Electric Motor Secrets

[gyula]

...
I just tried it and they both agree on 2.7amps

Thanks

Robert

Hi Robert,

Because the bulb was connected to the right hand side, A1 measured the total input (1.8A) and did not measure the sum of A2 and A1 (2.7A), right? So the input power can be calculated as 50V*1.8A=90W in this test. So I agree with your test.

HOWEVER, in your post #1584 you wrote this:

"If I put the bulb wire before A1 , I get 3.55a on A1 which means the effective power going in the motor is 130v x 3.55a =461 watts ??"

And this is the input power calculation what I do not agree with because in that setup you considered the SUM of A1 and A2 when figuring out input power.
First the bulb was connected to the right hand side of A1 and input current was A1=2.2A from V1=130V and A2=1.35A from V2=230V. So in this latter case input power was 130V*2.2A=286W, ok?

And when you put the bulb wire before A1 (i.e. to the left hand side of A1) then A1 showed 3.55A which is okay that it showed the sum of the earlier values A1 and A2 but you cannot use this summed current for calculating input power: this is what I do not think is correct. How would input power change from 286W to 461W by merely connecting the bulb wire before a current meter? A good current meter must be a short circuit for the current, it cannot cause such a huge change in power.
Is this okay? Maybe you meant something else with your question and I misunderstood.

Thanks, Gyula

[Robert49]

Hi gyula

I did not mean "input power" . I meant effective power inside the motor.

I think we agree on everything without knowing. To make an analogy, we are in the same car going in the same direction but one of us is looking through the windshield and the other through the rear window.

In any case I think I should do the torque test as Peter Lindemann showed in his video .

Thanks

Robert

[gyula]

Oh, indeed, you used effective power inside the motor and somehow my mind read it as "input power" ... my bad.

The power inside the motor indeed increases whenever you utilize the collapsing field of the coils because the current from that field flows also through the coils, not only via the bulb load.

Earlier you wrote that "this motor has no dead time: there's always current flowing into it because the pulses are overlapping."
Was this so also in your latest tests too? I mean by manipulating this and the on time you could also increase efficiency.

Gyula

[Robert49]

Oh, indeed, you used effective power inside the motor and somehow my mind read it as "input power" ... my bad.

The power inside the motor indeed increases whenever you utilize the collapsing field of the coils because the current from that field flows also through the coils, not only via the bulb load.

Earlier you wrote that "this motor has no dead time: there's always current flowing into it because the pulses are overlapping."
Was this so also in your latest tests too? I mean by manipulating this and the on time you could also increase efficiency.

Gyula

Gyula

After I posted "no dead time" I modified the timing wheel and now there's a 5 degree dead time. I had to do this because the overlapping pulses were creating drag and the amperage was too high.
I intend to make a variable timing wheel so I can adjust it.

Thanks

Robert

[Peter Lindemann]

Great Work!!

Hi gyula

I did not mean "input power" . I meant effective power inside the motor.

I think we agree on everything without knowing. To make an analogy, we are in the same car going in the same direction but one of us is looking through the windshield and the other through the rear window.

In any case I think I should do the torque test as Peter Lindemann showed in his video .

Thanks

Robert

Robert,

Sorry I've been gone a few days. A lot has happened in this thread. 69% recovery in your tests is very good and on the high side of what I was saying is to be expected. As you move toward the dynamometer test, just realize that this motor has a completely different speed/power curve than a standard DC traction motor. Your best power measurement will NOT BE at 1/2 of idle speed, but will probably appear at a higher speed. Start by loading it to about 80% speed and see what you get. Then load to 70% speed, then 60% speed, etc. Mechanical energy production is inversely proportional to air gap, so cutting the air gap in half will double the torque (as a general rule).

Currently, your air gap is quite large for a motor operated on magnetic attraction, as .5mm is about 0.019 inches. Torque would be 4 times more if the total gap was about 0.005 inches. In the Flux Motor model we built in Santa Barbara in 1983, we were testing air gaps in the 0.002 range with extremely excellent results

I love your design and implementation, though.

Best regards,
Peter

[Robert49]

Robert,

Sorry I've been gone a few days. A lot has happened in this thread. 69% recovery in your tests is very good and on the high side of what I was saying is to be expected. As you move toward the dynamometer test, just realize that this motor has a completely different speed/power curve than a standard DC traction motor. Your best power measurement will NOT BE at 1/2 of idle speed, but will probably appear at a higher speed. Start by loading it to about 80% speed and see what you get. Then load to 70% speed, then 60% speed, etc. Mechanical energy production is inversely proportional to air gap, so cutting the air gap in half will double the torque (as a general rule).

Currently, your air gap is quite large for a motor operated on magnetic attraction, as .5mm is about 0.019 inches. Torque would be 4 times more if the total gap was about 0.005 inches. In the Flux Motor model we built in Santa Barbara in 1983, we were testing air gaps in the 0.002 range with extremely excellent results

I love your design and implementation, though.

