[MN Driver]

Originally Posted by Tesseract

You know exactly why I did this, don't you?

I do. It shows that the Warp 9, and likely pretty much any series DC requires a very high battery voltage to get the RPM needed for 1000 amps of battery current to be matched. It seems it could possibly be enough to add another to your count of motors versus Soliton's. I only recently realized that a large portion of the EV community doesn't understand the dynamics of back EMF and the relationship of motor voltage and motor current being different than battery current. Oh well, its something that will be understood soon enough.

I am very interested at what you find the amperage saturation point of the Warp 9 to be at. Would this be the most efficient point of operation, or would RPM/voltage be a larger factor? I'm looking to operate at the most efficient point possible when I complete my conversion and am thinking that the Warp 9 might be too big for a highway cruising ~2500 pound car to get maximum efficiency but I could be wrong, I'm not sure on the specifics of efficiency and its what I'm trying to figure out right now. I wouldn't mind flying down the road with a bigger motor than I need though and it seems the differences may be negligible.

[Tesseract]

Originally Posted by MN Driver

...I only recently realized that a large portion of the EV community doesn't understand the dynamics of back EMF and the relationship of motor voltage and motor current being different than battery current....

Yep - and it's depressing how vocal some are in their ignorance. Not that I'm the least bit bitter, angry or frustrated...

Originally Posted by MN Driver

I am very interested at what you find the amperage saturation point of the Warp 9 to be at. Would this be the most efficient point of operation, or would RPM/voltage be a larger factor?

I just posted this data around the same time as you posted, but saturation seems to occur around 400A, and it appears to be rather abrupt as major said, rather than a "gentle bend" as I predicted.

I'll defer to major on your second question, but I would *expect* for the RPM per volt ratio to go up quite a bit once in saturation because the ratio of flux per amp went down quite a bit.

[major]

Originally Posted by Tesseract

It appears that the WarP-9 goes into saturation (changes over from torque being proportional to the square of current to being linear) at 400A. The direct pressure readings from our hydraulic dyno:

200A 1014 rpm 24.7V <200 psi <- too low to read on gauge
250A 1010 rpm 28.7V 325 psi
300A 1010 rpm 31.6V 500 psi
350A 1015 rpm 34.4V 625 psi
400A 1008 rpm 37.6V 850 psi
450A 1015 rpm 39.0V 900 psi

I'll try this.


I guess I pasted it in here Ain't the smoothest sat curve I ever saw, but not bad. Looks like it just starts to saturate about 200A and continues to "knee" over thru 400A. Which one would expect with straight sided slots. Also, this isn't a true representation of the saturation characteristics because we have resistive voltage drops mixed into the voltage measurements. So we're not looking at the generated voltage alone. This is why saturation curves are most often taken at no load.

But interesting data Tess Not sure what good it does anybody

major

[major]

Originally Posted by MN Driver

I am very interested at what you find the amperage saturation point of the Warp 9 to be at. Would this be the most efficient point of operation, or would RPM/voltage be a larger factor? I'm looking to operate at the most efficient point possible when I complete my conversion and am thinking that the Warp 9 might be too big for a highway cruising ~2500 pound car to get maximum efficiency but I could be wrong,

Hi MN,

I don't think you need to worry about saturation when considering motor efficiency in your context. Maybe if you were over exciting a SepEx or something like that. But "cruise" efficiency will best be maximized by choosing the proper ratio so the loading and RPM are "right" for the motor. And if the "cruise" power level is under the rated motor power, it means the motor 1) should not overheat and 2) should run efficiently.

Until you're very close to no load, if your RPM is reasonable, the larger motor should be more efficient than a smaller motor because the iron is stressed less and winding resistance is less. There may be a little more friction and windage with the larger motor, but the previously mentioned factors should offset that. Also the larger motor should run cooler, and cooler copper has less resistivity.

Regards,

major

[major]

Originally Posted by MN Driver

It shows that the Warp 9, and likely pretty much any series DC requires a very high battery voltage to get the RPM needed for 1000 amps of battery current to be matched.

IMO, it is the opposite. When you have a high voltage battery, your controller will likely always be reducing voltage to the motor and therefore motor current will always be higher than battery current (except when = 0). Take Tesser's last test there. If you extrapolate you would see a 1000A at maybe 45 to 50 volts. So if one had a 48V or lower V battery, the controller would go 100% dc at the 1000A CL whereas if you had a much higher voltage battery, it would not. Of course that was at 1000 RPM. So you could double the voltage at 2000 RPM and see 100% dc at CL or triple V (to 144) and see 100% dc at CL for 3000RPM. But if you have a 300V battery, 100% dc at CL would occur at or above 6000RPM, so your battery current would never equal motor current (except at 0 dc) because you're well beyond the zorch limit

Well, that's pretty confusing I hope you can follow it.

major

Just a bit for all those who have decided to NEVER go back to see what's up with Jack.

