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G-2 ORIENTAL MOTOR GENERAL CATALOGUE
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.
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 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.
Determine the drive
mechanism component
Confirm the required specifications
(Equipment specifications)
Calculate the speed and load
Select motor type
Selection calculation
G-
Service Life
Motor and Fan Sizing
Standard AC Motors
Brushless DC Motors
Stepping Motors
Gearheads
Linear Motion
Fan Motors
Formulas for Calculating Load Torque
Calculate 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.)
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/s 2 ])
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/s^2 ] (9.807)
F B
D
Machine
Pulley
Spring Balance
T L = [Nm] y
F B D
F A m
D
F (^) F A m
D
F
T L = [Nm] r
F
2 =^
FD
2 i
D
i
F = F A mg (sin a cos a ) [N] t
F A m
D
T L =
D
i
F A mg
= [Nm] e
( F A mg ) D 2 i
F A m
Direct Coupling
F A
F m
T L = ( ) [Nm] q
FP B
i
0 F 0 P B
F = F A mg (sin a cos a ) [N] w
Formulas for Calculating Moment of Inertia
Inertia 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 =7.9 103 [kg/m^3 ]
Aluminum =2.8 103 [kg/m^3 ]
Bronze =8.5 103 [kg/m^3 ]
Nylon =1.1 103 [kg/m^3 ]
Jx =Inertia on x axis [kgm^2 ]
Jy =Inertia on y axis [kgm^2 ]
J 0 =Inertia on x 0 axis (passing through center
of gravity) [kgm 2 ]
m =Mass [kg]
D 1 =External diameter [m]
D 2 =Internal diameter [m]
=Density [kg/m^3 ]
L =Length [m]
A =Unit of movement [m/rev]
J = m ( 2 ) 2 [kgm^2 ]!
A
A (^) B
C
x
y
Jx = 12 m ( A^2 B^2 )= [kgm^2 ]!
12 ABC^ ( A^2 B^2 )
Jy = 12 m ( B^2 C^2 )= [kgm^2 ]!
12 ABC^ ( B^2 C^2 )
C
A B
x x^0
Jx = Jx 0 ml^2 = 12 m ( A^2 B^2 12 l^2 ) [kgm^2 ]!
r= Distance between x and x 0 axes [m]
L
D 1 D 2
x
y
Jx = m ( D 12 D 22 )= [kgm^2 ] o 8
L ( D 14 D 24 )
Jy = 4 m [kgm^2 ]!
D 12 D 22
L^2
D 1
L
x
y
Jx = 8 mD 12 [kgm^2 ] u
= 32 LD 14
Jy = 4 m [kgm^2 ] i
D 12
L^2
G-
Service Life
Motor and Fan Sizing
Standard AC Motors
Brushless DC Motors
Stepping Motors
Gearheads
Linear Motion
Fan Motors
Selection Considerations
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.
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 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 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.
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 Ratio =
Total Inertia of the Machine [kgm^2 ] Rotor Inertia of the Motor [kgm 2 ](Gear Ratio) 2
J L
J 0 i^2
Product Series Motor Frame Size
Stepping Motor and Driver Packages
Inertia Ratio 30 Maximum 5 Maximum 10 Maximum
Inertia Ratio =
Total Inertia of the Machine [kgm 2 ] Rotor Inertia of the Motor [kgm 2 ]
J L
J 0
Running Duty Cycle 100
Running Time Running Time
Stopping Time
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 ])
Torque [
Nm
]
Product Series
Stepping Motor and Driver Packages
Safety Factor (Reference Value) 1.5 2 2
Product Series Motor Frame Size Acceleration/Deceleration Rate T R[ms/kHz] 28, 42, 60, 85 42, 60 85, 90
0.1 Maximum✽ 20 Maximum 30 Maximum
Stepping Motor and Driver Packages
G-6 ORIENTAL MOTOR GENERAL CATALOGUE
Sizing Example
Ball Screw
Determine the Drive Mechanism
Total mass of the table and work ····································· m =40 kg
Frictional coefficient of sliding surfaces ······························· =0.
Ball screw efficiency ····························································· =0.
Internal frictional coefficient of pilot pressure nut ················· 0 =0.
Ball screw shaft diameter············································· DB =15 mm
Total length of ball screw ············································ LB =600 mm
Material of ball screw ·················Iron (density =7.9 103 [kg/m^3 ])
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.
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 t 1 =0.80.25=0.2 [s]
(3) Determine the Operating Pulse Speed f 2 [Hz]
t 1 t (^1) period [ s]
t 0 =0.
