
B–65142E/04
4. SELECTING A MOTORDESCRIPTIONS FOR THE α series
35
Example)
Try to perform linear acceleration/deceleration under the following
condition.
V
m
+3000 (min
–1
)+50 (s
–1
), ta+0.1 (s), ks+30 (s
–1
),
J
L
+0.0247 (kgf·cm·s
2
)
Select α2/3000, and calculate its acceleration torque.
J
M
motor inertia is 0.0056 (kgf·cm·s
2
) when α2/3000 is selected, so
the load inertia is calculated as follows:
Ta +50 2π 1/0.1 0.0056 (1*e
–30
0.1
)
)50 2π 1/0.1 0.0247 (1*e
–30
0.1
)B0.9
+98.6 (kgf·cm)+9.67 (Nm)
Torque (Nm)
Speed (min
–1
)
Speed–torque characteristics of
α2/3000
12.1 (Nm)
2049 (min
–1
)
Torque (Nm)
Speed (min
–1
)
Speed–torque characteristics of
α3/3000
The speed–torque characteristics of α2/3000 show that the acceleration
torque of 9.67 (Nm) is beyond the intermittent operating zone of α2/3000
(see the characteristic curve above and data sheet). (The torque is
insufficient for α2/3000.)
If the operation specifications of the shaft (for instance, the acceleration
time) cannot be changed, a larger motor must be selected. Select an
α3/3000 (J
M
is 0.014 (kgf⋅cm⋅s
2
)) and calculate the acceleration torque
again.
Ta = 123.7 (kg⋅cm) = 12.1 (Nm)
Vr = 2050 (min
–1
)
In acceleration, an acceleration torque of 12.1 (Nm) is required at 2050
(min
–1
). The speed–torque characteristic curve shown above shows that
the acceleration is possible with α3/3000.
As α2/3000 is changed to α3/3000, the size of the attachment flange is
increased from 90 mm 90 mm to 130 mm 130 mm. If the machine
tool does not allow a larger motor, the specifications must be changed.
For example, the acceleration time must lengthen.