Exemple de Calcul Motoare

8/8/2019 Exemple de Calcul Motoare

http://slidepdf.com/reader/full/exemple-de-calcul-motoare 1/10G-2 ORIENTAL MOTOR GENERAL CATALOGUE

T e ch ni c al R ef er en c e

Motor Sizing Calculations

Selecting a motor that sufficiently satisfies the specifications required by the equipment is an important key to ensuring the desired reliability

and economy of the equipment.

This section describes the procedure to select the optimum motor for a particular application, as well as the selection calculations, key points

of selection and examples.

Selection ProcedureAn overview of selection procedure is explained below.

First, determine the drive mechanism. Representative drive mechanisms include simple body of rotation, ball

screw, belt pulley, and rack and pinion. Along with the type of drive mechanism, you must also determine the

mass of transferred work, dimensions of each part, friction coefficient of the sliding surface, and so on.

Confirm the drive conditions such as the speed of movement, drive time, and also positioning distance and

period if positioning operation will be performed. Also confirm the stop accuracy, resolution, position holding,

operating voltage, operating environment, and so on.

Calculate the values for load torque and load inertia at the motor drive shaft. See the left column on page G-3 forthe calculation of load torque for representative mechanisms. See the right column on page G-3 for the

calculation of inertia for representative shapes.

Select a motor type from AC Motors, Brushless DC Motors or Stepping Motors based on the required

specifications.

Make a final determination of the motor after confirming that the specifications of the selected motor/gearhead

satisfy all of the requirements, such as mechanical strength, acceleration period and acceleration torque. Since

the specific items that must be checked will vary depending on the motor model, see the selection calculations

and selection points explained on page G-4 and subsequent pages.

Determine the drive

mechanism component

Confirm the required specifications

(Equipment specifications)

Calculate the speed and load

Select motor type

Selection calculation


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Service Life

Motor and

Fan Sizing

Standard

AC Motors

Brushless

DC Motors

Stepping

Motors

Gearheads

Linear

Motion

Fan Motors

Formulas for Calculating Load TorqueCalculate the friction torque for the applicable drive mechanism.

Ball Screw

Pulley

Wire Belt Mechanism, Rack and Pinion Mechanism

By Actual Measurement

F =Force of moving direction [N]

F 0 =Pilot pressure load [N] (1/3F)

0 =Internal friction coefficient of pilot pressure nut (0.10.3)

=Efficiency (0.850.95)

i =Gear ratio (This is the gear ratio of the mechanism and not

the gear ratio of the Oriental Motor gearhead you are

selecting.)

PB =Ball screw pitch [m/rev]

F A =External force [N]

F B =Force when main shaft begins to rotate [N]

( F B= [value for spring balance] (kg) g [m/s2])

m =Total mass of work and table [kg]

=Frictional coefficient of sliding surfaces (0.05)

=Angle of inclination [˚ ]

D =Final pulley diameter [m]

g =Gravitational acceleration [m/s2] (9.807)

F B

D

M a c h i n e

Pulley

Spring Balance

= [Nm] yT L F B D

2

m F A

D

F m F A

D

F

= [Nm] rT L F

2 =

FD

2 i

D

i

= F Am g (sin a cos a) [N] t F

m F A

D

= T L D

i

F Am g

2

= [Nm] e( F Am g) D

2i

F A m

Direct Coupling

F A

F

m

= ( ) [Nm] qT L FPB

2 2

1

i

0 F 0 PB

= F Am g (sin a cos a) [N] w F

Formulas for Calculating Moment of InertiaInertia of a Cylinder

Inertia of a Hollow Cylinder

Inertia for Off-center Axis of Rotation

Inertia of a Rectangular Pillar

Inertia of an Object in Linear Motion

Density

Iron =7.9103 [kg/m3]

Aluminum =2.8103 [kg/m3]

Bronze =8.5103

[kg/m3

]Nylon =1.1103 [kg/m3]

Jx =Inertia on x axis [kgm2]

Jy =Inertia on y axis [kgm2]

J 0 =Inertia on x0 axis (passing through center

of gravity) [kgm2]

m =Mass [kg]

D1 =External diameter [m]

D2 =Internal diameter [m]

=Density [kg/m3]

