Electric Motor Secrets

[Jetijs]

Peter, here is the schematic:

I used only one wire strand of the coil and only one transistor. Also the output was a 24V battery and not a light bulb as in the circuit.

[lighty]

You're quite right Peter. I really should be more cautious with my assumptions. The thing is- I cannot easily get rid of the habit of assuming that everybody know what they're measuring.

Not everybody has the appropriate technical expertise you idiot! (that should be my mantra before posting anything).

[lighty]

@Jetijs

Isn't a value of 0.5 Ohm a bit high regarding the total resistance of your coil. You're probably dissipating some energy at the resistor as well.

[Jetijs]

Lighty,
I know I should have used a lower ommage, but I could not get the resistance wire, so I used what I had, a high wattage pot turned to its lowest setting. In fact I did not notice any RPM drop maybe just a little tiny bit

[Peter Lindemann]

Next Step

Lighty,
I know I should have used a lower ommage, but I could not get the resistance wire, so I used what I had, a high wattage pot turned to its lowest setting. In fact I did not notice any RPM drop maybe just a little tiny bit

Jetijs,

Ok, with the resistor where you placed it, it IS measuring BOTH the input current AND the output current. If the resistor had been about .01 Ohms, the wave-form would have been more of a linear ramp than the arching form you see. Never the less, it tells me what I want to know.

The rise-time of the inductance is filling the entire ON-TIME of your commutator.

What we want to do is cut the inductance by at least 75% so we can get about 4 times the RPM. So here is what I recommend.

Unwind the remaining wire from your coil. Take the three strands of wire and fold them in half. Cut all three strands at the half length. Then, take one set of wires and wind them onto a spool and set them aside. Take the other set of three strands and fold them in half again. Then cut them at this new half-way point. Now take all six strands and twist them together along the length until you have at least one full twist per inch of length. Then wind this twisted six wire group back on your core.

Wire up three of the wires to the transistors as the power coils and use the other three as isolated output windings with a common negative and a diode on the positive of each of the three windings. Connect the dedicated output windings to your output battery or light bulb.

This should give you about 60 turns on 6 strands of wire. The wire lengths should be approximately 1/4 the length of your current windings.

Please draw this up and post so I can see that you understand what I am asking.

Also, please remove the 0.5 Ohm resistor from the circuit. It has served its purpose and is too high a value to leave in place as the frequency and power levels rise in your model.

Peter

[elias]

I started a new thread,

Peter,

I started a new thread as you asked (http://www.energeticforum.com/renewa...html#post12929) and deleted my previous posts, but I put the portion of it that in fact is closely related for producing excess mechanical torque in attraction motors here:

http://www.energeticforum.com/12844-post358.html

Any comments on this?

[Jetijs]

Peter, did you mean the circuit should look like this?


I understand about the wire size, we basically want to cut the lenght by 75% (That would be aprox 10 metres of wire) and use 6 strands of twisted wire.
Thanks,
Jetijs

[Jetijs]

Edit: I wound all the wire from the startor core and folded it twice so I got exactly 1/4 of the lenght. I wound 6 strands on a spool and put them aside and I twisted the remaining six strands with a hand drill. Then I wound this thick 6 strand wire on the startor core and the wire lenght was just long enough for 59 turns (almost 60 as you predicted) Here's a picture:


Now I only need you to verify I have drawed the circuit diagram correcltly

[Peter Lindemann]

Minor Modifications

Peter, did you mean the circuit should look like this?


I understand about the wire size, we basically want to cut the lenght by 75% (That would be aprox 10 metres of wire) and use 6 strands of twisted wire.
Thanks,
Jetijs

Jetijs,

Not quite right. Please note that the output coils will have the OPPOSITE polarity of what you have drawn. When the field collapses, the magnetic flux lines cut the wires in the opposite direction as when the electricity is applied. So, if the top of the power coils is POSITIVE, then the top of the output coils is the NEGATIVE and the bottom is the POSITIVE (opposite polarity as the power coils). Also, the NEGATIVE of the output coils goes to the NEGATIVE of the second battery, not the negative of the first battery. Also, the second battery is now ISOLATED and not connected to the positive of the first battery.

Also, did you ever lower the resistor values in the voltage divider run by the photo-transistor? Are they still 2K?

And finally, are you protecting your 2N3055's with neon lights across the emitter-collector junction?

Everything else looks good!

