Fagor DDS, Hardware Manual

Page: 208 / 386

   Selection criteria
232
5.
S
E
LECT
ING CRI
T
ERIA
Sele
cti
on of
the
synchro
nou
s motor and its associated
dri
ve
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
bal 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
Dp2
4
L2
    
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.