1000 RPM shaft speed

Linear to Rotary Conversion Velocity

Linear to Rotary Conversion Acceleration

Motion System Application Example13

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13

MOTION SYSTEMFORMULAS

Jin= m 2 πL 2

X η1 + JS= 9Kg

2 π002032 m 2

X 0861 + 388x10ndash 7Kg-m2

= 1095 X 10ndash 5 Kg-m2 + 388x10ndash 7 Kg-m2

= 1134 x 10ndash 5 Kg-m2

Τin= Τa+Τf +Τg+ΤD

Τa = Jin ain

Τg =sinOslashmgL

2 π η

= (1134 x 10ndash 5 Kg-m2)(2786 rads2)

= 00316 Nm

Τf =cosOslashmgμL

2 π η= (cos90)(9 Kg)(98ms2)(001)(002032m)

2 π (086)= 0 Nm

= (sin90)(9 Kg)(98ms2)(002032m)2 π (086)

=1792 Nm

5404 = 03316 Nm

ΤDSince a free-wheeling lead screw nut was used there is no preload force If an anti-backlash nut was used ΤD would be gt0 due to preload force Always reference manufacturerrsquos data for more information

= 0 Nm

Linear to Rotary Conversion Torque

Linear to Rotary Conversion Inertia

Motion System Application Example14

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14

MOTION SYSTEMFORMULAS

t 1 t 2 t 3 t dwell Time0333 s 0333 s 0333 s 05 s

Rotary Motion Profile

ωPK

ω

= 9276 rads a = 2786 rads2

Τ1= Τa+Τf +Τg+ΤD

= (00316 Nm) + (0 Nm) + (03316 Nm) + (0 Nm)

= 03632 Nm

Τ2= Τa+Τf +Τg+ΤD

= (0 Nm) + (0 Nm) + (03316 Nm) + (0 Nm)

= 03316 Nm

Τ3= Τa+Τf +Τg+ΤD

= (00316 Nm) + (0 Nm) + (03316 Nm) + (0 Nm)

= 03632 Nm

Τ1

Τ2

Τ3

Motion System Application Example15

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

15

MOTION SYSTEMAPPLICATION EXAMPLE

Lead Screw Input Requirements

= 9276 rads= 2786 rads2

= 03632 Nm= 03316 Nm= ndash 03632 Nm= 0333 s= 0333 s= 0333 s= 05 s= 3369 W= 3076 W

ωPKaΤ1Τ2Τ3

t1t2t3tdwell

PPKPCV

PPK = (Τ1)(ωPK)

= (03632 Nm)(9276 rads)= 3369 W

PCV = (Τ2)(ωPK)

= (03316 Nm)(9276 rads)= 3076 W

RMS Torque Requirement Lead Screw Input

=

0333 s + 0333 s + 0333 s + 05 s

(03632 Nm)2(0333 s)+(03316 Nm)2(0333 s)+(- 03632 Nm)2(0333 s)

ΤRMS = Τ12t 1 + Τ2

2t2 + Τ32t 3

t 1 + t 2 + t 3 + tdwell

15

00439 + 00366 + 00439 = 00829 =

ΤRMS = 02879 Nm

Motion System Application Example16

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

16

MOTION SYSTEMAPPLICATION EXAMPLE

Load Parameters at the Lead Screw Shaft

= 9276 rads= 2786 rads2

= 03632 Nm= 03316 Nm= ndash 03632 Nm= 02879 Nm= 3369 W= 3076 W

ωPKaΤ1Τ2Τ3ΤRMSPPKPCV

Motor SelectionA direct-drive motor can be used however it would be large and expensive For example a DC057B-3 brush motor

would easily meet continuous torque and continuous speed as well as a 181 load-to-rotor inertia This motor however has significantly more output power than needed at 128 watts

A smaller motor with a transmission would optimize the system Look for a motor starting at a rated power around 40 watts Pittman DC040B-6

DC040B-6Reference VoltageContinuous TorqueRated CurrentRated PowerTorque ConstantVoltage ConstantTerminal ResistanceMax Winding TemperatureRotor Inertia

240 V00812 Nm236 A37 W0042 NmA0042 Vrads185 Ω155degC847 x 10ndash6 Kg-m2

Motor TransmissionLead Screw Support Rails

ΤRMS

PPK

ωPK

= 02879 Nm= 3369 W= 9276 rads

M

Other Motor Considerationsndash Type stepper brush BLDCndash Input speed (high input speed will result in high noise levels)ndash Total motor footprintndash Ambient temperaturendash Environmental conditionsndash Load-to-rotor inertiandash Encoder ready needed

Motion System Application Example17

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

17

MOTION SYSTEMAPPLICATION EXAMPLE

Gearbox Transmission Selection

When selecting a reduction ratio keep in mind that the RMS required torque at the motor shaft needs to fall below the continuous rated output torque of the motor

G30 A - Planetary GearboxMaximum output load 247 Nm

TRMS( Lead Screw)

02879 Nm

02879 Nm02879 Nm

Reduction

41

5161

Efficiency

090

090090

TRMS( Motor)

00798 Nm

00640 Nm00533 Nm

A 51 gearbox would allow about a 27 safety margin between what the system requires and the continuous torque output of the motor

System Summary 51η = 090

M

= 00807 Nm= 00737 Nm= ndash 00807 Nm= 00640 Nm= 4638 rads= 504 x 10 ndash 6 Kg-m2

= 3743 W

Τ1Τ2Τ3ΤRMS

PPK

ωPK

J

= 03632 Nm= 03316 Nm= ndash 03632 Nm= 02879 Nm= 9276 rads= 1134 x 10 ndash 5 Kg-m2

= 3369 W

Τ1Τ2Τ3ΤRMS

PPK

ωPK

J

Motion System Application Example18

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

18

MOTION SYSTEMAPPLICATION EXAMPLE

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

19

Motion Profile at the Motor

t 1 t 2 t 3 t dwell Time0333 s 0333 s 0333 s 05 s

ωΤ1

Τ2

Τ3= 4638 rads= 4429 RPM= ΤPK = 00807 Nm

ωPK

Τ1

nPK

Drive and Power Supply Requirements

Peak Current Required

IPK =TPK

KT+ IO =

00807 Nm0042 NmA

+ 0180 A = 210 A

RMS Current Required

IRMS =TRMS

KT+ IO =

00640 Nm0042 NmA

+ 0180 A = 170 A

Minimum bus Voltage Required

Vbus= IPKRmt+ ωPKKE = (210A)(185Ω) + (4638 rads)(0042 Vrads)= 389 V + 1948 V= 2337 V

0100 Nm 0200 Nm 0300 Nm 0400 Nm 0500 Nm 0600 Nm 0700 Nm

30 VDC

0100 Nm 0200 Nm 0300 Nm 0500 Nm 0600 Nm 0700 Nm

24 VDC

700060005000

RPM 4000300020001000

0400 Nm

TRMS

TPK

700060005000

RPM 4000300020001000

TRMS

TPK

Performance using 24V reference voltage

Performance using a 30V bus voltage

A 30 VDC bus will meet the application requirements as wells supply a margin of safety

MOTION SYSTEMAPPLICATION EXAMPLE

0100 Nm 0200 Nm 0300 Nm 0400 Nm 0500 Nm 0600 Nm 0700 Nm

30 VDC

0100 Nm 0200 Nm 0300 Nm 0500 Nm 0600 Nm 0700 Nm

24 VDC

700060005000

RPM 4000300020001000

0400 Nm

TRMS

TPK

700060005000

RPM 4000300020001000

TRMS

TPK

Ambient temperature Rated motor temperature Temperature rise Motor temperature

= 30degC= 155degC= 7638degC= 10638degC

Θa

Θm

ΘrΘrated

=Rth x IRMS

2 x Rmt

1 ndash (Rth x IRMS2 x Rmt x 000392degC)

Θr

=11degCω x 170A2

x 185Ω 1 ndash (11degCω x 170A2

x 185Ω x 000392degC)

= 5881 1 ndash (023)

= 5881 077

=

Θm = Θr + Θa = 7638degC + 30degC = 10638degC

7638degC

Performance using 24V reference voltage

Performance using a 30V bus voltage

Will the motor meet the temperature rise caused by the application

Motion System Application Example20

A 30 VDC bus will meet the application requirements as wells supply a margin of safety

203 756 7441 1500 Meriden Road Waterbury CT USA 06705

267 933 2105 534 Godshall DriveHarleysville PA USA 19438

wwwHaydonKerkPittmancom

copy All rights reserved AMETEK Advanced Motion Solutions No part of this doc-ument or technical information can be used reproduced or altered in any form without approval or proper authorization from AMETEK Inc and global affiliates

NOW ASK us the tough motion control questions ONLINE

ASK THE EXPERTS

HaydonKerkPittmancomwww

MOTION SYSTEMAPPLICATION EXAMPLE

Page 10: FORMULAS FOR MOTORIZED LINEAR MOTION SYSTEMS

Motion System Formulas9

For PMDC motors (brush type) brush drop can be approximated as a constant resistance connected in series with the armature winding This series resistance is included in the Rmt factor found on manufacturerrsquos data sheet

Copper graphiteSilver graphiteElectrographite

02 to 04 Ω 02 to 04 Ω 08 to 10 Ω

Motor Brush Resistance

Brush Material Resistance

DC Motor Thermal Resistance and Temperature Rise

ndash Thermal resistance

ndash Motor temperature rise

Θm = Θr + Θa

ndash OR ndash

Temperature Rise degCRth =

Machine Losses W

Θr = Rth x I2 x Rmt

Θr = Rth x I

2 x Rmt

1 ndash (Rth x I2 x Rmt x α)

ndash OR ndash

Θr = Rth x(TC

Km)

2

ndash Final motor temperature

Use caution when applying these formulas Depending on the conditions used to determine motor specifications only one formula should be used Contact Appications Engineering for questions

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

9

MOTION SYSTEMFORMULAS

Stepper Motor Formulas

Required ldquoTime Offrdquo using an LR drive (constant voltage drive)

180Step Oslash = (phases)(rotor teeth)

ndash Motor step angle

ndash RPM (revolutionsminute) to SPS (stepssecond)

Maximum accuracy of a step motor system has nothing to do with maximum resolution of a step motor system Maximum accuracy is always based on the accuracy of 1 full step but maximum resolution is based on all system components including lead screw encoder stepper motor and step mode on the controller

ndash SPS (stepssecond) to RPM (revolutionsminute)(SPS) x (Step Oslash)

RPM = 6

(RPM) x 6 SPS = (Step Oslash)

ndash Travel per step

mstep =L

positionsrev

ndash Overdriving capability

Required ldquoTime Offrdquo using a chopper drive (constant current drive)

Time Off = (Time On) (Applied Voltage)2

(Rated Voltage)ndash Time On

Time Off = (Time On) (Applied Current)2

(Rated Current)ndash Time On

It is not unusual for a customer to drive Haydon Kerk stepper motors beyond their rated power to obtain the most force in the smallest package size Precautions must be taken to prevent the motor from exceeding its maximum temperature The ldquoonrdquo time should not exceed 2 - 3 minutes As general rule of thumb driving a motor at 25 to 3x its rated voltage or current may result in wasted energy and erratic behavior due to saturation

Special notes for can-stack motors bull more easily saturated due to less active material (steel) bull more iron losses due to unlaminated steel

Motion System Formulas10

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

10

MOTION SYSTEMFORMULAS

Lead Screw Performance Characteristics

End Mounting Rigidity

Impact Impact

ScrewLength

Increase In Affects Increase In Affects

Critical speed

Compression load

Critical speed

Inertia

Compression load

Stiffness

Spring Rate

Load capacity

Drive torque

Angular velocity

Load cpacity

Positioning accuracy

ScrewDiameter

Lead

Load

Preload

Nut length

SI Unit Systems

LengthMassTimeElectric currentThermodynamic temperatureLuminous intensityAmount of substance

meterkilogramsecondamperekelvincandelamole

mKgsAKcdmol

Base Units

Derived Unitsareavolumefrequencymass density (density)speed velocityangular velocityaccelerationangular accelerationforcepressure (mechanical stress)kinematic viscositydynamic viscositywork energy quantity of heatpowerentropyspecific heat capacitythermal conductivity

square metercubic meterhertzkilogram per cubic metermeter per secondradian per secondmeter per second squaredradian per second squarednewtonpascalsquare meter per secondnewton-second per square meter joulewattjoule per kelvinjoule per kilogram kelvinwatt per meter kelvin

m2

m3

HzKgm3

msradsms2

rads2

NPam2sN-sm2

JWJKJKg-K)W(m-K)

Quantity Name of Unit Symbol

Motion System Reference Tables11

Critical speed

Compression load

System stiffness

Life

Positioning accuracy

System stiffness

Drag torque

Load capacity

Stiffness

sndash1

Kg-ms2

Nm2

N-mJs or N-ms

ExamplesAn INCREASE ( ) in screw length results in a DECREASE ( ) in critical speed

An INCREASE in screw diameter results in an INCREASE in critical speed

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

11

MOTION SYSTEMREFERENCE TABLES

t 1 t 2 t 3 t dwell

vTime

vpk

a -a

MassMove distanceMove timeDwell timeMotion profileScrew speed limitLoad supportOrientationRotary to linear conversionAmbient temperature

