How to convert a hybrid

Technical discussion on converting internal combustion to electric
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Post by T2 »

- Coulomb You have opened rebuttal on so many fronts that I don't have a lot of time to answer but I'll try.

First glad you caught that two pole/four pole error, we all have those moments ! I should point out since the phases are in 120 deg seperation the interphase voltage needs to be multiplied by root3/2 which reduces the expected 250Vac to 206Vac lowering the effective base speed to just 3090 rpm.

You brought up slip and for me at the present time this has to be a second order thing. Then this :

So really, you want a motor with higher V/Hz, which sounds bad, so you run out of volts earlier, which sounds worse, so the torque drops off earlier, which definitely sounds bad. But the size of the motor is proportional to its power, and the lower the base speed, the higher the torque you get to the wheels.

I disagree with one thing there - the bit about the size of a motor being proportional to its power. Not true.

The size of a motor is proportional to its torque

I do agree more iron and copper = more torque but not more power. Because that's not how power works. And Coulomb knows this, but for others who might be reading this -

Power is the product of speed and torque.

Ever wonder why the Tesla sets its base speed at 5000rpm and not 3000 rpm ? It's because it needs to be developing full power at around 40+mph in order to accomplish its 4.0sec to sixty mandate and it gets that power from rpms. If it could only reach full power beyond 60mph I think you would agree that would be somewhat unproductive. Most kinetic energy lies between 40 and 60mph so you must drop your power peak in there. {Prius -51.2mph EV1 42mph for example}

A high V/F is favoured by drive manufacturers. Lots of N-m for the amp.

Off the shelf motors in North America 460Vac 60Hz have a V/F of 8, for Japan, 100Vac 50Hz have a V/F of 2.0. Contrast with Aircraft duty 100Vac 400Hz V/F =0.25.
For aircraft power and lightness are important so high rpms are the norm. I`m just saying that shouldn`t this be the case for EVs as well.

But for the systems integrator - that's you Coulomb - it's a royal pain when torque starts dropping off early. Then you'll get someone crawling out the wood work with "that's why I use a four speed manual transmission, I just change into a higher gear the motor slows down and most of my torque comes back." Which kinda' defeats the idea of an electric transmission in my mind.

the V/f ratio is probably a bit crude when applied to a proper motor controller Hmmm Sounds like controller designer speke to me.

In fact use of a fractional V/F motor allows you, at one end, to avoid gearchanging by allowing more of those useful rpms and at the other, saves you a possible tasering while dealing with dangerous pack voltages in cramped quarters. Not to mention all those enthusiastic high torque couplings I've seen which could probably beef up an automotive transmission with enough strength to beach an oil tanker.

- Coulomb I'm pretty sure most of this stuff you already knew. I know that. And you didn't get this far without it. I do know that when we are trying out something new in a less familiar area it is important not to let the facts get in the way of a good idea. Know the rules. Break the rules. Is my motto. Besides some facts aren't even facts at all. That's what I've discovered. They are just convention which may have been true in the past, but the future will eventually overrun convention. In short, the truth is only temporary.

Thanks for putting up those graphs and responding with other stuff BTW, I may post again if you`re up for some friendly ribbing.
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Post by woody »

My take on slip was that it remains fairly proportional to torque, assuming motor magnetism is the same.

i.e. if nominal slip is 50rpm, at max torque (3 x nominal) the slip would be 150rpm.

As you get to the breakdown point, the ABB formulas suggest you use more current (~4x current at 3x torque), perhaps the slip is greater at this point, e.g. 200rpm.

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Post by T2 »

-Coulomb I've been forgetting that this is a 4-pole motor, so at 50 Hz, synchronous speed is 1500 rpm, not 3000 rpm. But that doesn't explain all the difference; considering only the V/f ratio, the base speed would be just under 4000 rpm (265/100 x 1500 rpm minus slip speed).


As I pointed out since the phases are in 120 deg seperation the interphase voltage needs to be multiplied by root3/2 which reduces the expected 250Vac to 206Vac lowering the effective base speed to just 3090 rpm.
..........or 2900rpm if we include slip, and could be even lower depending on the bus voltage sag with FLC.

