Motor Selection

Motor sizing & selection

    1. To move an object, the motor needs to overcome friction and opposing force. For example, to lift an object, the gravitational force needs to be overcome. To push an object or another robot, the object static friction force and pushing force needs to be overcome. Once any opposing force is overcome, the motor's output force will accelerate the load's inertia.
    2. Power requirement is proportional to acceleration and speed. Hence it is important to set realistic requirements to avoid getting a motor that is oversized.
    3. There is a wide selection of motor to select from to suit one’s application. Normally, we aim to choose one with the lowest cost that meets our criteria.
    4. Power efficiency and long service life are some of the factors to consider, especially in industrial applications. For competition robotics, we are probably more concern with performance rather than long service life or energy saving.
    5. To size a motor, we need to estimate the maximum power requirement of the robots. It varies from application to application
    6. Select a motor at least 1.5 to 2.0 times the power required.
    7. Select the right gear ratio to match the motor output to the load. DC motors prefer to operate nearer to the no load speed as the power efficiency is higher there. To operate the motor at the higher speed, we need a higher gear ratio. This has the advantage of having a higher output torque, which is necessary for high acceleration. I would operate the motor at around 70% of no load speed for the worst case condition. If we were to operate at a higher speed, say 80% of noload, where the power efficiency is higher, we need a bigger motor as the output power is quite low at this point.
    8. At 70% noload speed, output power = 84% of max power. Since we started off with a motor that's >1.5 times output power, it is safe to operate at this speed. At 80% no load speed, output power = 64% of max power, hence we need a motor that is at least twice the output power.
    9. We can get higher motor power by supplying a higher operating voltage. Just make sure that the motor is not overheated. The motor power is approximately proportional to the supply voltage squared. For example, if you have a 6-volt motor and your battery supply is 2 Lithium Polymer cells in series, you get approximately 7.6 volts. Minus the MOSFET motor driver voltage drop, you get around 7 volts supply for the motor. Hence the power increase is (7/6)*(7/6) = 1.36, a big increase of over 30%!

Example on motor selection

A 1 kg robot has a 2 wheels differential drive. It's desired speed profile is as shown below. Its’ wheel diameter is 5cm. Assume that the robot is not pushing any external objects and that it's drive transmission is rather efficient.

    1. Determine maximum force required
        • Force = mass*acceleration
              • = 1 kg * 3m/s 2
              • = 3 N
    2. Determine maximum power required to move robot
        • Power = force * speed
        • Maximum power required would be near the top speed of 2m/s
        • Max power required = 3N * 2m/s
                  • = 6 watts
        • With 2 drivemotors, power per motor = 3 watts
    3. Choose a motor power at least 1.5 times or 4.5 watts -> 2224SR
    4. Find the gearing ratio required
        • Wheel circumference = Pi*diameter = 0.157 m.
        • Wheel speed = 60 * robot speed/circumference
          1. = 60*2/0.157 rpm
          2. = 764 rpm
      • Motor no-load speed = 8,200 rpm
      • Therefore, 70% no-load speed = 5,740 rpm
      • Operating at this point, the gear ratio would be 5,740/764 = 7.51
    5. Double check torque value
        • Wheel torque = force per wheel * wheel radius
              1. = 3/2 * 0.05/2 N-m
              2. = 0.0375 N-m
        • At 70% no-load speed, motor output torque = 30% stalled torque = 0.0212 * 0.3= 0.00636 N-m
        • Required motor torque = wheel torque /gear ratio
            1. = 0.0375/7.51 N-m
            2. = 0.005 Nm (less than the 0.0636 N-m motor output torque)
        • Even assuming gear loss of 20%, there still enough torque. (Well, actually about just nice. But we can increased the motor voltage to bump up the motor output power)
        • Spur gear efficiency is approximately 90%. To take that into account, the output torque of gear needs to be reduced by 10%
    6. Note that we may not be able to achieve exactly 7.51:1 gear ratio. Based on gears available, we would end up with slightly higher or lower gear ratio. For most applications, I would try to select a higher gear ratio so that the motor can operate in the higher speed and hence higher efficiency region.
    7. The above calculation does not take into account the motor's rotor's inertia, which is assumed to be small.
    8. If the motor Torque-Speed-Power graph is available, it might be easier to use the graph to select the right gear ratio.

Motor selection, other consideration

    1. Voltage available. Normally between 6 to 24 volts. Lower voltage, less battery cells required, but current increased. Drive electronic will typically have better efficiency with higher operating voltage.
    2. Do not operate near stalled torque continuously or over no-load speed. Near stalled torque, the brush or coil may be damaged by overheating.
    3. Some manufacturers specify maximum permissible speed, which is near or higher than no-load speed. Operating near this speed increased mechanical wear and probably sparking at the brush to commutator interface.
    4. The motor cannot be driven at high current continuously as this heat up the motor and cause damage to the coil. This maximum continuous current limits the maximum continuous torque, which is normally less than the stall torque.
    5. The maximum permissible speed and maximum permissible torque limit the continuous operating range of the motor. However, for optimum motor service life, the manufacturer would normally recommend an operating speed and torque below these limits.
    6. Note that for short term operation, it is permissible to operate beyond these limits
    7. Maximum power occurs at half stall torque and half no-load speed. But the efficiency at this operating point is rather low and will result in heating the motor if operated continuously. For maximum efficiency, the operating speed should be between 70-90% and the operating torque should be between 10-30% stalled torque. In competition robotic, we are probably more concerned with maximum performance rather than power efficiency.
    8. Motor power is limited by the motor's ability to dissipate heat. Hence, if the motor is operating near it's limit, it's important that the motor is well ventilated so that it's winding does not burnt. It also means that we can operate a motor temporarily above it's power rating provided we ensure that it does not get heated up. It also means that we can supply a voltage higher than the nominal voltage to the motor. Of course, operating a motor at too high a speed or power region will likely shorten the life of the motor.

Gearing

    1. Gearing is often necessary as the motor output torque and speed normally does not match the required load torque and speed.
    2. Most motor has high output speed and low output torque. Hence gear reduction is often used to lower the speed and increase the torque. Without gear reduction, you would need an oversized motor in order to meet the requirements.

Basic physics for robotics

There are 2 kinds of movement, translational and rotational. All robots’ movements can be resolved into these 2 components. Not surprisingly, for every physic equation defining translational movement, there is an equivalent equation for rotational movement.

Moment of inertia is proportional to distance squared from centre of mass. Hence, to reduce moment of inertia, heavy mass should be placed as close to centre of robot as possible.

The inertias for translational and rotational movements are different. Hence it is better to have to different PID loop to control them separately, so that their gains can be tuned separately.