Jack Rickard said...
My my my my my. Such a tremendous number of very strongly held opinions. Well, well, well, well....

I'm having a very happy end to a marvelous day thank you. And truly delighted to meet the EV Whisperer, Mr. Sebastien Bourgouis. We had a lovely day, and thanks to his very rapid analysis, we were drinking whiskey and partying lock Rock Stars by two in the afternoon.

Hate to give away this week's show, but I can't read any more of this battery amp/motor amp theory nonsense that of course never was and never could have been the issue. The motor loading issue was interesting, but as I noted, rather paints us into a corner of never getting well even with a nuclear power plant driving a 12000 lb motor.

Apparently, if the calibration of the hall effect sensor INSDIE the Soliton is off, it is quite possible for the Soliton to be entirely convinced it IS putting out 1000 amps, while the rest of the real world, including the fat kid, is kind of convinced it's more like 722. In that event, the device has a cunning strategy for self preservation, it simply limits itself to 1000 amps, even if they are not real.

If you replace that unit with a DIFFERENT Soliton1 from the magic bag, that does NOT have this calibration issue, guess what you get on the very first test drive?

A TOTALLY SMOKED Stage 2 clutch and pressure plate assembly. I mean NASTY SMELLING and pretty much destroyed.

Throught the mircacle of time lapse photography, we should have a new CB Performance kevlar double sided ceramic trick double throw me down stage 4 unit with 4000 lb pressure plate by AM tomorrow.

0-60 times to follow.

Quite a relief. And quite an impressive show on the first run.

Jack Rickard
March 31, 2011 11:10 PM


More is posted on the blog for you all to read.

[jddcircuit]

Originally Posted by Tesseract

200A 1014 rpm 24.7V <200 psi <- too low to read on gauge
250A 1010 rpm 28.7V 325 psi
300A 1010 rpm 31.6V 500 psi
350A 1015 rpm 34.4V 625 psi
400A 1008 rpm 37.6V 850 psi
450A 1015 rpm 39.0V 900 psi

(HP = RPM x PSI x 0.0000182)
.

Does this mean that you are spending about 1kW per HP? Would that equate to a little over 60% conversion efficiency? I am interested in trying to gather similar data for another motor-controller combination to compare efficiencies.

[major]

Originally Posted by jddcircuit

Does this mean that you are spending about 1kW per HP? Would that equate to a little over 60% conversion efficiency?

Hi jdd,

One HP = 746 watts. So Power out / Power in = 746 / 1000 = 74.6 %. Not too shabby for efficiency of controller, motor and pump at these loads, RPM and voltage.

I did not calculate the power out, just commenting on your 1kw/HP statement.

Regards,

major

[jddcircuit]

Originally Posted by major

Hi jdd,

One HP = 746 watts. So Power out / Power in = 746 / 1000 = 74.6 %. Not too shabby for efficiency of controller, motor and pump at these loads, RPM and voltage.

I did not calculate the power out, just commenting on your 1kw/HP statement.

Regards,

major

thanks. I was calculating wrong.

[Tesseract]

Originally Posted by major

IMO, it is the opposite. When you have a high voltage battery, your controller will likely always be reducing voltage to the motor and therefore motor current will always be higher than battery current...

Loathe as I am to disagree with you, I think that it's worth pointing out that few battery packs appreciate having 1000A sucked out of them (much less the 2000A the Z2K could do), but most cars would benefit from having full torque beyond 3000 rpm, so I'd say it is definitely worthwhile to have more than 200V on tap even with an advanced motor like a WarP-9.

'Course, I am somewhat biased here...

Originally Posted by gottdi

Just a bit for all those who have decided to NEVER go back to see what's up with Jack.....

While I obtained the data reported in this thread to support the argument that Jack Rickard had run out of voltage, I didn't start this thread as a rebuttal to Jack - it was strictly to share the data because I figured others would find it valuable.

It turns out Jack was right all along, but he never did tell me what the motor amps were. Seb had to fly up there to make that measurement in person, but then he was able to diagnose the likely culprit immediately.

We don't claim to be perfect, and we don't claim our products are perfect, but we are honorable people so when something goes wrong we try our damnedest to make things right.