I know I have been flogging this high rpm strategy using a downsized motor but there is a further reason. High speed machines are also more efficient since the copper loss is constant with torque. When a 1500rpm motor which has been changed from a V/Hz of 2 to 0.25 and then spun to 12000rpm, it will have the same copper loss as before but with eight times the power. Therefore it logically follows that if it were 92% effc at 1500rpm it will now be 99% effc at 12000rpm. Neglecting friction windage, though iron loss will continue to be a factor.

In practice base speed could be brought down to 6000rpm yielding only four times the power but even then the effcy improvement would be at 98%. So there are actually two good reasons to consider downsizing.
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Post by coulomb »

T2 wrote: As I pointed out since the phases are in 120 deg seperation the interphase voltage needs to be multiplied by root3/2 which reduces the expected 250Vac to ......
I believe that it is well accepted that you can just divide the bus voltage by sqrt(2) to get the highest three phase sinusoidal line-to-line AC voltage; usually a trick like neutral wobble is needed to achieve this. This may be slightly different with trapezoidal output, but I doubt that the factor would be as high as sqrt(3)/2.
When a 1500rpm motor which has been changed from a V/Hz of 2 to 0.25 and then spun to 12000rpm, it will have the same copper loss as before but with eight times the power. Therefore it logically follows that if it were 92% effc at 1500rpm it will now be 99% effc at 12000rpm. Neglecting friction windage, though iron loss will continue to be a factor.
Since iron losses increase quickly with frequency, it's not clear to me that efficiency will increase at all. Also, friction and windage are often neglected because they're messy and not instructive of how a motor works, but at 12 000 rpm, I suspect they are not negligible.
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Post by T2 »

I am prepared to drop this but let's try one time and forget neutral wobble for now !

I am Proposing a nominal 141.4Vdc bus.

a single dual mosfet leg will generate an ac rms voltage v
where 2root(2) x v = 141.4

Therefore   v = 141.4/(2root(2)   =   50 Vac rms

A similar leg will also generate 50Vac rms

IF the two voltages at these legs are in antiphase by 180 degrees then YES, we can add them vectorially to give 100Vac rms.

But the legs are 120 deg apart, this is less than optimal but nevertheless when we add vectorially the factor root(3) which originates from twice the sine of 60 deg, since the 120 deg phase difference exists not 180 deg.

Therefore instead of adding 50Vac to 50Vac i.e. 50Vac TIMES TWO
                  in this case we have to mutiply by root(3) or 1.732
    
Therefore we get 86.6Vac rms instead of 100Vac rms and that's the unexpected loss that you've been seeing.

What's this got to do with the price of rice ? Well in your system it has effectively lowered the base speed of your motor to below 3000 rpm.
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Post by T2 »

I am glad you concur that the fractional V/Hz motor when spun beyond 50/60Hz frequencies becomes more efficient as far as copper loss is concerned. It is not commonly recognized.

Above 6000 rpm I agree windage is no longer trivial. The usual recommendation is to remove the integral fan and mount a seperately excited blower.

Iron loss does increase with rpm but initially it starts out with a very small coefficient. Iron loss = copper loss by design but a fractional V/HZ wound motor has extremely low copper loss when cruising so there is plenty of extra capacity to absorb more iron loss.

Taking the extreme case, how do 400Hz aircraft duty machines seem to survive ?

Usually when the acceleration phase is over the controller should drop the applied voltage. Weakening the excitation significantly reduces iron loss. It is my understanding that it is only the older V/Hz controllers which will overheat motors and then only beyond three times overspeed.

Any motor should be rotor balanced for 8000 rpm or at least checked on no load for vibration.

Personally the answer is that MBA's rather than engineers have sway and the thickest laminations (with lower cost) that do not impair 120Hz operation are specified.
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Post by Johny »

T2 wrote:Therefore   v = 141.4/(2root(2)   =   50 Vac rms
Doesn't the typical bridge output topology double this? Output would be DC/root(2).
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Post by T2 »

-Johny
Doesn't the typical bridge output topology double this?

Still Nope.

