Help, any gurus with alternator experience or knowledge?

Sure that's what a mechanic would do with a car - where one has lots of room and can make a mount or switch pulley sizes etc to make it work.

I, on the other hand, have a crankshaft mounted alternator that must fit in a certain space to keep it out of the rain - and I'd much rather put in permanent magnets and redesign the regulator than rewind the rotor - but I don't own a metal lathe, and I'm not too sure if wood would work - not to mention the hassle of getting the inside taper on the rotor right with the tools I have. This seems like the best option to me.

I have thought of just putting a sheave on the end of the crankshaft, kludging in a mount to hold a John Deere permanent magnet alternator out there in the elements - but there ain't no good way to do it.

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Well I calculated for three sizes of wire, in the same physical space. The coil resistance in going to larger gauges went from 5 ohm to 1 ohm, Power dissipation went from 28 watts to 144 watts. Magnetic force went from 1,415 Gilbert's to 1,417 Gilbert's.

Always the chance I made a mistake somewhere.

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I think so... if you show your working I might be able to spot it.

Bye. Jasen

I don't have my data - it was from the original winding five years ago. The formula to get from ampere turns to Gilbert's is simple (current *number of turns*1.257).

If my memory serves, I used something like the number of turns that could fit in a cross sectional area times the mean length per turn to get wire length and resistance, from a 1914 book on solenoid construction . . . .

But logically I see your point and it seems likely I went wrong somewhere in there . . . For the sake of argument:

If each turn of wire has one ohm resistance, and I have a four turn coil and 8 volts to drive it I have a current of 2 amps and 8 ampere turns dissipating 16 watts total.

If I halve the diameter of the wire the cross section drops by a factor of four so the resistance should increase by a factor of four. So now I have 16 turns fitting where 4 where, and the resistance is 64 ohms. Four times the turns, with four times the resistance per turn. With the same 8 volt supply that's 0.125 amps for 2 ampere turns dissipating 1 watt.

Conclusion: I was full of shit to state that wire size didn't matter.

Efficiency: To produce the same 8 ampere turns with a 1/2 size wire will take only 4 watts - so it becomes four times more efficient to decrease the wire size by one half (keeping the volume the same), or I could produce 32 amp/turns of field strength for the same 16 watts that produced 8 A/T with larger wire. (if the volume were to increase)

Practically speaking there's something like a theoretical increase of

7% or so when wires lay in the interstices created by the layer below. Increase the turns by a factor of four and that 7% becomes significant too.

See any flaws in the logic?

I'm glad we had this chat.

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Huh? It comes out exactly the same either way:

8V, 4 ohm, 4T = 16W, 8A-T 32V, 64 ohm, 16T = 16W, 8A-T

The only issues as far as efficiency goes are the insulation thickness as a percentage of the winding, and the wasted space due to the imperfect packing between the wires. You can write equations for each, and calculate the best wire size for an 8A-T coil, if you want.

-Chuck

If a single coil will give me two ampere turns with 1 watt of dissipation won't four parallel coils give me the equivalent of 8 ampere turns at 4 watts dissipation? All running at 8 volts in parallel.

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Of course, but you will have twice as much wire volume as the original coil.

TANFL!

-Chuck