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MDS-D-SVJ3/SPJ3 Series Instruction Manual
5-2 Adjustment procedures for each control
[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) x 1.2 = 5500 x 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]
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 spindle drive unit 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.
CAUTION
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]