I think you and Coulomb are doing scalar addition, you need to be doing vector addition which takes account of the 120 deg phase displacement between two legs. It's not double but root3 in fact.

So your AC motor will run out of volts earlier than anticipated which is depicted by the torque curves peaking early.

Hopefully you will straighten out on this.

On this thread I am pursuing a design using the Prius as a donor since it presents a unique opportunity with its 10.7 ratio which is midway between 1st and 2nd of a conventional transmission. Unfortunately both of the controllers under consideration limit output frequency to 300Hz. This will spin a 4 pole motor synchronously to 9000 rpm which works out to be 57 MPH. Accounting for induction motor slip maybe 55 MPH.    
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Post by Johny »

Your calculations don't pass the reasonable test.

A typical 3 phase VFD rectifies 3 phase to create a DC bus - typically around 1.4 times peak phase to phase - lets use 400V P-P and 560 VDC.

The typical 3 phase drive then chops the DC up using a FULL IGBT bridge to re-create the 3 phase almost volt for volt accurate (after inductive smoothing) - then gets the same torque at the same speed out of the motor in comparison to running Direct Online.

I gather you disagree with this?
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Post by coulomb »

T2 wrote: I am prepared to drop this but let's try one time and forget neutral wobble for now !

T2, please take it that I know how to add voltages vectorially.

I agree that if you don't use neutral wobble, or similar, you can't get the "required" 1/sqrt(2) AC/DC voltage ratio. That would fail the reasonableness test, as Johny pointed out. It would mean that with 415 V in, you could never get 415 V out. Using neutral wobble, you can get a metric smidge more than 415 V out, even with a couple of volts drop across each IGBT or MOSFET.

Here is a post I did on this some time ago: Neutral Wobble Theory. I have the feeling that I didn't finish it off properly, but at least the basics are there.

As a result, every reasonable inverter uses this technique. So I don't think we can "forget neutral wobble for now".
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Post by Johny »

I think T2 is eluding to a lot more than neutral wobble.

Let me put this another way.
An IGBT bridge can apply the full DC Bus, in either polarity, from any phase to any phase. Other than semiconductor losses, phase to phase voltage out = DC Bus voltage - then shape as desired.
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Post by weber »

T2 wrote: I am prepared to drop this but let's try one time and forget neutral wobble for now !

But forgetting neutral wobble is exactly where you are going wrong. All 3-phase inverters do it in one way or another, which is why they all can have a line to line sine-wave RMS voltage that is Vdc/sqrt(2) as Coulomb and others say.

[Edit: Added "sine-wave RMS" for clarification]
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Post by T2 »

Neutral Wobble is a new term to me, here in North America.

I am well aware that unlike UPS's, motor controllers don't synthesise
sine waves. I have never seen any math writeup on it. So you're saying
it makes as much as 15% difference. Thanks, I did not know that.

Let's add this to the thread for other ignoramuses like me :

Generally it is the fundamental sine plus the addition of one sixth amplitude third harmonic sine. This yields a double hump on the sine peak which increases the effective modulation depth. The effect at the motor is the inverter is running at a higher voltage bus than actual. And we accept the third harmonic cancels in the motor. And it makes a 15% difference !!
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Post by T2 »

- Weber Thank you for your input, obviously if anyone here needed straightening out just then, it was me.

-Coulomb please accept my apology for the misunderstanding if I was appearing to cast doubt on math skills over in the Aus.

That aside I have looked up your ref and see that Neutral Wobble is not just a problem with Holden manual transmissions of which you know quite well, but can also be a colloquialism for space vector modulation. The drawings were good by the way, thank you again for clearing that up.
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Post by T2 »

-Johny Your calculations don't pass the reasonable test.

A typical 3 phase VFD rectifies 3 phase to create a DC bus - typically around 1.4 times peak phase to phase - lets use 400V P-P and 560 VDC.

I gather you disagree with this?


This supposition will drag us way off topic ! Perhaps this should go to a think tank style thread.

Do I disagree ? Yes I do.

I think you have mounted an erroneous supposition right out the gate here, namely that you can run an inverter from the rectified and filtered input from a 3-Phase diode bridge and then turn around and reconstitute that same input AC waveform minus a few semiconductor volt drops here and there.

