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I know that electric vehicles have different performances depending on battery and motor, but it's not clear how electrical and mechanical units are related.

Can anybody please help?

Will a 100V motor raise against slopes better than a 50V motor?

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Possibly, but it's impossible to say without knowing the current which can be supplied or the efficiency, allowable loading (thermal, magnet damage if a PM design, etc) for the duration of the time it takes to climb the hill, or even the gearing. Consider that a 120v electric eraser is most likely inferior at drilling holes when compared to a 12v cordless drill. –  Chris Stratton Sep 3 '12 at 12:46

4 Answers 4

The relationship between a motor's electrical characteristics and mechanical performance can be calculated as such (note: this is the analysis for an ideal brushed DC motor, but some of it should still apply to a non-ideal brushless DC motor).

A DC motor can be approximated as a circuit with a resistor, and voltage back-emf source. The resistor models the intrinsic resistance of the motor windings. The back-emf models the voltage generated by the moving electric current in the magnetic field (basically a DC electric motor can function as a generator). It's also possible to model the inherent inductance of the motor by adding an inductor in series, however for the most part I've ignored this and assumed the motor is at quasi steady state electrically, or the motor's time response is dominated by the time response of the mechanical systems instead of the time response of the electrical systems. This is usually true, but not necessarily always true.

The generator produces a back EMF proportional to speed of the motor:

$$ V_{emf} = k_i * \omega $$

Where:

$$k_i~is~a~constant.$$ $$\omega~is~the~motor~speed~in~rad/sec$$

Ideally at stall speed there is no back emf, and at no the no-load speed the back emf is equal to the driving source voltage.

The current flowing through the motor can then be calculated:

$$ I = (V_S - V_{emf}) / R = (V_S - k_i * \omega) / R $$ $$V_S = source~voltage$$ $$R~is~the~motor~electrical~resistance$$

Now let's consider the mechanical side of the motor. The torque generated by the motor is proportional to the amount of current flowing through the motor:

$$ \tau = k_t * I $$

$$k_t~is~a~constant$$ $$\tau~is~the~torque$$

Using the above electrical model you can verify that at the stall speed the motor has the maximum current flowing through it, and thus the maximum torque. Also, at the no load speed the motor has no torque and no current flowing through it.

When does the motor produce the most power? Well, power can be calculated one of two ways:

Electrical Power: $$ P_e = V_S * I $$

Mechanical Power: $$ P_m = \tau * \omega $$

If you plot these, you'll find that for an ideal DC motor the maximum power comes at half the no-load speed.

So all things considered, how does the motor voltage stack up?

For the same motor, ideally if you apply double the voltage you'll double the no-load speed, double the torque, and quadruple the power. This is assuming of course the DC motor doesn't burn up, reach a state which violates this simplistic ideal motor model, etc.

However, between different motors it's impossible to tell how two motors will perform compared to each other based only on the voltage rating. So what do you need to compare two different motors?

Ideally you'd want to know the voltage rating and stall current so you can design your electronics appropriately and you'd want to know the no-load speed and stall torque so you can calculate the mechanical performance of your motor. You may also want to see the current rating of the motor (some motors can be damaged if you stall them for too long!). This analysis also somewhat neglects the efficiency aspect of the motor. For a perfectly efficient motor, ki = kt, or rather Pe = Pm. This would cause the power calculations using the two equations to be equal (i.e. electrical power equals mechanical power). However, real motors aren't perfectly efficient. Some are close, some aren't.

p.s. In my calculations I used motor speed as rad/sec. This can be converted to Hz or rev/sec by multiplying by 2*pi.

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For many of us non-experts, if we automatically understood what each of the symbols in the formula stood for, we'd have no need for your otherwise fascinating explanation. Perhaps you could improve your answer with a bit of "where τ = fill_in_the_blank" and "where ω = fill_in_the_blank" ? –  mickeyf Sep 4 '12 at 1:56
    
So in brief, V is proportional to ... V :-) and I is proportional to torque. –  jumpjack Sep 4 '12 at 19:26
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I also think that a higher voltage motor is "better" as it involves lower currents, hence lower energy loss due to Joule effect. –  jumpjack Sep 4 '12 at 19:27

Electric motors can be designed over a fairly wide range of voltage and current for the same speed and torque out. Just comparing the intended operating voltage of two motors doesn't tell you much about what those motors can ultimately do. Motors designed for high power do tend to work at higher voltages, but that is mostly so that the current can be within a reasonable limit.

To compare two motors for a particular job, you have to look at the output parameters. These will be the torque, speed range, and power.

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The mechanical performance of a motor will of course depend mainly on it's physical build, not necessarily its nominal voltage. High power motors will operate on higher voltages, but that does not tell you much.

I won't elaborate on the specifics, but there is a good rule of thumb to use when you want to estimate the parameters of a motor by look. A long motor will achieve higher rpms, and a wide motor will be able to deliver more torque. You can perhaps imagine how this works - a wide motor will have a wide rotor, so the forces of the magnetic fields inside will create a larger torque.

So, if you have two motors of identical length, but one of them is wider, you can expect the wider one to be able to generate higher torque.

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In very basic terms (helloworld's answer has the science bit covered):

Power is Volts * Amps (P=IV). For a given power, say 1000 Watts / 1kW, You can design a 10v motor that uses 100A or a 100v motor that uses 10A for the same nominal power:

10V * 100A = 1000 Watts    
100V * 10A = 1000 Watts

Your next consideration is how the various efficiencies stack up - for each part of the power train there will be some optimum way of building each part that gives best efficiency for the price. For example, if you went for the 10v option you need a lot of big heavy wires (or bus bars) to handle 100A, whereas 10A will flow happily down quite skinny little wires.

However, maybe it's harder to build a control unit / charger that works at 100v than at 10v (it's certainly safer for the average user if there's no high voltages kicking around for them to stick their fingers in).

So, there's a juggling act to be done to work out how the system stacks up - for each watt of power you put in, how much useful energy can you get out the other end?

It's a bit like the difference between a big lazy V8 and a screaming turbo motor, both can make the same power but each is a very different answer to the problem.

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