9 Kg200 mm10 s05 s13 13 13 Trapezoidal1000 RPMμ = 001VerticalTFE lead screw 8 mm diameter 275 mm long30degC

ControlDrive Power Supply

4 quadrant with encoder32 VDC 35 Arms50 A peak

0333 s 0333 s 0333 s 05 s

vPK =3S2t =

(3)(02m)(2)(10 s) =

06m2 s =

03ms

Linear Velocity

Linear Acceleration

a = vf - vi

t = 03ms - 0 ms 0333 s 0901ms2 =

1st step is to thoroughly understand system constraints and motion profile requirements

Linear Motion ExampleData

Motion System Application Example12

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

12

MOTION SYSTEMAPPLICATION EXAMPLE

Lmin =60vPK

n =(60)(03ms)1000 revmin = 0018mrev = 18mmrev

Refer to Kerk screw chart ndash closest lead in an 8mm screw diameter is

2032mmrev = 002032 mrevη = 086 free wheeling nut TFE coated screw

JS = 388 x 10ndash 7 Kg-m2 Other critical lead screw considerations bull Column loading bull Critical speed

Jm aa

ωPKvPK

TF

= ωPK = 2 π vPK

L=

(2π)(03ms)002032 mrev

9276 rads

a = =2πaL

=(2π)(0901ms2)002032 mrev

2786 rads2

Linear to Rotary Conversion ndash Velocity Acceleration Inertia Torque

Minimum screw lead to maintain lt 1000 RPM shaft speed

Linear to Rotary Conversion Velocity

Linear to Rotary Conversion Acceleration

Motion System Application Example13

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

13

MOTION SYSTEMFORMULAS

Jin= m 2 πL 2

X η1 + JS= 9Kg

2 π002032 m 2

X 0861 + 388x10ndash 7Kg-m2

= 1095 X 10ndash 5 Kg-m2 + 388x10ndash 7 Kg-m2

= 1134 x 10ndash 5 Kg-m2

Τin= Τa+Τf +Τg+ΤD

Τa = Jin ain

Τg =sinOslashmgL

2 π η

= (1134 x 10ndash 5 Kg-m2)(2786 rads2)

= 00316 Nm

Τf =cosOslashmgμL

2 π η= (cos90)(9 Kg)(98ms2)(001)(002032m)

2 π (086)= 0 Nm

= (sin90)(9 Kg)(98ms2)(002032m)2 π (086)

=1792 Nm

5404 = 03316 Nm

ΤDSince a free-wheeling lead screw nut was used there is no preload force If an anti-backlash nut was used ΤD would be gt0 due to preload force Always reference manufacturerrsquos data for more information

= 0 Nm

Linear to Rotary Conversion Torque

Linear to Rotary Conversion Inertia

Motion System Application Example14

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

14

MOTION SYSTEMFORMULAS

t 1 t 2 t 3 t dwell Time0333 s 0333 s 0333 s 05 s

Rotary Motion Profile

ωPK

ω

= 9276 rads a = 2786 rads2

Τ1= Τa+Τf +Τg+ΤD

= (00316 Nm) + (0 Nm) + (03316 Nm) + (0 Nm)

= 03632 Nm

Τ2= Τa+Τf +Τg+ΤD

= (0 Nm) + (0 Nm) + (03316 Nm) + (0 Nm)

= 03316 Nm

Τ3= Τa+Τf +Τg+ΤD

= (00316 Nm) + (0 Nm) + (03316 Nm) + (0 Nm)

= 03632 Nm

Τ1

Τ2

Τ3

Motion System Application Example15

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

15

MOTION SYSTEMAPPLICATION EXAMPLE

Lead Screw Input Requirements

= 9276 rads= 2786 rads2

= 03632 Nm= 03316 Nm= ndash 03632 Nm= 0333 s= 0333 s= 0333 s= 05 s= 3369 W= 3076 W

ωPKaΤ1Τ2Τ3

t1t2t3tdwell

PPKPCV

PPK = (Τ1)(ωPK)

= (03632 Nm)(9276 rads)= 3369 W

PCV = (Τ2)(ωPK)

= (03316 Nm)(9276 rads)= 3076 W

RMS Torque Requirement Lead Screw Input

=

0333 s + 0333 s + 0333 s + 05 s

(03632 Nm)2(0333 s)+(03316 Nm)2(0333 s)+(- 03632 Nm)2(0333 s)

ΤRMS = Τ12t 1 + Τ2

2t2 + Τ32t 3

t 1 + t 2 + t 3 + tdwell

15

00439 + 00366 + 00439 = 00829 =

ΤRMS = 02879 Nm

Motion System Application Example16

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

16

MOTION SYSTEMAPPLICATION EXAMPLE

Load Parameters at the Lead Screw Shaft

= 9276 rads= 2786 rads2

= 03632 Nm= 03316 Nm= ndash 03632 Nm= 02879 Nm= 3369 W= 3076 W

ωPKaΤ1Τ2Τ3ΤRMSPPKPCV

Motor SelectionA direct-drive motor can be used however it would be large and expensive For example a DC057B-3 brush motor

would easily meet continuous torque and continuous speed as well as a 181 load-to-rotor inertia This motor however has significantly more output power than needed at 128 watts

A smaller motor with a transmission would optimize the system Look for a motor starting at a rated power around 40 watts Pittman DC040B-6

DC040B-6Reference VoltageContinuous TorqueRated CurrentRated PowerTorque ConstantVoltage ConstantTerminal ResistanceMax Winding TemperatureRotor Inertia

240 V00812 Nm236 A37 W0042 NmA0042 Vrads185 Ω155degC847 x 10ndash6 Kg-m2

Motor TransmissionLead Screw Support Rails

ΤRMS

PPK

ωPK

= 02879 Nm= 3369 W= 9276 rads

M

Other Motor Considerationsndash Type stepper brush BLDCndash Input speed (high input speed will result in high noise levels)ndash Total motor footprintndash Ambient temperaturendash Environmental conditionsndash Load-to-rotor inertiandash Encoder ready needed

Motion System Application Example17

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

17

MOTION SYSTEMAPPLICATION EXAMPLE

Gearbox Transmission Selection

When selecting a reduction ratio keep in mind that the RMS required torque at the motor shaft needs to fall below the continuous rated output torque of the motor

G30 A - Planetary GearboxMaximum output load 247 Nm

TRMS( Lead Screw)

02879 Nm

02879 Nm02879 Nm

Reduction

41

5161

Efficiency

090

090090

TRMS( Motor)

00798 Nm

00640 Nm00533 Nm

A 51 gearbox would allow about a 27 safety margin between what the system requires and the continuous torque output of the motor

System Summary 51η = 090

M

= 00807 Nm= 00737 Nm= ndash 00807 Nm= 00640 Nm= 4638 rads= 504 x 10 ndash 6 Kg-m2

= 3743 W

Τ1Τ2Τ3ΤRMS

PPK

ωPK

J

= 03632 Nm= 03316 Nm= ndash 03632 Nm= 02879 Nm= 9276 rads= 1134 x 10 ndash 5 Kg-m2

= 3369 W

Τ1Τ2Τ3ΤRMS

PPK

ωPK

J

Motion System Application Example18

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

18

MOTION SYSTEMAPPLICATION EXAMPLE

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

19

Motion Profile at the Motor

t 1 t 2 t 3 t dwell Time0333 s 0333 s 0333 s 05 s

ωΤ1

Τ2

Τ3= 4638 rads= 4429 RPM= ΤPK = 00807 Nm

ωPK

Τ1

nPK

Drive and Power Supply Requirements

Peak Current Required

IPK =TPK

KT+ IO =

00807 Nm0042 NmA

+ 0180 A = 210 A

RMS Current Required

IRMS =TRMS

KT+ IO =

00640 Nm0042 NmA

+ 0180 A = 170 A

Minimum bus Voltage Required

Vbus= IPKRmt+ ωPKKE = (210A)(185Ω) + (4638 rads)(0042 Vrads)= 389 V + 1948 V= 2337 V

0100 Nm 0200 Nm 0300 Nm 0400 Nm 0500 Nm 0600 Nm 0700 Nm

30 VDC

0100 Nm 0200 Nm 0300 Nm 0500 Nm 0600 Nm 0700 Nm

24 VDC

700060005000

RPM 4000300020001000

0400 Nm

TRMS

TPK

700060005000

RPM 4000300020001000

TRMS

TPK

Performance using 24V reference voltage

Performance using a 30V bus voltage

A 30 VDC bus will meet the application requirements as wells supply a margin of safety

MOTION SYSTEMAPPLICATION EXAMPLE

0100 Nm 0200 Nm 0300 Nm 0400 Nm 0500 Nm 0600 Nm 0700 Nm

30 VDC

0100 Nm 0200 Nm 0300 Nm 0500 Nm 0600 Nm 0700 Nm

24 VDC

700060005000

RPM 4000300020001000

0400 Nm

TRMS

TPK

700060005000

RPM 4000300020001000

TRMS

TPK

Ambient temperature Rated motor temperature Temperature rise Motor temperature

= 30degC= 155degC= 7638degC= 10638degC

Θa

Θm

ΘrΘrated

=Rth x IRMS

2 x Rmt

1 ndash (Rth x IRMS2 x Rmt x 000392degC)

Θr

=11degCω x 170A2

x 185Ω 1 ndash (11degCω x 170A2

x 185Ω x 000392degC)

= 5881 1 ndash (023)

= 5881 077

=

Θm = Θr + Θa = 7638degC + 30degC = 10638degC

7638degC

Performance using 24V reference voltage

Performance using a 30V bus voltage

Will the motor meet the temperature rise caused by the application

Motion System Application Example20

A 30 VDC bus will meet the application requirements as wells supply a margin of safety

203 756 7441 1500 Meriden Road Waterbury CT USA 06705

267 933 2105 534 Godshall DriveHarleysville PA USA 19438

wwwHaydonKerkPittmancom

copy All rights reserved AMETEK Advanced Motion Solutions No part of this doc-ument or technical information can be used reproduced or altered in any form without approval or proper authorization from AMETEK Inc and global affiliates

NOW ASK us the tough motion control questions ONLINE

ASK THE EXPERTS

HaydonKerkPittmancomwww

MOTION SYSTEMAPPLICATION EXAMPLE

Page 11: FORMULAS FOR MOTORIZED LINEAR MOTION SYSTEMS

Stepper Motor Formulas

Required ldquoTime Offrdquo using an LR drive (constant voltage drive)

180Step Oslash = (phases)(rotor teeth)

ndash Motor step angle

ndash RPM (revolutionsminute) to SPS (stepssecond)

Maximum accuracy of a step motor system has nothing to do with maximum resolution of a step motor system Maximum accuracy is always based on the accuracy of 1 full step but maximum resolution is based on all system components including lead screw encoder stepper motor and step mode on the controller

ndash SPS (stepssecond) to RPM (revolutionsminute)(SPS) x (Step Oslash)

RPM = 6

(RPM) x 6 SPS = (Step Oslash)

ndash Travel per step

mstep =L

positionsrev

ndash Overdriving capability

Required ldquoTime Offrdquo using a chopper drive (constant current drive)

Time Off = (Time On) (Applied Voltage)2

(Rated Voltage)ndash Time On

Time Off = (Time On) (Applied Current)2

(Rated Current)ndash Time On

It is not unusual for a customer to drive Haydon Kerk stepper motors beyond their rated power to obtain the most force in the smallest package size Precautions must be taken to prevent the motor from exceeding its maximum temperature The ldquoonrdquo time should not exceed 2 - 3 minutes As general rule of thumb driving a motor at 25 to 3x its rated voltage or current may result in wasted energy and erratic behavior due to saturation

Special notes for can-stack motors bull more easily saturated due to less active material (steel) bull more iron losses due to unlaminated steel

Motion System Formulas10

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

10

MOTION SYSTEMFORMULAS

Lead Screw Performance Characteristics

End Mounting Rigidity

Impact Impact

ScrewLength

Increase In Affects Increase In Affects

Critical speed

Compression load

Critical speed

Inertia

Compression load

Stiffness

Spring Rate

Load capacity

Drive torque

Angular velocity

Load cpacity

Positioning accuracy

ScrewDiameter

Lead

Load

Preload

Nut length

SI Unit Systems

LengthMassTimeElectric currentThermodynamic temperatureLuminous intensityAmount of substance

meterkilogramsecondamperekelvincandelamole

mKgsAKcdmol

Base Units

Derived Unitsareavolumefrequencymass density (density)speed velocityangular velocityaccelerationangular accelerationforcepressure (mechanical stress)kinematic viscositydynamic viscositywork energy quantity of heatpowerentropyspecific heat capacitythermal conductivity

square metercubic meterhertzkilogram per cubic metermeter per secondradian per secondmeter per second squaredradian per second squarednewtonpascalsquare meter per secondnewton-second per square meter joulewattjoule per kelvinjoule per kilogram kelvinwatt per meter kelvin

m2

m3

HzKgm3

msradsms2

rads2

NPam2sN-sm2

JWJKJKg-K)W(m-K)

Quantity Name of Unit Symbol

Motion System Reference Tables11

Critical speed

Compression load

System stiffness

Life

Positioning accuracy

System stiffness

Drag torque

Load capacity

Stiffness

sndash1

Kg-ms2

Nm2

N-mJs or N-ms

ExamplesAn INCREASE ( ) in screw length results in a DECREASE ( ) in critical speed

