Analysis and Control of Electric Drives. Ned Mohan

have the following disadvantages:

       Additional power loss.

       Introduction of nonlinearity due to a phenomenon called backlash.

       Wear and tear.

      2‐7‐1 Conversion Between Linear and Rotary Motion

In many systems, a linear motion is achieved by using a rotating‐type motor, as shown in Fig. 2-14.

Fig. 2-14 Combination of rotary and linear motion.

      In such a system, the angular and the linear speeds are related by the radius r of the drum:

(2-35)

      To accelerate the mass M in Fig. 2-14, in the presence of an opposing force f_L, the force f applied to the mass, from Eq. (2-1), must be

      (2-36)

This force is delivered by the motor in the form of a torque T, which is related to f, using Eq. (2-35), as

      (2-37)

      Therefore, the electromagnetic torque required from the motor is

      (2-38)

EXAMPLE 2‐8

In the vehicle of Example 2-6 with C_w = 0.5, assume that each wheel is powered by its own electric motor that is directly coupled to it. If the wheel diameter is 60 cm, calculate the torque and the power required from each motor to overcome the drag force when the vehicle is traveling at a speed of 100 km/h.

       Solution

In Example 2-6, the vehicle with C_w = 0.5 presented a drag force f_L = 413.7 N at the speed u = 100 km/h. The force required from each of the four motors is . Therefore, the torque required from each motor is

      From Eq. (2-35),

      Therefore, the power required from each motor is

      2‐7‐2 Gears

For matching speeds, Fig. 2-15 shows a gear mechanism where the shafts are assumed to be of infinite stiffness, and the masses of the gears are ignored. We will further assume that there is no power loss in the gears. Both gears must have the same linear speed at the point of contact. Therefore, their angular speeds are related by their respective radii r₁ and r₂ such that

(2-39)

      and

(2-40)

Combining Eqs. (2-39) and (2-40),

(2-41)

where T₁ and T₂ are the torques at the ends of the gear mechanism, as shown in Fig. 2-15. Expressing T₁ and T₂ in terms of T_em and T_L in Eq. (2-41),

(2-42)

From Eq. (2-42), the electromagnetic torque required from the motor is

(2-43)

      where the equivalent inertia at the motor side is

(2-44)

Fig. 2-15 Gear mechanism for coupling the motor to the load.

       Optimum Gear Ratio

Equation (2-43) shows that the electromagnetic torque required from the motor to accelerate a motor‐load combination depends on the gear ratio. In a basically inertial load where T_L can be assumed to be negligible, T_em can be minimized, for a given load‐acceleration , by selecting an optimum gear ratio (r₁/r₂)_opt.. The derivation of the optimum gear ratio shows that the load inertia “seen” by the motor should equal the motor inertia, that is, in Eq. (2-44),

      (2-45a)