L =Length [m]

A=Unit of movement [m/rev]

[kgm2] !4 J = m ( ) 2

2

A

A B

C

x

y

= [kgm2] !2 Jx m ( A2 B2)=12

1 ABC ( A2 B2)

12

1

= [kgm2] !3 Jy m ( B2C2)=12

1 ABC ( B2C2)

12

1

C

A B

x x0

= Jx0 [kgm2] !1 Jx ml2= m ( A2 B212l2)12

1

r= Distance between x and x0 axes [m]

L

D1

D2

x

y

= [kgm2] o Jx m ( D12 D2

2)=8

1 L ( D1

4 D24)

32

= [kgm2] !0 Jy m4

1( )

4

D12 D2

2

3

L2

D1

L

x

y

= [kgm2] u Jx mD12

8

1= LD1

4

32

= [kgm2] i Jy m4

1( )

4

D12

3

L2




Motor Sizing CalculationsThe following explains the calculation for selecting a stepping

motor based on pulse control:

Operating PatternsThere are 2 basic motion profiles. One is a start/stop operation

and the other is an acceleration/deceleration operation.

Acceleration/deceleration operation is the most common. When

load inertia is small, start/stop operation can be used.

Calculate the Number of Operating Pulses AThe number of operating pulses is expressed as the number of

pulse signals that adds up to the angle that the motor must move

to get the work from point A to point B.

Calculate the Operating Pulse Speed f 2

The operating pulse speed can be found from the number of

operating pulses, the positioning period and the

acceleration/deceleration period.

qFor Acceleration/Deceleration Operation

The level of acceleration (deceleration) period is an important

point in the selection. The acceleration (deceleration) period

cannot be set hastily, because it correlates with the acceleration

torque and acceleration/deceleration rate.Initially, set the acceleration (deceleration) period at roughly 25%

of the positioning period. (The setting must be fine-tuned before

the final decision can be made.)

Acceleration (Deceleration) Period [s] =Positioning Period [s]0.25

wFor Start-Stop Operation

Calculate the Acceleration/Deceleration Rate T R The values represent the specifications of Oriental Motor

controllers.

The acceleration/deceleration rate indicates the degree of

acceleration of pulse speed and is calculated using the equation

shown below:

t1

T R

PulseSpeed

[ k

H z]

=

Acceleration/ deceleration

rate T R

[ms/kHz]

✽Calculate the pulse speed in full-step equivalents.

Operating Pulse

Speed[kHz]

Starting Pulse

Speed [kHz]

Acceleration (Deceleration)Period[ms]

= f 2 f 1

t1

=Operating PulseSpeed f 2[Hz]

Number of Operating Pulses

[Pulses]

Positioning Period [s]

= A

t0

=Operating Pulse

Speed f 2[Hz]

Number of Operating Pulses

[Pulses]Starting Pulse

Speed [Hz]

Acceleration(Deceleration)

Period [s]

Positioning Period[s]

Acceleration (Deceleration)Period [s]

=A f 1t1

t0t1

=

No. of PulsesRequired for1 Motor Rotation

Number of OperatingPulses A[Pulses]

Distance per MovementDistance per Motor Rotation

= : Step Anglel

lrev

360˚θ s

θ s

Positioning Period

Start/Stop Operation

Operating PulseSpeed

Number ofOperating Pulses

A

t 0

f 2

Positioning Period

Acceleration/Deceleration Operation

Acceleration Period t 1 t 1

t 0 DecelerationPeriod

Operating PulseSpeedf 2

f 1

Starting PulseSpeed

Number ofOperating Pulses

A

Calculate the Operating Speed from Operating Pulse speed

Calculate the Load TorqueSee basic equations on page G-3.

Calculate the Acceleration TorqueRegardless of the motor type, the acceleration/deceleration torque

must always be set if the speed is to be varied.

The basic equation is the same for all motors. However, different

equations apply to stepping motors, as shown below, because the

specifications of stepping motors are often calculated on the basis

of pulse speed.

<Standard AC Motors> (The same equation applies to brushless

DC motors.)

<Stepping Motors>

qFor Acceleration/Deceleration Operation

wFor Start-Stop Operation

Calculate the Required Torque T M

The required torque is calculated by multiplying the sum of load

torque and acceleration torque by the safety factor.