An INCREASE in screw diameter results in an INCREASE in critical speed

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

11

MOTION SYSTEMREFERENCE TABLES

t 1 t 2 t 3 t dwell

vTime

vpk

a -a

MassMove distanceMove timeDwell timeMotion profileScrew speed limitLoad supportOrientationRotary to linear conversionAmbient temperature

9 Kg200 mm10 s05 s13 13 13 Trapezoidal1000 RPMμ = 001VerticalTFE lead screw 8 mm diameter 275 mm long30degC

ControlDrive Power Supply

4 quadrant with encoder32 VDC 35 Arms50 A peak

0333 s 0333 s 0333 s 05 s

vPK =3S2t =

(3)(02m)(2)(10 s) =

06m2 s =

03ms

Linear Velocity

Linear Acceleration

a = vf - vi

t = 03ms - 0 ms 0333 s 0901ms2 =

1st step is to thoroughly understand system constraints and motion profile requirements

Linear Motion ExampleData

Motion System Application Example12

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

12

MOTION SYSTEMAPPLICATION EXAMPLE

Lmin =60vPK

n =(60)(03ms)1000 revmin = 0018mrev = 18mmrev

Refer to Kerk screw chart ndash closest lead in an 8mm screw diameter is

2032mmrev = 002032 mrevη = 086 free wheeling nut TFE coated screw

JS = 388 x 10ndash 7 Kg-m2 Other critical lead screw considerations bull Column loading bull Critical speed

Jm aa

ωPKvPK

TF

= ωPK = 2 π vPK

L=

(2π)(03ms)002032 mrev

9276 rads

a = =2πaL

=(2π)(0901ms2)002032 mrev

2786 rads2

Linear to Rotary Conversion ndash Velocity Acceleration Inertia Torque

Minimum screw lead to maintain lt 1000 RPM shaft speed

Linear to Rotary Conversion Velocity

Linear to Rotary Conversion Acceleration

Motion System Application Example13

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

13

MOTION SYSTEMFORMULAS

Jin= m 2 πL 2

X η1 + JS= 9Kg

2 π002032 m 2

X 0861 + 388x10ndash 7Kg-m2

= 1095 X 10ndash 5 Kg-m2 + 388x10ndash 7 Kg-m2

= 1134 x 10ndash 5 Kg-m2

Τin= Τa+Τf +Τg+ΤD

Τa = Jin ain

Τg =sinOslashmgL

2 π η

= (1134 x 10ndash 5 Kg-m2)(2786 rads2)

= 00316 Nm

Τf =cosOslashmgμL

2 π η= (cos90)(9 Kg)(98ms2)(001)(002032m)

2 π (086)= 0 Nm

= (sin90)(9 Kg)(98ms2)(002032m)2 π (086)

=1792 Nm

5404 = 03316 Nm

ΤDSince a free-wheeling lead screw nut was used there is no preload force If an anti-backlash nut was used ΤD would be gt0 due to preload force Always reference manufacturerrsquos data for more information

= 0 Nm

Linear to Rotary Conversion Torque

Linear to Rotary Conversion Inertia

Motion System Application Example14

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

14

MOTION SYSTEMFORMULAS

t 1 t 2 t 3 t dwell Time0333 s 0333 s 0333 s 05 s

Rotary Motion Profile

ωPK

ω

= 9276 rads a = 2786 rads2

Τ1= Τa+Τf +Τg+ΤD

= (00316 Nm) + (0 Nm) + (03316 Nm) + (0 Nm)

= 03632 Nm

Τ2= Τa+Τf +Τg+ΤD

= (0 Nm) + (0 Nm) + (03316 Nm) + (0 Nm)

= 03316 Nm

Τ3= Τa+Τf +Τg+ΤD

= (00316 Nm) + (0 Nm) + (03316 Nm) + (0 Nm)

= 03632 Nm

Τ1

Τ2

Τ3

Motion System Application Example15

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

15

MOTION SYSTEMAPPLICATION EXAMPLE

Lead Screw Input Requirements

= 9276 rads= 2786 rads2

= 03632 Nm= 03316 Nm= ndash 03632 Nm= 0333 s= 0333 s= 0333 s= 05 s= 3369 W= 3076 W

ωPKaΤ1Τ2Τ3

t1t2t3tdwell

PPKPCV

PPK = (Τ1)(ωPK)

= (03632 Nm)(9276 rads)= 3369 W

PCV = (Τ2)(ωPK)

= (03316 Nm)(9276 rads)= 3076 W

RMS Torque Requirement Lead Screw Input

=

0333 s + 0333 s + 0333 s + 05 s

(03632 Nm)2(0333 s)+(03316 Nm)2(0333 s)+(- 03632 Nm)2(0333 s)

ΤRMS = Τ12t 1 + Τ2

2t2 + Τ32t 3

t 1 + t 2 + t 3 + tdwell

15

00439 + 00366 + 00439 = 00829 =

ΤRMS = 02879 Nm

Motion System Application Example16

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

16

MOTION SYSTEMAPPLICATION EXAMPLE

Load Parameters at the Lead Screw Shaft

= 9276 rads= 2786 rads2

= 03632 Nm= 03316 Nm= ndash 03632 Nm= 02879 Nm= 3369 W= 3076 W

ωPKaΤ1Τ2Τ3ΤRMSPPKPCV

Motor SelectionA direct-drive motor can be used however it would be large and expensive For example a DC057B-3 brush motor

would easily meet continuous torque and continuous speed as well as a 181 load-to-rotor inertia This motor however has significantly more output power than needed at 128 watts

A smaller motor with a transmission would optimize the system Look for a motor starting at a rated power around 40 watts Pittman DC040B-6

DC040B-6Reference VoltageContinuous TorqueRated CurrentRated PowerTorque ConstantVoltage ConstantTerminal ResistanceMax Winding TemperatureRotor Inertia

240 V00812 Nm236 A37 W0042 NmA0042 Vrads185 Ω155degC847 x 10ndash6 Kg-m2

Motor TransmissionLead Screw Support Rails

ΤRMS

PPK

ωPK

= 02879 Nm= 3369 W= 9276 rads

M

Other Motor Considerationsndash Type stepper brush BLDCndash Input speed (high input speed will result in high noise levels)ndash Total motor footprintndash Ambient temperaturendash Environmental conditionsndash Load-to-rotor inertiandash Encoder ready needed

Motion System Application Example17

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

17

MOTION SYSTEMAPPLICATION EXAMPLE

Gearbox Transmission Selection

When selecting a reduction ratio keep in mind that the RMS required torque at the motor shaft needs to fall below the continuous rated output torque of the motor

G30 A - Planetary GearboxMaximum output load 247 Nm

TRMS( Lead Screw)

02879 Nm

02879 Nm02879 Nm

Reduction

41

5161

Efficiency

090

090090

TRMS( Motor)

00798 Nm

00640 Nm00533 Nm

A 51 gearbox would allow about a 27 safety margin between what the system requires and the continuous torque output of the motor

System Summary 51η = 090

M

= 00807 Nm= 00737 Nm= ndash 00807 Nm= 00640 Nm= 4638 rads= 504 x 10 ndash 6 Kg-m2

= 3743 W

Τ1Τ2Τ3ΤRMS

PPK

ωPK

J

= 03632 Nm= 03316 Nm= ndash 03632 Nm= 02879 Nm= 9276 rads= 1134 x 10 ndash 5 Kg-m2

= 3369 W

Τ1Τ2Τ3ΤRMS

PPK

ωPK

J

Motion System Application Example18

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

18

MOTION SYSTEMAPPLICATION EXAMPLE

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

19

Motion Profile at the Motor

t 1 t 2 t 3 t dwell Time0333 s 0333 s 0333 s 05 s

ωΤ1

Τ2

Τ3= 4638 rads= 4429 RPM= ΤPK = 00807 Nm

ωPK

Τ1

nPK

Drive and Power Supply Requirements

Peak Current Required

IPK =TPK

KT+ IO =

00807 Nm0042 NmA

+ 0180 A = 210 A

RMS Current Required

IRMS =TRMS

KT+ IO =

00640 Nm0042 NmA

+ 0180 A = 170 A

Minimum bus Voltage Required

Vbus= IPKRmt+ ωPKKE = (210A)(185Ω) + (4638 rads)(0042 Vrads)= 389 V + 1948 V= 2337 V

0100 Nm 0200 Nm 0300 Nm 0400 Nm 0500 Nm 0600 Nm 0700 Nm

30 VDC

0100 Nm 0200 Nm 0300 Nm 0500 Nm 0600 Nm 0700 Nm

24 VDC

700060005000

RPM 4000300020001000

0400 Nm

TRMS

TPK

700060005000

RPM 4000300020001000

TRMS

TPK

Performance using 24V reference voltage

Performance using a 30V bus voltage

A 30 VDC bus will meet the application requirements as wells supply a margin of safety

MOTION SYSTEMAPPLICATION EXAMPLE

0100 Nm 0200 Nm 0300 Nm 0400 Nm 0500 Nm 0600 Nm 0700 Nm

30 VDC

0100 Nm 0200 Nm 0300 Nm 0500 Nm 0600 Nm 0700 Nm

24 VDC

700060005000

RPM 4000300020001000

0400 Nm

TRMS

TPK

700060005000

RPM 4000300020001000

TRMS

TPK

Ambient temperature Rated motor temperature Temperature rise Motor temperature

= 30degC= 155degC= 7638degC= 10638degC

Θa

Θm

ΘrΘrated

=Rth x IRMS

2 x Rmt

1 ndash (Rth x IRMS2 x Rmt x 000392degC)

Θr

=11degCω x 170A2

x 185Ω 1 ndash (11degCω x 170A2

x 185Ω x 000392degC)

= 5881 1 ndash (023)

= 5881 077

=

Θm = Θr + Θa = 7638degC + 30degC = 10638degC

7638degC

Performance using 24V reference voltage

Performance using a 30V bus voltage

Will the motor meet the temperature rise caused by the application

Motion System Application Example20

A 30 VDC bus will meet the application requirements as wells supply a margin of safety

203 756 7441 1500 Meriden Road Waterbury CT USA 06705

267 933 2105 534 Godshall DriveHarleysville PA USA 19438

wwwHaydonKerkPittmancom

copy All rights reserved AMETEK Advanced Motion Solutions No part of this doc-ument or technical information can be used reproduced or altered in any form without approval or proper authorization from AMETEK Inc and global affiliates

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MOTION SYSTEMAPPLICATION EXAMPLE

Page 12: FORMULAS FOR MOTORIZED LINEAR MOTION SYSTEMS

Lead Screw Performance Characteristics

End Mounting Rigidity

Impact Impact

ScrewLength

Increase In Affects Increase In Affects

Critical speed

Compression load

Critical speed

Inertia

Compression load

Stiffness

Spring Rate

Load capacity

Drive torque

Angular velocity

Load cpacity

Positioning accuracy

ScrewDiameter

Lead

Load

Preload

Nut length

SI Unit Systems

LengthMassTimeElectric currentThermodynamic temperatureLuminous intensityAmount of substance

meterkilogramsecondamperekelvincandelamole

mKgsAKcdmol

Base Units

Derived Unitsareavolumefrequencymass density (density)speed velocityangular velocityaccelerationangular accelerationforcepressure (mechanical stress)kinematic viscositydynamic viscositywork energy quantity of heatpowerentropyspecific heat capacitythermal conductivity

square metercubic meterhertzkilogram per cubic metermeter per secondradian per secondmeter per second squaredradian per second squarednewtonpascalsquare meter per secondnewton-second per square meter joulewattjoule per kelvinjoule per kilogram kelvinwatt per meter kelvin

m2

m3

HzKgm3

msradsms2

rads2

NPam2sN-sm2

JWJKJKg-K)W(m-K)

Quantity Name of Unit Symbol

Motion System Reference Tables11

Critical speed

Compression load

System stiffness

Life

Positioning accuracy

System stiffness

Drag torque

Load capacity

Stiffness

sndash1

Kg-ms2

Nm2

N-mJs or N-ms

ExamplesAn INCREASE ( ) in screw length results in a DECREASE ( ) in critical speed

An INCREASE in screw diameter results in an INCREASE in critical speed

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11

MOTION SYSTEMREFERENCE TABLES

t 1 t 2 t 3 t dwell

vTime

vpk

a -a

MassMove distanceMove timeDwell timeMotion profileScrew speed limitLoad supportOrientationRotary to linear conversionAmbient temperature

9 Kg200 mm10 s05 s13 13 13 Trapezoidal1000 RPMμ = 001VerticalTFE lead screw 8 mm diameter 275 mm long30degC

ControlDrive Power Supply

4 quadrant with encoder32 VDC 35 Arms50 A peak

0333 s 0333 s 0333 s 05 s

vPK =3S2t =

(3)(02m)(2)(10 s) =

06m2 s =

03ms

Linear Velocity

Linear Acceleration

a = vf - vi

t = 03ms - 0 ms 0333 s 0901ms2 =

1st step is to thoroughly understand system constraints and motion profile requirements

Linear Motion ExampleData

Motion System Application Example12

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12

MOTION SYSTEMAPPLICATION EXAMPLE

Lmin =60vPK

n =(60)(03ms)1000 revmin = 0018mrev = 18mmrev

Refer to Kerk screw chart ndash closest lead in an 8mm screw diameter is

2032mmrev = 002032 mrevη = 086 free wheeling nut TFE coated screw

JS = 388 x 10ndash 7 Kg-m2 Other critical lead screw considerations bull Column loading bull Critical speed

