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6 Spindle Adjustment
MITSUBISHI CNC
[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 0.0148 [kg•m
2
], and when the motor
shaft conversion load inertia is 0.05 [kg•m
2
].
Po = (Short-time rated output) × 1.2 = 5500 × 1.2 = 6600 [W]
J
all
= (Motor inertia) + (load inertia)
= 0.0148 + 0.05 =0.0648 [kg•m
2
]
Thus,
t = t1 + t2 + t3 = 0.242 + 1.818 + 4.691 = 6.751 [s]
CAUTION
1. Note that the inertia (J) is a quarter of "GD
2
".
Ex.) When "GD
2
" is 0.2 [kg•m
2
], the inertia is "0.2 / 4 = 0.05 [kg•m
2
]".
2. If the AC input power voltage to the power supply is low, or if the input power impedance is high, the acceleration/
deceleration time may be long. (Especially, the acceleration/deceleration time of the deceleration output range may be
long.)
3. For the actual measurement in comparison with the theoretical value, perform under the same condition as the
calculated load inertia of Jall. The acceleration/deceleration time differs according to the inertia. When performing the
measurement with a workpiece or tool installed to the spindle, confirm that the acceleration/deceleration time has been
calculated when the total inertia is included in the installed workpiece and tool.
0
1500
6000
0
2.0
6.0
8.0
4.0
Out
p
ut
[
kW
]
10000
15-minute
rating
3.7
5.5
4.1
Continuous
ratin
g
Rotation s
p
eed
[
r/min
]
Spindle motor characteristics
2.8
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 =
2Po
=
26600
= 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]