Oriental Motor Motor Sizing Calculations 6g5f56
This document was ed by and they confirmed that they have the permission to share it. If you are author or own the copyright of this book, please report to us by using this report form. Report 3i3n4
Overview 26281t
& View Oriental Motor Motor Sizing Calculations as PDF for free.
More details 6y5l6z
- Words: 4,844
- Pages: 10
Technical Reference
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 Procedure An overview of selection procedure is explained below. Determine the drive mechanism component
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 required specifications (Equipment specifications)
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 speed and load
Select motor type
Selection calculation
G-2
ORIENTAL MOTOR GENERAL CATALOGUE
Calculate the values for load torque and load inertia at the motor drive shaft. See the left column on page G-3 for the 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.
Formulas for Calculating Moment of Inertia
Calculate the friction torque for the applicable drive mechanism.
Inertia of a Cylinder 1 mD12 =
LD14 [kgm2] 8 32 L2 1 D12 Jy = m( ) [kgm2] 3 4 4
u
Jx =
Ball Screw TL = (
FPB 0F0PB 1 ) [Nm] i 2 2
q
i
Technical Reference
Formulas for Calculating Load Torque
D1
F =FA mg (sin a cos a) [N]
x
w
Direct Coupling F
m
FA
L
y
m
F
A
Inertia of a Hollow Cylinder
Pulley FA mg D i 2 ( FA mg) D = [Nm] 2i
1 m (D12 D22)=
L (D14D24) [kgm2] 8 32 1 D12 D22 L2 Jy = m ( ) [kgm2] 4 4 3
Jx =
TL =
e
D1
o !0
x D2
D
Service Life
L
y FA
F D FD TL = = [Nm] 2 i 2 i
r
F =FA mg (sin a cos a) [N]
t
F
m
FA
Jx =Jx0 ml2=
By Actual Measurement
x
!1 x0
A
B
1 m (A2 B2)= 12 1 Jy = m (B2 C2)= 12
y Spring Balance
1
ABC (A2 B2) [kgm2] 12 1
ABC (B2 C2) [kgm2] 12 x A B
!2
hine
Stepping Motors
y
Gearheads
!3 Linear Motion
FB Mac
Brushless DC Motors
C
Jx =
[Nm]
1 m (A2 B2 12l2) [kgm2] 12
Inertia of a Rectangular Pillar
D
FBD 2
Inertia for Off-center Axis of Rotation
r=Distance between x and x0 axes [m]
F
m
D
TL =
Standard AC Motors
m
Wire Belt Mechanism, Rack and Pinion Mechanism
FA
Motor and Fan Sizing
C
D Fan Motors
Pulley
F F0 0 i
=Force of moving direction [N] =Pilot pressure load [N] (1/3F) =Internal friction coefficient of pilot pressure nut (0.10.3) =Efficiency (0.850.95) =Gear ratio (This is the gear ratio of the mechanism and not the gear ratio of the Oriental Motor gearhead you are selecting.) P B =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)
Inertia of an Object in Linear Motion A J= m ( 2 ) 2 [kgm2]
!4 A=Unit of movement [m/rev]
Density
=7.9103 [kg/m3] Iron Aluminum =2.8103 [kg/m3]
=8.5103 [kg/m3] Bronze
=1.1103 [kg/m3] Nylon Jx =Inertia on x axis [kgm2] Jy =Inertia on y axis [kgm2] J0 =Inertia on x0 axis (ing through center of gravity) [kgm2] m =Mass [kg] D1 =External diameter [m] D2 =Internal diameter [m]
=Density [kg/m3] L =Length [m]
G-3
Technical Reference
Motor Sizing Calculations
Calculate the Operating Speed from Operating Pulse speed
The following explains the calculation for selecting a stepping motor based on pulse control:
Operating =Operating Pulse Speed [Hz] Step Angle 60 Speed[r/min] 360
Operating Patterns
Calculate the Load Torque
There 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 Acceleration Torque
Operating Pulse Speed f2
Operating Pulse Speed f2
Number of Operating Pulses A
Number of Operating Pulses A
Starting Pulse Speed f1 Acceleration Period t1
Positioning Period t0
See basic equations on page G-3.
Regardless 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.
Positioning Period t0
Start/Stop Operation
Deceleration Period t1
Acceleration/Deceleration Operation
Calculate the Number of Operating Pulses A The 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.
<Standard AC Motors> (The same equation applies to brushless DC motors.) Rotor Inertia Total Inertia [kgm2] [kgm2] Operating Speed [r/min] coefficient Acceleration (Deceleration) Period [s] NM (J0 JL) = 9.55 t1
Acceleration Torque = Ta[Nm]
Operating Speed NM [r/min]
No. of Pulses Distance per Movement Number of Operating = Required for Pulses A[Pulses] Distance per Motor Rotation 1 Motor Rotation l 360˚ : Step Angle θ s = lrev θs
Movement
Calculate the Operating Pulse Speed f 2
Acceleration Period t 1
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 Number of Acceleration Operating Pulses Starting Pulse (Deceleration) Operating Pulse Speed [Hz] [Pulses] Period [s] Speed f2[Hz] = Positioning Period Acceleration (Deceleration) [s] Period [s] =
Af1t1 t0t1
Number of Operating Pulses [Pulses] Operating Pulse = Positioning Period [s] Speed f2[Hz] A t0
Calculate the Acceleration/Deceleration Rate TR
=
t1 f2f1
✽Calculate the pulse speed in full-step equivalents.