Peter

[Jetijs]

Yes, I lowered the resistor value to 500 Ohm in the phototrigger voltage divider.
There are no neon bulbs right now acros the transistor emmiter and collector. I will add them now. As for the direction of the electricity, I put two small dots on the core in the diagramm, the dots are showing the top of the power and output coils. So If all coils are wound in the same direction should it now look like this?

[Peter Lindemann]

Almost..

Yes, I lowered the resistor value to 500 Ohm in the phototrigger voltage divider.
There are no neon bulbs right now acros the transistor emmiter and collector. I will add them now. As for the direction of the electricity, I put two small dots on the core in the diagramm, the dots are showing the top of the power and output coils. So If all coils are wound in the same direction should it now look like this?

Jetijs,

Did you see the last edit of my previous post?

Peter

[Jetijs]

I was acting too fast and did not see your edited post. I edited the diagram, here it is:

Is it right now?

[Peter Lindemann]

Correct!

I was acting too fast and did not see your edited post. I edited the diagram, here it is:

Is it right now?

Jetijs,

That looks right now. Wire it up and see what it does.

Great work!!!

Peter

[elias]

No one seems to be interested in what I am trying to point out, so ...

Hello

I am trying to point out something very important for increasing motor torque in attraction motors, but nobody seems to be interested in it so I decided to perform some primitive calculations to verify my claim and I found interesting results, very interesting. I performed these calculations at a position where the rotor is halfway slided across the stator, and to obtain more complete results, the value of the mean effective force must be found by integrating it all over the period of the "on state" of the coil. But at least this may be a crude estimation for the mean value of F_effective along its attraction cycle.

I calculated the effective force exerted on the Iron rotor to produce torque in two rotor configurations: The first rotor resembled Jetijs' rotor, which is about 5 units long (each unit is the width of the stator) and the second one only 2 units long.

Consider these:
- alpha is the amount of displacement of the rotor from its alignment position with the stator in radians.
- F_pull = beta * F_slide
- F_effective = F_pull * sin(alpha) + F_slide * cos(alpha)
- F_slide = F
Here is what I found out by simple math:

In the first rotor size (5 units of length):
sin(alpha) = 0.2
F_effective = (0.97 + 0.2*beta)F
beta = 1 then F_effective = 1.17F
beta = 2 then F_effective = 1.37F
beta = 3 then F_effective = 1.57F

In the second rotor size (2 units of length):
sin(alpha) = 0.5
F_effective = (0.85 + 0.5*beta)F
beta = 1 then F_effective = 1.35F
beta = 2 then F_effective = 1.85F
beta = 3 then F_effective = 2.35F

As it is demonstrated by these calculations, by increasing beta (by decreasing the airgap) the effective force increases much more in the second geometry.

beta = 1 then F_effective_2/F_effective_1 = 1.15
beta = 2 then F_effective_2/F_effective_1 = 1.37
beta = 3 then F_effective_2/F_effective_1 = 1.50

I think that the value of beta can be about 3 with a small airgap and thus changing the rotor length from 5 units to 2 units can result in 1.5 times increase in the effective force.

The most interesting observation was that even when beta = 1 which means that F_pull = F_slide, a slight increase in the effective force was resulted (1.15 times).

I hope that this makes someone interested !
Or at least corrects me if I am in error.
I have attached the calculations.


Elias

[Jetijs]

Peter, I hooked everything up. First I tried to run the motor on only one transistor and power coil strand. The motor ran very slow, about 60-70 RPM and the current draw was 0.8A. Then I tried to run the motor with two transistors and power coil strands. This time the motor reached 2000RPM and more, drawing 1.6A. I figured that the first transistor was dead because it could not be that with one coil I have only 70 RPM but adding the second coil made such an improvement. So I tried to run the motor again with only one transistor, but this time I used the second transistor. Also this time I got only 70-80 RPM and the current draw was 0.8A. I left it running for a minute or so and then the transistor blew. I touched ir and almost burned my fingers, it was HOT. So I disconnected it and hooked up the two remaining transistors and two power coils. I started the motor and it went on fast and the RPMs increased slowly. Also I checked the transistors for temperature. They started to get hot very fast, so I stopped the motor, because I did not want to burn them out. But still I could see that the charging battery on the output went up from 24.9V to 25.1V in that short time the motor was running at high speed. I did not notice any neon flash at all in the whole operating time, even when the transistor blew. What could be the problem?