12 Appendix
12.1.2 Selection Example for Position Control
12-4
12.1.2 Selection Example for Position Control
(1) Speed Diagram
(2) Rotation Speed
• Load axis rotation speed
• Motor shaft rotation speed with direct coupling: Gear ratio 1/R = 1/1
Therefore,
(3) Load Torque
(4) Load Moment of Inertia
• Liner motion section
• Ball screw
• Coupling
• Load moment of inertia at the motor shaft
(5) Load Moving Power
• Load speed: V = 15 m/min
• Positioning times: n = 40 times/min
• Linear motion section mass: M = 80 kg
• Positioning distance: = 0.25 m
• Ball screw length: L
B
= 0.8 m
• Positioning time: tm = Less than 1.2 s
• Ball screw diameter: D
B
= 0.016 m
• Electrical stop accuracy:
δ
=
±
0.01 mm
• Ball screw lead: P
B
= 0.005 m
• Friction coefficient:
μ
= 0.2
• Coupling mass: M
C
= 0.3 kg
• Mechanical efficiency:
η
= 0.9 (90%)
• Coupling outer diameter: D
C
=0 .03 m
Where ta = td, ts = 0.1 (s)
1
Ball screw
Servomotor
Linear motion
Coupling
Mechanical Specifications
Load
speed
Reference
pulse
s
t
d
t
ta
c
t
(m/min)
Speed
Time (s)
tm
t
V
15
t
60
n
------
60
40
------
1.5 s
( )
=
=
=
ta = tm
−
ts
−
= 1.2
−
0.1
−
= 0.1 (s)
60
60
×
0.25
V
15
tc
1.2 0.1
–
0.1 2
×
–
0.9 s
( )
=
=
N = = = 3000 (min )
V
1
5
P
B
-1
0.005
N = N R = 3000
×
1 = 3000 (min )
M
-1
T
L
=
= = 0.139 (N m)
9.8
μ
M P
B
2
π
R
η
9.8
×
0.2
×
80
×
0.005
2
π ×
1
×
0.9
J
L1
= M
(
)
2
= 80
×
(
)
2
= 0.507
×
10
-4
(kg m )
P
B
2
π
R
0.005
2
π ×
1
2
J
B
=
ρ
L
B
D
B
4
=
×
7.87
×
10
3
×
0.8
×
(0.016)
4
= 0.405
×
10
-4
(kg
m
2
)
π
32
π
32
J
C
= M
C
D
C
4
=
×
0.3
×
(0.03)
2
= 0.338
×
10
-4
(kg
m
2
)
8
1
1
8
J
L
= J
L1
J
B
J
C
=
1.25
×
10
-4
(kg
m
2
)
P
O
=
= = 43.7 (W)
2
π
N
M
T
L
60
2
π ×
3000
×
0.139
60
Содержание Sigma-II Series
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