Jm aa

ωPKvPK

TF

= ωPK = 2 π vPK

L=

(2π)(03ms)002032 mrev

9276 rads

a = =2πaL

=(2π)(0901ms2)002032 mrev

2786 rads2

Linear to Rotary Conversion ndash Velocity Acceleration Inertia Torque

Minimum screw lead to maintain lt 1000 RPM shaft speed

Linear to Rotary Conversion Velocity

Linear to Rotary Conversion Acceleration

Motion System Application Example13

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13

MOTION SYSTEMFORMULAS

Jin= m 2 πL 2

X η1 + JS= 9Kg

2 π002032 m 2

X 0861 + 388x10ndash 7Kg-m2

= 1095 X 10ndash 5 Kg-m2 + 388x10ndash 7 Kg-m2

= 1134 x 10ndash 5 Kg-m2

Τin= Τa+Τf +Τg+ΤD

Τa = Jin ain

Τg =sinOslashmgL

2 π η

= (1134 x 10ndash 5 Kg-m2)(2786 rads2)

= 00316 Nm

Τf =cosOslashmgμL

2 π η= (cos90)(9 Kg)(98ms2)(001)(002032m)

2 π (086)= 0 Nm

= (sin90)(9 Kg)(98ms2)(002032m)2 π (086)

=1792 Nm

5404 = 03316 Nm

ΤDSince a free-wheeling lead screw nut was used there is no preload force If an anti-backlash nut was used ΤD would be gt0 due to preload force Always reference manufacturerrsquos data for more information

= 0 Nm

Linear to Rotary Conversion Torque

Linear to Rotary Conversion Inertia

Motion System Application Example14

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14

MOTION SYSTEMFORMULAS

t 1 t 2 t 3 t dwell Time0333 s 0333 s 0333 s 05 s

Rotary Motion Profile

ωPK

ω

= 9276 rads a = 2786 rads2

Τ1= Τa+Τf +Τg+ΤD

= (00316 Nm) + (0 Nm) + (03316 Nm) + (0 Nm)

= 03632 Nm

Τ2= Τa+Τf +Τg+ΤD

= (0 Nm) + (0 Nm) + (03316 Nm) + (0 Nm)

= 03316 Nm

Τ3= Τa+Τf +Τg+ΤD

= (00316 Nm) + (0 Nm) + (03316 Nm) + (0 Nm)

= 03632 Nm

Τ1

Τ2

Τ3

Motion System Application Example15

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15

MOTION SYSTEMAPPLICATION EXAMPLE

Lead Screw Input Requirements

= 9276 rads= 2786 rads2

= 03632 Nm= 03316 Nm= ndash 03632 Nm= 0333 s= 0333 s= 0333 s= 05 s= 3369 W= 3076 W

ωPKaΤ1Τ2Τ3

t1t2t3tdwell

PPKPCV

PPK = (Τ1)(ωPK)

= (03632 Nm)(9276 rads)= 3369 W

PCV = (Τ2)(ωPK)

= (03316 Nm)(9276 rads)= 3076 W

RMS Torque Requirement Lead Screw Input

=

0333 s + 0333 s + 0333 s + 05 s

(03632 Nm)2(0333 s)+(03316 Nm)2(0333 s)+(- 03632 Nm)2(0333 s)

ΤRMS = Τ12t 1 + Τ2

2t2 + Τ32t 3

t 1 + t 2 + t 3 + tdwell

15

00439 + 00366 + 00439 = 00829 =

ΤRMS = 02879 Nm

Motion System Application Example16

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16

MOTION SYSTEMAPPLICATION EXAMPLE

Load Parameters at the Lead Screw Shaft

= 9276 rads= 2786 rads2

= 03632 Nm= 03316 Nm= ndash 03632 Nm= 02879 Nm= 3369 W= 3076 W

ωPKaΤ1Τ2Τ3ΤRMSPPKPCV

Motor SelectionA direct-drive motor can be used however it would be large and expensive For example a DC057B-3 brush motor

would easily meet continuous torque and continuous speed as well as a 181 load-to-rotor inertia This motor however has significantly more output power than needed at 128 watts

A smaller motor with a transmission would optimize the system Look for a motor starting at a rated power around 40 watts Pittman DC040B-6

DC040B-6Reference VoltageContinuous TorqueRated CurrentRated PowerTorque ConstantVoltage ConstantTerminal ResistanceMax Winding TemperatureRotor Inertia

240 V00812 Nm236 A37 W0042 NmA0042 Vrads185 Ω155degC847 x 10ndash6 Kg-m2

Motor TransmissionLead Screw Support Rails

ΤRMS

PPK

ωPK

= 02879 Nm= 3369 W= 9276 rads

M

Other Motor Considerationsndash Type stepper brush BLDCndash Input speed (high input speed will result in high noise levels)ndash Total motor footprintndash Ambient temperaturendash Environmental conditionsndash Load-to-rotor inertiandash Encoder ready needed

Motion System Application Example17

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17

MOTION SYSTEMAPPLICATION EXAMPLE

Gearbox Transmission Selection

When selecting a reduction ratio keep in mind that the RMS required torque at the motor shaft needs to fall below the continuous rated output torque of the motor

G30 A - Planetary GearboxMaximum output load 247 Nm

TRMS( Lead Screw)

02879 Nm

02879 Nm02879 Nm

Reduction

41

5161

Efficiency

090

090090

TRMS( Motor)

00798 Nm

00640 Nm00533 Nm

A 51 gearbox would allow about a 27 safety margin between what the system requires and the continuous torque output of the motor

System Summary 51η = 090

M

= 00807 Nm= 00737 Nm= ndash 00807 Nm= 00640 Nm= 4638 rads= 504 x 10 ndash 6 Kg-m2

= 3743 W

Τ1Τ2Τ3ΤRMS

PPK

ωPK

J

= 03632 Nm= 03316 Nm= ndash 03632 Nm= 02879 Nm= 9276 rads= 1134 x 10 ndash 5 Kg-m2

= 3369 W

Τ1Τ2Τ3ΤRMS

PPK

ωPK

J

Motion System Application Example18

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18

MOTION SYSTEMAPPLICATION EXAMPLE

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19

Motion Profile at the Motor

t 1 t 2 t 3 t dwell Time0333 s 0333 s 0333 s 05 s

ωΤ1

Τ2

Τ3= 4638 rads= 4429 RPM= ΤPK = 00807 Nm

ωPK

Τ1

nPK

Drive and Power Supply Requirements

Peak Current Required

IPK =TPK

KT+ IO =

00807 Nm0042 NmA

+ 0180 A = 210 A

RMS Current Required

IRMS =TRMS

KT+ IO =

00640 Nm0042 NmA

+ 0180 A = 170 A

Minimum bus Voltage Required

Vbus= IPKRmt+ ωPKKE = (210A)(185Ω) + (4638 rads)(0042 Vrads)= 389 V + 1948 V= 2337 V

0100 Nm 0200 Nm 0300 Nm 0400 Nm 0500 Nm 0600 Nm 0700 Nm

30 VDC

0100 Nm 0200 Nm 0300 Nm 0500 Nm 0600 Nm 0700 Nm

24 VDC

700060005000

RPM 4000300020001000

0400 Nm

TRMS

TPK

700060005000

RPM 4000300020001000

TRMS

TPK

Performance using 24V reference voltage

Performance using a 30V bus voltage

A 30 VDC bus will meet the application requirements as wells supply a margin of safety

MOTION SYSTEMAPPLICATION EXAMPLE

0100 Nm 0200 Nm 0300 Nm 0400 Nm 0500 Nm 0600 Nm 0700 Nm

30 VDC

0100 Nm 0200 Nm 0300 Nm 0500 Nm 0600 Nm 0700 Nm

24 VDC

700060005000

RPM 4000300020001000

0400 Nm

TRMS

TPK

700060005000

RPM 4000300020001000

TRMS

TPK

Ambient temperature Rated motor temperature Temperature rise Motor temperature

= 30degC= 155degC= 7638degC= 10638degC

Θa

Θm

ΘrΘrated

=Rth x IRMS

2 x Rmt

1 ndash (Rth x IRMS2 x Rmt x 000392degC)

Θr

=11degCω x 170A2

x 185Ω 1 ndash (11degCω x 170A2

x 185Ω x 000392degC)

= 5881 1 ndash (023)

= 5881 077

=

Θm = Θr + Θa = 7638degC + 30degC = 10638degC

7638degC

Performance using 24V reference voltage

Performance using a 30V bus voltage

Will the motor meet the temperature rise caused by the application

Motion System Application Example20

A 30 VDC bus will meet the application requirements as wells supply a margin of safety

203 756 7441 1500 Meriden Road Waterbury CT USA 06705

267 933 2105 534 Godshall DriveHarleysville PA USA 19438

wwwHaydonKerkPittmancom

copy All rights reserved AMETEK Advanced Motion Solutions No part of this doc-ument or technical information can be used reproduced or altered in any form without approval or proper authorization from AMETEK Inc and global affiliates

NOW ASK us the tough motion control questions ONLINE

ASK THE EXPERTS

HaydonKerkPittmancomwww

MOTION SYSTEMAPPLICATION EXAMPLE

Page 13: FORMULAS FOR MOTORIZED LINEAR MOTION SYSTEMS

t 1 t 2 t 3 t dwell

vTime

vpk

a -a

MassMove distanceMove timeDwell timeMotion profileScrew speed limitLoad supportOrientationRotary to linear conversionAmbient temperature

9 Kg200 mm10 s05 s13 13 13 Trapezoidal1000 RPMμ = 001VerticalTFE lead screw 8 mm diameter 275 mm long30degC

ControlDrive Power Supply

4 quadrant with encoder32 VDC 35 Arms50 A peak

0333 s 0333 s 0333 s 05 s

vPK =3S2t =

(3)(02m)(2)(10 s) =

06m2 s =

03ms

Linear Velocity

Linear Acceleration

a = vf - vi

t = 03ms - 0 ms 0333 s 0901ms2 =

1st step is to thoroughly understand system constraints and motion profile requirements

Linear Motion ExampleData

Motion System Application Example12

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

12

MOTION SYSTEMAPPLICATION EXAMPLE

Lmin =60vPK

n =(60)(03ms)1000 revmin = 0018mrev = 18mmrev

Refer to Kerk screw chart ndash closest lead in an 8mm screw diameter is

2032mmrev = 002032 mrevη = 086 free wheeling nut TFE coated screw

JS = 388 x 10ndash 7 Kg-m2 Other critical lead screw considerations bull Column loading bull Critical speed

Jm aa

ωPKvPK

TF

= ωPK = 2 π vPK

L=

(2π)(03ms)002032 mrev

9276 rads

a = =2πaL

=(2π)(0901ms2)002032 mrev

2786 rads2

Linear to Rotary Conversion ndash Velocity Acceleration Inertia Torque

Minimum screw lead to maintain lt 1000 RPM shaft speed

Linear to Rotary Conversion Velocity

Linear to Rotary Conversion Acceleration

Motion System Application Example13

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13

MOTION SYSTEMFORMULAS

Jin= m 2 πL 2

X η1 + JS= 9Kg

2 π002032 m 2

X 0861 + 388x10ndash 7Kg-m2

= 1095 X 10ndash 5 Kg-m2 + 388x10ndash 7 Kg-m2

= 1134 x 10ndash 5 Kg-m2

Τin= Τa+Τf +Τg+ΤD

Τa = Jin ain

Τg =sinOslashmgL

2 π η

= (1134 x 10ndash 5 Kg-m2)(2786 rads2)

= 00316 Nm

Τf =cosOslashmgμL

2 π η= (cos90)(9 Kg)(98ms2)(001)(002032m)

2 π (086)= 0 Nm

= (sin90)(9 Kg)(98ms2)(002032m)2 π (086)

=1792 Nm

5404 = 03316 Nm

ΤDSince a free-wheeling lead screw nut was used there is no preload force If an anti-backlash nut was used ΤD would be gt0 due to preload force Always reference manufacturerrsquos data for more information

= 0 Nm

Linear to Rotary Conversion Torque

Linear to Rotary Conversion Inertia

Motion System Application Example14

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14

MOTION SYSTEMFORMULAS

t 1 t 2 t 3 t dwell Time0333 s 0333 s 0333 s 05 s

Rotary Motion Profile

ωPK

ω

= 9276 rads a = 2786 rads2

Τ1= Τa+Τf +Τg+ΤD

= (00316 Nm) + (0 Nm) + (03316 Nm) + (0 Nm)

= 03632 Nm

Τ2= Τa+Τf +Τg+ΤD

= (0 Nm) + (0 Nm) + (03316 Nm) + (0 Nm)

= 03316 Nm

Τ3= Τa+Τf +Τg+ΤD

= (00316 Nm) + (0 Nm) + (03316 Nm) + (0 Nm)

= 03632 Nm

Τ1

Τ2

Τ3

Motion System Application Example15

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15

MOTION SYSTEMAPPLICATION EXAMPLE

Lead Screw Input Requirements

= 9276 rads= 2786 rads2

= 03632 Nm= 03316 Nm= ndash 03632 Nm= 0333 s= 0333 s= 0333 s= 05 s= 3369 W= 3076 W

ωPKaΤ1Τ2Τ3

t1t2t3tdwell

PPKPCV

PPK = (Τ1)(ωPK)

= (03632 Nm)(9276 rads)= 3369 W

PCV = (Τ2)(ωPK)