Required Torque T M= (Load Torque Acceleration Torque) Safety Factor

=(T LT a)S f

=

Acceleration TorqueT a [Nm]

Rotor Inertia[kgm2]

Total Inertia[kgm2]

Step Angle[˚ ](Operating Pulse Speed)2 [Hz]

180˚coefficient

= ( J 0 J L) s f 22

180n n: 3.6 ̊/ sθ

θ

=

Acceleration TorqueT a [Nm]

Rotor Inertia[kgm2]

Total Inertia[kgm2]

Operating Pulse

Speed [Hz]

Starting Pulse

Speed [Hz]


Step Angle[˚ ]

180˚

= ( J 0 J L) f 2 f 1

t1

s

180

θ

Operating

Speed[r/min]

N M

AccelerationPeriod t 1

Deceleration

Period t 1Positioning Period t 0

For AC Motors

Movement

AccelerationTorqueT a[Nm]

=coefficient

Rotor Inertia[kgm2]

Total Inertia[kgm2] Operating Speed [r/min]


=9.55

( J 0 J L)

t1

N M

60Operating

Speed[r/min]=Operating Pulse Speed [Hz]

Step Angle

360




Service Life

Motor and

Fan Sizing

Standard

AC Motors

Brushless

DC Motors

Stepping

Motors

Gearheads

Linear

Motion

Fan Motors

Selection ConsiderationsThere are differences in characteristics between standard AC

motors and stepping motors. Shown below are some of the points

you should know when sizing a motor.

Standard AC MotorsqSpeed Variation by Load

The speed of induction motors and reversible motors varies by

several percent with the size of the load torque. Therefore,when selecting an induction motor or reversible motor the

selection should take into account this possible speed variation

by load.

wRating

There can be a difference of continuous and short-term ratings,

due to the difference in motor specifications, despite the fact

that two motors have the same output power. Motor selection

should be based on the operating time (operating pattern).

ePermissible Load Inertia for Gearheads

If instantaneous stop (using a brake pack, etc.), frequent

intermittent operations or instantaneous reversing will be

performed using a gearhead, an excessive load inertia maydamage the gearhead. In these applications, therefore, the

selection must be made so the load inertia does not exceed the

permissible load inertia for the gearhead (see page A-13).

Stepping MotorsqChecking the Running Duty Cycle

A stepping motor is not intended to be run continuously with

rated current. Lower than 50% running duty cycle is

recommended.

wChecking the Inertia Ratio

Large inertia ratios cause large overshooting and undershooting

during starting and stopping, which can affect start-up times and

settling times. Depending on the conditions of usage, operation

may be impossible.

Calculate the inertia ratio with the following equation and check

that the values found are at or below the inertia ratios shown in

the table.

Inertia Ratio (Reference Values)

Except geared motor types

When these values are exceeded, we recommend a geared

motor.

Using a geared motor can increase the drivable inertia load.

=Inertia RatioTotal Inertia of the Machine [kgm2]

Rotor Inertia of the Motor [kgm2](Gear Ratio)2

= J L J 0 i2

Product Series Motor Frame Size

Stepping Motor and

Driver Packages

28, 42, 60, 85

20, 28

42, 60, 85

Inertia Ratio

30 Maximum

5 Maximum

10 Maximum

=Inertia RatioTotal Inertia of the Machine [kgm2]

Rotor Inertia of the Motor [kgm2]

= J L

J 0

Running Duty Cycle 100Running Time

RunningTime

StoppingTime

eCheck the Acceleration/Deceleration Rate

Most controllers, when set for acceleration or deceleration,

adjust the pulse speed in steps. For that reason, operation may

sometimes not be possible, even though it can be calculated.

Calculate the acceleration/deceleration rate from the following

equation and check that the value is at or above the

acceleration/deceleration rate in the table.

Acceleration Rate (Reference Values with XG9200, SG9200 Series)

✽This item need not be checked for . The value in the table represents the lower

limit of setting for the XG9200 and SG9200 Series.

rChecking the Required Torque

Check that the required torque falls within the pull-out torque of

the speed-torque characteristics.

