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I'm missing something fundamental in my understanding of the torque of a motor.

Consider the following analogy: I have a 10kg dumbell on the ground and a 1kg dumbell next to it. Let's say that I lift the 10kg dumbell off the ground with a certain force 'F'. Now let's say I use the same force 'F' to lift the 1kg dumbbell. Naturally, I would lift the 1kg dumbell much faster, right?

Now coming to electric motors, let's say I attach a 10kg load to the motor shaft. The motor would produce a torque T and rotate at a certain rpm. Now, I put 1kg load on the shaft. As per the dumbell analogy, wouldn't the 1kg load cause an increase in speed because it's lighter than 10kg?

But we know that a motor's torque keeps changing with the load. So for a 1kg load, does the motor produce just enough torque required to rotate a 1kg load? If yes, how would the motor know what the load is so as to produce the right amount of torque? If no, them wouldn't the motor produce the same amount of torque for all loads and only the rpm keeps changing?

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  • \$\begingroup\$ It depends on the motor and field winding configuration. \$\endgroup\$ – Andy aka Dec 11 '19 at 8:06
  • \$\begingroup\$ Could you please elucidate \$\endgroup\$ – noorav Dec 11 '19 at 8:43
  • \$\begingroup\$ Some DC motors with the appropriately connected field windings can increase speed with an increasing load. Go google differential compound DC motors. \$\endgroup\$ – Andy aka Dec 11 '19 at 10:12
  • \$\begingroup\$ Additional to other input: Simplistically, in your 1kg/10kg load scenario, at a given power level the 1kg load will stabilise at ten times the RPM of the 10 kg load. | A useful rule of thumb is Watts ~= RPM x kg.m torque. (It's out by about 2%). || The "real formula is Watts = force x distance per time = kg.m torque x g x 2.Pi x RPM/60. [g = 9.8 m/s/s] It just happens that the various factors cancel out to give the useful approximation above. \$\endgroup\$ – Russell McMahon Dec 11 '19 at 12:10
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Assuming we are talking about a shunt wound or permanent magnet brushed DC or BLDC motor, torque is proportional to current, so the more torque load you put on the motor the more current it will draw to match it. However the current causes a voltage drop in the winding resistance, which reduces rpm. How much the rpm drops depends on the resistance. A large motor with thick low resistance wire could lift a heavy weight without its rpm dropping significantly, while a small motor with thinner wire might drop down to stall speed and not draw enough current to lift the heavy weight at all.

To understand the relationships between torque, current, rpm and voltage, consider this simplified model of a PMDC motor:-

schematic

simulate this circuit – Schematic created using CircuitLab

When the motor spins it generates a voltage proportional to its speed. This is a result of Faraday's Law, which says that moving a wire through a magnetic field induces a voltage in it proportional to its velocity, and occurs whether it is driven by an external force or by its own power. Depending on how it is constructed (number of turns, size, stator or armature slots, magnet poles etc.) each motor has a characteristic 'Kv' (Velocity constant), commonly expressed in rad/s/Volt or rpm/Volt.

Output torque is proportional to current according the Lorentz force law, which says that the magnetic force on a wire is proportional to current x magnetic field strength (which is constant in a permanent magnet or shunt wound DC motor). For a particular motor this relationship is a constant, called Kt (Torque constant), often expressed in N.m/A.

By symmetry, Kt is the inverse of Kv. So if you know one then you know the other. If Kv = 10.5 rad/s/V (100 rpm/V) then Kt must be 0.095 N.m/A.

For this simplified model we will ignore any internal losses due to magnetic hysteresis, eddy currents, friction and air drag. If the motor has no load then torque and current are zero, and no voltage is dropped across Rm. That means the generator voltage must equal the supply voltage, and the motor must be spinning at 100 rpm/V * 10V = 1000 rpm.

Now put an external torque load on the motor. It will slow down and produce less voltage, causing voltage to be dropped across Rm. By Ohms Law a current must then flow, producing torque. Speed continues to drop and current rises until output torque matches the externally applied torque.