I don't doubt that the inverter will indeed run from a DC voltage defined as the rms voltage input times root2 at least, that is, until the inverter begins to pull current. At that point the bus voltage will simply collapse.

The input bus capacitors will draw huge currents at the crest of the input waveform which could be destructive to the supply structure. The addition of a series inductor in the input filter will help extend the conduction angle of the rectifying diodes and smooth out the high current peaks being drawn by the capacitors but the average power will then be supplied only at the RMS value of the input voltage - not its peak.
Sorry Johny but I am not with you here.
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Post by coulomb »

T2 wrote: I think you have mounted an erroneous supposition right out the gate here, namely that you can run an inverter from the rectified and filtered input from a 3-Phase diode bridge and then turn around and reconstitute that same input AC waveform minus a few semiconductor volt drops here and there.
Arrgh - lost reply with fat fingers. Briefly, yes, customers do expect this, and they get it. You even get the metric smidge extra voltage, which sounds implausible but it's true. (At least for sine wave outputs.)
The input bus capacitors will draw huge currents at the crest of the input waveform which could be destructive to the supply structure.

This is only true for single phase input inverters, which here only go up to about 2.4 kW. The voltage ripple on a three-phase smoothing capacitor is only about 6%, and the line inductors take care of that.

In Europe and Australia, all VFDs operating above 240 V and about 2.4 kW (maybe 3 kW occasionally) are three phase. In North America, they might use 480 VAC, something we don't have here. Where this is used, is this typically 3-phase or single phase? If single phase, it would have the problems you mentioned.
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Post by T2 »


The metric smidge extra voltage ?

Respectfully Coulomb, but are you on something ?

I accept the fact that you were able to flumox me the other day re Neutral Wobble

And I have to appreciate the thinking that "we got him to buy into that so let's try him with something like the er..umm...metric smidge extra voltage "

But it isn't going to work this time, buddy, you can keep your 'modified sinewave' sorcery. I'm not buying !!
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Post by coulomb »

T2 wrote: The metric smidge extra voltage ?

Respectfully Coulomb, but are you on something ?

Sorry for the colourful language, T2. It's of the order of 2%, less IGBT drops, from memory.

Ok, while I've got you in denial, consider this further sorcery: at some parts of the output waveform, the voltage is actually outside (above the positive rail, or below the negative rail) by a diode drop. This gives a kind of "crossover distortion", because the output amplitude one one side changes from an IGBT drop below the rail to a diode drop above. This distortion apparently can somewhat be corrected for in software; the distortion occurs as the current (not voltage) crosses zero.

This happens more often during regeneration.
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Gosh T2 haven't you seen his picture? Image
Image
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Post by T2 »

And now, and now, it's time for us to take our leave and abandon the world of sorcery for the moment. Now is the time to enter a universe, a place to stand and observe, while he, T2, weaves a little uninterrupted wizardry. No, we are not about to have a Harry Potter moment, this is just the calm before the storm, a moment in time, if you will, marking, nay, heralding the prelude to the opening of a such a splendiferous array of technical sophistry that some say could only spring from the fertile mind of onesuch, T2.

Whoever would have guessed that despite his academic record at University being unexpectedly cut short, (being shown the door twice will do that) thereby depriving his mind of that final coating in the form of a patina of diseducation that academia manages to impart to so many others, that he would be, himself, able to come so far.    
It's time. I would now ask you people to quieten down. Cell phones off please. He's about to speak :

Welcome ! Welcome Everybody. Glad you could make it. WOW !! That was some intro wasn't it ! Hopefully what we are going to get ourselves into next will be worth it.

So here we go. And to start we'll begin by discussing the subject of base speed. What is base speed ? Let us take a look at how some in the industry define it. And who better than the makers of that famous roadster - TeslaMotors.

TESLA defined base speed in a recent patent application as the speed at which the torque drops to 95% of the flat peak torque and will continue to drop beyond base speed proceeding towards top speed under constant power source limits. For myself, I prefer to define it as the centre of the peak of the power curve. And the centre of the peak of the power curve is subjectively a lot easier to identify visually and a lot less long winded to describe I'm sure you'll agree.