G-4
ORIENTAL MOTOR GENERAL CATALOGUE
Pulse Speed [kHz]
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: Acceleration (Deceleration) Acceleration/ Period[ms] deceleration = rate TR Operating Pulse Starting Pulse [ms/kHz] Speed[kHz] Speed [kHz]
Deceleration Period t 1
For AC Motors
<Stepping Motors> qFor Acceleration/Deceleration Operation Acceleration Torque Ta [Nm] =
Rotor Inertia Total Inertia Step Angle[˚ ] [kgm2] [kgm2] 180˚
= (J0 JL)
θ s 180
f2f1 t1
Operating Pulse Starting Pulse Speed [Hz] Speed [Hz] Acceleration (Deceleration) Period [s]
wFor Start-Stop Operation Acceleration Torque Ta [Nm] =
Step Angle[˚ ](Operating Pulse Speed)2 [Hz] Rotor Inertia Total Inertia [kgm2] [kgm2] 180˚coefficient
= (J0 JL)
θ s f22 180n
n: 3.6˚/ θ s
Calculate the Required Torque TM 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 L T a)S f
wFor Start-Stop Operation
=
Positioning Period t 0
TR
t1
Standard AC Motors qSpeed 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 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 may damage 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 Motors qChecking 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. Running Duty Cycle
Running Time Running Stopping Time Time
100
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.
Product Series
Motor Frame Size Acceleration/Deceleration Rate TR[ms/kHz] 28, 42, 60, 85 0.1 Maximum✽ Stepping Motor and 20 Maximum 42, 60 Driver Packages 30 Maximum 85, 90 ✽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.
Service Life
Safety Factor: Sf (Reference Value) Product Series Stepping Motor and Driver Packages
Safety Factor (Reference Value) 1.52 2
Motor and Fan Sizing
Standard AC Motors
Brushless DC Motors
Required Torque
Speed [r/min] (Pulse Speed [kHz])
Stepping Motors
Gearheads
Linear Motion
Fan Motors
Total Inertia of the Machine [kgm2] Rotor Inertia of the Motor [kgm2]
Inertia Ratio =
=
Acceleration Rate (Reference Values with XG9200, SG9200 Series)
Torque [Nm]
There 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.
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.
Technical Reference
Selection Considerations
JL J0
Inertia Ratio (Reference Values) Product Series
Motor Frame Size 28, 42, 60, 85 Stepping Motor and 20, 28 Driver Packages 42, 60, 85
Inertia Ratio 30 Maximum 5 Maximum 10 Maximum
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 Ratio =
=
Total Inertia of the Machine [kgm2] Rotor Inertia of the Motor [kgm2](Gear Ratio)2 JL J0 i 2
G-5
Technical Reference
Sizing Example
(4) Calculate the Operating Speed N [r/min]
Ball Screw
Operating Speed =f2
Using Stepping Motors ( Controller
Stepping Motor
)
θS 60 360
=10000
0.72 60 360
=1200 [r/min] m Coupling
Calculate the Required Torque T M [Nm] (see page G-4) (1) Calculate the Load Torque T L [Nm]
Direct Connection
Force of moving direction F =FA mg (sin a cos a)
Driver
=0 409.807 (sin0 0.05cos0) =19.6 [N] DB
PB
Pilot Pressure Load F0 =
F 19.6 = =6.53 [N] 3 3
Load Torque TL =
F PB 0 F0 PB 2 2
Programmable Controller
=
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=0.3 Ball screw shaft diameter·············································DB =15 mm Total length of ball screw ············································LB=600 mm Material of ball screw ·················Iron (density =7.9103 [kg/m3]) Ball screw pitch ····························································PB=15 mm Desired resolution···············································∆l=0.03 mm/step (feed per pulse) Feed ·············································································l=180 mm Positioning period ·············································t 0=Within 0.8 sec.
(2) Calculate the Acceleration Torque T a [Nm] qCalculate the total moment of inertia JL [kgm2] (See page G-3 for basic equations) LB DB4 32 = 7.9103600103 (15103)4 32
Inertia of Ball Screw JB =
=0.236104 [kgm2] Inertia of Table and Work JT =m (
wCalculate the acceleration torque Ta [Nm]
360˚0.03 =0.72 [˚ ] 15
Acceleration torque Ta =(J0 JL)
180 360˚ =6000 [pulses] 15 0.72˚
(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 f2 [Hz]
(3) Calculate the Required Torque T M [Nm] Required torque T M=(T L Ta)2 ={0.0567 (628J0 0.158)}2 =1256J0 0.429 [Nm]
Select a Motor (1) Provisional Motor Selection
AS66AA
6000 pulses
t1
t1 t0=0.8
0.2
period [s]
20
2.0
15
1.5
10
Torque [Nm]
10000
0.2
Required Torque [Nm] 0.48
(2) Determine the Motor from the Speed-Torque Characteristics AS66AA
Torque [kgfcm]
Operating pulse speed [Hz]
60000 =10000 [Hz] 0.80.2
Rotor Inertia [kgm2] 405107
Model
Operating pulse Operating pulses (A)Starting pulse speed (f1)Acceleration (Deceleration) Period (t1) = speed f2 Positioning Period (t0)Acceleration (Deceleration) Period (t1) =
0.72 100000 180˚ 0.2
=628J0 0.158 [Nm]
(1) Finding the Number of Operating Pulses A [pulses] 360˚ Feed per Unit (l) Step Angle (θ s) Ball Screw Pitch (PB)
θ s f2f1 180˚ t1
=(J0 2.52104)
Determine the Operating Pattern (See page G-4 for basic equations)
=
15103 2 ) =2.28104 [kgm2] 2
=0.236104 2.28104=2.52104 [kgm2]
can be connected directly to the application.