Operating pulse speed [
Hz
]
6000 pulses
Operating pulse speed f 2
Operating pulses ( A ) Starting pulse speed ( f 1 )Acceleration (Deceleration) Period ( t 1 ) Positioning Period ( t 0 ) Acceleration (Deceleration) Period ( t 1 )
=
=10000 [Hz]
Operating pulses A =
Feed per Unit ( l ) Ball Screw Pitch ( P B )
Step Angle ( s)
=6000 [pulses]
Required Resolution s =
360 ˚Desired Resolution (∆ l ) Ball Screw Pitch ( P B )
= =0.72 [˚ ]
P B
DB
Stepping Motor
Controller
Driver
Programmable Controller
Coupling
Direct Connection
m
(4) Calculate the Operating Speed N [r/min]
Calculate the Required Torque T M [N m] (see page G-4) (1) Calculate the Load Torque T L [N m]
(2) Calculate the Acceleration Torque T a [N m]
qCalculate the total moment of inertia J L [kgm^2 ]
(See page G-3 for basic equations)
wCalculate the acceleration torque Ta [Nm]
(3) Calculate the Required Torque T M [N m]
Required torque T M =( T L Ta ) 2
={0.0567 (628 J 0 0.158)} 2
=1256 J 0 0.429 [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.
(^0 1000 2000 3000 )
Torque [kgfcm]Torque [N
m]
Speed [r/min]
Pulse Speed [kHz]
0 10 20 30 40 50 60 (Resolution Setting: 1000 P/R)
15
20
10
5
0
AS66AA 405 10 7
Rotor Inertia
[kgm 2 ]
Required Torque
Model [Nm]
Acceleration torque T a J 0 J L
f 2 f 1 t 1
s 180 ˚
J 0 2.52
628 J 0 0.158 [Nm]
Inertia of Ball Screw J B =
32 ^ ^ L B^ ^ D B
4
=0.236 10 4 [kgm 2 ]
32 7.9^10
= m (
Inertia of Table and Work J T )^2
= J B J T
=0.236 10 4 2.28 10 4 =2.52 10 4 [kgm 2 ]
Total Inertia J L
P B
2 =2.28^10
15 (^10) ) 2 4 [kgm 2 ] 3
=F A mg (sin a cos a ) =0 40 9.807 (sin0 0.05cos0) =19.6 [N]
Force of moving direction F
Load Torque TL =
F P B
0 F 0 P B
Pilot Pressure Load F 0 =
F
3 = =6.53 [N]
=0.0567 [Nm]
Operating Speed = f 2 60
S 360
=10000 60
=1200 [r/min]
Using Stepping Motors ( )
G-8 ORIENTAL MOTOR GENERAL CATALOGUE
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.
Total mass of belt and work ············································· m 1 =20kg
Frictional coefficient of sliding surfaces ································· =0.
Drum radius ·································································· D =100mm
Mass of drum····································································· m 2 =1kg
Belt roller efficiency ······························································· =0.
Belt speed························································· V =140mm/s 10%
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
1450 1550 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 =0.3 20 9.807=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 TM , then:
Look for a margin of safety of 2 times, taking into consideration
commercial power voltage fluctuation.
82.6 2 165 [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.
TM =
TL
i G =^ =0.0826 [Nm]=82.6 [mNm]
(Gearhead transmission efficiency G =0.66)
Load torque TL = =3.27 [Nm]
F D
2 =^
the gear ratio ( i ) is calculated as follows:
i
N G
Speed at the gearhead output shaft: NG = =
V 60
D
=26.7 2.7 [r/min]
Belt Conveyor
Gearhead
Motor
D
V
Using Standard AC Motors
Load Inertia Check
Gearhead Shaft Load Inertia
J =500 10 4 12.5 10 4 2=525 10 4 [kgm^2 ]
Here, the 5GN60K permissible load inertia is (see page A-13):
JG =0.75 10 4 602
=2700 10 4 [kgm^2 ]
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 =^
60 60 =152.7 [mm/s]
(Where NM is the motor speed)
Roller Moment of Inertia Jm2 =
8 m^2 D
2
=12.5 10 4 [kgm^2 ]
8 ^1 ^ (100^10
= m 1 ( )
2 Belt and Work Moment of Inertia Jm
D
=20 ( )
100 10 3^2
=500 10 4 [kgm^2 ]
G-
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 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 ················································· m 2 =1kg
Total mass of belt and work····························· m 1 =15kg
Frictional coefficient of sliding surfaces ················ =0.