In a previous post I prodded Coulomb to remember that frame size is proportional to torque and not power.
Any emphasis towards torque therefore has the disadvantage of requiring a heavy and expensive motor.

However stipulating the base speed to be lower, at whatever horsepower, similarly implies that you are voting for a motor with a correspondingly higher torque and cost. Whether a high or low base speed is preferable I begin with the following statements :

As Coulomb previously indicated graphically, the motor should start accelerating from rest at maximum torque and proceed towards its base speed which, in his case, he would like to attain as early as possible since this is where the motor is able to reach its greatest power.
During this short acceleration ramp, power would have increased steadily on its way towards base speed thus describing a triangle with the area enclosed representing the power delivered. The triangular power profile produced by constant torque.

Then continuing on with maximum power on tap the motor continues to accelerate notwithstanding that acceleration will fall since torque will decline inversely with speed. This is the rectangular power profile produced by constant power. Being limited in this case by the ampacity of the battery pack.

On paper this looks a good strategy for the controller. It is, however not so good for the motor because the low base speed has forced the use of a much larger frame size motor. And not so much for the battery either which will not come to full current until base speed is reached. Then there's going to be a further problem with the motor upstream in the expectation of constant power from the motor. There is a strong likelihood the motor will not be able to provide full power beyond twice base speed. A point where power may be needed the most.

Introspection of of just the Kinetic Energy alone which is the highest load in a 0-60mph ramp up.
0 mph K.E = 0
                          DIF = 1   1/36= 3%

10 mph K.E.= 1
                          DIF = 3   3/36= 8%

20 mph K.E.= 4
                          DIF = 5   5/36=15%

30 mph K.E.= 9
                          DIF = 7   7/36=19%

40 mph K.E.= 16
                          DIF = 9    9/36=25%

50 mph K.E.= 25
                          DIF = 11 11/36=30%
60 mph K.E.= 36

The problem is that as any dragster knows a great launch trumps an average finish.
So ideally it would seem that high power is just as important at low speed during the initial launch as it is at high speed. Therefore the (imagined) need and motive for a low base speed motor to accomplish that bottom end launch.

That is where we differ. At low speeds I say we certainly need to go for high torque yes, but high power no. Consequently my preference is to set the base speed as high as possible but not compromise performance in the low speed region. I maintain that it is possible to have our cake and eat it too. In which case we first need to ascertain the optimal position to be placing base speed.

From the Kinetic Energy speed chart above, for a 60mph terminal velocity the following observations can be made.
That by 20mph only 3+8 or just 11% of the total K.E. has been invested in the vehicle.
That between 40mph and 60mph, a total of 25+30 or 55% of K.E will need investing.

This introspection demonstrates the importance of achieving full power by 40mph and following thru to 60mph.
Therefore, despite what others might think, it appears that setting base speed around 40mph will be the most propitious.
It happens that the half way point for K.E. is 60/root2 mph = 42mph. There is no math equation that could place base speed at the most propitious point since some variables have already been decided. For example, reducer gear ratio already set to 10.7 . Available motors have limited V/Hz values.

It may be of interest, and perhaps of no little coincidence, that 42 mph was also specifically chosen by the designers of the IMPACT (later the EV1).

So, What about the launch ? The selection of a higher base speed will undoubtably enlarge the triangular part of the overall power profile and that certainly, is Not Good. However this is where we can "compensate" by specifying a higher current controller with the high current rating specifically for use when operating in the triangular section.

Then, when the voltage on the motor is only a percentage of the max voltage we can draw an equivalent percentage of current extra, in order to compensate. It is open to speculation whether Curtis supplies a 650A 96Vdc AC controller with the underlying intention to allow the battery to come to full current well before base speed is reached.

It occurred to me that there may be those reading the previous paragraph who are unaware that it is perfectly possible for the motor running below base speed to receive 650 amps under certain conditions while the controller draws the limit of, say, 200 amps from the battery, since the controller has a battery current limiter in addition to a motor current limiter. I have been attempting to find the curtis programming manuals but they don't appear available on the web so far.
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Post by T2 »

- Johny NICE GRAPHIC damn, wish I had your computer skills

- Coulomb just saw your post also. And yes I appreciate the conduction of the anti-parallel diode must give that effect.