Safety Factor: Sf (Reference Value)

Required Torque

Speed [r/min]

(Pulse Speed [kHz])

T o r q u e

[ N

m ]

Product Series

Stepping Motor andDriver Packages

Safety Factor(Reference Value)

1.52

2

Product Series Motor Frame Size Acceleration/Deceleration RateT R[ms/kHz]28, 42, 60, 85

42, 60

85, 90

0.1 Maximum✽20 Maximum

30 Maximum

Stepping Motor and

Driver Packages




Sizing Example

Ball Screw

Determine the Drive Mechanism

Total mass of the table and work ·····································m=40 kg

Frictional coefficient of sliding surfaces ·······························=0.05

Ball screw efficiency ····························································· =0.9

Internal frictional coefficient of pilot pressure nut ·················0=0.3Ball screw shaft diameter············································· D B=15 mm

Total length of ball screw ············································ L B=600 mm

Material of ball screw ·················Iron (density =7.9103 [kg/m3])

Ball screw pitch ···························································· P B=15 mm

Desired resolution···············································∆l=0.03 mm/step

(feed per pulse)

Feed ·············································································l=180 mm

Positioning period ·············································t0=Within 0.8 sec.

Calculate the Required Resolution

can be connected directly to the application.

Determine the Operating Pattern (See page G-4 for basic equations)

(1) Finding the Number of Operating Pulses A [pulses]

(2) Determine the Acceleration (Deceleration) Period t1 [s]

An acceleration (deceleration) period of 25% of the positioning

period is appropriate.

Acceleration (Deceleration) Period t1=0.80.25=0.2 [s]

(3) Determine the Operating Pulse Speed f 2 [Hz]

period [s]t 1 t 1

t 0=0.8

Ope

rating

pulse

speed

[ H

z]

10000

0.2 0.2

6000 pulses

=Operating pulse

speed f 2

Operating pulses ( A)Starting pulse speed ( f 1) Acceleration (Deceleration) Period (t1)

Positioning Period (t0) Acceleration (Deceleration) Period (t1)

=60000

0.80.2=10000 [Hz]

= Operating pulses AFeed per Unit (l)

Ball Screw Pitch ( PB)

360˚

Step Angle ( s)

= 180

15=6000 [pulses]

360˚

0.72˚

θ

=Required Resolution s360˚Desired Resolution (∆l)

Ball Screw Pitch ( PB)

= =0.72 [˚ ]360˚0.03

15

θ

P B

D B

SteppingMotor

Controller

Driver

Programmable

Controller

Coupling

DirectConnection

m

(4) Calculate the Operating Speed N [r/min]

Calculate the Required Torque T M [Nm] (see page G-4)

(1) Calculate the Load Torque T L [Nm]

(2) Calculate the Acceleration Torque T a [Nm]

qCalculate the total moment of inertia J L [kgm2

](See page G-3 for basic equations)

wCalculate the acceleration torque Ta [Nm]

(3) Calculate the Required Torque T M [Nm]

Required torque T M =(T LTa)2

={0.0567(628 J 00.158)}2

=1256 J 00.429 [Nm]

Select a Motor

(1) Provisional Motor Selection

(2) Determine the Motor from the Speed-Torque Characteristics

AS66AA

Select a motor for which the required torque falls within the pull-

out torque of the speed-torque characteristics.

01000 2000 3000 4000

Torque[kgfcm]

Torque[N

m]

Speed [r/min]

Pulse Speed [kHz]100 20 30 40 50 60

(Resolution Setting: 1000 P/R)

1.5

2.0

1.0

0.5

15

20

10

5

0

405107 AS66AA

Rotor Inertia[kgm2]

0.48

Required Torque[Nm]Model

=( J 0 J L ) Acceleration torque T a f 2 f 1

t1

s180˚

=( J 02.52104 )

=628 J 00.158 [Nm]

100000

0.2

0.72180˚

θ

=Inertia of Ball Screw J B

32 LB DB

4

=

=0.236104 [kgm2]

327.9103600103 (15103)4

=m (

=40(

)2

Inertia of Table and Work J T

= J B J T

=0.2361042.28104=2.52104 [kgm2]

Total Inertia J L

PB

2

2 =2.28104 [kgm2])