Operating pulses A =
PB 2 ) 2
Total Inertia JL =JB JT
360˚Desired Resolution (∆l) Ball Screw Pitch (PB)
=
=0.0567 [Nm]
=40(
Calculate the Required Resolution Required Resolution θ s =
19.615103 0.36.5315103 2 0.9 2
1.0
5
0.5
0
0
0
1000 10
2000 3000 Speed [r/min]
4000
20
30 40 50 60 Pulse Speed [kHz] (Resolution Setting: 1000 P/R)
Select a motor for which the required torque falls within the pullout torque of the speed-torque characteristics.
G-6
ORIENTAL MOTOR GENERAL CATALOGUE
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.
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 (TM) is given by: TM =
TL 0.283 = =0.0388 [Nm]=38.8 [mNm] i G 90.81
(Gearhead transmission efficiency G =0.81)
Technical Reference
Using Standard AC Motors
Look for a margin of safety of 2 times.
Motor Gearhead Coupling
Ball Screw FA Slide Guide v
m
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.
Motor and Fan Sizing
Load Inertia Check Ball Screw Moment of Inertia JB =
Total mass of the table and work ···································m=30 [kg] Table speed ························································V=15 2 [mm/s] External force·····································································FA=0[N] Ball screw tilt angle ····························································=90 [˚ ] Total length of ball screw···········································LB=800 [mm] Ball screw shaft diameter············································DB=20 [mm] Ball screw pitch ·····························································PB=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 Speed at the gearhead output shaft: NG =
V 60 (15 2)60 = P 5
=180 24 [r/min]
Because the rated speed for a 4-pole motor at 60 Hz is 14501550 r/min,
=
LB DB4 32 7.9103800103 (20103)4 32
=0.993104 [kgm2] A 2 Table and Work Moment of Inertia Jm =m ( ) 2 =30 (
5103 2 ) 2
=0.190104 [kgm2]
Gearhead shaft total load inertia J=0.993104 0.190104=1.18104 [kgm2] Here, the 3GN9K permissible load inertia is (see page A-13): JG=0.1410492 JG=1.13103 [kgm2] Therefore, J < JG, the load inertia is less than the permissible inertia, so there is no problem. There is margin for the torque, so the rotation rate is checked with the no-load rotation rate (about 1750 r/min). V=
NM PB 17505 = =16.2 [mm/s] 60 i 609
Service Life
Standard AC Motors
Brushless DC Motors
Stepping Motors
Gearheads
Linear Motion
Fan Motors
(where NM is the motor speed)
This confirms that the motor meets the specifications.
the gear ratio ( i ) is calculated as follows: i=
14501550 14501550 = =7.19.9 NG 180 24
From within this range a gear ratio of i =9 is selected.
Calculate the Required Torque Load weight in the direction =FA mg (sin cos ) of the ball screw shaft F =0 309.807 (sin90˚ 0.05cos90˚) =294 [N] Pilot pressure load F0 = Load torque TL = =
F =98 [N] 3 FPB 0F0PB 2 2 2945103 0.3985103 2 20.9
=0.283 [Nm]
G-7
Technical Reference
Belt and Pulley
Load Inertia Check
Using Standard AC Motors Belt and Work Moment of Inertia Jm1 =m1
(
D 2
=20
(
100103 2 ) 2
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.
=500104 [kgm2] Roller Moment of Inertia Jm2 =
2
)
V
=
1 m2D2 8 1 1 (100103)2 8
=12.5104 [kgm2]
Belt Conveyor
D
Motor Gearhead
Total mass of belt and work ·············································m1=20kg Frictional coefficient of sliding surfaces ·································=0.3 Drum radius ··································································D=100mm Mass of drum·····································································m2=1kg Belt roller efficiency ·······························································=0.9 Belt speed·························································V=140mm/s 10% Motor power supply·························Single-Phase 110 VAC 60 Hz Movement time ···························································8 hours/day
Determine the Gear Ratio Speed at the gearhead output shaft: NG =
V 60 (140 14) 60 = D 100
=26.7 2.7 [r/min]
Because the rated speed for a 4-pole motor at 60 Hz is 14501550 r/min, the gear ratio (i) is calculated as follows: 14501550 14501550 = =49.364.6 i= 26.7 2.7 NG
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=mg=0.3209.807=58.8 [N] Load torque TL =
FD 58.8100103 = =3.27 [Nm] 20.9 2
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 TM, then: TM =
TL 3.27 = =0.0826 [Nm]=82.6 [mNm] i G 600.66
(Gearhead transmission efficiency G=0.66)
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.
G-8
ORIENTAL MOTOR GENERAL CATALOGUE
Gearhead Shaft Load Inertia J=500104 12.51042=525104 [kgm2] Here, the 5GN60K permissible load inertia is (see page A-13): JG=0.75104602 =2700104 [kgm2] Therefore, J < JG, 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=
NM D 60 i
=
1750 100 =152.7 [mm/s] 6060
(Where NM is the motor speed)
Using Low-Speed Synchronous Motors (SMK Series) 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.
Work m
Technical Reference
Using Brushless DC Motors Here is an example of how to select a brushless DC motor to drive a belt conveyor.