= (03316 Nm)(9276 rads)= 3076 W

RMS Torque Requirement Lead Screw Input

=

0333 s + 0333 s + 0333 s + 05 s

(03632 Nm)2(0333 s)+(03316 Nm)2(0333 s)+(- 03632 Nm)2(0333 s)

ΤRMS = Τ12t 1 + Τ2

2t2 + Τ32t 3

t 1 + t 2 + t 3 + tdwell

15

00439 + 00366 + 00439 = 00829 =

ΤRMS = 02879 Nm

Motion System Application Example16

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16

MOTION SYSTEMAPPLICATION EXAMPLE

Load Parameters at the Lead Screw Shaft

= 9276 rads= 2786 rads2

= 03632 Nm= 03316 Nm= ndash 03632 Nm= 02879 Nm= 3369 W= 3076 W

ωPKaΤ1Τ2Τ3ΤRMSPPKPCV

Motor SelectionA direct-drive motor can be used however it would be large and expensive For example a DC057B-3 brush motor

would easily meet continuous torque and continuous speed as well as a 181 load-to-rotor inertia This motor however has significantly more output power than needed at 128 watts

A smaller motor with a transmission would optimize the system Look for a motor starting at a rated power around 40 watts Pittman DC040B-6

DC040B-6Reference VoltageContinuous TorqueRated CurrentRated PowerTorque ConstantVoltage ConstantTerminal ResistanceMax Winding TemperatureRotor Inertia

240 V00812 Nm236 A37 W0042 NmA0042 Vrads185 Ω155degC847 x 10ndash6 Kg-m2

Motor TransmissionLead Screw Support Rails

ΤRMS

PPK

ωPK

= 02879 Nm= 3369 W= 9276 rads

M

Other Motor Considerationsndash Type stepper brush BLDCndash Input speed (high input speed will result in high noise levels)ndash Total motor footprintndash Ambient temperaturendash Environmental conditionsndash Load-to-rotor inertiandash Encoder ready needed

Motion System Application Example17

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

17

MOTION SYSTEMAPPLICATION EXAMPLE

Gearbox Transmission Selection

When selecting a reduction ratio keep in mind that the RMS required torque at the motor shaft needs to fall below the continuous rated output torque of the motor

G30 A - Planetary GearboxMaximum output load 247 Nm

TRMS( Lead Screw)

02879 Nm

02879 Nm02879 Nm

Reduction

41

5161

Efficiency

090

090090

TRMS( Motor)

00798 Nm

00640 Nm00533 Nm

A 51 gearbox would allow about a 27 safety margin between what the system requires and the continuous torque output of the motor

System Summary 51η = 090

M

= 00807 Nm= 00737 Nm= ndash 00807 Nm= 00640 Nm= 4638 rads= 504 x 10 ndash 6 Kg-m2

= 3743 W

Τ1Τ2Τ3ΤRMS

PPK

ωPK

J

= 03632 Nm= 03316 Nm= ndash 03632 Nm= 02879 Nm= 9276 rads= 1134 x 10 ndash 5 Kg-m2

= 3369 W

Τ1Τ2Τ3ΤRMS

PPK

ωPK

J

Motion System Application Example18

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18

MOTION SYSTEMAPPLICATION EXAMPLE

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19

Motion Profile at the Motor

t 1 t 2 t 3 t dwell Time0333 s 0333 s 0333 s 05 s

ωΤ1

Τ2

Τ3= 4638 rads= 4429 RPM= ΤPK = 00807 Nm

ωPK

Τ1

nPK

Drive and Power Supply Requirements

Peak Current Required

IPK =TPK

KT+ IO =

00807 Nm0042 NmA

+ 0180 A = 210 A

RMS Current Required

IRMS =TRMS

KT+ IO =

00640 Nm0042 NmA

+ 0180 A = 170 A

Minimum bus Voltage Required

Vbus= IPKRmt+ ωPKKE = (210A)(185Ω) + (4638 rads)(0042 Vrads)= 389 V + 1948 V= 2337 V

0100 Nm 0200 Nm 0300 Nm 0400 Nm 0500 Nm 0600 Nm 0700 Nm

30 VDC

0100 Nm 0200 Nm 0300 Nm 0500 Nm 0600 Nm 0700 Nm

24 VDC

700060005000

RPM 4000300020001000

0400 Nm

TRMS

TPK

700060005000

RPM 4000300020001000

TRMS

TPK

Performance using 24V reference voltage

Performance using a 30V bus voltage

A 30 VDC bus will meet the application requirements as wells supply a margin of safety

MOTION SYSTEMAPPLICATION EXAMPLE

0100 Nm 0200 Nm 0300 Nm 0400 Nm 0500 Nm 0600 Nm 0700 Nm

30 VDC

0100 Nm 0200 Nm 0300 Nm 0500 Nm 0600 Nm 0700 Nm

24 VDC

700060005000

RPM 4000300020001000

0400 Nm

TRMS

TPK

700060005000

RPM 4000300020001000

TRMS

TPK

Ambient temperature Rated motor temperature Temperature rise Motor temperature

= 30degC= 155degC= 7638degC= 10638degC

Θa

Θm

ΘrΘrated

=Rth x IRMS

2 x Rmt

1 ndash (Rth x IRMS2 x Rmt x 000392degC)

Θr

=11degCω x 170A2

x 185Ω 1 ndash (11degCω x 170A2

x 185Ω x 000392degC)

= 5881 1 ndash (023)

= 5881 077

=

Θm = Θr + Θa = 7638degC + 30degC = 10638degC

7638degC

Performance using 24V reference voltage

Performance using a 30V bus voltage

Will the motor meet the temperature rise caused by the application

Motion System Application Example20

A 30 VDC bus will meet the application requirements as wells supply a margin of safety

203 756 7441 1500 Meriden Road Waterbury CT USA 06705

267 933 2105 534 Godshall DriveHarleysville PA USA 19438

wwwHaydonKerkPittmancom

copy All rights reserved AMETEK Advanced Motion Solutions No part of this doc-ument or technical information can be used reproduced or altered in any form without approval or proper authorization from AMETEK Inc and global affiliates

NOW ASK us the tough motion control questions ONLINE

ASK THE EXPERTS

HaydonKerkPittmancomwww

MOTION SYSTEMAPPLICATION EXAMPLE

Page 14: FORMULAS FOR MOTORIZED LINEAR MOTION SYSTEMS

Lmin =60vPK

n =(60)(03ms)1000 revmin = 0018mrev = 18mmrev

Refer to Kerk screw chart ndash closest lead in an 8mm screw diameter is

2032mmrev = 002032 mrevη = 086 free wheeling nut TFE coated screw

JS = 388 x 10ndash 7 Kg-m2 Other critical lead screw considerations bull Column loading bull Critical speed

Jm aa

ωPKvPK

TF

= ωPK = 2 π vPK

L=

(2π)(03ms)002032 mrev

9276 rads

a = =2πaL

=(2π)(0901ms2)002032 mrev

2786 rads2

Linear to Rotary Conversion ndash Velocity Acceleration Inertia Torque

Minimum screw lead to maintain lt 1000 RPM shaft speed

Linear to Rotary Conversion Velocity

Linear to Rotary Conversion Acceleration

Motion System Application Example13

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

13

MOTION SYSTEMFORMULAS

Jin= m 2 πL 2

X η1 + JS= 9Kg

2 π002032 m 2

X 0861 + 388x10ndash 7Kg-m2

= 1095 X 10ndash 5 Kg-m2 + 388x10ndash 7 Kg-m2

= 1134 x 10ndash 5 Kg-m2

Τin= Τa+Τf +Τg+ΤD

Τa = Jin ain

Τg =sinOslashmgL

2 π η

= (1134 x 10ndash 5 Kg-m2)(2786 rads2)

= 00316 Nm

Τf =cosOslashmgμL

2 π η= (cos90)(9 Kg)(98ms2)(001)(002032m)

2 π (086)= 0 Nm

= (sin90)(9 Kg)(98ms2)(002032m)2 π (086)

=1792 Nm

5404 = 03316 Nm

ΤDSince a free-wheeling lead screw nut was used there is no preload force If an anti-backlash nut was used ΤD would be gt0 due to preload force Always reference manufacturerrsquos data for more information

= 0 Nm

Linear to Rotary Conversion Torque

Linear to Rotary Conversion Inertia

Motion System Application Example14

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14

MOTION SYSTEMFORMULAS

t 1 t 2 t 3 t dwell Time0333 s 0333 s 0333 s 05 s

Rotary Motion Profile

ωPK

ω

= 9276 rads a = 2786 rads2

Τ1= Τa+Τf +Τg+ΤD

= (00316 Nm) + (0 Nm) + (03316 Nm) + (0 Nm)

= 03632 Nm

Τ2= Τa+Τf +Τg+ΤD

= (0 Nm) + (0 Nm) + (03316 Nm) + (0 Nm)

= 03316 Nm

Τ3= Τa+Τf +Τg+ΤD

= (00316 Nm) + (0 Nm) + (03316 Nm) + (0 Nm)

= 03632 Nm

Τ1

Τ2

Τ3

Motion System Application Example15

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15

MOTION SYSTEMAPPLICATION EXAMPLE

Lead Screw Input Requirements

= 9276 rads= 2786 rads2

= 03632 Nm= 03316 Nm= ndash 03632 Nm= 0333 s= 0333 s= 0333 s= 05 s= 3369 W= 3076 W

ωPKaΤ1Τ2Τ3

t1t2t3tdwell

PPKPCV

PPK = (Τ1)(ωPK)

= (03632 Nm)(9276 rads)= 3369 W

PCV = (Τ2)(ωPK)

= (03316 Nm)(9276 rads)= 3076 W

RMS Torque Requirement Lead Screw Input

=

0333 s + 0333 s + 0333 s + 05 s

(03632 Nm)2(0333 s)+(03316 Nm)2(0333 s)+(- 03632 Nm)2(0333 s)

ΤRMS = Τ12t 1 + Τ2

2t2 + Τ32t 3

t 1 + t 2 + t 3 + tdwell

15

00439 + 00366 + 00439 = 00829 =

ΤRMS = 02879 Nm

Motion System Application Example16

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16

MOTION SYSTEMAPPLICATION EXAMPLE

Load Parameters at the Lead Screw Shaft

= 9276 rads= 2786 rads2

= 03632 Nm= 03316 Nm= ndash 03632 Nm= 02879 Nm= 3369 W= 3076 W

ωPKaΤ1Τ2Τ3ΤRMSPPKPCV

Motor SelectionA direct-drive motor can be used however it would be large and expensive For example a DC057B-3 brush motor

would easily meet continuous torque and continuous speed as well as a 181 load-to-rotor inertia This motor however has significantly more output power than needed at 128 watts

A smaller motor with a transmission would optimize the system Look for a motor starting at a rated power around 40 watts Pittman DC040B-6

DC040B-6Reference VoltageContinuous TorqueRated CurrentRated PowerTorque ConstantVoltage ConstantTerminal ResistanceMax Winding TemperatureRotor Inertia

240 V00812 Nm236 A37 W0042 NmA0042 Vrads185 Ω155degC847 x 10ndash6 Kg-m2

Motor TransmissionLead Screw Support Rails

ΤRMS

PPK

ωPK

= 02879 Nm= 3369 W= 9276 rads

M

Other Motor Considerationsndash Type stepper brush BLDCndash Input speed (high input speed will result in high noise levels)ndash Total motor footprintndash Ambient temperaturendash Environmental conditionsndash Load-to-rotor inertiandash Encoder ready needed

Motion System Application Example17

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17

MOTION SYSTEMAPPLICATION EXAMPLE

Gearbox Transmission Selection

When selecting a reduction ratio keep in mind that the RMS required torque at the motor shaft needs to fall below the continuous rated output torque of the motor

G30 A - Planetary GearboxMaximum output load 247 Nm

TRMS( Lead Screw)

02879 Nm

02879 Nm02879 Nm

Reduction

41

5161

Efficiency

090

090090

TRMS( Motor)

00798 Nm

00640 Nm00533 Nm

A 51 gearbox would allow about a 27 safety margin between what the system requires and the continuous torque output of the motor

System Summary 51η = 090

M

= 00807 Nm= 00737 Nm= ndash 00807 Nm= 00640 Nm= 4638 rads= 504 x 10 ndash 6 Kg-m2

= 3743 W

Τ1Τ2Τ3ΤRMS

PPK

ωPK

J

= 03632 Nm= 03316 Nm= ndash 03632 Nm= 02879 Nm= 9276 rads= 1134 x 10 ndash 5 Kg-m2

= 3369 W

Τ1Τ2Τ3ΤRMS

PPK

ωPK

J

Motion System Application Example18

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18

MOTION SYSTEMAPPLICATION EXAMPLE

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19

Motion Profile at the Motor

t 1 t 2 t 3 t dwell Time0333 s 0333 s 0333 s 05 s

ωΤ1

Τ2

Τ3= 4638 rads= 4429 RPM= ΤPK = 00807 Nm

ωPK

Τ1

nPK

Drive and Power Supply Requirements

Peak Current Required

IPK =TPK

KT+ IO =

00807 Nm0042 NmA

+ 0180 A = 210 A

RMS Current Required

IRMS =TRMS

KT+ IO =

00640 Nm0042 NmA

+ 0180 A = 170 A

Minimum bus Voltage Required

Vbus= IPKRmt+ ωPKKE = (210A)(185Ω) + (4638 rads)(0042 Vrads)= 389 V + 1948 V= 2337 V

