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Overunity Machines Forum



Dia. Mag. Alternator

Started by z.monkey, May 27, 2010, 07:34:19 AM

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0 Members and 3 Guests are viewing this topic.

z.monkey

I've been trying to find low voltage bulbs that fit the Edison base.  The lowest voltage bulb that I could find local was 24 Volts.  It does light the bulb but not to full intensity.  I have some other 6.3 Volt bulbs but they are a bayonet base, and I would have to take the cool Edison base off my load jig...   Wah...

Video of the Light Bulb Test...
http://www.youtube.com/watch?v=O4iHdNWo86k
Goodwill to All, for All is One!

gyulasun

Hi z.monkey,

What is the DC resistance of your present coils? Also, if you have an L meter at hand, would be good to know the coils average inductance (on average I mean its change when rotor is slowly rotated).
I ask these because knowing these data a good optimum load impedance could be estimated that could insure a power match Zgen=Zload to get the highest output at a given rpm. The 24V light bulb gave a light load at random but if you use a 6.3V bulb its brightness will still depend on its wattage rate.
I guess you surely know these.
Excellent job by the way!

Gyula

z.monkey

Thanks Gyula,

Um, lets see, the coils are ~2.5 Ohms each, connected in series, out of phase 180 degrees.

I don't have an inductance meter, but I do have the magnet wire spec sheet...

http://www.belden.com/techdatas/english/8051.pdf

It give us the specific resistance of 22 AWG wire which is 16.2 Ohms per 1000 feet.

16.2 Ohms divided by 1000 feet is 0.0162 Ohms per foot...

5 Ohm coil divided by 0.0162 Ohms per foot is ~308 feet of wire divided by 2 coils = 154 feet per coil...

The data sheet also gives us linear turns per inch = 37.5, and turns per square inch = 1406...

The cross sectional area of the winding space is 0.5 inch x 0.75 inch = 0.375 square inches...

So, the number of windings we can fit in that space is ~ 527 windings per coil...

The core is Steel (Hard Iron) so its permeability is 8.75×10âˆ'4 uH/M (microHenries per Meter)

The core is an inch square, but the corners are rounded off, so I give the core area 0.95 Square Inches...

Convert that to metric and we have 2.413 Square Centimeters...

We plug all of this into an inductance equation...

Inductance = (permeability x number of turns squared x cross sectional area) divided by length...

Inductance = (0.00875 uH x 259091 turns2 x 2.413 cm2) / 1.27 cm = 4307 microHenries, 4.397 milliHenries, or 0.004397 Henries...

Of course my equation is not dynamic like a meter would be...

I may have to put that L-Meter on my Christmas list, good idea, thanks again...
Goodwill to All, for All is One!

z.monkey

Doh!

I even Forked Up the Equation...

"The core is Steel (Hard Iron) so its permeability is 8.75×10âˆ'4 uH/M (microHenries per Meter)"

s/b 8.75×10âˆ'4 uH/M, and...

"Inductance = (0.00875 uH x 259091 turns2 x 2.413 cm2) / 1.27 cm = 4307 microHenries"

s/b Inductance = (0.000875 uH x 5272 turns x 2.413 cm2) / 1.27 cm = 430.7 microHenries...

There are a lot of caveats, nuances and idiosyncrasies to this induction stuffz...
Goodwill to All, for All is One!

nievesoliveras

That generator is giving you a hard time.