Roller 2
F Roller 1
Fig. 1 Example of Belt Drive
Performance Belt speed VL 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
Structural Specifications Total mass of belt and work Roller diameter Mass of roller Frictional coefficient of sliding surfaces Belt and pulley efficiency Frequency of power supply
m1=1.5 [kg] D=30 [mm] m2=0.1 [kg] =0.04 =0.9 60 Hz (Motor speed: 72 r/min)
Motor and Fan Sizing
Service Life
CW 15 (sec) 5
10
Standard AC Motors
CCW Fig. 2 Operating Pattern
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. NG =
60VL D
NG : Speed at the gearhead output shaft
Belt Speed 600.015 0.015 m/s =2.86 r/min (Minimum Speed) 0.1 1 m/s
601 =191 r/min (Maximum Speed) 0.1
Calculate the Load Inertia JG Load Inertia of Roller: Jm2 Jm2 =
1 1 2 m2D = 10.12=12.5104 kgm2 8 8
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 V= D N = 3072 =113 [mm/s] 60 60
0.58930103 =9.82103 [Nm] Load Torque TL= F D = 20.9 2
Jm1= m1( D )2=1.5( 2
30103 )2=3.38104 [kgm2] 2
Load Inertia of Roller: Jm2 Jm2= 1 m2D2= 1 0.1(30103)2=0.113104 [kgm2] 8 8
The load inertia JG is calculated as follows: JG=Jm22 Jm1=212.5104 375104 =400104kgm2
The load inertia JL is calculated as follows: JL=Jm1 Jm22=3.38104 0.1131042=3.5104 [kgm2] rCalculate the Acceleration Torque
Friction Coefficient of = m1 g=0.3159.807=44.1N the Sliding Surface: F Load Torque TL =
FD 44.10.1 2 20.9
=2.45 Nm
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.
Gearheads
Linear Motion
eCalculate the Load Inertia Load inertia of belt and work: Jm1
D 2 0.1 2 Jm1 =m1 ( ) =15 ( ) =375104 kgm2 2 2
Calculate the Load Torque T L
Stepping Motors
wCalculate the Required Torque T L Frictional coefficient of sliding surfaces: F= m1 g=0.041.59.807=0.589 [N]
Load inertia of belt and work: Jm1
From the specifications on page B-10, the permissible load inertia for BX5120A-15 is 420104 kgm2
Brushless DC Motors
Fan Motors
2 2 Ta=(J0 JL) θ s f =(J0 3.5104) 7.260 180 n 1800.5
=905 J0 0.32 [Nm]
Here, θ s=7.2˚, f=60 Hz, n=3.6˚/θ s=0.5 J0 =Rotor Inertia tCalculate the Required Torque (look for a margin of safety of 2 times) Required Torque T M=(T L Ta)2 =(9.82103 905J0 0.32)2 =1810J0 0.66 [Nm] ySelect a Motor Select a motor that satisfies both the required torque and the permissible load inertia. Motor SMK5100A-A
Rotor Inertia [kgm2] 1.4104
Permissible Load Inertia Output Torque [kgm2] [Nm] 1.12 7104
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 in e is also below the permissible load inertia, SMK5100A-A can be used in this application.
G-9
Technical Reference
Index Table
(4) Determine the Operating Pulse Speed f 2 [Hz]
Geared stepping motors are suitable for systems with high inertia, such as index tables.
Operating Pulse Speed f2 =
Determine the Drive Mechanism
A 600 = t0t1 0.250.1 =6667 [Hz]
DT=300 mm =125 mm Operating Pulse Speed [Hz]
6667
Driver
Controller
600 Pulses
0.1
t1
t1 t0=0.25
Programmable Controller
Geared Stepping Motor
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 maximum starting/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] θ Operating pulses A = θs =
36˚ 0.036˚
=1000 [Pulses]
(2) Determine the Acceleration (Deceleration) Period t1 [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] Operation Speed N =
θ 60 36 60 = 360 0.250.1 360 t0t1
=40 [r/min]
✽The permissible speed range for a PN geared motor with a gear ratio of 1:10 is 0 to 300 r/min.
G-10
ORIENTAL MOTOR GENERAL CATALOGUE
0.1
Period [s]
Select a Motor
(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]
(1) Provisional Motor Selection
(2) Calculate the Acceleration Torque T a [Nm] qCalculate the total moment of inertia JL [kgm2] (See page G-3 for basic equations) LTDT4 32
250
25
7.9103(10103)(300103)4 32
200
20
=6.28102 [kgm2]
Inertia of Work JW1 = LWDW4 32 (Center of gravity) = 7.9103(30103)(40103)4 32 =0.596104 [kgm2] Mass of Work mW = =
LWDW2 4
150
=0.3 [kg]
Inertia of work JW [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) Jw1 [kgm2] .
TM=9.55
100
15 10
50
5
0
0
Motor and Fan Sizing
Permissible Torque
0
7.9103(30103)(40103)2 4
Required Torque [Nm]
(2) Determine the Motor from the Speed-Torque Characteristics AS66AA-N10
Torque [Nm]
=
AS66AA-N10
Torque [kgfcm]
Inertia of Table JT =
Rotor Inertia [kgm2] J0=405107
Model
Technical Reference
Calculate the Required Torque T M [Nm] (see page G-4)
50 5
10
100 15
150 200 Speed [r/min]
250
20 25 30 35 40 Pulse Speed [kHz] (Resolution Setting: 1000P/R)
300 45
Service Life
50
Select a motor for which the required torque falls within the pullout 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.
Standard AC Motors
Brushless DC Motors
Stepping Motors
Since the number of works, n, is 10 [pcs], Inertia of Work JW = n(JW1 mWL2) (Center of gravity)
Gearheads
= 10[(0.596104) 0.3(125103)2] =4.71102 [kgm2]
Linear Motion
Total Inertia JL = JT JW = (6.28 4.71)102
Fan Motors
=11102 [kgm2]
wCalculate the acceleration torque T a [Nm] θ s Acceleration =(J0i2 JL) 180 Torque Ta =(J0102 11102)
f2f1 t1
66670 0.036 180 0.1
=4.19103J0 4.61 [Nm]
(3) Calculate the Required Torque T M Safety Factor S f =2.0 Required Torque T M=(T L Ta)S f ={0 (4.19103 J 0 4.61)}2.0 =8.38103 J 0 9.2 [Nm]
G-11
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 Procedure An overview of selection procedure is explained below. Determine the drive mechanism component
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 required specifications (Equipment specifications)
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 speed and load
Select motor type
Selection calculation
G-2
ORIENTAL MOTOR GENERAL CATALOGUE
Calculate the values for load torque and load inertia at the motor drive shaft. See the left column on page G-3 for the 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.
