76
AX
Series
Selection guide (1)
(working conditions)
STEP 1
Calculation of moment
of inertia
STEP 2
Max. rotation speed
STEP 3
Load torque
STEP 5
Selection guide
STEP 4
Regenerative
electric power
* Only for AX4300, AX4500
a) Table J
1
=
= 6.32
2
W
t
R
2
2
79 0.4
2
b) Jig and workpiece (kg·m
2
)
(kg·m
2
)
(kg·m
2
)
J
2
= N W
j
R
e
2
=4 10 0.325
2
= 4.225
c) Total sum of moment of inertia
N
max
= V
m
·
Confirm that Nmax does not exceed the direct drive actuator's max. rotation speed.
Load total sum of moment of inertia
Max. rotation speed
Max. load torque
Effective load torque
Regenerative electric power
10.545
18
33
100
231.3
300
70.7
100
15.68
40
(kg·m
2
)
(rpm)
(N·m)
(N·m)
(w)
AX4300 is used.
Study whether the temporarily selected AX4300 is used.
T
m
= [ A
m
· (J + J
M
) ·
T
rms
=
= [
5.53
(
10.545 + 0.326
)
=
231.3
(N·m)
= 70.7
(N·m)
+ 0 + 0 ] 1.5 + 10
+ T
F
+ T
W
] · fc
+ T
MF
(rpm) =
1.76
=
33
6
· t
1
6 0.8
90
J = J
1
+ J
2
= 6.32 + 4.225 = 10.545
180
·
t
1
2
180
0.8
2
90
t
1
t
0
T
rms
=
0.8
4
W
=
(
W 40 (W)
(W)
)
2
)
2
V
m
(J + J
M
)
t
1
·
180
2 · t
0
=
(
1.76
90
0.8
180
= 16.23
10.871
2
4
Table radius : R = 0.4 (m)
Table weight : Wt = 79 (kg)
Jig radius of rotation : Re = 0.325 (m)
Jig weight : Wj = 10 (kg/pc.)
(Including workpiece weight)
Jig number : N = 4
(operation conditions)
Movement angle : = 90 (˚)
Moving time : t
1
= 0.8 (s)
Cycle time : t
0
= 4 (s)
Load friction torque : T
F
= 0 (N·m)
Working torque : T
W
= 0 (N·m)
Output shaft friction torque : T
MF
(N·m)
Follows actuator specifications
Cam curve : MS (modified sine)
=
Calculate the smallest model that can tolerate the load moment of inertia.
The AX4300 allowable moment of inertia is 18.0 (kg·m
2
) or over, so this load is allowable.
Max. load torque
·
··
180
·
t
1
2
·
180
0.8
2
90
Effective load torque
· [r · A
m
· (J + J
M
) ·
· fc ]
2
+ ( T
F
· fc + T
W
·fc + T
MF
)
2
[0.707 5.53 10.871
1.5 ]
2
+ (0 1.5 + 0 1.5 + 10)
2
·
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