215103

=F Am g (sin a cos a)

=0409.807 (sin00.05cos0)

=19.6 [N]

Force of moving direction F

=Load Torque T L F PB

2

0 F 0 PB

2

=Pilot Pressure Load F 0 F

3=6.53 [N]=

19.6

3

=

=0.0567 [Nm]

19.615103

2 0.9

0.36.5315103

2

= f 2 60Operating SpeedS

360

=10000 600.72360

=1200 [r/min]

θ

Using Stepping Motors ( )




Service Life

Motor and

Fan Sizing

Standard

AC Motors

Brushless

DC Motors

Stepping

Motors

Gearheads

Linear

Motion

Fan Motors

This example demonstrates how to select an AC motor with an

electromagnetic brake for use on a tabletop moving vertically on a

ball screw. In this case, a motor must be selected that meets the

following basic specifications.

<Required and Structural Specifications>

Total mass of the table and work ···································m=30 [kg]

Table speed ························································V =152 [mm/s]

External force····································································· F A=0[N]Ball screw tilt angle ····························································=90 [˚ ]

Total length of ball screw··········································· L B=800 [mm]

Ball screw shaft diameter············································ D B=20 [mm]

Ball screw pitch ····························································· P B=5 [mm]

Distance moved for one rotation of ball screw ················ A=5 [mm]

Ball screw efficiency ······························································· =0.9

Material of ball screw ·····················Iron (density =7.9103kg/m3)

Internal frictional coefficient of pilot pressure nut ·················0=0.3

Frictional coefficient of sliding surfaces································=0.05

Motor power supply ·························Single-Phase 110 VAC 60 Hz

Movement time ·····················Intermittent operation, 5 hours/day

Load with repeated starts and stops

Required load holding

Determine the Gear Ratio

Because the rated speed for a 4-pole motor at 60 Hz is

14501550 r/min,

From within this range a gear ratio of i =9 is selected.

Calculate the Required Torque

= F Am g (sin cos )

=0309.807 (sin90˚0.05cos90˚)

=294 [N]

Load weight in the directionof the ball screw shaft F

=Load torque T L F P B

2 2

2

0 F 0 P B

=Pilot pressure load F 0 F

3=98 [N]

=

=0.283 [Nm]

2945103

2 0.9

0.3985103

= =7.19.9

the gear ratio ( i ) is calculated as follows:

i14501550

N G=

14501550

18024

= =Speed at the gearhead output shaft: N GV 60

P

(152)60

5

=18024 [r/min]

Motor

Gearhead

Coupling

Ball Screw

Slide Guide

mv

FA

Using Standard AC Motors This value is the load torque at the gearhead drive shaft, and must

be converted into load torque at the motor output shaft.

The required torque at the motor output shaft (T M ) is given by:

Look for a margin of safety of 2 times.

38.82=77.6 [mNm]

To find a motor with a start-up torque of 77.6 mNm or more,

select motor 3RK15GN-AWMU.

This motor is equipped with an electromagnetic brake to hold a load.

A gearhead with a gear ratio of 1:9 that can be connected to the

motor 3RK15GN-AWMU is 3GN9K.

Load Inertia Check

Gearhead shaft total load inertia

J =0.9931040.190104=1.18104 [kgm2]

Here, the 3GN9K permissible load inertia is (see page A-13):

J G=0.1410492

J G=1.13103 [kgm2]

Therefore, J < J G, the load inertia is less than the permissible

inertia, so there is no problem. There is margin for the torque, sothe rotation rate is checked with the no-load rotation rate (about

1750 r/min).

This confirms that the motor meets the specifications.

=V N M P B

60 i17505609

=16.2 [mm/s]=

(where N M is the motor speed)

=Ball Screw Moment of Inertia J B

32 L B D B

4

=

=0.993104 [kg m2]

327.9103800103 (20103)4

=m ( ) 2Table and Work Moment of Inertia J m A

2

2=30 ( ) 2

=0.190104 [kg m2]

5103

=T M T L

i G= =0.0388 [Nm]=38.8 [mNm]

0.28390.81

(Gearhead transmission efficiency =0.81) G




Belt and Pulley

Here is an example of how to select an induction motor to drive a

belt conveyor.