0100 Nm 0200 Nm 0300 Nm 0400 Nm 0500 Nm 0600 Nm 0700 Nm

30 VDC

0100 Nm 0200 Nm 0300 Nm 0500 Nm 0600 Nm 0700 Nm

24 VDC

700060005000

RPM 4000300020001000

0400 Nm

TRMS

TPK

700060005000

RPM 4000300020001000

TRMS

TPK

Performance using 24V reference voltage

Performance using a 30V bus voltage

A 30 VDC bus will meet the application requirements as wells supply a margin of safety

MOTION SYSTEMAPPLICATION EXAMPLE

0100 Nm 0200 Nm 0300 Nm 0400 Nm 0500 Nm 0600 Nm 0700 Nm

30 VDC

0100 Nm 0200 Nm 0300 Nm 0500 Nm 0600 Nm 0700 Nm

24 VDC

700060005000

RPM 4000300020001000

0400 Nm

TRMS

TPK

700060005000

RPM 4000300020001000

TRMS

TPK

Ambient temperature Rated motor temperature Temperature rise Motor temperature

= 30degC= 155degC= 7638degC= 10638degC

Θa

Θm

ΘrΘrated

=Rth x IRMS

2 x Rmt

1 ndash (Rth x IRMS2 x Rmt x 000392degC)

Θr

=11degCω x 170A2

x 185Ω 1 ndash (11degCω x 170A2

x 185Ω x 000392degC)

= 5881 1 ndash (023)

= 5881 077

=

Θm = Θr + Θa = 7638degC + 30degC = 10638degC

7638degC

Performance using 24V reference voltage

Performance using a 30V bus voltage

Will the motor meet the temperature rise caused by the application

Motion System Application Example20

A 30 VDC bus will meet the application requirements as wells supply a margin of safety

203 756 7441 1500 Meriden Road Waterbury CT USA 06705

267 933 2105 534 Godshall DriveHarleysville PA USA 19438

wwwHaydonKerkPittmancom

copy All rights reserved AMETEK Advanced Motion Solutions No part of this doc-ument or technical information can be used reproduced or altered in any form without approval or proper authorization from AMETEK Inc and global affiliates

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MOTION SYSTEMAPPLICATION EXAMPLE

Page 15: FORMULAS FOR MOTORIZED LINEAR MOTION SYSTEMS

Jin= m 2 πL 2

X η1 + JS= 9Kg

2 π002032 m 2

X 0861 + 388x10ndash 7Kg-m2

= 1095 X 10ndash 5 Kg-m2 + 388x10ndash 7 Kg-m2

= 1134 x 10ndash 5 Kg-m2

Τin= Τa+Τf +Τg+ΤD

Τa = Jin ain

Τg =sinOslashmgL

2 π η

= (1134 x 10ndash 5 Kg-m2)(2786 rads2)

= 00316 Nm

Τf =cosOslashmgμL

2 π η= (cos90)(9 Kg)(98ms2)(001)(002032m)

2 π (086)= 0 Nm

= (sin90)(9 Kg)(98ms2)(002032m)2 π (086)

=1792 Nm

5404 = 03316 Nm

ΤDSince a free-wheeling lead screw nut was used there is no preload force If an anti-backlash nut was used ΤD would be gt0 due to preload force Always reference manufacturerrsquos data for more information

= 0 Nm

Linear to Rotary Conversion Torque

Linear to Rotary Conversion Inertia

Motion System Application Example14

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

14

MOTION SYSTEMFORMULAS

t 1 t 2 t 3 t dwell Time0333 s 0333 s 0333 s 05 s

Rotary Motion Profile

ωPK

ω

= 9276 rads a = 2786 rads2

Τ1= Τa+Τf +Τg+ΤD

= (00316 Nm) + (0 Nm) + (03316 Nm) + (0 Nm)

= 03632 Nm

Τ2= Τa+Τf +Τg+ΤD

= (0 Nm) + (0 Nm) + (03316 Nm) + (0 Nm)

= 03316 Nm

Τ3= Τa+Τf +Τg+ΤD

= (00316 Nm) + (0 Nm) + (03316 Nm) + (0 Nm)

= 03632 Nm

Τ1

Τ2

Τ3

Motion System Application Example15

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15

MOTION SYSTEMAPPLICATION EXAMPLE

Lead Screw Input Requirements

= 9276 rads= 2786 rads2

= 03632 Nm= 03316 Nm= ndash 03632 Nm= 0333 s= 0333 s= 0333 s= 05 s= 3369 W= 3076 W

ωPKaΤ1Τ2Τ3

t1t2t3tdwell

PPKPCV

PPK = (Τ1)(ωPK)

= (03632 Nm)(9276 rads)= 3369 W

PCV = (Τ2)(ωPK)

= (03316 Nm)(9276 rads)= 3076 W

RMS Torque Requirement Lead Screw Input

=

0333 s + 0333 s + 0333 s + 05 s

(03632 Nm)2(0333 s)+(03316 Nm)2(0333 s)+(- 03632 Nm)2(0333 s)

ΤRMS = Τ12t 1 + Τ2

2t2 + Τ32t 3

t 1 + t 2 + t 3 + tdwell

15

00439 + 00366 + 00439 = 00829 =

ΤRMS = 02879 Nm

Motion System Application Example16

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16

MOTION SYSTEMAPPLICATION EXAMPLE

Load Parameters at the Lead Screw Shaft

= 9276 rads= 2786 rads2

= 03632 Nm= 03316 Nm= ndash 03632 Nm= 02879 Nm= 3369 W= 3076 W

ωPKaΤ1Τ2Τ3ΤRMSPPKPCV

Motor SelectionA direct-drive motor can be used however it would be large and expensive For example a DC057B-3 brush motor

would easily meet continuous torque and continuous speed as well as a 181 load-to-rotor inertia This motor however has significantly more output power than needed at 128 watts

A smaller motor with a transmission would optimize the system Look for a motor starting at a rated power around 40 watts Pittman DC040B-6

DC040B-6Reference VoltageContinuous TorqueRated CurrentRated PowerTorque ConstantVoltage ConstantTerminal ResistanceMax Winding TemperatureRotor Inertia

240 V00812 Nm236 A37 W0042 NmA0042 Vrads185 Ω155degC847 x 10ndash6 Kg-m2

Motor TransmissionLead Screw Support Rails

ΤRMS

PPK

ωPK

= 02879 Nm= 3369 W= 9276 rads

M

Other Motor Considerationsndash Type stepper brush BLDCndash Input speed (high input speed will result in high noise levels)ndash Total motor footprintndash Ambient temperaturendash Environmental conditionsndash Load-to-rotor inertiandash Encoder ready needed

Motion System Application Example17

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

17

MOTION SYSTEMAPPLICATION EXAMPLE

Gearbox Transmission Selection

When selecting a reduction ratio keep in mind that the RMS required torque at the motor shaft needs to fall below the continuous rated output torque of the motor

G30 A - Planetary GearboxMaximum output load 247 Nm

TRMS( Lead Screw)

02879 Nm

02879 Nm02879 Nm

Reduction

41

5161

Efficiency

090

090090

TRMS( Motor)

00798 Nm

00640 Nm00533 Nm

A 51 gearbox would allow about a 27 safety margin between what the system requires and the continuous torque output of the motor

System Summary 51η = 090

M

= 00807 Nm= 00737 Nm= ndash 00807 Nm= 00640 Nm= 4638 rads= 504 x 10 ndash 6 Kg-m2

= 3743 W

Τ1Τ2Τ3ΤRMS

PPK

ωPK

J

= 03632 Nm= 03316 Nm= ndash 03632 Nm= 02879 Nm= 9276 rads= 1134 x 10 ndash 5 Kg-m2

= 3369 W

Τ1Τ2Τ3ΤRMS

PPK

ωPK

J

Motion System Application Example18

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18

MOTION SYSTEMAPPLICATION EXAMPLE

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19

Motion Profile at the Motor

t 1 t 2 t 3 t dwell Time0333 s 0333 s 0333 s 05 s

ωΤ1

Τ2

Τ3= 4638 rads= 4429 RPM= ΤPK = 00807 Nm

ωPK

Τ1

nPK

Drive and Power Supply Requirements

Peak Current Required

IPK =TPK

KT+ IO =

00807 Nm0042 NmA

+ 0180 A = 210 A

RMS Current Required

IRMS =TRMS

KT+ IO =

00640 Nm0042 NmA

+ 0180 A = 170 A

Minimum bus Voltage Required

Vbus= IPKRmt+ ωPKKE = (210A)(185Ω) + (4638 rads)(0042 Vrads)= 389 V + 1948 V= 2337 V

0100 Nm 0200 Nm 0300 Nm 0400 Nm 0500 Nm 0600 Nm 0700 Nm

30 VDC

0100 Nm 0200 Nm 0300 Nm 0500 Nm 0600 Nm 0700 Nm

24 VDC

700060005000

RPM 4000300020001000

0400 Nm

TRMS

TPK

700060005000

RPM 4000300020001000

TRMS

TPK

Performance using 24V reference voltage

Performance using a 30V bus voltage

A 30 VDC bus will meet the application requirements as wells supply a margin of safety

MOTION SYSTEMAPPLICATION EXAMPLE

0100 Nm 0200 Nm 0300 Nm 0400 Nm 0500 Nm 0600 Nm 0700 Nm

30 VDC

0100 Nm 0200 Nm 0300 Nm 0500 Nm 0600 Nm 0700 Nm

24 VDC

700060005000

RPM 4000300020001000

0400 Nm

TRMS

TPK

700060005000

RPM 4000300020001000

TRMS

TPK

Ambient temperature Rated motor temperature Temperature rise Motor temperature

= 30degC= 155degC= 7638degC= 10638degC

Θa

Θm

ΘrΘrated

=Rth x IRMS

2 x Rmt

1 ndash (Rth x IRMS2 x Rmt x 000392degC)

Θr

=11degCω x 170A2

x 185Ω 1 ndash (11degCω x 170A2

x 185Ω x 000392degC)

= 5881 1 ndash (023)

= 5881 077

=

Θm = Θr + Θa = 7638degC + 30degC = 10638degC

7638degC

Performance using 24V reference voltage

Performance using a 30V bus voltage

Will the motor meet the temperature rise caused by the application

Motion System Application Example20

A 30 VDC bus will meet the application requirements as wells supply a margin of safety

203 756 7441 1500 Meriden Road Waterbury CT USA 06705

267 933 2105 534 Godshall DriveHarleysville PA USA 19438

wwwHaydonKerkPittmancom

copy All rights reserved AMETEK Advanced Motion Solutions No part of this doc-ument or technical information can be used reproduced or altered in any form without approval or proper authorization from AMETEK Inc and global affiliates

NOW ASK us the tough motion control questions ONLINE

ASK THE EXPERTS

HaydonKerkPittmancomwww

MOTION SYSTEMAPPLICATION EXAMPLE

Page 16: FORMULAS FOR MOTORIZED LINEAR MOTION SYSTEMS

t 1 t 2 t 3 t dwell Time0333 s 0333 s 0333 s 05 s

Rotary Motion Profile

ωPK

ω

= 9276 rads a = 2786 rads2

Τ1= Τa+Τf +Τg+ΤD

= (00316 Nm) + (0 Nm) + (03316 Nm) + (0 Nm)

= 03632 Nm

Τ2= Τa+Τf +Τg+ΤD

= (0 Nm) + (0 Nm) + (03316 Nm) + (0 Nm)

= 03316 Nm

Τ3= Τa+Τf +Τg+ΤD

= (00316 Nm) + (0 Nm) + (03316 Nm) + (0 Nm)

= 03632 Nm

Τ1

Τ2

Τ3

Motion System Application Example15

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15

MOTION SYSTEMAPPLICATION EXAMPLE

Lead Screw Input Requirements

= 9276 rads= 2786 rads2

= 03632 Nm= 03316 Nm= ndash 03632 Nm= 0333 s= 0333 s= 0333 s= 05 s= 3369 W= 3076 W

ωPKaΤ1Τ2Τ3

t1t2t3tdwell

PPKPCV

PPK = (Τ1)(ωPK)

= (03632 Nm)(9276 rads)= 3369 W

PCV = (Τ2)(ωPK)

= (03316 Nm)(9276 rads)= 3076 W

RMS Torque Requirement Lead Screw Input

=

0333 s + 0333 s + 0333 s + 05 s

(03632 Nm)2(0333 s)+(03316 Nm)2(0333 s)+(- 03632 Nm)2(0333 s)

ΤRMS = Τ12t 1 + Τ2

2t2 + Τ32t 3

t 1 + t 2 + t 3 + tdwell

15

00439 + 00366 + 00439 = 00829 =

ΤRMS = 02879 Nm

Motion System Application Example16

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16

MOTION SYSTEMAPPLICATION EXAMPLE

Load Parameters at the Lead Screw Shaft

= 9276 rads= 2786 rads2

= 03632 Nm= 03316 Nm= ndash 03632 Nm= 02879 Nm= 3369 W= 3076 W

ωPKaΤ1Τ2Τ3ΤRMSPPKPCV

Motor SelectionA direct-drive motor can be used however it would be large and expensive For example a DC057B-3 brush motor

would easily meet continuous torque and continuous speed as well as a 181 load-to-rotor inertia This motor however has significantly more output power than needed at 128 watts