Belt roller efficiency ··············································· =0.
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 JG
Load Inertia of Roller: Jm 2
Load inertia of belt and work: Jm 1
The load inertia JG is calculated as follows:
JG = J m 2 2 J m 1 =212.5 10 4 375 10 4
=400 10 4 kgm^2
From the specifications on page B-10, the permissible load inertia
for BX5120A-15 is 420 10 4 kgm^2
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 TL
Friction Coefficient of = m 1 g =0.3 15 9.807=44.1N the Sliding Surface: F F D
Jm 1 = m 1 (
D
) =15 ( ) =375 10 4 kgm^2
2 2
Jm 2 =^1 8
m 2 D^2 = 1 8
1 0.1^2 =12.5 10 4 kgm^2
NG = NG : Speed at the gearhead output shaft
Belt Speed
60 V L
D
0.015 m/s =2.86 r/min (Minimum Speed)
1 m/s =191 r/min (Maximum Speed)
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 m 1 =1.5 [kg]
Roller diameter D =30 [mm]
Mass of roller m 2 =0.1 [kg]
Frictional coefficient of sliding surfaces =0.
Belt and pulley efficiency =0.
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 = m 1 g =0.041.59.807=0.589 [N]
eCalculate the Load Inertia
Load inertia of belt and work: Jm 1
Load Inertia of Roller: Jm 2
The load inertia JL is calculated as follows:
JL = J m 1 J m 2 2=3.38 10 4 0.113 10 4 2=3.5 10 4 [kgm^2 ]
rCalculate the Acceleration Torque
Here, θ s=7.2˚, f=60 Hz, n=3.6˚/ θ s=0.
J 0 =Rotor Inertia
tCalculate the Required Torque (look for a margin of safety of 2 times)
Required Torque T M =( T L Ta ) 2
=(9.82 10 3 905 J 0 0.32) 2
=1810 J 0 0.66 [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 in e is also below the permissible load inertia,
SMK5100A-A can be used in this application.
Rotor Inertia Motor [kgm 2 ] 1.4 10 4
Permissible Load Inertia [kgm 2 ]
Output Torque [Nm] SMK5100A-A 7 10 4 1.
Ta =( J 0 JL )
=905 J 0 0.32 [Nm]
s f =( J 0 3.5 10 4 )
2
180 n
J m2 = 1
m2 D^2 = 1
0.1(30 10 3 )^2 =0.113 10 4 [kgm^2 ]
J m1 = m 1 ( D
)^2 =1.5( ^30 ^10 )^2 =3.38 10 4 [kgm 2 ]
3
Load Torque TL = F^ ^ D
= 0.589^30 ^10 =9.82 10 3 [Nm]
3
V = D^ ^ N = =113 [mm/s]
CW
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)
G-
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 [N m] (see page G-4)
(1) Calculate the Load Torque T L [N m]
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 [N m]
qCalculate the total moment of inertia J L [kgm 2 ]
(See page G-3 for basic equations)
Inertia of work J W [kgm^2 ] 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 [kgm^2 ].
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.
Required Torque T M=( T L T a) S f
={0 (4.19 103 J 0 4.61)}2.
=8.38 103 J 0 9.2 [Nm]
Acceleration =( J 0 i (^2) J L ) Torque T a
f 2 f 1 t 1
s 180
=( J 0 102 11 10 2 )
=4.19 103 J 0 4.61 [Nm]
= J T J W
Total Inertia J L
=11 10 2 [kgm 2 ]
= n ( J W1 m W L 2 )
= 10 [(0.596 10 4 ) 0.3(125 10 3 ) 2 ]
Inertia of Work J W
=4.71 10 2 [kgm 2 ]
(Center of gravity)
Mass of Work m W =
4 L^ W D^ W
2
=0.3 [kg]
4 7.9^10
Inertia of Table J T =
32 L T D T
4
32 L W D W
4
=6.28 10 2 [kgm 2 ]
32 7.9^10
32 7.9^10
Inertia of Work J W
=0.596 10 4 [kgm 2 ]
(Center of gravity)
Select a Motor
(1) Provisional Motor Selection
(2) Determine the Motor from the Speed-Torque Characteristics
AS66AA-N
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)
Speed [r/min]
Torque [N
m]
Torque [kgfcm]
Permissible Torque
AS66AA-N10 J 0=405 10 7
Rotor Inertia [kgm 2 ] TM=9.
Required Torque Model [Nm]
