Why 1/f²?

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Why 1/f²?

Post by coulomb » Sun, 09 May 2010, 22:22

From this thread, it seems that the conventional wisdom is that peak torque falls off with the square of speed over base speed.

However, the Tesla Roadster torque/speed curve does not show this:

Image

[Edit: the above image was originally from http://www.teslamotors.com/display_data ... aph_v2.gif, rescued via archive.org]

Was from here.

Peak torque is almost 300 ft.lb (ick, about 400 Nm) at 5000 RPM, for a respectable 215 kW of power. But at 10,000 RPM, the torque is about 120 ft.lb, just under half the base torque, and way more than a quarter (which would be just under 75 ft.lb).

If the peak torque was falling off at 1/f², then the power would be falling off at 1/f, but it's only down about 20% (170 vs 215 kW).

So what's going on there? Is this marketing hype? Surely not. Is the Roadster controller doing something unusual? Probably, but if so, why can't we do the same?

My understanding has been that the 1/f² thing comes about as a result of how the various DOL torque vs speed curves changes as the electrical frequency changes, and has to do with breakdown torque. Now TJ is claiming that breakdown torque is almost irrelevant when a motor is VFD driven. So why is the received wisdom still that peak torque will fall as 1/f²?

Another datapoint:

Image

From http://metadope.com/Bus/pdf/AC55.pdf and many other places.

Base speed seems to be 2500 RPM, where the torque is 250 Nm. At 5000 RPM, torque is about 110 Nm, again a little under half the base torque, and a lot more than a quarter.

I'm assuming that these torque/speed curves are peak torque, not continuous.

This matters a lot in direct drive vehicles. If the power is relatively constant above base speed, then a gearbox won't help much at all above base speed. If the power decreases as 1/f above base speed, then a gearbox will improve high speed performance, despite its extra losses.

Edit: torque verse slip curve -> torque vs speed curve
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Why 1/f²?

Post by Tritium_James » Sun, 09 May 2010, 22:52

I think it might actually be 1/f, not 1/f². I'll check when I'm in the office tomorrow.

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Why 1/f²?

Post by coulomb » Sun, 09 May 2010, 23:36

Here is a Baldor article where they come frustratingly close to explaining what I want to know:

http://www.reliance.com/prodserv/motgen ... 0Operation

Indeed, in their graph, the peak power does come down to half at 2x base speed. But they don't say why.
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Why 1/f²?

Post by Tritium_James » Mon, 10 May 2010, 00:20

It's because past the base speed you're limited to constant power (can't put in more volts). Double the speed, power can't increase, therefore you must have halved the torque.

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Why 1/f²?

Post by coulomb » Mon, 10 May 2010, 01:38

Tritium_James wrote: It's because past the base speed you're limited to constant power (can't put in more volts). Double the speed, power can't increase, therefore you must have halved the torque.

Yes, but it's half the power, and about a quarter of the torque. Hence 1/f² for torque, 1/f for power.

Why do the books say this, and the torque curves show a far more gentle decrease in power?
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Why 1/f²?

Post by weber » Mon, 10 May 2010, 01:49

coulomb wrote: From this thread, it seems that the conventional wisdom is that peak torque falls off with the square of speed over base speed.
Not quite. The conventional wisdom is that, at constant voltage, peak torque falls off as the inverse square of the electrical frequency, not the mechanical speed. Subtle difference perhaps. And it follows that law somewhat below nominal frequency as well, thereby giving you more than Tmax. But this can't go on indefinitely. As you go lower in frequency, without reducing the voltage, the current must increase in proportion to the torque, and the stator iron begins to saturate, so the stator flux is no longer proportional to V/f.

If I remember rightly, in the linear (non-saturating) region, peak torque is proportional to stator flux squared and flux is proportional to V/f, so peak torque is proportional to V^2/f^2. So if V is constant, peak torque is proportional to 1/f^2.

We need to understand how breakdown torque is defined. It is defined as the maximum possible torque at a given voltage and frequency, with absolutely no consideration given to efficiency or temperature. You apply that voltage and frequency to the motor and you gradually load it up mechanically. As you do, the slip increases, which means the rotor sees increasing frequency. At first the rotor flux increases with increasing slip, providing a kind of positive feedback so the motor just draws more current to meet the increasing torque. But eventually the rotor flux levels off, at the critical slip frequency, and on entry into the negative feedback region the motor just folds back its torque and stops dead. This is the "breakdown".

Or something like that. Image

A properly functioning VFD will not let that happen. As breakdown approaches it will simply lower its output frequency.