My previous post had been on the backburner for several days and I spent most of the day putting a new front end on it to make the subject less dry.

I am attempting to give rhyme and reason for AC motor selection as a proactive decision rather than discussing the prospects of an unknown motor someone happened upon at a forklift breakers yard. Where single ratio reducers are concerned I don't see any useful guidance being given anywhere on this subject.

Mostly what I see is the captain of the Titanic being given helpful advice just after he hit the iceberg.

For the Prius conversion there will be a part two coming soon. After accomplishing that I will have time to field questions and provide answers for the missing parts of the puzzle. Your help would be welcomed.
For now the plan is to establish that reference design for pre 2004 Prii and perhaps some even newer than that which have sustained such damage to either the engine or electrical system that economic repair for restoration as hybrids is no longer possible.
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Post by coulomb »

T2 wrote: I would now ask you people to quieten down. Cell phones off please. He's about to speak :
Nice buildup, T2. Image Unfortunately, I don't think that the speech matches the buildup.
TESLA defined base speed in a recent patent application as the speed at which the torque drops to 95% of the flat peak torque ...
I prefer to define it as the centre of the peak of the power curve.
I see these as equivalent, although with an ideal motor, where the power really was constant in the "constant power" region, the peak power would be a band, not a point. But you could use the point where the power stopped rising as the base speed instead.

So: we're in agreement so far.
However stipulating the base speed to be lower, at whatever horsepower, similarly implies that you are voting for a motor with a correspondingly higher torque and cost.
Yes, I suppose so. But it seems to me that for successful direct drive, you need good low end torque, so you are stuck with a large, heavy motor, compared with what you could get away with for non-direct drive. (The Prius is direct drive, as far as MG2 is concerned; there is a fixed ratio between it and the wheels.)
During this short acceleration ramp, power would have increased steadily on its way towards base speed thus describing a triangle with the area enclosed representing the power delivered.
Actually, the area under the power-time curve represents energy. Power is just the height of the curve at a particular point.
... This is the rectangular power profile produced by constant power. Being limited in this case by the ampacity of the battery pack.
The limit may not be the battery; it could be that the battery can dish out the power, but the controller has run out of volts, so it's the current limit of the inverter that limits the power. The effect is the same; power is clamped at a limit.
It is, however not so good for the motor because the low base speed has forced the use of a much larger frame size motor. And not so much for the battery either which will not come to full current until base speed is reached.
It seems you agree that it's best for the battery to deliver as much power as possible as soon as possible. However, that means that a lower base speed would be better; you get to maximum power sooner, and spend less time in the low power region where the battery isn't delivering all that it could.
There is a strong likelihood the motor will not be able to provide full power beyond twice base speed. A point where power may be needed the most.
That depends entirely on the motor. IPM motors, as used in the Prius and most commercial EVs, seem to produce at least 80% of peak power at and beyond twice base speed.
That is where we differ. At low speeds I say we certainly need to go for high torque yes, but high power no.
But power is the product of torque and speed. If you are saying that we want high torque but we don't want high power in the first few seconds of launch, then you are saying we don't want much speed at launch.
This introspection demonstrates the importance of achieving full power by 40mph and following thru to 60mph.
Right. It is good to have full power by 40 mph or earlier, so we want base speed to be lower, so that we are at or near full power at that speed. Unfortunately, I think you meant to write that you want full torque at 40 mph, and that to have the highest torque at that speed, you would want base speed to be higher. But this is faulty thinking.

Assuming a constant power motor, (otherwise, it's not a fair comparison), varying the base speed will NOT affect the torque at 40 if the base speed is less than 40. If you push the base speed beyond 40, then torque will actually be lower at 40 than it would have been for a motor whose base speed is 40 or lower.

