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
tt
a
c
t
(m/min)
Speed
Time (s)
tm
t
V
15
t
60
n
------
60
40
------1.5s()===
ta = tm − ts − = 1.2 − 0.1 − = 0.1 (s)
60
60 × 0.25
V
15
tc 1.2 0.1–0.12×–0.9s()==
N = = = 3000 (min )
V
15
P
B
-1
0.005
N = N R = 3000 × 1 = 3000 (min )
M
-1
T
L
= = = 0.139 (Nm)
9.8μ
M P
B
2πR η
9.8 × 0.2 × 80 × 0.005
2π × 1 × 0.9
J
L1
= M
(
)
2
= 80 ×
(
)
2
= 0.507 × 10
-4
(kg
m )
P
B
2πR
0.005
2π × 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)
2πN
M
T
L
60
2π × 3000 × 0.139
60
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