I note that continuous torque depends on considerations of heat dissipation and tends to fall off as 1/f.

The efficiency vs. torque curves in this thread may be relevant.
forum_posts.asp?TID=1420&PID=25234#25234
However, the Tesla Roadster torque/speed curve does not show this:
...
So what's going on there? Is this marketing hype? Surely not. Is the Roadster controller doing something unusual? Probably, but if so, why can't we do the same?
Nothing unusual. Try looking at the far right of the graph. Torque at 13500 rpm is 50 dram-cubits and torque at half that (6750 rpm) is very nearly 4 times, at 200 hogshead-furlongs.
Now TJ is claiming that breakdown torque is almost irrelevant when a motor is VFD driven. So why is the received wisdom still that peak torque will fall as 1/f²?
The breakdown torque is still the limiting torque at the nominal frequency and voltage, even with a VFD. But it is not the limiting torque overall.
I'm assuming that these torque/speed curves are peak torque, not continuous.
Therein lies the source of your quandary.

They are not peak (i.e. instantaneous), except at high speeds, but they may not be continuous either, they may be based on operation within temperature limits for a few seconds or minutes.

[Edit: "stator flux levels off" -> "rotor flux levels off". Thanks Coulomb]

[Edit: I may be confusing positive and negative feedback above. The latter may be "positive feedback with negative consequences", but hopefully you get the idea, even if the details are not correct.]
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Why 1/f²?

Post by coulomb » Mon, 10 May 2010, 02:15

weber wrote: I note that continuous torque depends on considerations of heat dissipation and tends to fall off as 1/f.
And this has relevance to the answer, I think?
Nothing unusual. Try looking at the far right of the graph. Torque at 13500 rpm is 50 dram-cubits and torque at half that (6750 rpm) is very nearly 4 times, at 200 hogshead-furlongs.
An excellent point, and one that I missed. So both motors (let's say, I haven't checked the AC55) obey the laws of physics at high speed, but what allows them to exceed the limits just above base speed?
The breakdown torque is still the limiting torque at the nominal frequency and voltage, even with a VFD.
Actually, I thought that was to do with the DOL torque verses speed curve, which depends on the ratio of resistance to reactance, so my understanding is that a VFD, by changing the power factor (which a DOL connected motor can't), could in fact exceed breakdown torque at nominal voltage and speed. The power factor in an induction motor is of course essentially the ratio of motoring (real) current to magnetising (imaginary) current.
Therein lies the source of your quandary.
So the answer is here... I wish I could recognise it!   Image
They are not peak (i.e. instantaneous), except at high speeds,
Ok, I'll bite. How does high verses low speed affect whether the torque is peak or continuous? Are you saying that at low speeds (say below 2000 RPM) the motor has a chance to cool off between mechanical revolutions (or electrical cycles), but doesn't get a chance to do this at higher speeds? Surely thermal time constants are all way longer than any mechanical or electrical cycle, except perhaps below about 6 RPM (0.1 Hz)?

Edit: had motoring and magnetising round the wrong way (ratios have an implied order of operands, I believe)
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Why 1/f²?

Post by coulomb » Mon, 10 May 2010, 02:24

coulomb wrote: So both motors (let's say, I haven't checked the AC55) obey the laws of physics at high speed,

Actually, at 6500 and 3250 RPM, I read about 60 and 180 Nm for the AC55, so that's 3:1, not 4:1. Perhaps it would get closer to 4:1 if the test was continued at higher speed? I'll accept that it may be heading in that direction, but I maintain that peak torque shouldn't be allowed, with the current theory, to fall less sharply than 1/f².
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Why 1/f²?

Post by Tritium_James » Mon, 10 May 2010, 02:44

coulomb wrote: Yes, but it's half the power, and about a quarter of the torque. Hence 1/f² for torque, 1/f for power.

Why do the books say this, and the torque curves show a far more gentle decrease in power?
Why do you say it's half the power? I don't see it - the ideal (perfect) situation is constant power above the base speed, as shown in "Figure 9" in the Baldor link you posted.

The only reason the graphs you've posted from Tesla et al are not constant power above the base speed is the non-ideal things start to limit your power throughput: batteries fall over, motor losses increase, controller losses increase, motor power factor gets too high, etc, etc

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Why 1/f²?

Post by woody » Mon, 10 May 2010, 04:15

The Skateboard Ramp torque curve & Ski Jump power curve of an AC motor assume you have enough current to keep the motor at full torque.