A smaller motor with a transmission would optimize the system Look for a motor starting at a rated power around 40 watts Pittman DC040B-6

DC040B-6Reference VoltageContinuous TorqueRated CurrentRated PowerTorque ConstantVoltage ConstantTerminal ResistanceMax Winding TemperatureRotor Inertia

240 V00812 Nm236 A37 W0042 NmA0042 Vrads185 Ω155degC847 x 10ndash6 Kg-m2

Motor TransmissionLead Screw Support Rails

ΤRMS

PPK

ωPK

= 02879 Nm= 3369 W= 9276 rads

M

Other Motor Considerationsndash Type stepper brush BLDCndash Input speed (high input speed will result in high noise levels)ndash Total motor footprintndash Ambient temperaturendash Environmental conditionsndash Load-to-rotor inertiandash Encoder ready needed

Motion System Application Example17

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

17

MOTION SYSTEMAPPLICATION EXAMPLE

Gearbox Transmission Selection

When selecting a reduction ratio keep in mind that the RMS required torque at the motor shaft needs to fall below the continuous rated output torque of the motor

G30 A - Planetary GearboxMaximum output load 247 Nm

TRMS( Lead Screw)

02879 Nm

02879 Nm02879 Nm

Reduction

41

5161

Efficiency

090

090090

TRMS( Motor)

00798 Nm

00640 Nm00533 Nm

A 51 gearbox would allow about a 27 safety margin between what the system requires and the continuous torque output of the motor

System Summary 51η = 090

M

= 00807 Nm= 00737 Nm= ndash 00807 Nm= 00640 Nm= 4638 rads= 504 x 10 ndash 6 Kg-m2

= 3743 W

Τ1Τ2Τ3ΤRMS

PPK

ωPK

J

= 03632 Nm= 03316 Nm= ndash 03632 Nm= 02879 Nm= 9276 rads= 1134 x 10 ndash 5 Kg-m2

= 3369 W

Τ1Τ2Τ3ΤRMS

PPK

ωPK

J

Motion System Application Example18

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

18

MOTION SYSTEMAPPLICATION EXAMPLE

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

19

Motion Profile at the Motor

t 1 t 2 t 3 t dwell Time0333 s 0333 s 0333 s 05 s

ωΤ1

Τ2

Τ3= 4638 rads= 4429 RPM= ΤPK = 00807 Nm

ωPK

Τ1

nPK

Drive and Power Supply Requirements

Peak Current Required

IPK =TPK

KT+ IO =

00807 Nm0042 NmA

+ 0180 A = 210 A

RMS Current Required

IRMS =TRMS

KT+ IO =

00640 Nm0042 NmA

+ 0180 A = 170 A

Minimum bus Voltage Required

Vbus= IPKRmt+ ωPKKE = (210A)(185Ω) + (4638 rads)(0042 Vrads)= 389 V + 1948 V= 2337 V

0100 Nm 0200 Nm 0300 Nm 0400 Nm 0500 Nm 0600 Nm 0700 Nm

30 VDC

0100 Nm 0200 Nm 0300 Nm 0500 Nm 0600 Nm 0700 Nm

24 VDC

700060005000

RPM 4000300020001000

0400 Nm

TRMS

TPK

700060005000

RPM 4000300020001000

TRMS

TPK

Performance using 24V reference voltage

Performance using a 30V bus voltage

A 30 VDC bus will meet the application requirements as wells supply a margin of safety

MOTION SYSTEMAPPLICATION EXAMPLE

0100 Nm 0200 Nm 0300 Nm 0400 Nm 0500 Nm 0600 Nm 0700 Nm

30 VDC

0100 Nm 0200 Nm 0300 Nm 0500 Nm 0600 Nm 0700 Nm

24 VDC

700060005000

RPM 4000300020001000

0400 Nm

TRMS

TPK

700060005000

RPM 4000300020001000

TRMS

TPK

Ambient temperature Rated motor temperature Temperature rise Motor temperature

= 30degC= 155degC= 7638degC= 10638degC

Θa

Θm

ΘrΘrated

=Rth x IRMS

2 x Rmt

1 ndash (Rth x IRMS2 x Rmt x 000392degC)

Θr

=11degCω x 170A2

x 185Ω 1 ndash (11degCω x 170A2

x 185Ω x 000392degC)

= 5881 1 ndash (023)

= 5881 077

=

Θm = Θr + Θa = 7638degC + 30degC = 10638degC

7638degC

Performance using 24V reference voltage

Performance using a 30V bus voltage

Will the motor meet the temperature rise caused by the application

Motion System Application Example20

A 30 VDC bus will meet the application requirements as wells supply a margin of safety

203 756 7441 1500 Meriden Road Waterbury CT USA 06705

267 933 2105 534 Godshall DriveHarleysville PA USA 19438

wwwHaydonKerkPittmancom

copy All rights reserved AMETEK Advanced Motion Solutions No part of this doc-ument or technical information can be used reproduced or altered in any form without approval or proper authorization from AMETEK Inc and global affiliates

NOW ASK us the tough motion control questions ONLINE

ASK THE EXPERTS

HaydonKerkPittmancomwww

MOTION SYSTEMAPPLICATION EXAMPLE

Page 17: FORMULAS FOR MOTORIZED LINEAR MOTION SYSTEMS

Lead Screw Input Requirements

= 9276 rads= 2786 rads2

= 03632 Nm= 03316 Nm= ndash 03632 Nm= 0333 s= 0333 s= 0333 s= 05 s= 3369 W= 3076 W

ωPKaΤ1Τ2Τ3

t1t2t3tdwell

PPKPCV

PPK = (Τ1)(ωPK)

= (03632 Nm)(9276 rads)= 3369 W

PCV = (Τ2)(ωPK)

= (03316 Nm)(9276 rads)= 3076 W

RMS Torque Requirement Lead Screw Input

=

0333 s + 0333 s + 0333 s + 05 s

(03632 Nm)2(0333 s)+(03316 Nm)2(0333 s)+(- 03632 Nm)2(0333 s)

ΤRMS = Τ12t 1 + Τ2

2t2 + Τ32t 3

t 1 + t 2 + t 3 + tdwell

15

00439 + 00366 + 00439 = 00829 =

ΤRMS = 02879 Nm

Motion System Application Example16

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

16

MOTION SYSTEMAPPLICATION EXAMPLE

Load Parameters at the Lead Screw Shaft

= 9276 rads= 2786 rads2

= 03632 Nm= 03316 Nm= ndash 03632 Nm= 02879 Nm= 3369 W= 3076 W

ωPKaΤ1Τ2Τ3ΤRMSPPKPCV

Motor SelectionA direct-drive motor can be used however it would be large and expensive For example a DC057B-3 brush motor

would easily meet continuous torque and continuous speed as well as a 181 load-to-rotor inertia This motor however has significantly more output power than needed at 128 watts

A smaller motor with a transmission would optimize the system Look for a motor starting at a rated power around 40 watts Pittman DC040B-6

DC040B-6Reference VoltageContinuous TorqueRated CurrentRated PowerTorque ConstantVoltage ConstantTerminal ResistanceMax Winding TemperatureRotor Inertia

240 V00812 Nm236 A37 W0042 NmA0042 Vrads185 Ω155degC847 x 10ndash6 Kg-m2

Motor TransmissionLead Screw Support Rails

ΤRMS

PPK

ωPK

= 02879 Nm= 3369 W= 9276 rads

M

Other Motor Considerationsndash Type stepper brush BLDCndash Input speed (high input speed will result in high noise levels)ndash Total motor footprintndash Ambient temperaturendash Environmental conditionsndash Load-to-rotor inertiandash Encoder ready needed

Motion System Application Example17

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17

MOTION SYSTEMAPPLICATION EXAMPLE

Gearbox Transmission Selection

When selecting a reduction ratio keep in mind that the RMS required torque at the motor shaft needs to fall below the continuous rated output torque of the motor

G30 A - Planetary GearboxMaximum output load 247 Nm

TRMS( Lead Screw)

02879 Nm

02879 Nm02879 Nm

Reduction

41

5161

Efficiency

090

090090

TRMS( Motor)

00798 Nm

00640 Nm00533 Nm

A 51 gearbox would allow about a 27 safety margin between what the system requires and the continuous torque output of the motor

System Summary 51η = 090

M

= 00807 Nm= 00737 Nm= ndash 00807 Nm= 00640 Nm= 4638 rads= 504 x 10 ndash 6 Kg-m2

= 3743 W

Τ1Τ2Τ3ΤRMS

PPK

ωPK

J

= 03632 Nm= 03316 Nm= ndash 03632 Nm= 02879 Nm= 9276 rads= 1134 x 10 ndash 5 Kg-m2

= 3369 W

Τ1Τ2Τ3ΤRMS

PPK

ωPK

J

Motion System Application Example18

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

18

MOTION SYSTEMAPPLICATION EXAMPLE

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

19

Motion Profile at the Motor

t 1 t 2 t 3 t dwell Time0333 s 0333 s 0333 s 05 s

ωΤ1

Τ2

Τ3= 4638 rads= 4429 RPM= ΤPK = 00807 Nm

ωPK

Τ1

nPK

Drive and Power Supply Requirements

Peak Current Required

IPK =TPK

KT+ IO =

00807 Nm0042 NmA

+ 0180 A = 210 A

RMS Current Required

IRMS =TRMS

KT+ IO =

00640 Nm0042 NmA

+ 0180 A = 170 A

Minimum bus Voltage Required

Vbus= IPKRmt+ ωPKKE = (210A)(185Ω) + (4638 rads)(0042 Vrads)= 389 V + 1948 V= 2337 V

0100 Nm 0200 Nm 0300 Nm 0400 Nm 0500 Nm 0600 Nm 0700 Nm

30 VDC

0100 Nm 0200 Nm 0300 Nm 0500 Nm 0600 Nm 0700 Nm

24 VDC

700060005000

RPM 4000300020001000

0400 Nm

TRMS

TPK

700060005000

RPM 4000300020001000

TRMS

TPK

Performance using 24V reference voltage

Performance using a 30V bus voltage

A 30 VDC bus will meet the application requirements as wells supply a margin of safety

MOTION SYSTEMAPPLICATION EXAMPLE

0100 Nm 0200 Nm 0300 Nm 0400 Nm 0500 Nm 0600 Nm 0700 Nm

30 VDC

0100 Nm 0200 Nm 0300 Nm 0500 Nm 0600 Nm 0700 Nm

24 VDC

700060005000

RPM 4000300020001000

0400 Nm

TRMS

TPK

700060005000

RPM 4000300020001000

TRMS

TPK

Ambient temperature Rated motor temperature Temperature rise Motor temperature

= 30degC= 155degC= 7638degC= 10638degC

Θa

Θm

ΘrΘrated

=Rth x IRMS

2 x Rmt

1 ndash (Rth x IRMS2 x Rmt x 000392degC)

Θr

=11degCω x 170A2

x 185Ω 1 ndash (11degCω x 170A2

x 185Ω x 000392degC)

= 5881 1 ndash (023)

= 5881 077

=

Θm = Θr + Θa = 7638degC + 30degC = 10638degC

7638degC

Performance using 24V reference voltage

Performance using a 30V bus voltage

Will the motor meet the temperature rise caused by the application

Motion System Application Example20

A 30 VDC bus will meet the application requirements as wells supply a margin of safety

203 756 7441 1500 Meriden Road Waterbury CT USA 06705

267 933 2105 534 Godshall DriveHarleysville PA USA 19438

wwwHaydonKerkPittmancom

copy All rights reserved AMETEK Advanced Motion Solutions No part of this doc-ument or technical information can be used reproduced or altered in any form without approval or proper authorization from AMETEK Inc and global affiliates

NOW ASK us the tough motion control questions ONLINE

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MOTION SYSTEMAPPLICATION EXAMPLE

Page 18: FORMULAS FOR MOTORIZED LINEAR MOTION SYSTEMS

Load Parameters at the Lead Screw Shaft

= 9276 rads= 2786 rads2

= 03632 Nm= 03316 Nm= ndash 03632 Nm= 02879 Nm= 3369 W= 3076 W

ωPKaΤ1Τ2Τ3ΤRMSPPKPCV

Motor SelectionA direct-drive motor can be used however it would be large and expensive For example a DC057B-3 brush motor

would easily meet continuous torque and continuous speed as well as a 181 load-to-rotor inertia This motor however has significantly more output power than needed at 128 watts

A smaller motor with a transmission would optimize the system Look for a motor starting at a rated power around 40 watts Pittman DC040B-6

DC040B-6Reference VoltageContinuous TorqueRated CurrentRated PowerTorque ConstantVoltage ConstantTerminal ResistanceMax Winding TemperatureRotor Inertia

240 V00812 Nm236 A37 W0042 NmA0042 Vrads185 Ω155degC847 x 10ndash6 Kg-m2

Motor TransmissionLead Screw Support Rails

ΤRMS

PPK

ωPK

= 02879 Nm= 3369 W= 9276 rads

M

Other Motor Considerationsndash Type stepper brush BLDCndash Input speed (high input speed will result in high noise levels)ndash Total motor footprintndash Ambient temperaturendash Environmental conditionsndash Load-to-rotor inertiandash Encoder ready needed

Motion System Application Example17

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

17

MOTION SYSTEMAPPLICATION EXAMPLE

Gearbox Transmission Selection

When selecting a reduction ratio keep in mind that the RMS required torque at the motor shaft needs to fall below the continuous rated output torque of the motor

