Fagor QC-PDS - 5.1.2 Second motor pre-selection

 Selection criteria
SELECTING CRITERIA
Synchronous motor selection
5.
· 177 ·
Ref.2003
QC-PDS
HARDWARE
5.1.2 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 for the rotation axis (from now on referred to as, inertia) J
TOTAL
is due to load J
LOAD
and the rotor of the motor 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 reduction factor
i as shown by the following equation.
The inertia due to the pulley that turns with the motor (pulley 2) is not
affected by the i factor.
The inertia of each element is:
The resulting inertia values are in kg·m².
The inertia of the motor J
MOTOR
is:
this data may be obtained from the characteristics table of the
corresponding 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. If this requisite is not met, a new motor must be selected
which meets the conditions of the first and second selections.