Fagor QC-PDS, Руководство по оборудованию

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  
Selection criteria
5.
S
E
LECT
ING CRI
T
ERIA
Synchr
onous motor
selection
· 176 ·
198
Ref.2003
QC-PDS
HARDWARE
is due to:
• the friction between table with its ways and with the ballscrew M
F
,
• the weight of the table when not moving horizontally M
W
,
• the cutting force of the tool M
C
.
 Friction torque
M
F
:
where:
 Torque due to the weight of the table
M
w
:
When the table does not move horizontally, but at an angle

as in figure
F.
H5/1
, the torque due to the weight of the table must also be considered:
If the total table weight is compensated for by means of some sort of
hydraulic system or counterweights so the motor makes the same effort to
move the table up as to move it down, the
% factor will be 0. If no
compensation is applied, % will be 1.
 Torque due to the needed cutting force
Mc
:
There is a cutting force between the tool and the part and this means a
hindrance for moving the table. The torque necessary at the motor to make
this movement is calculated as follows:
Motor speed calculation · rpm ·
The machine will need a maximum speed (rpm motor) in a linear
movement of the table. Therefore, the motor must have a maximum speed
of:
V
max
is the maximum linear speed required by the table.
In the characteristics table of the FAGOR synchronous motors
·
manual
man_fxm_fkm_motors.pdf
·
, select a motor that has:
• A stall torque equal to or greater than the calculated continuous torque M
s
.
• A maximum turning speed equal to or greater than the calculated value
RPM motor.
M
F
Torque due to friction in (N·m).
m Table mass in kg.
d
Leadscrew diameter in m.
g
Gravitational acceleration, 9.81 in m/s².
h
Leadscrew pitch per turn in m.
µ The friction coefficient between the table and the ways it moves on:
typical µ values depending on material:
Iron 0.1
÷
0.2
Turcite 0.05
Roller bearings 0.01
÷
0.02
M
W
Torque due to the weight of the table in N·m.

Inclination angle of the ballscrew with respect to the horizontal axis.
% Table weight compensation factor between 0 and 1.
M
C
Torque due to the cutting force of the tool in N·m.
F Cutting force of the tool in kg-force.
g Gravitational acceleration, 9.81 in m/s².
MF MF
TABLE –
MF
BALLSCREW –
+
 
1
i
---

m g

h
  
2

----------------------------
d
10
------ +
1
i
---

= =
MW
m g sen

h
  
2

--------------------------------------
%
i
-----

=
MC
F g h
 
2

-------------------
1
i
---

=
RPMmotor
Vmax
h
----------- i

=