Best regards,
Peter

Peter

I agree with you on the power curve being better at higher speed because I feel it when I slow it down.
I have measured the air gap at .010 inch. This was a 220 volt 1400watt ac conveyer motor which I modified .
If it has a too big air gap, I will have a new rotor made by my good friend Markus who has a machine shop. I repair his machines and he makes my stuff: nice exchange!
I'll watch your video again to repeat the dynamometer test and give you resuts when I have them.

Thanks for your help Peter

Robert

[Peter Lindemann]

New Rotor

Peter

I agree with you on the power curve being better at higher speed because I feel it when I slow it down.
I have measured the air gap at .010 inch. This was a 220 volt 1400watt ac conveyer motor which I modified .
If it has a too big air gap, I will have a new rotor made by my good friend Markus who has a machine shop. I repair his machines and he makes my stuff: nice exchange!
I'll watch your video again to repeat the dynamometer test and give you results when I have them.

Thanks for your help Peter

Robert

Robert,

If you are going to build a new rotor to close the gap (an excellent idea) then you may also wish to take a good look at the bearings in the system and consider replacing them with precision bearings. Production AC motors use large air gaps and cheap bearings to make sure the rotor never rubs the stator while keeping the costs at an absolute minimum.

As you close the gap, the attraction forces get extremely strong and will eat bearings that have any play in them. You may also have trouble running tests because the rotor may seem centered when the magnetic fields are off, but still rub during operation.

Start with a good SS shaft and an over-sized section of iron for the rotor. Use the full round shape. Mount the iron on the shaft and switch to the lathe. Shave down to the diameter that gives about 0.002 air gap and turns without touching when mounted in the stator. As you do this process, you may also start to identify if there are any eccentricities in the stator fingers.

[Obviously, doing a precision boring operation on the mating surface of the stator is to be avoided if possible. The reasons for this are many, but that operation is best accomplished with either a rotating grinder bit or a diamond cutter in a boring tool, taking extremely small slices at a time, keeping in mind that this is an intermittent cut on a silicon covered steel material.]

Next, switch to the mill and shave off the side sections to produce the "bar shape" the rotor needs to be. This will leave you with an issue of critical balance that will need to be addressed, but precision machining can get you close enough for early tests. The alternative is to make the bar shape first and mount it on the shaft, but that leaves you with an intermittent cut operation on the lathe, which is much more problematic and not recommended.

Once you solve that set of problems, you may also want to add back two plastic side pieces to the iron section so you don't have to push all of that air around. You'd be surprised how much drag turbulent air can cause at 4000 rpm.

So you see, "replacing the rotor" is going to be a process, not simply a one-time event.

I hope that helps.

Best regards,
Peter

[Robert49]

Peter

Thank you again.
My machinist is quite busy right now and i'm leaving for florida in a week or so for two months vacation. So you are not about to see the results of that.
But I will get to it.

By the way, my german friend Heinrich Liebner who also fiddles with bedini motors was in germany during the war and saw the original "Lochridge device" inside an underground bunker. they cranked the thing with a pull chord and after a few minutes they flipped a switch and lights on.
It really worked. He was about 12years old.
Thought I'd share that with you.

Best regards


Robert

[Peter Lindemann]

Life Interrupts Science....LOL

Robert,

I was just sharing some ideas with you, based on some of my past experiences. I'm sure your model will come together in its own time.

By the way, thanks for the report of the "Lockridge" unit working back in Germany, all those years ago. It has always been obvious that it was a real machine that worked, but its nice to hear of an eye-witness report.

Have a great vacation!

Best regards,
Peter

[Robert49]

Thank you Peter

More of the same.
As per diagram on post 1574,

V1 = 170v
A1 = 2.7a

V2 = 308v
A2 = 1.55a

V3 = 138v
A = 1.55a

rpm = 5000

It has a beautifull clean high pitch sound. I love it !!

Robert

[Robert49]

unexpected result

Hi friends

I have just finished winding another stator for the motor. It now has 150 turns #22 wire per set of coils instead of 120. nothing else has changed.

The motor takes less power to run and returns more power than before.
I think I should try to modify the first stator so I can get it to 200 turns per set of coils.

Robert

[gyula]

Hi Robert,

The 150 turns (instead of the 120) has certainly increased the coils L inductance, this involved an increase in their inductive reactance (XL) so this explains the less input power draw.

The L/R time constant for the coils has also increased and depending on how the previous L/R time constant and the previous switch ON time were related, the increased time constant may have shifted the ON time to the (more favorable) initial steeper section of the exponential charging curve of the coils so that the captured energy from the collapsing field has been increased. This latter is a speculation from me because I do not know yet whether V2 voltage has also increased or stayed about at the same level? You may already have some measured V1 A1, V2 A2 and V3 A numbers with the new stator coils.

