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In all of this you seem to be labouring under the assumption that batteries have power limits (which they do) and motors don't. The motor can't just take as much power as the battery will give it without overheating. I'm not sure what you're not understanding about this.
Motors have phase current and rpm limits not power input limits as inputting the same power necessary for max torque @ max rpm close to stall/0rpm would exceed the phase current limit. That’s because electrical power required for constant phase current/torque/resistive losses in motor increases with rpm.

That means for a short time the motor can handle the same current (same torque) at any rpm up till close to max rpm (til the speed, current and torque is voltage limited). It takes the same current/torque to magnetically saturate the stator regardless of rpm.

Obtaining the peak torque close to peak rpm uses lots of power from the battery and generates lots of heat in the motor, which is a symptom of the motor operating inefficiently in this condition. So, close to max motor rpm it’s more efficient to use 2 motors operating at half their rated phase current/torque limit than 1 motor operating at the phase current/torque limit.
 

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But why does the speed controller limit the current (actually the power, since a constant current won't give a flat power curve as the cells discharge)
Most commercially available controllers limit the battery and motor current not the power, and as a result they do lose performance as the voltage of the battery sags, because the same battery current at lower voltage means less power.
 

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When staying within the power dissipation envelope of the motor it naturally follows that the slower you go the more torque you can put out with the same power dissipation
No, torque is proportional to the square root of the resistive losses in the motor not the electrical power. For the same torque at twice the rpm the resistive losses in the motor (&phase current) remain constant and the electrical power (battery current) roughly doubles.

Please give up on the fantasy that EV motors can and should produce full torque all the way to maximum RPM.
They can but shouldn’t for the efficiency reasons stated previously... 1 motor at high torque output is less efficient than 2 motors at half torque output.

To design a motor that can produce full torque at full RPM essentially just means you are purposely sacrificing the low speed torque and this would not result in a satisfactory car.
False, all motors “can” put out full torque (according to the phase current limit) up to the voltage limited speed.

When staying within the power dissipation envelope of the motor it naturally follows that the slower you go the more torque you can put out with the same power dissipation.
Resistive power dissipation in the motor remains constant with constant torque regardless of speed.
 

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Motors have phase current and rpm limits not power input limits as inputting the same power necessary for max torque at max rpm close to stall/0rpm would exceed the phase current limit. That’s because electrical power required for constant phase current/torque increases with rpm.
Sorry but if you think electric motors don't have power input limits then I'm not sure why I'm still here trying to debate with you. :rolleyes: You're making this way too complicated and failing to see the wood for the trees.

Forget about phase current, voltage and all that stuff and boil it back to basics.

You have a black box which you put electrical power into, most of that power comes out as rotational energy, say 96% at a given operating condition, the rest generates heat.

That heat causes a temperature rise inside the motor which follows a time dependent curve that depends on factors like thermal mass and thermal resistance to ambient. (In practice more complicated than that as there are multiple different thermal resistance jumps and time constants between say a rotor/stator and ambient air)

The materials that make up the motor can only withstand a certain temperature or they will degrade or fail outright. Mainly the windings and magnetic materials, but also things like lubrication oil.

Therefore to keep the motor in its safe operating region power dissipation must be limited to some maximum figure regardless of speed, and indirectly a maximum input power established, assuming efficiency doesn't change much with power, which is the case with an electric motor. (But not a combustion engine)

This dissipation can't be exceeded for more than a short time (how short depends on each and every thermal mass/resistance step inside the motor) and this is what sets the power limit of the constant power region of the motor, assuming that the battery is not the limiting factor, which I don't believe it is in EV's with decently large batteries and "normal" performance levels.

Of course an AC motor is a reactive device which means the load current can be quite reactive under some conditions, this increases I^2R losses compared to the purely resistive case, however all that really does when you boil it down is reduce the overall motor efficiently slightly and hence increase heating for the same power input. This means the power input must be further de-rated based on the power factor.
That means for a short time the motor can handle the same current (same torque) at any rpm up till close to max rpm (til the speed, current and torque is voltage limited). It takes the same current/torque to magnetically saturate the stator regardless of rpm.
Yeah, and I could disable the rev limiter on my petrol car and rev it past it's 6500 rpm red line to 8000 rpm for a short time and it probably wouldn't blow up immediately - so what ?