In this case, a motor must be selected that meets the following

basic specifications.

<Required and Structural Specifications>

Total mass of belt and work ·············································m1=20kg

Frictional coefficient of sliding surfaces·································=0.3

Drum radius ·································································· D=100mm

Mass of drum·····································································m 2=1kg

Belt roller efficiency ······························································· =0.9Belt speed·························································V =140mm/s10%

Motor power supply·························Single-Phase 110 VAC 60 Hz

Movement time ··························································· 8 hours/day

Determine the Gear Ratio

Because the rated speed for a 4-pole motor at 60 Hz is

14501550 r/min,

From within this range a gear ratio of i =60 is selected.

Calculate the Required Torque

On a belt conveyor, the greatest torque is needed when starting

the belt. To calculate the torque needed for start-up, the friction

coefficient ( F ) of the sliding surface is first determined:

F =m g=0.3209.807=58.8 [N]

The load torque obtained is actually the load torque at the

gearhead drive shaft, so this value must be converted into load

torque at the motor output shaft. If the required torque at the motor

output shaft is T M , then:

Look for a margin of safety of 2 times, taking into consideration

commercial power voltage fluctuation.

82.62165 [mNm]

The suitable motor is one with a starting torque of 165 mNm or

more. Therefore, motor 5IK40GN-AWU is the best choice.

Since a gear ratio of 1:60 is required, select the gearhead

5GN60K which may be connected to the 5IK40GN-AWU

motor.

=T M T Li G

= =0.0826 [Nm]=82.6 [mNm]3.27

600.66

(Gearhead transmission efficiency G=0.66)

= =3.27 [Nm]Load torque T L F D

2 =

58.8100103

20.9

= =49.364.6

the gear ratio (i) is calculated as follows:

i

14501550

N G =

14501550

26.72.7

= =Speed at the gearhead output shaft: N GV 60

D

(14014) 60

100

=26.72.7 [r/min]

Belt Conveyor

Gearhead

Motor

D

V

Using Standard AC Motors

Load Inertia Check

Gearhead Shaft Load Inertia

J =50010412.51042=525104 [kgm2]

Here, the 5GN60K permissible load inertia is (see page A-13):

J G=0.75104602

=2700104 [kgm2]

Therefore, J < J G, the load inertia is less than the permissible

inertia, so there is no problem.

Since the motor selected has a rated torque of 260 mNm, which is

somewhat larger than the actual load torque, the motor will run at

a higher speed than the rated speed.

Therefore the speed is used under no-load conditions

(approximately 1750 r/min) to calculate belt speed, and thus

determine whether the selected product meets the required

specifications.

=V N M D

60 i=

1750 1006060

=152.7 [mm/s]

(Where N M is the motor speed)

=Roller Moment of Inertia J m218

m 2 D 2

=

=12.5104 [kgm2]

1

81 (100103)2

=m1 ( ) 2

Belt and Work Moment of Inertia J m1

D

2

=20 ( ) 2100103

2

=500104 [kgm2]




Service Life

Motor and

Fan Sizing

Standard

AC Motors

Brushless

DC Motors

Stepping

Motors

Gearheads

Linear

Motion

Fan Motors

Here is an example of how to select a brushless DC motor to drive

a belt conveyor.

Performance

Belt speed V L is 0.015 m/s1 m/s

Specifications for belt and work

Condition: Motor power supply···················Single-Phase 100 VAC

Belt conveyor drive

Roller diameter ·············································· D=0.1m

Mass of roller ·················································m2=1kg

Total mass of belt and work·····························m1=15kg

Frictional coefficient of sliding surfaces ················=0.3

Belt roller efficiency ··············································· =0.9

Find the Required Speed Range

For the gear ratio, select 1:15 (speed range: 2200) from the

permissible torque table for combination type on page B-12 so that

the minimum/maximum speeds fall within the speed range.

Calculate the Load Inertia J G

Load Inertia of Roller: Jm2

Load inertia of belt and work: Jm1

The load inertia J G is calculated as follows:

J G= J m22 J m1=212.5104375104

=400104kgm2

From the specifications on page B-10, the permissible load inertia

for BX5120A-15 is 420104 kgm2

Calculate the Load Torque T L

Select BX5120A-15 from the permissible torque table on page

B-12.