Of course, the ON time could be seen on a scope and be compared to the L/R time constant, this would involve to check the average L inductance of the new coils [R includes not only the coils DC resistance (two coils in series at a time) but the ON resistance of a switch and the inner resistance of the V1 power supply, these latter two resistances may be neglected if they are much less than that of any two series coils].

EDIT: I did not consider your words: "and returns more power than before" when I answered I had not known whether V2 voltage increased, now it is sure V2 increased and this means my 'speculation' with the time constant - ON time relation has more merit. If this proves to be so, then the importance of finding the optimum ON time for an actual setup is justified i.e. making the ON time adjustable is a must.

rgds, Gyula

[Robert49]

Hi Robert,

The 150 turns (instead of the 120) has certainly increased the coils L inductance, this involved an increase in their inductive reactance (XL) so this explains the less input power draw.

The L/R time constant for the coils has also increased and depending on how the previous L/R time constant and the previous switch ON time were related, the increased time constant may have shifted the ON time to the (more favorable) initial steeper section of the exponential charging curve of the coils so that the captured energy from the collapsing field has been increased. This latter is a speculation from me because I do not know yet whether V2 voltage has also increased or stayed about at the same level? You may already have some measured V1 A1, V2 A2 and V3 A numbers with the new stator coils.

Of course, the ON time could be seen on a scope and be compared to the L/R time constant, this would involve to check the average L inductance of the new coils [R includes not only the coils DC resistance (two coils in series at a time) but the ON resistance of a switch and the inner resistance of the V1 power supply, these latter two resistances may be neglected if they are much less than that of any two series coils].

EDIT: I did not consider your words: "and returns more power than before" when I answered I had not known whether V2 voltage increased, now it is sure V2 increased and this means my 'speculation' with the time constant - ON time relation has more merit. If this proves to be so, then the importance of finding the optimum ON time for an actual setup is justified i.e. making the ON time adjustable is a must.

rgds, Gyula

Gyula

Here are some numbers for you.
Take into consideration that I play a lot with the timing and that changes my voltages but
look at the difference between power in on V1 and power out on V2 line.
That's where the change is and the motor is a lot stronger.
V1 = 178v
A1 = 1.8a
V2 = 285v
A2 = 1.4a
V3 = 112v
The inductance varies from 4.65mh to 8.85mh over all but the pulse comes on at 7.0mh and rises to 8.0mh where the pulse stops.
The rotor is wide enough to catch 3 coils at a time.
I agree with you the on time has to be adjustable and it will be.

Thanks Gyula. I appreciate all the help I can get on this forum.

Robert

[gyula]

Hi Robert,

Thanks for the details.

You mention the difference between power in on V1 and power out on V2 line but in fact your bulb draws 112V*1.4A=156.8W while power input is 320W and you have had similar out/in ratios earlier?
(It is okay that the motor is a lot stronger while the input current is less.)

Would like to show you a link to an interesting explanation on how to recover the most energy from a pulsed coil:
STEORN DEMO LIVE & STREAM in Dublin, December 15th, 10 AM

Of course, one would need to adopt the ON time to a particular setup, with minimal trade-offs, to approach the suggested 1/8*Tau i.e. L/8R time. (I did not check this with measurements but it makes sense to me.)

Gyula

[Robert49]

Hi Robert,

Thanks for the details.

You mention the difference between power in on V1 and power out on V2 line but in fact your bulb draws 112V*1.4A=156.8W while power input is 320W and you have had similar out/in ratios earlier?
(It is okay that the motor is a lot stronger while the input current is less.)

Would like to show you a link to an interesting explanation on how to recover the most energy from a pulsed coil:
STEORN DEMO LIVE & STREAM in Dublin, December 15th, 10 AM

Of course, one would need to adopt the ON time to a particular setup, with minimal trade-offs, to approach the suggested 1/8*Tau i.e. L/8R time. (I did not check this with measurements but it makes sense to me.)

Gyula

Hi Gyula



The power dissipated in the bulb varies with the timing of the motor.
But if you look at the overall picture , I may be wrong but I think there is a lot more power in the system than I put in.
I adjusted the timing as to get the least input amps.
I have now put three 200w light bulbs in parallel and the result is:

V1 = 173v A1 = 1.65a = 285.45w
V2 = 220v A2 = 2.7a = 594w of which 173v x 2.7a = 467.1w goes back to the motor.It is now very strong compared to the 285w input.
V3 = 47v
I also added an ampmeter directly on the motor input and it reads 4.7a at these settings.

Now in this next test I have connected one 200w bulb across the cap bank.
I got thesenumbers
V1=96v A1=3.2a = 307.2w input.
V2=135v A2=1.6a = 216w on light bulb.
Since V2 is higher than V1 the cap bank only gets its energy from the hv spikes .
This means I get 70.3% back.

ps: My knowledge in electronics is very good but very poor in magnetism and all my life I always thought everything was possible. I do not think "inside or outside the box" because for me there is no box exept for the limits of my imagination. I do everything with an open mind. I may be right or I may be wrong but nevertheless I try.

Thank you for your help

Robert