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Motor Sizing Guideline for Treadmills

When designing a treadmill with a given horsepower motor (H, horsepower), the engineer is faced with trading off top treadmill speed (S, mph) for available track force (F, pounds). Available track force is defined as the force that the motor has available to move the track and overcome loading by friction and the runner. This force is measured at the surface of the track. The following formulas illustrate the relationship between top speed and available track force. Note that parasitic losses such as friction and windage are ignored in the following discussions. Figure 1 graphs the relationship between top treadmill speed and available track force for a range of motor powers. The x-axis is speed in mph, the y-axis is force in pounds.

F = H * 375/S is derived from: Conversions:
F= H hp *33000 ft-lbs/min*1/hp
*60min/hr*1/(S mile/hr)*
1 mile/5280 ft
1 hp = 33000 ft-lbs/min
1 mile = 5280 ft
1 hour = 60 min

 

speed_vs_force
Figure 1

 

Don't Worry About Torque

When choosing a motor, the engineer is given a choice of rated horsepower as well as rated speed. "Rated" is defined as the value that can be sustained with a 100% duty cycle. Since horsepower is the product of torque and speed, for a given horsepower, a motor with a rated speed of 3000 rpm would have only half the torque of a motor rated at 1500 rpm. At first glance, the novice might conclude that the best choice for a motor is a low speed one because it has more available torque. If the motor was directly connected to the track without any mechanical speed changers (belts/pulleys or gears) and the top speed was of no concern, this would be a correct conclusion. However, in the real world, the engineer designs the treadmill to have a specific top speed. Once a top track speed is established and a motor is chosen, the engineer can determine what ratio of mechanical speed reduction is required. For example, the top speed of a treadmill is 10 mph and it has a 1 hp, 3000 rpm motor. The required gear reduction ratio might be 10:1. If a 1 hp, 1500 rpm motor is substituted, the gear ratio must be reduced to 5:1. The 1500 rpm motor has twice the torque of the 3000 rpm, but because the gearing ratio is half of that needed with the 3000 rpm motor, the net force at the track is the same. The conclusion, therefore, is that you don't need to worry about how much torque a motor has because the horsepower and top track speed determine the available track force.

So, should the engineer choose a high speed motor with a lot of mechanical speed reduction, or a lower speed motor with lower reduction? A high speed motor has the advantages of smaller size, less construction material, and therefore lower cost. On the other hand, a slower motor will tend to have longer brush life, less commutation problems, and less mechanical problems due to components running at a high speed. In addition, the lower gear ratio required with the slower motor may be less expensive to implement.

 

Relationship Between Torque and Armature Current

For a given motor, the available torque can be increased by increasing the armature current. Generally, a motor can deliver twice its rated torque (and thus horsepower) for transient time periods if the armature current is doubled. However, this linear relationship starts to drop off for most motors above 250% of rated current due to armature reaction and other affects. The incremental torque/ampere ratio rapidly falls off as the armature current increases. See Figure 4 for a typical torque - current curve.

 

SCR vs. Gemini PWM IGBT Controls

A "side effect" of this torque - current relationship is noticed when comparing a typical SCR-based motor control with Gemini's pulse-width modulated (PWM) controls. The Gemini PWM controls use high speed IGBTs at a switching frequency of 16 kHz. These IGBT's switch on and off a dc voltage bus and the pulse width, or duty cycle, is varied with motor speed. The combined benefit of switching a dc voltage and the filtering effect of the largely inductive motor at high frequencies filters out virtually all of the ripple, providing pure dc current to the motor. The SCR control typically switches at 120 Hz. At this relatively low frequency, the filtering by the motor is significantly less, resulting in a significant ripple component. Furthermore, the SCR control switches its power from unfiltered, rectified ac. This compounds the ripple problem. In order to supply a given average torque, the motor armature must be provided with a proportional average current. For example, if the control is to try to get 200% torque from the motor, the average current in the motor must be 200% of rated current. Figures 2 and 3 show the current waveforms for this example. Note the large ripple content of the SCR waveform. It has peaks of over 400% of rated current in order to maintain the 200% average current.

figure2
Figure 2

 

figure3
Figure 3

In the previous paragraph the nonlinearity of the torque-current relationship was shown. Figure 4 shows that the linear relationship breaks down above 250% of rated current (typical motor).

figure4

Figure 4 - Motor Torque vs. Armature Current

Thus, the instantaneous 400% current can not produce an instantaneous 400% torque, and the 200% average current from the SCR control can not provide the desired 200% average torque. Translated more simply, for a given motor and a given average current, a Gemini PWM control can provide more torque and power than can a SCR control. Furthermore, SCR controls draw a higher average line current because the motor current comes directly from the ac line. As the motor speed decreases, the line current drawn by a Gemini PWM control also decreases, whereas with the SCR control, the power factor just gets worse.

 

 

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