I'll attempt a diagram to show this:

Image

So the red line represents the peak power of the motor on a torque-speed diagram. I'll consider two motors, both with the same peak power (equal to the red line), one with a base speed of 25 and another with a base speed of 50. The units of speed don't matter; you can consider them to equate to miles per hour if you want; we stopped using miles per hour over 40 years ago. So the first motor, with base speed at 25, produces 100 units of torque up to speed 25, then the torque drops off. The other motor makes only 50 units of torque, but that torque continues out to speed 50. Which is better for acceleration at 40? It's the one with the lower base speed. The other motor has torque that doesn't start decreasing till later, but until that speed, its torque is lower all the time than the other motor.
... Then, when the voltage on the motor is only a percentage of the max voltage we can draw an equivalent percentage of current extra, in order to compensate. It is open to speculation whether Curtis supplies a 650A 96Vdc AC controller with the underlying intention to allow the battery to come to full current well before base speed is reached.
I think you may be mixing up motor and battery current here. When the motor speed is low and the battery current is low, if the controller is going flat out, the motor current will be at maximum. It's the motor current that is the controller's limit; all the motor current has to go through the IGBTs. So there is no "current extra" at low speed. You might be able to force more torque from an induction motor, by just increasing slip at low speed, but I don't think you can do that sort of trick with an IPM motor. I'm on shaky ground here, so experts, feel free to remind me how "torque multiplication" works in an industrial controller.

Edit: Ah! I've just realised that I'm assuming that it is fair to compare motors of the same power; you will no doubt claim that since peak torque is proportional to motor size, I should be comparing motors with the same peak torque. I'll consider what would be a fair comparison and get back.

Certainly, if the two motors start with the same constant torque, then the higher the base speed, the higher the peak power of the motor. I don't think you can get power "for free" like that.
Last edited by coulomb on Thu, 28 Jun 2012, 09:13, edited 1 time in total.
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How to convert a hybrid

Post by coulomb »

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In defence of my position, I noted the other day that the motor constant which defines volts per hertz also defines torque per amp. In fact, in the metric system (if you use radians per second for speed), the constant is the same value (volts per radian/s or newton-metres per amp).

So a motor with a higher base speed, and everything else the same, has a lower motor constant: so less back emf at a given speed, so at a given speed, you need fewer volts to overcome the back emf. The lower motor constant would then imply proportionally lower torque per amp, hence the lower starting torque for the same current controller.

But then again, if you really could get more power from the motor by "just changing its base speed", you would need a higher current controller, so you could get the higher starting torque after all.

So the question comes down to this: for the same motor frame, if you "just rewind it", do you merely change the motor constant, or can you somehow get more power from the motor by lowering its constant?

I note that if you double the base speed, you halve the motor constant, so you'd double the current needed for the same torque. That would increase the I^2.R losses by a factor of 4, while only increasing the motor power by a factor of 2. So you double the copper losses (per unit power output) this way.
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How to convert a hybrid

Post by weber »

I'm completely lost. What are we trying to optimise? What are we allowed to vary? What must we keep constant? Are we talking about induction motors made from unobtainium, with a 1/f torque dropoff region (constant power) at constant voltage, or induction motors we can actually buy, which go straight to 1/f^2 torque dropoff (1/f power dropoff)?

Edit: What's with the "mph"? I thought Canada went metric in the early 1970's like Australia.
Last edited by weber on Thu, 28 Jun 2012, 18:45, edited 1 time in total.
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How to convert a hybrid

Post by T2 »

- Coulomb I've composed a post for your 11.28pm entry which I'm holding back since your 11.06pm post really needs to be answered in a couple of places, not to do so would be impolite.

I think a lot of where we differ is in the engineering tradeoffs. Some of the things I have written I can't say I was initially comfortable with either but that is because I was unaccustomed to those ideas and most other people including yourself will undoubtably feel the same.

It's probably not always obvious but I try hard to stay on subject and not ramble but I am reaching a stage here where making succint and accurate statements is again becoming time consuming to the point where I am beginning to display symptoms of OCD. I know going forward we will have to pick our battles but the fact I glossed over that energy, for example, is in fact the integral of power is not going to be a showstopper in the grand scheme of things but thanks for being on the sidelines anyway. So I am going to review everything you wrote (11.06pm) and get back to you later next week. In the meantime we are not excluding anybody who wishes to join in the fray. Happy Holidays.
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