Fagor DDS Series Hardware Manual page 203

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
rotor of the motor itself J
MOTOR
J TOTAL
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.
J + J
TABLE BALLSCREW
J = ---------------------------------------------------------------------------------------------------------------------------------------------------------------------
LOAD
The inertia of each element is:
2
h
 ------
J = m
TABLE
2
4
    
D L
p1 1
J = ------------------------------------
PULLEY1
32
The resulting inertia will be in kg·m².
L Leadscrew length in m.
L Width of pulley 1 in m.
1
L Width of pulley 2 in m.
2
D Diameter of pulley 1 in m.
p1
D Diameter of pulley 2 in m.
p2

Material density:
7700 kg/m³ for iron/steel
2700 kg/m³ for aluminum
i, h are data used earlier.
See previous sections.
The inertia of the motor J
MOTOR
J
MOTOR
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
J
MOTOR
where k is a factor whose value depends on the application given to the
motor.
The ideal will be to obtain a J
For a positioning axis, the typical value of "K" will be between 1 and 3.
WARNING. Note that if this requisite is not met, a new motor must be
selected which meets the conditions of the 1st selection and the 2nd one.
) J is due to the load J
TOTAL
.
= J LOAD + J MOTOR
+ J + J
PULLEY1 COMPENSATION
2
i
d
J = ----------------------------- -
BALLSCREW
D
J = ------------------------------------
PULLEY2
is:
= J + J
ROTOR BRAKE
that meets the following condition:
   
J K
LOAD
= J
MOTOR LOAD
  Selection criteria
and to the
LOAD
+ J
PULLEY2
4
 L    
32
4
    
L
p2 2
32
HARDWARE
5.
DDS
Ref.1912
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