Formulas for Calculating Moment of Inertia
Calculate the friction torque for the applicable drive mechanism.
Inertia of a Cylinder 1 mD12 =
LD14 [kgm2] 8 32 L2 1 D12 Jy = m( ) [kgm2] 3 4 4
u
Jx =
Ball Screw TL = (
FPB 0F0PB 1 ) [Nm] i 2 2
q
i
Technical Reference
Formulas for Calculating Load Torque
D1
F =FA mg (sin a cos a) [N]
x
w
Direct Coupling F
m
FA
L
y
m
F
A
Inertia of a Hollow Cylinder
Pulley FA mg D i 2 ( FA mg) D = [Nm] 2i
1 m (D12 D22)=
L (D14D24) [kgm2] 8 32 1 D12 D22 L2 Jy = m ( ) [kgm2] 4 4 3
Jx =
TL =
e
D1
o !0
x D2
D
Service Life
L
y FA
F D FD TL = = [Nm] 2 i 2 i
r
F =FA mg (sin a cos a) [N]
t
F
m
FA
Jx =Jx0 ml2=
By Actual Measurement
x
!1 x0
A
B
1 m (A2 B2)= 12 1 Jy = m (B2 C2)= 12
y Spring Balance
1
ABC (A2 B2) [kgm2] 12 1
ABC (B2 C2) [kgm2] 12 x A B
!2
hine
Stepping Motors
y
Gearheads
!3 Linear Motion
FB Mac
Brushless DC Motors
C
Jx =
[Nm]
1 m (A2 B2 12l2) [kgm2] 12
Inertia of a Rectangular Pillar
D
FBD 2
Inertia for Off-center Axis of Rotation
r=Distance between x and x0 axes [m]
F
m
D
TL =
Standard AC Motors
m
Wire Belt Mechanism, Rack and Pinion Mechanism
FA
Motor and Fan Sizing
C
D Fan Motors
Pulley
F F0 0 i
=Force of moving direction [N] =Pilot pressure load [N] (1/3F) =Internal friction coefficient of pilot pressure nut (0.10.3) =Efficiency (0.850.95) =Gear ratio (This is the gear ratio of the mechanism and not the gear ratio of the Oriental Motor gearhead you are selecting.) P B =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)
Inertia of an Object in Linear Motion A J= m ( 2 ) 2 [kgm2]
!4 A=Unit of movement [m/rev]
Density
=7.9103 [kg/m3] Iron Aluminum =2.8103 [kg/m3]
=8.5103 [kg/m3] Bronze
=1.1103 [kg/m3] Nylon Jx =Inertia on x axis [kgm2] Jy =Inertia on y axis [kgm2] J0 =Inertia on x0 axis (ing through center of gravity) [kgm2] m =Mass [kg] D1 =External diameter [m] D2 =Internal diameter [m]
=Density [kg/m3] L =Length [m]
G-3
Technical Reference
Motor Sizing Calculations
Calculate the Operating Speed from Operating Pulse speed
The following explains the calculation for selecting a stepping motor based on pulse control:
Operating =Operating Pulse Speed [Hz] Step Angle 60 Speed[r/min] 360
Operating Patterns
Calculate the Load Torque
There 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 Acceleration Torque
Operating Pulse Speed f2
Operating Pulse Speed f2
Number of Operating Pulses A
Number of Operating Pulses A
Starting Pulse Speed f1 Acceleration Period t1
Positioning Period t0
See basic equations on page G-3.
Regardless 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.
Positioning Period t0
Start/Stop Operation
Deceleration Period t1
Acceleration/Deceleration Operation
Calculate the Number of Operating Pulses A The 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.
<Standard AC Motors> (The same equation applies to brushless DC motors.) Rotor Inertia Total Inertia [kgm2] [kgm2] Operating Speed [r/min] coefficient Acceleration (Deceleration) Period [s] NM (J0 JL) = 9.55 t1
Acceleration Torque = Ta[Nm]
Operating Speed NM [r/min]
No. of Pulses Distance per Movement Number of Operating = Required for Pulses A[Pulses] Distance per Motor Rotation 1 Motor Rotation l 360˚ : Step Angle θ s = lrev θs
Movement
Calculate the Operating Pulse Speed f 2
Acceleration Period t 1
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 Number of Acceleration Operating Pulses Starting Pulse (Deceleration) Operating Pulse Speed [Hz] [Pulses] Period [s] Speed f2[Hz] = Positioning Period Acceleration (Deceleration) [s] Period [s] =
Af1t1 t0t1
Number of Operating Pulses [Pulses] Operating Pulse = Positioning Period [s] Speed f2[Hz] A t0
Calculate the Acceleration/Deceleration Rate TR
=
t1 f2f1
✽Calculate the pulse speed in full-step equivalents.
