AX
Series
47
Selection guide (1)
(working conditions)
Table radius
: R = 0.4 (m)
Table weight : Wt = 79 (kg)
Jig radius of rotation : Re = 0.25 (m)
Jig weight
: Wj = 10 (kg/pc.)
(Including workpiece weight)
Jig number
: N = 4
(operation conditions)
Moving angle
:
ψ=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)
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
a) Table
J1 =
=
=6.32
2
Wt×R2
2
79×0.42
b) Jig and workpiece
(kg•m
2
)
(kg•m
2
)
(kg•m
2
)
J2 = N x Wj x Re2 = 4 x 10 x 0.3252 = 4.225
c) Total sum of moment of inertia
N•max. =
Confirm that Nmax does not exceed the direct drive actuator’s maximum rotation speed.
Calculate the smallest model that can tolerate the load moment of inertia.
The AX allowable moment of inertia is 18.0 (kg•m
2
) or over, so this load is allowable.
Max load torque
Effective load torque
Total sum of load moment of inertia
Max. rotation speed
Max. load torque
Effective load torque
Regenerative electric power
10.545 180
33 100
231.3 300
70.7 100
16.23 40
(kg•m
2
)
(rpm)
(N•m)
(N•m)
(w)
Thus, AX4300T can be used.
Determine if the selected AX4300T can be used.
Tm = [Am•(J+J
M
)•
Trms=
= [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
Vm/ψ
6×0.8
90
J = J1+J2=6.32+4.225=10.545
180•t
12
ψ•π
180×0.82
90×π
•fc]
2
+(T
F
•fc+T
W
•fc+T
MF
)
2
•[r•Am•(J+J
M
)•
t
1
t
0
ψ•π
180•t
12
Trms=
×1.5]
2
+(0×1.5+0×1.5+10)
2
×[0.707×5.53×10.871×
0.8
4
90×π
180×0.82
W=
W 40 (W)
(W)
)
2
•
Vm•ψ•π
(J+J
M
)
t
1
•180
2•t
0
= 1.76×90×π
0.8×180
=16.23
10.871
2×4
(
)
2
•
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