38
n
60
60
40
60
×
0.5
30
Mechanical Specifications
Time (s)
Reference pulse
Load speed
Speed
(m/min)
30
t
1.2
ta
tc
td ts
V
•
Load axis speed
2
π
R
η
µ
9.8
M
P
B
+ 2
π
T
30
0.201
•
Motor speed
Where acceleration time (
t
a
) = deceleration time (
t
d
) and
setting time (
t
c
) = 0.1 s when the filter setting of the FIL rotary switch is 0.
Cycle time
N
M
=
N
R=149
×
5= 745 (min
-1
)
2
π
×
5
×
0.9
0.2
×
9.8
×
4
×
0.201 + 2
π
×
0.05
•
Load speed
•
Mass of linear-motion unit
•
Pulley diameter
•
Pulley thickness
•
Coupling mass
•
Coupling outer diameter
•
Gear A outer diameter
•
Gear A thickness
•
Gear B outer diameter
•
Gear B thickness
•
Gear density
•
Gear ratio
•
Positioning frequency
•
Traveling distance
•
Positioning interval
•
Friction coefficient
•
Effective load torque
at motor shaft
•
Combined efficiency
:
:
:
:
:
:
:
:
:
:
:
:
:
:
:
:
:
:
= 30m/min
= 4kg
= 0.064m
= 0.02m
= 2690kg/m
3
= 0.03m
= 0.02m
= 0.02m
= 0.1m
= 0.02m
= 7870kg/m
3
= 5
= 40 times/min
= 0.5m
= 1.2
S max.
= 0.2
= 0.05N
m
= 0.9
(90%)
V
M
D
P
L
P
M
C
D
C
D
A
L
A
D
B
L
B
A
,
B
R
n
t
m
µ
T
η
t = = = 1.5 (s)
t
a
=
t
d
=
t
m
–
t
s
–
= 1.2
–
0.1
–
= 0.1 (s)
t
c
=
t
m
–
t
s
–
t
a
–
t
d
= 1.2
–
0.1
–
0.1
–
0.1 = 0.9 (s)
60
×
V
N
=
=
= 149 (min
-1
)
P
B
=
π
d
=
π
×
0.064 = 0.201
J
L
=
J
L1
+
J
L2
+
J
L3
=
(
1.639
+
0.687
+
0.362
)
×
10
-4
= 2.69
×
10
-4
(
kg
m
2
)
P
B
V
T
L
=
= 0.0669 (N
m)
=
32
π
J
i
5
2
1
8
1
•
Linear motion
•
Load-shaft motion: Pulley
×
2 + Gear B
•
Motor-shaft motion: Gear A + Coupling
2
π
R
2
π
×
5
0.201
2
2
J
L1
=
M
=1.639
×
10
-4
(kg
m
2
)
= 4
×
P
B
×
×
(2690
×
0.02
×
(0.064)
4
×
2 + 7870
×
0.02
×
(0.1)
4
)
= 0.687
×
10
-4
(kg
m
2
)
=
J
L2
=
32
π
R
2
×
0.3
×
(0.03)
2
= 0.362
×
10
-4
(kg
m
2
)
×
7870
×
0.02
×
(0.02)
4
+
J
L3
=
ρ ρ
Σ
Constant-speed time :
Acceleration time :
Gear B
Load
Servomotor
Pulley
Gear A
Coupling
Selection of Servomotor Size
Selection of Servomotor Size
Servomotor Selection Example
Speed Diagram
Speed
Effective Torque at Motor Shaft
Effective Load Moment of Inertia at Motor Shaft