G-4
ORIENTAL MOTOR GENERAL CATALOGUE
Pulse Speed [kHz]
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: Acceleration (Deceleration) Acceleration/ Period[ms] deceleration = rate TR Operating Pulse Starting Pulse [ms/kHz] Speed[kHz] Speed [kHz]
Deceleration Period t 1
For AC Motors
<Stepping Motors> qFor Acceleration/Deceleration Operation Acceleration Torque Ta [Nm] =
Rotor Inertia Total Inertia Step Angle[˚ ] [kgm2] [kgm2] 180˚
= (J0 JL)
θ s 180
f2f1 t1
Operating Pulse Starting Pulse Speed [Hz] Speed [Hz] Acceleration (Deceleration) Period [s]
wFor Start-Stop Operation Acceleration Torque Ta [Nm] =
Step Angle[˚ ](Operating Pulse Speed)2 [Hz] Rotor Inertia Total Inertia [kgm2] [kgm2] 180˚coefficient
= (J0 JL)
θ s f22 180n
n: 3.6˚/ θ s
Calculate the Required Torque TM 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 L T a)S f
wFor Start-Stop Operation
=
Positioning Period t 0
TR
t1
Standard AC Motors qSpeed 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 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 may damage 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 Motors qChecking 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. Running Duty Cycle
Running Time Running Stopping Time Time
100
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.
Product Series
Motor Frame Size Acceleration/Deceleration Rate TR[ms/kHz] 28, 42, 60, 85 0.1 Maximum✽ Stepping Motor and 20 Maximum 42, 60 Driver Packages 30 Maximum 85, 90 ✽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.
Service Life
Safety Factor: Sf (Reference Value) Product Series Stepping Motor and Driver Packages
Safety Factor (Reference Value) 1.52 2
Motor and Fan Sizing
Standard AC Motors
Brushless DC Motors
Required Torque
Speed [r/min] (Pulse Speed [kHz])
Stepping Motors
Gearheads
Linear Motion
Fan Motors
Total Inertia of the Machine [kgm2] Rotor Inertia of the Motor [kgm2]
Inertia Ratio =
=
Acceleration Rate (Reference Values with XG9200, SG9200 Series)
Torque [Nm]
There 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.
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.
Technical Reference
Selection Considerations
JL J0
Inertia Ratio (Reference Values) Product Series
Motor Frame Size 28, 42, 60, 85 Stepping Motor and 20, 28 Driver Packages 42, 60, 85
Inertia Ratio 30 Maximum 5 Maximum 10 Maximum
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 Ratio =
=
Total Inertia of the Machine [kgm2] Rotor Inertia of the Motor [kgm2](Gear Ratio)2 JL J0 i 2
G-5
Technical Reference
Sizing Example
(4) Calculate the Operating Speed N [r/min]
Ball Screw
Operating Speed =f2
Using Stepping Motors ( Controller
Stepping Motor
)
θS 60 360
=10000
0.72 60 360
=1200 [r/min] m Coupling
Calculate the Required Torque T M [Nm] (see page G-4) (1) Calculate the Load Torque T L [Nm]
Direct Connection
Force of moving direction F =FA mg (sin a cos a)
Driver
=0 409.807 (sin0 0.05cos0) =19.6 [N] DB
PB
Pilot Pressure Load F0 =
F 19.6 = =6.53 [N] 3 3
Load Torque TL =
F PB 0 F0 PB 2 2
Programmable Controller
=
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=0.3 Ball screw shaft diameter·············································DB =15 mm Total length of ball screw ············································LB=600 mm Material of ball screw ·················Iron (density =7.9103 [kg/m3]) Ball screw pitch ····························································PB=15 mm Desired resolution···············································∆l=0.03 mm/step (feed per pulse) Feed ·············································································l=180 mm Positioning period ·············································t 0=Within 0.8 sec.
(2) Calculate the Acceleration Torque T a [Nm] qCalculate the total moment of inertia JL [kgm2] (See page G-3 for basic equations) LB DB4 32 = 7.9103600103 (15103)4 32
Inertia of Ball Screw JB =
=0.236104 [kgm2] Inertia of Table and Work JT =m (
wCalculate the acceleration torque Ta [Nm]
360˚0.03 =0.72 [˚ ] 15
Acceleration torque Ta =(J0 JL)
180 360˚ =6000 [pulses] 15 0.72˚
(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 f2 [Hz]
(3) Calculate the Required Torque T M [Nm] Required torque T M=(T L Ta)2 ={0.0567 (628J0 0.158)}2 =1256J0 0.429 [Nm]
Select a Motor (1) Provisional Motor Selection
AS66AA
6000 pulses
t1
t1 t0=0.8
0.2
period [s]
20
2.0
15
1.5
10
Torque [Nm]
10000
0.2
Required Torque [Nm] 0.48
(2) Determine the Motor from the Speed-Torque Characteristics AS66AA
Torque [kgfcm]
Operating pulse speed [Hz]
60000 =10000 [Hz] 0.80.2
Rotor Inertia [kgm2] 405107
Model
Operating pulse Operating pulses (A)Starting pulse speed (f1)Acceleration (Deceleration) Period (t1) = speed f2 Positioning Period (t0)Acceleration (Deceleration) Period (t1) =
0.72 100000 180˚ 0.2
=628J0 0.158 [Nm]
(1) Finding the Number of Operating Pulses A [pulses] 360˚ Feed per Unit (l) Step Angle (θ s) Ball Screw Pitch (PB)
θ s f2f1 180˚ t1
=(J0 2.52104)
Determine the Operating Pattern (See page G-4 for basic equations)
=
15103 2 ) =2.28104 [kgm2] 2
=0.236104 2.28104=2.52104 [kgm2]
can be connected directly to the application.
Operating pulses A =
PB 2 ) 2
Total Inertia JL =JB JT
360˚Desired Resolution (∆l) Ball Screw Pitch (PB)
=
=0.0567 [Nm]
=40(
Calculate the Required Resolution Required Resolution θ s =
19.615103 0.36.5315103 2 0.9 2
1.0
5
0.5
0
0
0
1000 10
2000 3000 Speed [r/min]
4000
20
30 40 50 60 Pulse Speed [kHz] (Resolution Setting: 1000 P/R)
Select a motor for which the required torque falls within the pullout torque of the speed-torque characteristics.
G-6
ORIENTAL MOTOR GENERAL CATALOGUE
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.
