Fuji Electric FRENIC MEGA G2 Series - 10.3 Equations for Selections

10.3 Equations for Selections
10-6
10.3 Equations for Selections
10.3.1 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  (m/s) is F (N) and the motor speed for driving this is
N
M
(r/min), the required motor output torque 
M
(N·m) is shown in the following equation (Equation 10.3-1):
τ
M
=
60・υ
・
F
(N·m)
(Equation 10.3-1)
2π・N
Ｍ
η
Ｇ
where, 
G
is reduction-gear efficiency.
When the inverter brakes the motor, efficiency works inversely, so the required motor torque should be calculated
as follows:
τ
M
=
60・υ
・
Ｆ・η
Ｇ
(N·m)
(Equation 10.3-2)
2π∙N
M
(60·) / (2·N
M
) in the above equation is an equivalent turning radius corresponding to speed  (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 Fig. 10.3-1. 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  , then the friction force F (N) is expressed
as follows (Equation (Equation 10.3-3), which is equal to a required force for driving the load:
F =
（Wo＋W）・ g・μ
(N)
(Equation 10.3-3)
where, g is the gravity acceleration (≒ 9.8 (m/s
2
)).
Then, the driving torque around the motor shaft is expressed with the following equation ((Equation 10.3-4):
τ
M
=
60・υ
・
（Wo＋W）・ g・μ
(N·m)
(Equation 10.3-4)
2π・N
Ｍ
η
Ｇ
Fig. 10.3-1 Moving a load horizontally