If you don't (e.g. using a 37kW motor with a 37kW VFD), then you can't get full breakdown torque at any speed, and when you run out of voltage, you can still pump in current as you're well below peak anyway.

e.g. plug the ABB formulas into a 230V 131-008 (22kW 2 pole) using a much too small Danfoss 6032:
Image

You only have enough current to get about 90Nm of torque (purple curve - nowhere near the full 285Nm) but after you run out of voltage you have plenty of headroom to keep the current up, giving you a flat power curve (orange curve).

With a bit too small controller (Danfoss 5042):
Image
You get a flattened top power curve which gives you a flatish power curve which drops off.

Much better overall is a big controller - your power and torque drop off, but not as low as either of these.

It's possible that the Tesla's controller hasn't got the amps for absolute maximum torque? Sacreligious I know...
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Post by weber » Mon, 10 May 2010, 04:31

I didn't mean to be cryptic. All I'm saying is that you're mistaken in assuming those curves show peak torque. They show continuous torque or something close to it. i.e. They show something based on temperature considerations which is a bit fuzzy, not the sharply defined limit of peak or breakdown torque which depends only on flux (i.e. on V/f and saturation).

Temperature related torque limits fall off as 1/f. The ultimate flux-related torque limit starts from a higher point but falls off as 1/f^2. However eventually the temperature related 1/f curve hits the 1/f^2 curve and must follow it down from then on. This is shown in the graph below. I note that what they call maximum torque (T) here is in fact maximum continuous torque.

Image

The graph came from this old post. Thanks Richo.
forum_posts.asp?TID=585&PID=10811#10811

The transition from 1/f to 1/f^2 is smoothed out in the Tesla's torque curve.

But those curves (immediately above) still don't tell the whole story because they assume constant V/f below the base frequency. If you don't lower the voltage as you go below base frequency the 1/f^2 curve of breakdown torque just keeps on going up at 1/f^2 until saturation gradually starts to kick in. There doesn't seem to be much written about exactly what happens here.

Another point you may be missing is that you can't really change phase independently of frequency, since they are derivative and integral of each other.

I also note that the terms active and reactive are often used instead of real and imaginary.
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Post by woody » Mon, 10 May 2010, 04:46

Good topic BTW Coulomb.

What I didn't say before was my graphs are all based on fixed V/F (below field weakening point).
Weber wrote:If you don't lower the voltage as you go below base frequency the 1/f^2 curve of breakdown torque just keeps on going up at 1/f^2 until saturation gradually starts to kick in. There doesn't seem to be much written about exactly what happens here.
There's a hint on the bottom of this:
Image
Running your 500V motor at 550V (20% more V/F) you get 19% more torque - i.e. saturation isn't really a big problem yet.

The old textbook I borrowed said that saturation is a shallow curve, and that motors are generally designed to only go into saturation but not by much - for efficiency reasons.
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Why 1/f²?

Post by coulomb » Mon, 10 May 2010, 04:49

Tritium_James wrote: Why do you say it's half the power? I don't see it -

Ah, my mistake. I was referring to figures 10 and especially 11 (torque and power respectively, so the latter is the former times speed). In figure 11, the peak power is clearly halving at 200% of base speed. The nearest anchor point on the page was a bit high up, sorry.

Figure 9 is the sort of idealised continuous power curve, and I agree, it is the best you can hope for: constant maximum torque until you run out of voltage, and then the most torque you can get without breaking the law of conservation of energy (and hence power).
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Post by coulomb » Mon, 10 May 2010, 05:21

weber wrote: I didn't mean to be cryptic.
That's all right, I didn't mean to be thick!    Image
Temperature related torque limits fall off as 1/f. The ultimate flux-related torque limit starts from a higher point but falls off as 1/f^2. However eventually the temperature related 1/f curve hits the 1/f^2 curve and must follow it down from then on.
Ah! So my mistake was believing that marketing can be completely trusted to always show the best possible results for a product - I thought that this was surely law!   Image

So the torque curves shown, while not continuous, are likely not instantaneous either. So the motors are capable of more than what the manufacturers claim - for very short periods of time. Presumably, these might be half hour or possibly 10 or even 1 minute ratings. I truly learn something every day.

That possibly explains the bump in the AC55 torque curve, where it may be transitioning from 1/f to 1/f²... but then I should still see a 4:1 decrease in torque for a doubling of speed. The controller could be doing odd things; there may be saturation, windage, or other non-linear losses involved.