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 (TM) is given by: TM =
TL 0.283 = =0.0388 [Nm]=38.8 [mNm] i G 90.81
(Gearhead transmission efficiency G =0.81)
Technical Reference
Using Standard AC Motors
Look for a margin of safety of 2 times.
Motor Gearhead Coupling
Ball Screw FA Slide Guide v
m
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.
Motor and Fan Sizing
Load Inertia Check Ball Screw Moment of Inertia JB =
Determine the Gear Ratio Speed at the gearhead output shaft: NG =
V 60 (15 2)60 = P 5
=180 24 [r/min]
Because the rated speed for a 4-pole motor at 60 Hz is 14501550 r/min,
=
LB DB4 32 7.9103800103 (20103)4 32
=0.993104 [kgm2] A 2 Table and Work Moment of Inertia Jm =m ( ) 2 =30 (
5103 2 ) 2
=0.190104 [kgm2]
Gearhead shaft total load inertia J=0.993104 0.190104=1.18104 [kgm2] Here, the 3GN9K permissible load inertia is (see page A-13): JG=0.1410492 JG=1.13103 [kgm2] Therefore, J < JG, the load inertia is less than the permissible inertia, so there is no problem. There is margin for the torque, so the rotation rate is checked with the no-load rotation rate (about 1750 r/min). V=
NM PB 17505 = =16.2 [mm/s] 60 i 609
Service Life
Standard AC Motors
Brushless DC Motors
Stepping Motors
Gearheads
Linear Motion
Fan Motors
(where NM is the motor speed)
This confirms that the motor meets the specifications.
the gear ratio ( i ) is calculated as follows: i=
14501550 14501550 = =7.19.9 NG 180 24
From within this range a gear ratio of i =9 is selected.
Calculate the Required Torque Load weight in the direction =FA mg (sin cos ) of the ball screw shaft F =0 309.807 (sin90˚ 0.05cos90˚) =294 [N] Pilot pressure load F0 = Load torque TL = =
F =98 [N] 3 FPB 0F0PB 2 2 2945103 0.3985103 2 20.9
=0.283 [Nm]
G-7
Technical Reference
Belt and Pulley
Load Inertia Check
Using Standard AC Motors Belt and Work Moment of Inertia Jm1 =m1
(
D 2
=20
(
100103 2 ) 2
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.
=500104 [kgm2] Roller Moment of Inertia Jm2 =
2
)
V
=
1 m2D2 8 1 1 (100103)2 8
=12.5104 [kgm2]
Belt Conveyor
D
Motor Gearhead
Total mass of belt and work ·············································m1=20kg Frictional coefficient of sliding surfaces ·································=0.3 Drum radius ··································································D=100mm Mass of drum·····································································m2=1kg Belt roller efficiency ·······························································=0.9 Belt speed·························································V=140mm/s 10% Motor power supply·························Single-Phase 110 VAC 60 Hz Movement time ···························································8 hours/day
Determine the Gear Ratio Speed at the gearhead output shaft: NG =
V 60 (140 14) 60 = D 100
=26.7 2.7 [r/min]
Because the rated speed for a 4-pole motor at 60 Hz is 14501550 r/min, the gear ratio (i) is calculated as follows: 14501550 14501550 = =49.364.6 i= 26.7 2.7 NG
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=mg=0.3209.807=58.8 [N] Load torque TL =
FD 58.8100103 = =3.27 [Nm] 20.9 2
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 TM, then: TM =
TL 3.27 = =0.0826 [Nm]=82.6 [mNm] i G 600.66
(Gearhead transmission efficiency G=0.66)
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.
G-8
ORIENTAL MOTOR GENERAL CATALOGUE
Gearhead Shaft Load Inertia J=500104 12.51042=525104 [kgm2] Here, the 5GN60K permissible load inertia is (see page A-13): JG=0.75104602 =2700104 [kgm2] Therefore, J < JG, 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=
NM D 60 i
=
1750 100 =152.7 [mm/s] 6060
(Where NM is the motor speed)
Using Low-Speed Synchronous Motors (SMK Series) 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.
Work m
Technical Reference
Using Brushless DC Motors Here is an example of how to select a brushless DC motor to drive a belt conveyor.
Roller 2
F Roller 1
Fig. 1 Example of Belt Drive
Performance Belt speed VL 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
Structural Specifications Total mass of belt and work Roller diameter Mass of roller Frictional coefficient of sliding surfaces Belt and pulley efficiency Frequency of power supply
m1=1.5 [kg] D=30 [mm] m2=0.1 [kg] =0.04 =0.9 60 Hz (Motor speed: 72 r/min)
Motor and Fan Sizing
Service Life
CW 15 (sec) 5
10
Standard AC Motors
CCW Fig. 2 Operating Pattern
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. NG =
60VL D
NG : Speed at the gearhead output shaft
Belt Speed 600.015 0.015 m/s =2.86 r/min (Minimum Speed) 0.1 1 m/s
601 =191 r/min (Maximum Speed) 0.1
Calculate the Load Inertia JG Load Inertia of Roller: Jm2 Jm2 =
1 1 2 m2D = 10.12=12.5104 kgm2 8 8
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 V= D N = 3072 =113 [mm/s] 60 60
0.58930103 =9.82103 [Nm] Load Torque TL= F D = 20.9 2
Jm1= m1( D )2=1.5( 2
30103 )2=3.38104 [kgm2] 2
Load Inertia of Roller: Jm2 Jm2= 1 m2D2= 1 0.1(30103)2=0.113104 [kgm2] 8 8
The load inertia JG is calculated as follows: JG=Jm22 Jm1=212.5104 375104 =400104kgm2
The load inertia JL is calculated as follows: JL=Jm1 Jm22=3.38104 0.1131042=3.5104 [kgm2] rCalculate the Acceleration Torque
Friction Coefficient of = m1 g=0.3159.807=44.1N the Sliding Surface: F Load Torque TL =
FD 44.10.1 2 20.9
=2.45 Nm
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.
