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5. Spindle Adjustment
5 - 28
[Calculation example]
Calculate the acceleration/deceleration time from 0 to
10000[r/min] for an spindle motor having the output
characteristics shown on the right when the motor inertia is
0.059 [kg･m
2
], and when the motor shaft conversion load
inertia is 0.2 [kg
•m
2
].
Po = (Short-time rated output) x 1.2
= 5500 x 1.2 = 6600 [W]
J
all
= (Motor inertia) + (load inertia)
= 0.0148 + 0.05 =0.0648 [kg･m
2
]
1.097 x 10
-2
x J
all
x N1
2
1.097 x 10
-2
x 0.0648 x 1500
2
t1 =
Po
=
6600
= 0.242 [s]
1.097 x 10
-2
x J
all
x (N2
2
- N1
2
) 1.097 x 10
-2
x 0.0648 x (6000
2
- 1500
2
)
t2 =
2×Po
=
2×6600
= 1.818 [s]
1.097 x 10
-2
x J
all
x (N3
3
- N2
3
) 1.097 x 10
-2
x 0.0648 x (10000
3
- 6000
3
)
t3 =
3 x Po x N2
=
3 x 6600 x 6000
= 4.691 [s]
Thus,
t = t1 + t2 + t3 = 0.242 + 1.818 + 4.691 = 6.751 [s]
0
1500
6000
0
2.0
6.0
8.0
4.0
Output [kW]
10000
15-minute
rating
3.7
5.5
4.1
Continuous
rating
Rotation speed [r/min]
Spindle motor characteristics
2.8