If the torque load is high enough to stop the motor spinning then there is no generator voltage and the full 10V is dropped across Rm, causing a current of 10V/1Ω = 10A. With Kt = 0.095 N.m/A and current = 10A, the 'stall' torque must be 0.95 N.m.

Now imagine an otherwise identical motor which has thicker wire, reducing Rm to 0.1Ω. The same 10A current (producing the same 0.95 N.m torque) now only drops 1V across Rm, so the generator must be producing 9V. This motor must be spinning at 9*100 = 900 rpm (only 10% less than the no-load speed) while lifting the weight that stalled the other motor.

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  • \$\begingroup\$ Bruce, thank you for the answer. But I still have a few doubts and I would appreciate it if you could answer them. 1) You mentioned that the motor draws more current when load is increased to match it. What do you mean by matching? Moreover, how does the motor "know" how much current to draw? 2)If the motor has no load, how is torque zero? The shaft would still be rotating even at no load, hence there must be a torque right? 3)How is less voltage produced cause a voltage to be "dropped" across Rm? 4) If the output torque matches the external torque, would the motor shaft still rotate? \$\endgroup\$ – noorav Dec 11 '19 at 15:21
  • \$\begingroup\$ 1) By 'matching' I mean when the torques are equal. If motor torque is less than load torque it must slow down, since there is a force imbalance. If it's greater then the motor will speed up. If they are equal then then rpm and current don't change. At any instant the motor 'knows' how much current to draw due to Ohms law, as I explained (Irm = Vrm/Rm). 2) No load is zero torque by definition. how fast the shaft is rotating is irrelevant, since with no load it is not delivering any torque (we are not considering internal torque load due to friction and windage). \$\endgroup\$ – Bruce Abbott Dec 11 '19 at 18:38
  • \$\begingroup\$ 3) 4) If output torque matches load torque the motor must be drawing current (determined by Kt). That current produces a voltage drop across Rm. If the voltage drop across Rm is less than the supply voltage there must be generated voltage (since the only components in the series circuit are the resistor and the generator, and their voltages must add up to the supply voltage). If the generator is producing voltage the motor must be spinning, (at a speed determined by Kv). \$\endgroup\$ – Bruce Abbott Dec 11 '19 at 18:46
  • \$\begingroup\$ Wanted to stress that this is the behavior when a motor is hooked up to an ideal voltage source. If connected to an ideal current source, it will behave exactly as you described in your force/dumbell analogy \$\endgroup\$ – Ocanath Dec 11 '19 at 18:52
  • \$\begingroup\$ @Ocanath With a constant current supply the motor will put out constant torque for sure (assuming no internal losses), but this is an unusual situation. Motors are generally operated from a voltage supply which closely approximates constant voltage. A speed controller may regulate current to produce constant torque, but now we are talking about more than just a motor. \$\endgroup\$ – Bruce Abbott Dec 11 '19 at 19:13
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Motor torque is directly related to motor current, not with voltage. Ignoring motor resistance, friction etc, when you drive the motor with a current source that would give the same analogy as pulling a dumbell with force F.

But we know that a motor's torque keeps changing with the load.

This is only true when driving a motor with voltage source.

Using the electrical domain:
- With a current source, we know that the voltage keeps changing with the load.
- With a voltage source, we know that the current keeps changing with the load.

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Additional to other input:

Simplistically, in your 1kg/10kg load scenario, at a given power level the 1kg load will stabilise at ten times the RPM of the 10 kg load.

A useful rule of thumb is

  • Watts ~= RPM x kg.m torque.

Remember this for life!!!! (It's out by about 2%).

Because:

Watts = force x distance per time

= g x kg x m x 2.Pi x RPM/60.
= [Force]x[ distance ..........................]

[g = 9.8 m/s/s]

It just happens that the various factors cancel out to give the useful approximation above.
For this example assume we have 1 kg.m of torque.
Assume the torque acts at 1m radius.
You'll get the same result with say 2m radius and half the force or 0.1m radius and 10 x the force.