Gearheads
Linear Motion
eCalculate the Load Inertia Load inertia of belt and work: Jm1
D 2 0.1 2 Jm1 =m1 ( ) =15 ( ) =375104 kgm2 2 2
Calculate the Load Torque T L
Stepping Motors
wCalculate the Required Torque T L Frictional coefficient of sliding surfaces: F= m1 g=0.041.59.807=0.589 [N]
Load inertia of belt and work: Jm1
From the specifications on page B-10, the permissible load inertia for BX5120A-15 is 420104 kgm2
Brushless DC Motors
Fan Motors
2 2 Ta=(J0 JL) θ s f =(J0 3.5104) 7.260 180 n 1800.5
=905 J0 0.32 [Nm]
Here, θ s=7.2˚, f=60 Hz, n=3.6˚/θ s=0.5 J0 =Rotor Inertia tCalculate the Required Torque (look for a margin of safety of 2 times) Required Torque T M=(T L Ta)2 =(9.82103 905J0 0.32)2 =1810J0 0.66 [Nm] ySelect a Motor Select a motor that satisfies both the required torque and the permissible load inertia. Motor SMK5100A-A
Rotor Inertia [kgm2] 1.4104
Permissible Load Inertia Output Torque [kgm2] [Nm] 1.12 7104
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 in e is also below the permissible load inertia, SMK5100A-A can be used in this application.
G-9
Technical Reference
Index Table
(4) Determine the Operating Pulse Speed f 2 [Hz]
Geared stepping motors are suitable for systems with high inertia, such as index tables.
Operating Pulse Speed f2 =
Determine the Drive Mechanism
A 600 = t0t1 0.250.1 =6667 [Hz]
DT=300 mm =125 mm Operating Pulse Speed [Hz]
6667
Driver
Controller
600 Pulses
0.1
t1
t1 t0=0.25
Programmable Controller
Geared Stepping Motor
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 maximum starting/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] θ Operating pulses A = θs =
36˚ 0.036˚
=1000 [Pulses]
(2) Determine the Acceleration (Deceleration) Period t1 [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] Operation Speed N =
θ 60 36 60 = 360 0.250.1 360 t0t1
=40 [r/min]
✽The permissible speed range for a PN geared motor with a gear ratio of 1:10 is 0 to 300 r/min.
G-10
ORIENTAL MOTOR GENERAL CATALOGUE
0.1
Period [s]
Select a Motor
(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]
(1) Provisional Motor Selection
(2) Calculate the Acceleration Torque T a [Nm] qCalculate the total moment of inertia JL [kgm2] (See page G-3 for basic equations) LTDT4 32
250
25
7.9103(10103)(300103)4 32
200
20
=6.28102 [kgm2]
Inertia of Work JW1 = LWDW4 32 (Center of gravity) = 7.9103(30103)(40103)4 32 =0.596104 [kgm2] Mass of Work mW = =
LWDW2 4
150
=0.3 [kg]
Inertia of work JW [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) Jw1 [kgm2] .
TM=9.55
100
15 10
50
5
0
0
Motor and Fan Sizing
Permissible Torque
0
7.9103(30103)(40103)2 4
Required Torque [Nm]
(2) Determine the Motor from the Speed-Torque Characteristics AS66AA-N10
Torque [Nm]
=
AS66AA-N10
Torque [kgfcm]
Inertia of Table JT =
Rotor Inertia [kgm2] J0=405107
Model
Technical Reference
Calculate the Required Torque T M [Nm] (see page G-4)
50 5
10
100 15
150 200 Speed [r/min]
250
20 25 30 35 40 Pulse Speed [kHz] (Resolution Setting: 1000P/R)
300 45
Service Life
50
Select a motor for which the required torque falls within the pullout 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.
Standard AC Motors
Brushless DC Motors
Stepping Motors
Since the number of works, n, is 10 [pcs], Inertia of Work JW = n(JW1 mWL2) (Center of gravity)
Gearheads
= 10[(0.596104) 0.3(125103)2] =4.71102 [kgm2]
Linear Motion
Total Inertia JL = JT JW = (6.28 4.71)102
Fan Motors
=11102 [kgm2]
wCalculate the acceleration torque T a [Nm] θ s Acceleration =(J0i2 JL) 180 Torque Ta =(J0102 11102)
f2f1 t1
66670 0.036 180 0.1
=4.19103J0 4.61 [Nm]
(3) Calculate the Required Torque T M Safety Factor S f =2.0 Required Torque T M=(T L Ta)S f ={0 (4.19103 J 0 4.61)}2.0 =8.38103 J 0 9.2 [Nm]
G-11
Related Documents 3h463d
Oriental Motor Motor Sizing Calculations 6g5f56
October 2022 0
Motor Sizing Calculation 64v40
December 2019 54
Abb Motor-vsd Sizing 5n1731
August 2021 0
Dc Motor Calculations 3p283o
December 2019 37
Nec Motor Calculations Article 430 5l2d65
January 2022 0
Main Hoist Motor Power Calculations z245
October 2021 0More Documents from "Kha Veen" 14526s
Oriental Motor Motor Sizing Calculations 6g5f56
October 2022 0
Renault Dp0 Al4 Fault Finding Mr341laguna2 6d593p
November 2019 96
October 2020 0
Etabs-example-rc Building Seismic Load _design Skill 6h2a26
November 2020 0
23.000 Afrique j3e34
January 2021 0