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Overunity Machines Forum



GENERATOR- YOU DO THE IN/OUT POWER MATH

Started by magnetman12003, April 19, 2010, 09:16:15 AM

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0 Members and 10 Guests are viewing this topic.

DeepCut

Ah brilliant Mark, thanks :)

I will try that and post back results.

*EDIT* I was wondering if there is a limit to this setup. Tom has 3,200 feet of wire in his inductive coil but surely there is a critical threshold where a certain amount of input voltage an't excite beyond a certain length of wire ?

I've googled 'limits of electromagnetic induction' and the like but with no joy as yet.


Gary.

mscoffman

Quote from: DeepCut on May 09, 2010, 05:26:31 PM

...
*EDIT* I was wondering if there is a limit to this setup. Tom has 3,200 feet of wire in his inductive coil but surely there is a critical threshold where a certain amount of input voltage an't excite beyond a certain length of wire ?

I've googled 'limits of electromagnetic induction' and the like but with no joy as yet.


Gary.

Oh please, you're begining to sound like Joe Newman!  :)
Next you'll be saying that scotch tape can't be a fuel. ;)

DeepCut

Quote from: mscoffman on May 09, 2010, 09:25:11 PM
Oh please, you're begining to sound like Joe Newman!  :)
Next you'll be saying that scotch tape can't be a fuel. ;)

::W H O O S H:: ... that's the sound of what you just said going right over my head ... ;+}

gyulasun

Quote from: DeepCut on May 09, 2010, 05:26:31 PM
...
*EDIT* I was wondering if there is a limit to this setup. Tom has 3,200 feet of wire in his inductive coil but surely there is a critical threshold where a certain amount of input voltage an't excite beyond a certain length of wire ?
...

Hi Gary,

As long as the magnetic field is able to penetrate the coil's volume and 'cut' the wires, you will get induction hence induced voltage. Of course higher volume coils need stronger fields to penetrate because of the higher coil sizes, both in diameter and in length. And stronger fields need higher input power, this is one side of the coin.

The problem is not really what you ask but the increasing lenght of wire...   Here is a link to copper wire resistances in the function of wire diameter and length: http://www.thelenchannel.com/1wire.php

for instance, choosing AWG 30 Solid wire, the chart says this gauge has 113 Ohm resistance for 1000 feet. You mentioned Tom's 3200 feet long wire and if he uses this AWG 30 wire (a diameter of 0.25mm) he will have about 113 x 3.2=361.6 Ohm DC resistance for the total coil. 

This means that if he wishes to load this generator coil and the load would take 0.1 Amper as the load current, then the voltage drop just due to the long wire would be V=I x R= 0.1 x 361.6= 36.16V, ok? 
This is just the voltage loss that would never reach the load, it dissipates into heat in the wire  (the heat loss in Watts is I2 x R, so 0.1 x 0.1 x 361.6= 3.616 W). 

So the longer the wire you wish to use for the generator coil, the higher diameter you wish to use to reduce copper loss but this involves higher volume coils and higher cost and this is an ever-increasing problem: you are forced for a trade-off.  Conventional generators surely use as thick wires as the output power demand dictates, to keep the physical sizes of the generator and the heat losses at a compromised and still acceptable level. (Of course I am not a generator designer, I just think on this problem.)

You could say: what is the problem once Tom has more than a 100V output voltage, so 36V loss from it may not be a big problem. 

Well if this nearly one/third voltage loss would be indeed THE loss, than I do not think this loss would be a big problem,  however you have to consider Lenz law which is the Number 1 enemy here, as you have already experienced.  All you wanted to do is to take out at least as much current as you invested at the input (about 75mA in your case @ 9.5V or so) and your magnet did not even start to rotate or came to a halt as soon as you used a REAL load of a 100 Ohm resistor.  On the word REAL I mean a load which is present across the voltage source in every and each moment of the AC cycle.
LED diodes do not behave in the same way as a load in the sense as a normal resistor does.  For instance LEDs are diodes and as such they conduct current when they sense forward bias polarity (any moment when their anode is positive with respect to their cathode). This can only happen in the half cycle part of a full AC cycle, ok? 
Next problem with the LEDS is they have a certain forward voltage threshold under which they do not conduct even if the polarity is correct: for instance a white LED needs any voltage between from 3 to 3.4V forward voltage to start conducting and give out light.  IF you connect them in series, than you have to multiply this 3 to 3.4V with the number of diodes in series, like 20 pieces of LED do not draw any current under 20 x 3V=60V. In case of a sinusoidal output voltage that suppose has 120V peak to peak value, the 20 LEDs in series would not conduct whenever the instantenous voltage value is under 60V across them even if the voltage polarity is correctly forward biases them in the appropiate half cycle of the full AC waveform. And you can calculate power consumption only when the LEDS conduct!
So you see what difference LEDs make as a load wrt a resistor and how decisive it can be.

rgds,  Gyula

DeepCut

Thank you Gyula, that clears a lot up in my mind.

I am reading two fantastic books at the moment :

Introduction to Magnetic Materials, by B.D. Cullity and C.D Graham,

and,

Practical Electronics for Inventors, by Paul Scherz.

I hope to know a lot more by the end of this year than i did at the beginning :)


Gary.