Fuji Electric FRENIC-AQUA series, User Manual

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3-6
3.1.3 Equations for selections
3.1.3.1 Load torque during constant speed running
[ 1 ] General equation
The frictional force acting on a horizontally moved load must be calculated. Calculation for driving a
load along a straight line with the motor is shown below.
Where the force to move a load linearly at constant speed
υ
(m/s) is F (N) and the motor speed for
driving this is N
M
(r/min), the required motor output torque
τ
M
(N·m) is as follows:
) m N (
F
N
2
60
G
M
M
• •
•
•
η
τ
π
υ
=
(3.1)
where,
η
G
is Reduction-gear efficiency.
When the inverter brakes the motor, efficiency works inversely, so the required motor torque should
be calculated as follows:
) m N ( F
N
2
60
G
M
M
• • •
•
• η
τ
π
υ
=
(3.2)
(60·
υ
) / (2
π
·N
M
) in the above equation is an equivalent turning radius corresponding to speed
υ
(m/s) around the motor shaft.
The value F (N) in the above equations depends on the load type.
[ 2 ] Obtaining the required force F
Moving a load horizontally
A simplified mechanical configuration is assumed as shown in Figure 3.7. If the mass of the carrier
table is W
0
(kg), the load is W (kg), and the friction coefficient of the ball screw is
μ
, then the friction
force F (N) is expressed as follows, which is equal to a required force for driving the load:
) N ( g ) W W ( F
0
μ + =
• •
(3.3)
where, g is the gravity acceleration (
≈
9.8 (m/s
2
)).
Then, the driving torque around the motor shaft is expressed as follows:
) m N (
g ) W W (
N
2
60
G
0
M
M
•
• •
•
•
•
η
τ
μ +
π
υ
=
(3.4)
Figure 3.7 Moving a Load Horizontally