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Fagor AXD 1.35 - Page 208

Fagor AXD 1.35
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Selection criteria
232
5.
SELECTING CRITERIA
Selection of the synchronous motor and its associated drive
208
DDS
HARDWARE
Ref.1310
Second motor pre-selection
Calculation of inertia (J)
The next step is to calculate the load that the motor has to move when
accelerating; that is the moment of inertia of all the elements it moves.
Total inertia (from now on
inertia) J
TOTAL
is due to the load J
LOAD
and to
the rotor of the motor itself
J
MOTOR
.
The inertia due to load may be divided into that of the table + that of the
ballscrew + that of the system used to compensate for non - horizontal
axes + that of the pulley or gear used for transmission and which turns
with the ballscrew (pulley 1). All these elements are affected by the reduc-
tion factor
i as shown by the following equation.
The inertia due to the pulley that turns with the motor (pulley 2) is not af-
fected by the
i factor.
The inertia of each element is:
The resulting inertia will be in kg·m².
See previous sections.
The inertia of the motor
J
MOTOR
is:
this data may be obtained from the characteristics table of the corre-
sponding motor manual.
Verify that in the characteristics table the rotor of the motor chosen in the
1st selection has an inertia J
MOTOR
that meets the following condition:
where
k is a factor whose value depends on the application given to the
motor.
The ideal will be to obtain a
J
MOTOR
= J
LOAD
For a positioning axis, the typical value of “K” will be between 1 and 3.
L
Leadscrew length in m.
L
1
Width of pulley 1 in m.
L
2
Width of pulley 2 in m.
D
p1
Diameter of pulley 1 in m.
D
p2
Diameter of pulley 2 in m.
Material density:
7700 kg/m³
for iron/steel
2700 kg/m³ for aluminum
i, h are data used earlier.
J
TOTAL
J
LOAD
J
MOTOR
+=
J
LOAD
J
TABLE
J
BALLSCREW
J
PULLEY1
J
COMPENSATION
+++
2
i
------------------------------------------------------------------------------------------------------------------------------------------------ J
PULLEY2
+=
J
TABLE
m
h
2
------
2
=
J
PULLEY1
D
p1
4
L
1

32
------------------------------------=
J
BALLSCREW
d
4
L 
32
------------------------------=
J
PULLEY2
D
p2
4
L
2

32
---------------------------------------=
J
MOTOR
= J
ROTOR
+ J
BRAKE
J
MOTOR
J
LOAD
K
WARNING. Note that if this requisite is not met, a new motor must be se-
lected which meets the conditions of the 1st selection and the 2nd one.

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