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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 7 Guests are viewing this topic.

DeepCut

Hi.

Am ordering a second magnet tomorrow, same as the current one.

There is just enough room inside my coil for two.

Reading the equations for DC motors it looks to me like RPM and magnetic field strength are important to output so i'm hoping for a generous improvement.


Gary.

gyulasun

Gary,

The question rather is what happens to any scope if a large voltage spike hits its input?
Well, most likely the scope gets damaged, its input circuits probably get 'fried' etc. Normally, USB scopes are designed to "bear" 20-30V maximum (either DC or peak AC), beyond that you have to use 10:1 probes, these lift up the max limit to about 300V, and beyond that you have to use a 100:1 probe, designed for handling max 2500-3000V.

See this simple explanation on probes: http://www.hobbyprojects.com/oscilloscope_tutorial/oscilloscope_probes.html


DeepCut

Ah nice one.

That makes me feel safer, i didn't realise the protection was in the probe :)


Thanks again.


DeepCut

I added the second magnet to the axle but it doesn't turn.

At first i thought that it was the extra mass that was the problem then i noticed it would turn if only one magnet was directly above it.

I've realised there is a kind of a dead-spot where the two magnets meet on the shaft. They are cylinder magnets, diametrically magnetised with a 6mm hole through length for the shaft.

I'm thinking if i had one, long magnet this would work.


Gary.

DeepCut

Someone (i can't remember who) suggested a double-pole double-throw switch to be used to test whether or not this device can self-power.

Would this be the right thing to buy :

http://maplin.co.uk/Module.aspx?ModuleNo=37518

Also, check these test results, it seems an air-core is better than a ferrite one :

INPUT : 9.57 VDC @ 0.83 A = 0.79431 W

OUTPUT : 19.8 VDC @ 0.22 A = 4.356 W

I didn't measure the current but calculated it using Ohm's Law.

The resistance of the coil is 90 ohms therefore the current (I = V/R) is 19.8/90 = 0.22 A.

Am i doing this right, because it's an apparent COP of 5.48 ?


Thanks.