Why is there a rev limiter then ? To keep the engine in a safe operating region which is going to let it last for years. Just like the maximum power limit on an EV motor. They want these things to last for years without blowing up.

I think I'm going to bow out at this point because you seem extremely fixated on a single small technical point to the exclusion of all else. Are you sure you're not @donald 's cousin ? ;)
 

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You have a black box which you put electrical power into, most of that power comes out as rotational energy, say 96% at a given operating condition, the rest generates heat.

That heat causes a temperature rise inside the motor which follows a time dependent curve that depends on factors like thermal mass and thermal resistance to ambient.
That heat is directly proportional to the square of the torque not power input... which means the same torque generates the same resistive heat, regardless of rpm.
 

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It's very simple... since torque is directly proportional to the square root of resistive heating losses in the motor, doubling the torque with 1 motor requires quadrupling the resistive heating while doubling the torque by adding a second motor doubles the resistive heating.

Doubling torque w/ 1 motor = 4x resistive heating
Doubling torque adding 2nd motor = 2x resistive heating
 

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Per motor. But what exactly is your point.
I have to say I would not base my car purchasing decisions on anything you have said or posted in this thread.
 

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Doubling torque w/ 1 motor = 4x resistive heating
Doubling torque adding 2nd motor = 2x resistive heating
Per motor. But what exactly is your point.
No, doubling torque adding 2nd motor (coupled output shaft) doesn't change resistive heating per motor but doubles it overall because there are 2 motors. Doubling torque with 1 motor quadruples resistive heating per motor (doubling torque requires doubling phase current, but resistive heating is proportional to phase current squared, so doubling torque from one motor quadruples resistive heating, reducing conversion efficiency). So 2 identical motors can output twice as much power at the same peak efficiency as 1 motor.
 

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Indeed, but it would be more helpful if you just get it right first time.
see post #14

each doubling of motors halves the total joule heating losses and gives 1/4th the heating per motor at constant speed, so 4 motors gives 1/4th the total heating losses and 1/16th the heating per motor at the same constant speed.
& post #2

They get much lower efficiency while accelerating through the lower rpms at full throttle, or towing very heavy trailers up very steep slopes at low speeds. That’s because those situations require high torque from the motor and torque is directly proportional to the motor current and heat (loss) increases at the square of the current. Doubling the motor torque quadruples the heating
 

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see post #14



& post #2
You said "each doubling of motors halves the total joule heating losses and gives 1/4th the heating per motor " before, and I burned you down in flames.

If you don't learn from your mistakes then we're basically struggling to have any sensible discussion here.
 

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Each doubling of motors does halve the total joule heating losses & give 1/4th the joule heating per motor at constant speed.

You can calculate the joule heating losses by multiplying I^2R=W where R is the winding resistance, I is the motor current and W is the watts of joule heating. With 2 motors one needs half the phase current per motor for the same torque output with a coupled output shaft, which is 1/4th the joule heating per motor at constant speed and half the joule heating overall.

For this reason, in the end, 2 motors can output twice the power at peak efficiency as 1 motor.
 

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But you need to expand a lot more and tell us what the point of this endless repetition is. What is the actual crux of your argument?
 

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But you need to expand a lot more and tell us what the point of this endless repetition is. What is the actual crux of your argument?
Suppose I am in a vehicle race cruising along on flat ground at peak efficiency.

I encounter an uphill section that doubles the torque requirement for constant speed.

1) Doubling the gear ratio for double the thrust reduces efficiency because If I was already at peak efficiency on flat ground, doubling the motor rpm at the same torque on the hill reduces efficiency.

2) Doubling the torque from the single motor without changing gears is another option, but it also reduces the efficiency.

3) Engaging a second motor, both producing the same torque for double the total torque is a third option which doesn't reduce efficiency.

