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Overunity Machines Forum



Tesla's Charging Circuit and it's Application to Pulse Motors

Started by Farmhand, June 01, 2013, 05:39:16 AM

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0 Members and 3 Guests are viewing this topic.

Farmhand

Hows this look for a setup ?

http://www.youtube.com/watch?v=0HYCGAUF5mQ

The rotor as used weighs 626 grams, the power was disconnected completely and the generator coils were there but no loads.

Just a test of the test setup. Seems to work well, I'll find a better weight, fairly consistent results too.

Cheers

MileHigh

Farmhand:

That looks pretty good to me.  If I can make one comment it would be that this is a "pure" measurement of the moment of inertia of the rotor so things like pick-up coils and anything that might affect the rotor should be moved away (if possible.)

If you do work with this data and then start doing things like using the moment of inertia to determine rotor power at a given RPM like TK said, then you will be a pioneer among pulse motor testers blazing new trails for others (hopefully) to follow.

MileHigh

TinselKoala

Quote from: MileHigh on June 04, 2013, 05:51:40 PM
Farmhand:

That looks pretty good to me.  If I can make one comment it would be that this is a "pure" measurement of the moment of inertia of the rotor so things like pick-up coils and anything that might affect the rotor should be moved away (if possible.)

If you do work with this data and then start doing things like using the moment of inertia to determine rotor power at a given RPM like TK said, then you will be a pioneer among pulse motor testers blazing new trails for others (hopefully) to follow.

MileHigh

Yes, looks good to me too. Hopefully you can also compare to the straight calculation method too.

But.... it's not exactly completely unexplored territory!
http://www.youtube.com/watch?v=PJavCZX_-PI

Farmhand

Yeah MileHigh, I see, so do I leave the motor coils in place also ? I would remove the generator coils before testing but I would have left the motor coils there but maybe not connected, not really sure, it's a lot to try to take in for someone in my position. I'm understanding better as I go along, keep coming with the suggestions if you think of any gentlemen.  I need all the help I can get, but please be patient also. I think this is something important to learn and useful, I get the feeling it is anyway.

Tinsel, I hope you wear those "dislike tags" as badges of Honor  ;) That's how I think of them, every one means I peeved someone or spoke truth enough to upset some faker or some such other dishonorable person.  :)  You ought to get a lot more badges than me considering your high work rate.  :D

I'll be going back and rereading posts for a while as I work through things, but I'll speak up when I get stuck on something.

Cheers


Farmhand

Ok using the formula  I = 2MGh/w2, with 0.120 Kg falling 0.750 meters producing 404 RPM. I recorded 5 runs and averaged three run results after disregarding the highest and lowest. Not much point to the averaging because after disregarding the high and low result the others were the same save for fractions so I just disregarded those as well.

Frequency in Hz meaning RPM/60 I guess.

I get a value of  I = 0.00098655969

Question what does the number stand for ? :-[ Once one "gets" into the calculation it is actually easy when we have the computer type calculators to do the actual calculating for us. If I did the correct thing or not is another matter.

That is with no generator coils, the power disconnected and the motor coils negatives disconnected from the mosfets but with the return circuit in place, (as it would run down with no power and no switching). Side note is that with the generator coils in place the speed is higher because the cogging was less to begin the rotor to turn, I had to go to more weight when I removed them.  ;)

If I'm wrong I can show on my sketch pad what I did and find out why.

Now to pull out the rotor and carefully measure and weigh it while referring to the engineers toolbox pages.  ;) I'm a serious cheater, but the meaning is understood.

Getting there. The help is dearly appreciated, to get it for free is very fortunate.

Cheers