40
Servomotor Selection Example
Speed Diagram
Speed
Effective Torque at Motor Shaft
Effective Load Moment of Inertia at Motor Shaft
Load Running Power
Mechanical Specifications
Time (s)
Reference pulse
Load speed
Speed
(m/min)
15
t
1.2
ta tc td ts
V
Linear motion unit
Ball screw
Coupling
Servomotor
•
Load axis speed
15
0.005
2πR
η
µ
9.8
M
P
B
•
Motor speed
Because of direct coupling, the gear ratio is 1/R = 1/1.
Then,
NM
=
N
R=3000
×
1=3000
(
min
-1
)
2π
×
1
×
0.9
0.2
×
9.8
×
80
×
0.005
•
Positioning frequency
•
Traveling distance
•
Positioning interval
•
Friction coefficient
•
Combined efficiency
:
:
:
:
:
:
:
= 15m/min
= 80kg
= 0.8m
= 0.016m
= 0.005m
= 0.3kg
= 0.03m
=
40 times/min
= 0.25m
= 1.2
S max.
= 0.2
= 0.9
(
90%
)
V
M
L
B
DB
PB
MC
DC
n
t
m
µ
η
v
60 60
2π
×
3000
×
0.139
N
=
=
= 3000
(
min
-1
)
P
B
V
T
L
= = = 0.139
(
N
m
)
2
2
32
8
1
8
1
•
Linear motion
•
Ball screw
•
Coupling
J
L
=
J
L1
+
J
B
+
J
C
= 1.25
×
10
-4
(
kg
m
2
)
J
L1
=
M
= 80
×
= 0.507
×
10
-4
(
kg
m
2
)
2πR
P
B
2
×
1
0.005
π
J
B
= ρ
L
B
D
B
4
=
×
7.87
×
10
3
×
0.8
×
(
0.016
)
4
= 0.405
×
10
-4
(
kg
m
2
)
J
C
=
M
C
DC
2
=
×
0.3
×
(
0.03
)
2
= 0.338
×
10
-4
(
kg
m
2
)
2π
N
M
T
L
P
o
=
=
=
43.7
(
W
)
n
60 60
40
60
×
0.5
30
Where acceleration time (
ta
) = deceleration time (
td
) and
setting time (
tc
) = 0.1 s when the filter setting of the FIL rotary switch is 0.
Cycle time
t
= = = 1.5 (s)
ta
=
td
=
tm
—
ts
—
= 1.2
—
0.1
—
= 0.1 (s)
tc
=
tm
—
ts
—
ta
—
td
= 1.2
—
0.1
—
0.1
—
0.1 = 0.9 (s)
60
×
V
Constant-speed time :
Acceleration time :
Selection of Servomotor Size
Selection of Servomotor Size
32
ππ
•
Load speed
•
Mass of linear motion unit
•
Ball screw length
•
Ball screw diameter
•
Ball screw lead
•
Coupling mass
•
Coupling outer diameter
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