G30 A - Planetary GearboxMaximum output load 247 Nm

TRMS( Lead Screw)

02879 Nm

02879 Nm02879 Nm

Reduction

41

5161

Efficiency

090

090090

TRMS( Motor)

00798 Nm

00640 Nm00533 Nm

A 51 gearbox would allow about a 27 safety margin between what the system requires and the continuous torque output of the motor

System Summary 51η = 090

M

= 00807 Nm= 00737 Nm= ndash 00807 Nm= 00640 Nm= 4638 rads= 504 x 10 ndash 6 Kg-m2

= 3743 W

Τ1Τ2Τ3ΤRMS

PPK

ωPK

J

= 03632 Nm= 03316 Nm= ndash 03632 Nm= 02879 Nm= 9276 rads= 1134 x 10 ndash 5 Kg-m2

= 3369 W

Τ1Τ2Τ3ΤRMS

PPK

ωPK

J

Motion System Application Example18

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

18

MOTION SYSTEMAPPLICATION EXAMPLE

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

19

Motion Profile at the Motor

t 1 t 2 t 3 t dwell Time0333 s 0333 s 0333 s 05 s

ωΤ1

Τ2

Τ3= 4638 rads= 4429 RPM= ΤPK = 00807 Nm

ωPK

Τ1

nPK

Drive and Power Supply Requirements

Peak Current Required

IPK =TPK

KT+ IO =

00807 Nm0042 NmA

+ 0180 A = 210 A

RMS Current Required

IRMS =TRMS

KT+ IO =

00640 Nm0042 NmA

+ 0180 A = 170 A

Minimum bus Voltage Required

Vbus= IPKRmt+ ωPKKE = (210A)(185Ω) + (4638 rads)(0042 Vrads)= 389 V + 1948 V= 2337 V

0100 Nm 0200 Nm 0300 Nm 0400 Nm 0500 Nm 0600 Nm 0700 Nm

30 VDC

0100 Nm 0200 Nm 0300 Nm 0500 Nm 0600 Nm 0700 Nm

24 VDC

700060005000

RPM 4000300020001000

0400 Nm

TRMS

TPK

700060005000

RPM 4000300020001000

TRMS

TPK

Performance using 24V reference voltage

Performance using a 30V bus voltage

A 30 VDC bus will meet the application requirements as wells supply a margin of safety

MOTION SYSTEMAPPLICATION EXAMPLE

0100 Nm 0200 Nm 0300 Nm 0400 Nm 0500 Nm 0600 Nm 0700 Nm

30 VDC

0100 Nm 0200 Nm 0300 Nm 0500 Nm 0600 Nm 0700 Nm

24 VDC

700060005000

RPM 4000300020001000

0400 Nm

TRMS

TPK

700060005000

RPM 4000300020001000

TRMS

TPK

Ambient temperature Rated motor temperature Temperature rise Motor temperature

= 30degC= 155degC= 7638degC= 10638degC

Θa

Θm

ΘrΘrated

=Rth x IRMS

2 x Rmt

1 ndash (Rth x IRMS2 x Rmt x 000392degC)

Θr

=11degCω x 170A2

x 185Ω 1 ndash (11degCω x 170A2

x 185Ω x 000392degC)

= 5881 1 ndash (023)

= 5881 077

=

Θm = Θr + Θa = 7638degC + 30degC = 10638degC

7638degC

Performance using 24V reference voltage

Performance using a 30V bus voltage

Will the motor meet the temperature rise caused by the application

Motion System Application Example20

A 30 VDC bus will meet the application requirements as wells supply a margin of safety

203 756 7441 1500 Meriden Road Waterbury CT USA 06705

267 933 2105 534 Godshall DriveHarleysville PA USA 19438

wwwHaydonKerkPittmancom

copy All rights reserved AMETEK Advanced Motion Solutions No part of this doc-ument or technical information can be used reproduced or altered in any form without approval or proper authorization from AMETEK Inc and global affiliates

NOW ASK us the tough motion control questions ONLINE

ASK THE EXPERTS

HaydonKerkPittmancomwww

MOTION SYSTEMAPPLICATION EXAMPLE

Page 19: FORMULAS FOR MOTORIZED LINEAR MOTION SYSTEMS

Gearbox Transmission Selection

When selecting a reduction ratio keep in mind that the RMS required torque at the motor shaft needs to fall below the continuous rated output torque of the motor

G30 A - Planetary GearboxMaximum output load 247 Nm

TRMS( Lead Screw)

02879 Nm

02879 Nm02879 Nm

Reduction

41

5161

Efficiency

090

090090

TRMS( Motor)

00798 Nm

00640 Nm00533 Nm

A 51 gearbox would allow about a 27 safety margin between what the system requires and the continuous torque output of the motor

System Summary 51η = 090

M

= 00807 Nm= 00737 Nm= ndash 00807 Nm= 00640 Nm= 4638 rads= 504 x 10 ndash 6 Kg-m2

= 3743 W

Τ1Τ2Τ3ΤRMS

PPK

ωPK

J

= 03632 Nm= 03316 Nm= ndash 03632 Nm= 02879 Nm= 9276 rads= 1134 x 10 ndash 5 Kg-m2

= 3369 W

Τ1Τ2Τ3ΤRMS

PPK

ωPK

J

Motion System Application Example18

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

18

MOTION SYSTEMAPPLICATION EXAMPLE

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

19

Motion Profile at the Motor

t 1 t 2 t 3 t dwell Time0333 s 0333 s 0333 s 05 s

ωΤ1

Τ2

Τ3= 4638 rads= 4429 RPM= ΤPK = 00807 Nm

ωPK

Τ1

nPK

Drive and Power Supply Requirements

Peak Current Required

IPK =TPK

KT+ IO =

00807 Nm0042 NmA

+ 0180 A = 210 A

RMS Current Required

IRMS =TRMS

KT+ IO =

00640 Nm0042 NmA

+ 0180 A = 170 A

Minimum bus Voltage Required

Vbus= IPKRmt+ ωPKKE = (210A)(185Ω) + (4638 rads)(0042 Vrads)= 389 V + 1948 V= 2337 V

0100 Nm 0200 Nm 0300 Nm 0400 Nm 0500 Nm 0600 Nm 0700 Nm

30 VDC

0100 Nm 0200 Nm 0300 Nm 0500 Nm 0600 Nm 0700 Nm

24 VDC

700060005000

RPM 4000300020001000

0400 Nm

TRMS

TPK

700060005000

RPM 4000300020001000

TRMS

TPK

Performance using 24V reference voltage

Performance using a 30V bus voltage

A 30 VDC bus will meet the application requirements as wells supply a margin of safety

MOTION SYSTEMAPPLICATION EXAMPLE

0100 Nm 0200 Nm 0300 Nm 0400 Nm 0500 Nm 0600 Nm 0700 Nm

30 VDC

0100 Nm 0200 Nm 0300 Nm 0500 Nm 0600 Nm 0700 Nm

24 VDC

700060005000

RPM 4000300020001000

0400 Nm

TRMS

TPK

700060005000

RPM 4000300020001000

TRMS

TPK

Ambient temperature Rated motor temperature Temperature rise Motor temperature

= 30degC= 155degC= 7638degC= 10638degC

Θa

Θm

ΘrΘrated

=Rth x IRMS

2 x Rmt

1 ndash (Rth x IRMS2 x Rmt x 000392degC)

Θr

=11degCω x 170A2

x 185Ω 1 ndash (11degCω x 170A2

x 185Ω x 000392degC)

= 5881 1 ndash (023)

= 5881 077

=

Θm = Θr + Θa = 7638degC + 30degC = 10638degC

7638degC

Performance using 24V reference voltage

Performance using a 30V bus voltage

Will the motor meet the temperature rise caused by the application

Motion System Application Example20

A 30 VDC bus will meet the application requirements as wells supply a margin of safety

203 756 7441 1500 Meriden Road Waterbury CT USA 06705

267 933 2105 534 Godshall DriveHarleysville PA USA 19438

wwwHaydonKerkPittmancom

copy All rights reserved AMETEK Advanced Motion Solutions No part of this doc-ument or technical information can be used reproduced or altered in any form without approval or proper authorization from AMETEK Inc and global affiliates

NOW ASK us the tough motion control questions ONLINE

ASK THE EXPERTS

HaydonKerkPittmancomwww

MOTION SYSTEMAPPLICATION EXAMPLE

Page 20: FORMULAS FOR MOTORIZED LINEAR MOTION SYSTEMS

wwwHaydonKerkPittmancomHaydon Kerk Motion Solutions 203 756 7441Pittman Motors 267 933 2105

19

Motion Profile at the Motor

t 1 t 2 t 3 t dwell Time0333 s 0333 s 0333 s 05 s

ωΤ1

Τ2

Τ3= 4638 rads= 4429 RPM= ΤPK = 00807 Nm

ωPK

Τ1

nPK

Drive and Power Supply Requirements

Peak Current Required

IPK =TPK

KT+ IO =

00807 Nm0042 NmA

+ 0180 A = 210 A

RMS Current Required

IRMS =TRMS

KT+ IO =

00640 Nm0042 NmA

+ 0180 A = 170 A

Minimum bus Voltage Required

Vbus= IPKRmt+ ωPKKE = (210A)(185Ω) + (4638 rads)(0042 Vrads)= 389 V + 1948 V= 2337 V

0100 Nm 0200 Nm 0300 Nm 0400 Nm 0500 Nm 0600 Nm 0700 Nm

30 VDC

0100 Nm 0200 Nm 0300 Nm 0500 Nm 0600 Nm 0700 Nm

24 VDC

700060005000

RPM 4000300020001000

0400 Nm

TRMS

TPK

700060005000

RPM 4000300020001000

TRMS

TPK

Performance using 24V reference voltage

Performance using a 30V bus voltage

A 30 VDC bus will meet the application requirements as wells supply a margin of safety

MOTION SYSTEMAPPLICATION EXAMPLE

0100 Nm 0200 Nm 0300 Nm 0400 Nm 0500 Nm 0600 Nm 0700 Nm

30 VDC

0100 Nm 0200 Nm 0300 Nm 0500 Nm 0600 Nm 0700 Nm

24 VDC

700060005000

RPM 4000300020001000

0400 Nm

TRMS

TPK

700060005000

RPM 4000300020001000

TRMS

TPK

Ambient temperature Rated motor temperature Temperature rise Motor temperature

= 30degC= 155degC= 7638degC= 10638degC

Θa

Θm

ΘrΘrated

=Rth x IRMS

2 x Rmt

1 ndash (Rth x IRMS2 x Rmt x 000392degC)

Θr

=11degCω x 170A2

x 185Ω 1 ndash (11degCω x 170A2

x 185Ω x 000392degC)

= 5881 1 ndash (023)

= 5881 077

=

Θm = Θr + Θa = 7638degC + 30degC = 10638degC

7638degC

Performance using 24V reference voltage

Performance using a 30V bus voltage

Will the motor meet the temperature rise caused by the application

Motion System Application Example20

A 30 VDC bus will meet the application requirements as wells supply a margin of safety

203 756 7441 1500 Meriden Road Waterbury CT USA 06705

267 933 2105 534 Godshall DriveHarleysville PA USA 19438

wwwHaydonKerkPittmancom

copy All rights reserved AMETEK Advanced Motion Solutions No part of this doc-ument or technical information can be used reproduced or altered in any form without approval or proper authorization from AMETEK Inc and global affiliates

NOW ASK us the tough motion control questions ONLINE

ASK THE EXPERTS

HaydonKerkPittmancomwww

MOTION SYSTEMAPPLICATION EXAMPLE

Page 21: FORMULAS FOR MOTORIZED LINEAR MOTION SYSTEMS

0100 Nm 0200 Nm 0300 Nm 0400 Nm 0500 Nm 0600 Nm 0700 Nm

30 VDC

0100 Nm 0200 Nm 0300 Nm 0500 Nm 0600 Nm 0700 Nm

24 VDC

700060005000

RPM 4000300020001000

0400 Nm

TRMS

TPK

700060005000

RPM 4000300020001000

TRMS

TPK

Ambient temperature Rated motor temperature Temperature rise Motor temperature

= 30degC= 155degC= 7638degC= 10638degC

Θa

Θm

ΘrΘrated

=Rth x IRMS

2 x Rmt

1 ndash (Rth x IRMS2 x Rmt x 000392degC)

Θr

=11degCω x 170A2

x 185Ω 1 ndash (11degCω x 170A2

x 185Ω x 000392degC)

= 5881 1 ndash (023)

= 5881 077

=

Θm = Θr + Θa = 7638degC + 30degC = 10638degC

7638degC

Performance using 24V reference voltage

Performance using a 30V bus voltage

Will the motor meet the temperature rise caused by the application

Motion System Application Example20

A 30 VDC bus will meet the application requirements as wells supply a margin of safety

203 756 7441 1500 Meriden Road Waterbury CT USA 06705

267 933 2105 534 Godshall DriveHarleysville PA USA 19438

wwwHaydonKerkPittmancom

copy All rights reserved AMETEK Advanced Motion Solutions No part of this doc-ument or technical information can be used reproduced or altered in any form without approval or proper authorization from AMETEK Inc and global affiliates

NOW ASK us the tough motion control questions ONLINE

ASK THE EXPERTS

HaydonKerkPittmancomwww

MOTION SYSTEMAPPLICATION EXAMPLE