66 1018854 12/2015 V01
Load rotation speed n
2
= 40 [rpm]
Load torque (e. g. friction) T
L
= 5 [Nm]
Load inertia J
L
= 1.3 [kgm
2
]
Speed pattern
Acceleration; Deceleration t
1
= t
3
= 0.1 [s]
Operate with rated speed t
2
= 0.1 [s]
Stand still t
p
= 1 [s]
Total cycle time t
O
= 1.3 [s]
Load Conditions
Assume servo mechanism is used to cyclically position a mass with a horizontal axis of rotation.
Speed pattern
Torque pattern
n
2
= 40 rpm
Speed n [rpm]
T
1
T
2
T
3
t
1
=0,1
t
2
=0,1
t
3
=0,1
t
P
=1
Torque T [Nm]
t
0
= 1,3
Note t
1
= t
3
Time t [s]
Time t [s]
Please note: Each characteristic value should be converted to the value at the output shaft of the actuator.
Max. Torque T
max
= 151 [Nm]
Max. Speed n
max
= 90 [rpm]
Moment of inertia J
Out
= 0.86 [kgm
2
]
Actuator data FHA-25C-50-L
Example of actuator selection
Table 66.1
Illustration 66.2
Table 66.3
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