Force is the Newtons exerted at the load radius.
So for 1 kg.m of torque you get 1kg x g = 1 x 9.8 = 9.8N of force at 1m radius.

Distance is the distance travelled along the "load arc" in unit time.
Here radius is 1m so per revolution we travel 2 x Pi m.
In one second we get RPM/60s revolutions.
So distance = 2 x Pi x R x RPM/60
= 2.Pi.RPM//60 for 1m radius.

SO Watts = Force x distance per time
= kg.m x g x 2 x Pi x RPM / 60
= kg.m x RPM x (g x 2 x Pi /60)
= kg.m x RPM x (9.8 m/s/s x 2 x 3.14 /60) = kg.m x RPM x (9.8 x 2 x 3.14 /60)
= kg.m x RPM x 1.026

Which is 2.6% higher than the
Watts = kg.m torque x RPM rule of thumb.

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EXAMPLE:

Consider a motor that can deliver 100 mechanical watts to a pulley with 0.1 metre radius. How we cause the motor to provide 100 Watts depends on circumstances - we may vary voltage or current - either way we arrange things so V x I produces 100 Watts mechanical output (in this case).

Using the rule of thumb formula
Watts = kgm x RPM or P = T x R

(T = torque in kgm, P = Power in Watts, R = RPM of pulley). Rearranging P = T x R gives
R = P/T.

With 1 kg load torque in kg.m = 1kg x 0.1m = 0.1 kgm.
With 10 kg load T = 10 x 0.1 = 1 kgm.

So as R = P/T:
- For load = 1 kg, T = 0.1 kgm so
RPM = P/T = 100/0.1 = 1000 RPM.

  • For load = 10 kg, T = 1 kgm
    RPM = P/T = 100/1 = 100 RPM.

That is for 100 Watts mechanical at the pulley in each case.
So a 1 kg load will stabilise = operate steadily at 1000 RPM.
And a 10kg load stabilises at 100 RPM.

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  • \$\begingroup\$ What do you mean by "1kg load stabilising.."? \$\endgroup\$ – noorav Dec 11 '19 at 15:22
  • \$\begingroup\$ @noorav Say a motor can deliver 100 mechanical watts to a pulley with 0.1 metre radius. Using my formula Watts = kgm x RPM = T x R (T = torque in kgm, P = Power in Watts, R = RPM of pulley). Or rearranging R = P/T. || With 1 kg load torque = 1 x 0.1 = 0.1 kgm. With 10 kg load T = 10 x 0.1 = 1 kgm. || So if R = P/T for load = 1 kg, so T = 0.1 kgm RPM = P/T = 100/0.1 = 1000 RPM. For 10 kg load and 1 kgm torque - RPM = P/T = 100/1 = 100 RPM. That is for 100 Watts mechanical at the pulley in each case. So a 1 kg load will stabilise = operate steadily at 1000 RPM. 10kg stabilises at 100 RPM. \$\endgroup\$ – Russell McMahon Dec 11 '19 at 18:08
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    \$\begingroup\$ "at a given power level the 1kg load will stabilise at ten times the RPM of the 10 kg load." - this is true, but may be a bit too simplistic. The motor does not output constant power at any rpm. A PMDC motor has a parabolic power curve that peaks at half the no-load rpm and reduces to zero above and below it, so the rpm change depends on where on the power curve the different loads occur. \$\endgroup\$ – Bruce Abbott Dec 11 '19 at 19:00
  • \$\begingroup\$ @BruceAbbott "All models are wrong. Some models are useful". Yes. It's hard to cover all bases, as you know. One could also note that electric motors tend to produce around max torque at around zero speed. Or .... . I tried to cite a very specific set of conditions that gave useful insight. Inadequacy in covering all aspects is guaranteed. \$\endgroup\$ – Russell McMahon Dec 11 '19 at 19:50

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