And of course, I've seen the three-part curve before, and had forgotten it, probably because its significance never sunk in properly for me. Image

Woody - I'd seen that table on non-design voltages before too, and not understood the complete significance there. [Edit: BTW, running a 500 V motor at 550 V is 110% V/f.] Interesting that running a 400 V motor at 380 V (95% voltage) gives you 90% of breakdown torque. And 415 V (103.75% voltage) gives you 106% of breakdown torque. It's that f² term again, in this case it's (V/f)², so instead of 106% you'd actually expect 107.6%. So as you say, getting 119% of breakdown torque from 110% of voltage is doing pretty well (121% = 110%² is expected).
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Why 1/f²?

Post by coulomb » Mon, 10 May 2010, 05:41

Well, this is embarrassing. I'm recovering from a cold, yes, that's it - not going senile at all!

Somehow I thought TJ's motor was 200 V, when of course it's 100 V. So I'm afraid I'm going to remove the rest of this post. Those that received this message in email can have a good chortle at my expense.

Nothing to see here; move on.   Image

Edit: massive deletion to protect the, uh, recovering.
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Post by woody » Mon, 10 May 2010, 06:39

woody wrote:Running your 500V motor at 550V (20% more V/F) you get 19% more torque - i.e. saturation isn't really a big problem yet.
Err, Coulomb's cold is affecting me too:

550V is 10% more than 500V, i.e. 110% V/F => 121% of torque ignoring losses, but it's only 119%, which still isn't bad.
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Post by coulomb » Wed, 12 May 2010, 20:00

I've come across some further evidence of the worth of Weber's wise words:

Image

From http://corbinstreehouse.com/misc/M3-ac40_L.pdf. Unfortunately, this particular motor seems to be "unofficial", but I've not happened across information like this for an official motor. Other Electro Vehicles Europe motors here: http://www.electro-vehicles.eu/shop/bro ... ion_Motors

Note how the "peak" torque (60 second rating) merges with the other torque, marked "S21h", at around . I couldn't find any motor duty cycle S21, let alone the "h" version of this. Then it dawned on me that this is "S2 1h", that is IEC duty cycle S2, for 1 hour. IEC duty cycle are well documented; S2 is defined as:

"Short-time duty. The motor works at a constant load, but not long enough to reach temperature equilibrium. The rest periods are long enough for the motor to reach ambient temperature."

I'm not sure how the hour comes into this, I think it means the motor can take this level of torque for a whole hour, provided it starts at ambient and is allowed to cool to ambient before it is used again. Since a typical EV has a "time range" of about an hour (say 80 km @ 80 km/h), and needs hours to recharge and thus cool, this could be considered "real world". It might work a bit harder than this rating uphill, then rest at a traffic light.

For an EV, the S2 1h rating could be considered "continuous", and the 60 second rating could be considered "peak".

Here, the peak (1/f²) curve meets the S2 1h curve (1/f) at about 3000 RPM, while base speed for this motor is about 1800 RPM. Below base speed, continuous torque is about 2/3 of peak torque. I suspect that this is typical, so most motors can run at about 2/3 of peak power for about an hour before overheating. However, water verses air cooling may affect these numbers; I believe that this one is water cooled.

Edit: added link to Electro Vehicles Europe.
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Post by coulomb » Mon, 25 Oct 2010, 04:31

I noticed that the image wasn't working in the first post. When I went to check why, I found that the image on teslamotors.com no longer exists (it may have been moved, I haven't checked yet).

When I went to archive.org to pull out a copy, it seems that the image has changed somewhat. It used to be this:

Image

but later it became this:

Image

Interestingly, the first graph shows classic constant torque then approximately 1/f² fall (200 to 50 foot-lubs over a 2:1 speed range (7 000 to 14 000 RPM). They've drawn it as fairly linear, but the end points correspond to 1/f².

The second graph shows much more torque and power. Perhaps they changed the motor at this point, or radically changed the software in the inverter to get more power. Now the torque doesn't fall as 1/f² any more either.

All this just makes me more confused, but perhaps someone can enlighten me.

[Edit: archive.org makes it hard to change history Image )
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Why 1/f²?

Post by 7circle » Mon, 25 Oct 2010, 10:01

I'll try them side by side v2 new 4 cylinder on left and V2a 6 cylinder on right:
ImageImage

Apart from the ICE change from 6 cylinder to 4 cylinder.

I thought I'd play with the GIMP graphics software and do an overlay.
I mached up at 250kW and 13500 RPM
Image

Now I can see the controller current has been reduced lowering the torque.
it might just be controller parameters to give a softer torque-roll-off to give the driver a similar feel to ICE performance.

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