Since the permissible torque is 5.4 Nm, the safety margin is

T M / T L=5.4/2.452.2.

Usually, a motor can operate at the safety margin of 1.52 or more.

=

=2.45 Nm

Load Torque T L

= m1 g=0.3159.807=44.1NFriction Coefficient of the Sliding Surface: F

F D

2

44.10.1

20.9

=m1 ( Jm1 D

2

2

0.1) =15 ( ) =375104 kgm2 2 2

= Jm218

m2 D 2=

18

10.1 2=12.5104 kgm2

= N G: Speed at the gearhead output shaft N G

Belt Speed

60V L

D

=2.86 r/min (Minimum Speed)0.015 m/s 600.015

0.1

=191 r/min (Maximum Speed)1 m/s 601

0.1

Work

Using Brushless DC Motors

The mass of work is selected that can be driven with SMK5100A-A

when the belt-drive table shown in Fig. 1 is driven in the operation

pattern shown in Fig. 2.

Structural Specifications

Total mass of belt and work m1=1.5 [kg]

Roller diameter D=30 [mm]

Mass of roller m 2=0.1 [kg]

Frictional coefficient of sliding surfaces =0.04

Belt and pulley efficiency =0.9

Frequency of power supply 60 Hz (Motor speed: 72 r/min)

Low-speed synchronous motors share the same basic operating

principle with 2-phase stepping motors. Accordingly, the torque for

a low-speed synchronous motor is calculated in the same manner

as for a 2-phase stepping motor.

qBelt speed

Check the belt (work) speed

wCalculate the Required Torque T L

Frictional coefficient of sliding surfaces: F = m1 g=0.041.59.807=0.589 [N]

eCalculate the Load Inertia

Load inertia of belt and work: Jm1

Load Inertia of Roller: Jm 2

The load inertia J L is calculated as follows:

J L= J m1 J m 22=3.381040.1131042=3.5104 [kgm2]

rCalculate the Acceleration Torque

Here, θ s=7.2˚, f=60 Hz, n=3.6˚/ θ s=0.5

J 0=Rotor Inertia

tCalculate the Required Torque (look for a margin of safety of 2 times)

Required Torque T M =(T LTa)2

=(9.82103905 J 00.32)2

=1810 J 00.66 [Nm]

ySelect a Motor

Select a motor that satisfies both the required torque and the

permissible load inertia.

When the required torque is calculated by substituting the rotor

inertia, T M is obtained as 0.914 Nm, which is below the output

torque. Next, check the permissible load inertia. Since the load

inertia calculated ine is also below the permissible load inertia,

SMK5100A-A can be used in this application.

Rotor Inertia

[kgm

2

]

Motor

1.4104

Permissible Load Inertia

[kgm

2

]

Output Torque

[Nm]

7104 1.12SMK5100A-A

Ta=( J 0 J L)

=905 J 00.32 [Nm]

=( J 03.5104)

sf

2

180 n 7.260

2

1800.5θ

J m2=18

m 2 D2= 18

0.1(30103)2=0.113104 [kgm2]

J m1= m1( D

2 2)2=1.5( )2=3.38104 [kgm2]

30103

T L=Load Torque F D

2 = =9.82103 [Nm]

0.58930103

20.9

V = = =113 [mm/s] D N

60

3072

60

CW

5 10 15 (sec)

Fig. 2 Operating Pattern

CCW

Roller 2

Fig. 1 Example of Belt Drive

Roller 1

m

F

Using Low-Speed Synchronous Motors (SMK Series)




Index TableGeared stepping motors are suitable for systems with high inertia,

such as index tables.

Determine the Drive Mechanism

Diameter of index table ················································ DT =300mm

Index table thickness ··························································· LT =10mm

Diameter of work ·································································· Dw=40mm

Thickness of work································································· Lw=30mm

Material of table·································Iron (density =7.9103kg/m3)

Number of loads ······················································10 (one every 36˚)

Distance from center of index table

to center of load ······································································ l=125mm

Positioning angle ··········································································· θ =36˚

Positioning period································································· t0=0.25 sec

The PN geared type (gear ratio=1:10, resolution=0.036˚)

can be used.