^Option 3 is the best for efficiency

132441


if we also factor iron losses from the above chart (in addition to joule heating):

flat ground at peak efficiency (1 motor):

2000rpm @ 125Nm = 26.18kW output @ 93% efficiency (peak)

(26.18kw / 93) * (100 - 93) = 1.97kW losses

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doubling torque (1 motor)

2000rpm @ 250Nm = 52.36kW @ 90% efficiency

(52.36kW / 90) * (100 - 90) = 5.81kW losses

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doubling rpm / changing gears (1 motor)

4000rpm @ 125Nm = 52.36kW @ 85% efficiency

(52.36kW / 85) * (100 - 85) = 9.24kW losses

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doubling torque (adding 2nd motor w/ coupled output shaft)

2*(2000rpm @ 125Nm) = (2 * 26.18kW) = 52.36kW output @ 93% efficiency (peak)

(52.36kw / 93) * (100 - 93) = 3.94kW losses

^In the scenario, doubling torque by adding a 2nd motor w/ coupled output shaft has higher efficiency than doubling torque with a single motor, or doubling rpm by shifting gears
 

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Consider a vehicle similar to a Ford F-150 at max towing capacity:

132453


Flat Ground @ 65mph = 49384w power output required
5% slope @ 65mph = 2.78x power & wheel torque required
10% slope @65mph = 4.56x power & wheel torque required

Assume an electric motor in the vehicle is at peak efficiency outputting 49384w @ 65mph on flat ground...

a 5% slope requires nearly 3x as much wheel torque to maintain constant speed and a 10% slope requires nearly 5x as much wheel torque.

Engaging 2 additional motors to maintain 65mph on a 5% slope achieves greater efficiency than tripling the torque from the single motor or tripling the motor rpm by switching gears.
 

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7 motors—

The Mach-E 1400 has three electric motors that power the front wheels and four more powering the back wheels. Power can be sent to all four wheels, just the front wheels, just the back or it can be split between front and back wheels in any proportion.

 

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In the Ioniq, Hyundai have probably chosen a higher ratio single gear transmission than everyone else. Reviewers comment on a relatively poor acceleration from stopped and relatively good efficiency at motorway speed.

Plus great aerodynamics. Maybe they have a clever motor too, of course, but I suggest that you can get this result by careful choice of gear.
I've noticed this on mine. I was surprised the first time I drove it, initial pickup seems flat but after about 10mph it gets into its stride.

There is zero noise from it though, so whatever Hyundai are doing gearing wise, it is introducing no noticeable transmission noise. There was a review of the Audi e-Tron on Autogefuel's YouTube channel not long ago, it that sounds very noisy by comparison.
 

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Consider a vehicle similar to a Ford F-150 at max towing capacity:

View attachment 132453

Flat Ground @ 65mph = 49384w power output required
5% slope @ 65mph = 2.78x power & wheel torque required
10% slope @65mph = 4.56x power & wheel torque required

Assume an electric motor in the vehicle is at peak efficiency outputting 49384w @ 65mph on flat ground...

a 5% slope requires nearly 3x as much wheel torque to maintain constant speed and a 10% slope requires nearly 5x as much wheel torque.

Engaging 2 additional motors to maintain 65mph on a 5% slope achieves greater efficiency than tripling the torque from the single motor or tripling the motor rpm by switching gears.
So much pseudo-science and jumbling of technical terms. Just because the current/torque increases it doesn’t mean that the efficiency decreases. In fact, synchronous reluctance motors have high efficiencies at higher torque. Why not keep it simple and just talk about resistive losses (i squared r) in relation to heat? Instead we get ”joule heating” and any number of confusing terms as though they were copied at random from an electrical textbook.

The fact is that modern controllers and motors are capable of many different ways of operating by varying voltage, current input and frequency and by changing the motor design to suit the characteristics that the designer requires from the vehicle. Adding motors adds weight and cost but can be justified at higher powers for greater flexibility. This defies simple text book analysis of joule heating, wheel torque etc.
 

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Just because the current/torque increases it doesn’t mean that the efficiency decreases.
If a single motor is already operating at peak efficiency, and the current/torque increases, the efficiency decreases...

cruising along on flat ground at peak efficiency


...If the torque requirement for constant speed doubles on a hill when the motor is operating at peak efficiency on flat ground, the only way to supply the additional torque without decreasing the efficiency is engaging a 2nd motor.
 
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