Fuji Electric Frenic-Mini - Equations for Selections; 1 ] General Equation; 2 ] Obtaining the Required Force F; Load Torque During Constant Speed Running

7.1 Selecting Motors and Inverters
7-7
Chap. 7 SELECTING OPTIMAL INVERTER CAPACITIES
7.1.3 Equations for selections
7.1.3.1 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:
)mN(
F
N
2
60
G
M
M
••
•
•
η
τ
π
υ
=
(7.1)
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:
)mN(F
N
2
60
G
M
M
•••
•
•
η
τ
π
υ
=
(7.2)
(60·υ) / (2π·N
M
) in the above equation is an equivalent turning radius corresponding to speed υ 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 7.7. 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)WW(F
0
μ
+= ••
(7.3)
where, g is the gravity acceleration (≈ 9.8 m/s
2
).
Then, the required output torque around the motor shaft is expressed as follows:
)mN(
g)WW(
N
2
60
G
0
M
M
•
••
•
•
•
η
τ
μ+
π
υ
=
(7.4)
Figure 7.7 Moving a Load Horizontally