10.1 Selecting Motors and Inverters
10-8
Vertical Lift Load
A simplified mechanical configuration is assumed as shown in Figure 10.1-8. If the mass of the cage is W
0
(kg), the load is W (kg), and the balance weight is W
B
(kg), then the forces F (N) required for lifting the load
up and down are expressed as follows:
(For lifting up)
F= (W
0
+ W - W
B
) • g (N) (10.5)
(For lifting down)
F= (W
B
- W - W
0
) • g (N) (10.6)
Assuming the maximum load is W
max
, the mass of the balance weight W
B
(kg) is generally obtained with
the expression W
B
= W
O
+ Wmax /2. Depending on the mass of load W (kg), the values of F (N) may be
negative in both cases of lifting up and down, which means the lift is in braking mode. So, be careful in
motor and inverter selection.
For calculation of the required output torque τ around the motor shaft, apply the expression (10.1) or
(10.2) depending on the driving or braking mode of the lift, that is, apply the expression (10.1) if the value
of F (N) is positive, and the (10.2) if negative.
Figure 10.1-8 Vertical Lift Load
Inclined Lift Load
Although the mechanical configuration of an inclined lift load is similar to that of a vertical lift load,
unignorable friction force in the inclined lift makes a difference; in an inclined lift load, there is a distinct
difference between the expression to calculate the lift force F (N) for lifting up and that for lifting down. If
the incline angle is θ, and the friction coefficient is μ, as shown in the Figure 10.1-9, the driving force F (N)
is expressed as follows:
(For lifting up)
F =((W
O
+ W)(sinθ + μ • cosθ) - W
B
) • g (N) (10.7)
(For lifting down)
F = (W
B
- (W
O
+ W) (sinθ + μ • cosθ)) • g (N) (10.8)
The braking mode applies to both lifting up and down as in the vertical lift load. And the calculation of the
required output torque around the motor shaft is the same as in the vertical lift load;
apply the expression (10.1) if the value of F (N) is positive, and the (10.2) if negative.
Figure 10.1-9 Inclined Lift Load
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