✽The PN geared type can be used at the maximumstarting/stopping torque in the inertial drive mode.

Gear Ratio ··············································································· i=10

Resolution·············································································θ s=0.036˚

Determine the Operating Pattern (See page G-4 for basic equations)

(1) Find the Number of Operating Pulses A [pulses]

(2) Determine the Acceleration (Deceleration) Period t 1 [s]

Generally, an acceleration (deceleration) period should be set

approximately 25% of the positioning period.

Here, t1= 0.1[s] is provided as the accelation (deceleration) period.

(3) Calculate the Operation Speed N [r/min]

✽The permissible speed range for a PN geared motor with a gear

ratio of 1:10 is 0 to 300 r/min.

= = Operation Speed N 60

360

60

360

θ

t0t1

36

0.250.1

=40 [r/min]

=Operating pulses Aθ

θ s

36˚

0.036˚=

=1000 [Pulses]

DT=300 mm

=125 mm

Programmable

ControllerGeared Stepping Motor

ControllerDriver

(4) Determine the Operating Pulse Speed f 2 [Hz]

Period [s]t1 t1

t0=0.25

Operating

Pulse

Speed

[ H

z]

6667

0.1 0.1

600 Pulses

= =Operating Pulse Speed f 2 A

t0t1

600

0.250.1

=6667 [Hz]


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Service Life

Motor and

Fan Sizing

Standard

AC Motors

Brushless

DC Motors

Stepping

Motors

Gearheads

Linear

Motion

Fan Motors

Calculate the Required Torque T M [Nm] (see page G-4)

(1) Calculate the Load Torque T L [Nm]

Friction load is negligible and therefore omitted. The load torque is

assumed as zero.

Load Torque T L=0 [Nm]

(2) Calculate the Acceleration Torque T a [Nm]

qCalculate the total moment of inertia J L [kgm2]

(See page G-3 for basic equations)

Inertia of work J W [kgm2] relative to the center of rotation can be

obtained from distance L [mm] between the center of work and

center of rotation, mass of work mW [kg], and inertia of work

(center of gravity) J w1 [kgm2] .

Since the number of works, n, is 10 [pcs],

wCalculate the acceleration torque T a [Nm]

(3) Calculate the Required Torque T M

Safety Factor S f =2.0

Required Torque T M=(T LT a)S f

={0(4.19103 J 04.61)}2.0

=8.38103 J 09.2 [Nm]

=( J 0i2 J L) Acceleration

Torque T a

f 2 f 1

t1

s

180

=( J 010211102)

=4.19103 J 04.61 [Nm]

66670

0.1

0.036

180

θ

= J T J W

(6.284.71)102=

Total Inertia J L

=11102 [kgm2]

= n( J W1mW L 2)

10[(0.596104)0.3(125103)2]=

Inertia of Work J W

=4.71102 [kgm2]

(Center of gravity)

=Mass of Work mW

4 LW DW

2

=

=0.3 [kg]

47.9103(30103)(40103)2

=Inertia of Table J T

32 LT DT

4

=

32 LW DW

4

=

=6.28102 [kgm2]

327.9103(10103)(300103)4

=

327.9103(30103)(40103)4

Inertia of Work J W1

=0.596104 [kgm2]

(Center of gravity)

Select a Motor

(1) Provisional Motor Selection

(2) Determine the Motor from the Speed-Torque Characteristics

AS66AA-N10

Select a motor for which the required torque falls within the pull-

out torque of the speed-torque characteristics.

PN geared type can operate inertia load up to acceleration torque

less than Maximum torque.

If load torque is applied, the selection must be made so the

product of the load torque and the safety factor does not exceed

the permissible torque.

Pulse Speed [kHz]

(Resolution Setting: 1000P/R)

0 504540353020 2515105

15050 100 250200 300

Speed [r/min]

T o r q u e [ N

m ]

T o r q u e [ k g f c m ]

0

25

20

5

10

15

0

250

200

50

100

150

Permissible Torque

J0=405107 AS66AA-N10

Rotor Inertia[kgm2]

TM=9.55

Required Torque[Nm]Model

Exemple de Calcul Motoare

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