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illusionist wrote:Ok, i've begun scaling my design up, can't wait to get a resistor and try it out. Thanks, a lot!

No problem! Also, I can't definitively say that the maximum diameter is the best way to go. I haven't done enough testing yet. But for now that's where I'd start.

yup, i'm starting mine at 26cm, and i'll work from there. I think that the main variable i will change will be the angle of the blades. Any input on that?

That completely depends on the shape of the blades. But they way I try to think of it is, "If this spins really really fast, will the blades be cutting through the air, or just running into it?" You want the former, of course.

The rules state that the resistor should be between 5-7.5 ohms. At radioshack, they have resistors ranging from 10-1m ohms. Which one is right for wind power? Also, they say that there are 1-watt to .25 watt resitors. Which one should I get?

One more thing, what is the difference between 10 ohms, 10K ohms, and 1m ohms of resistance?

The K stands for kilo, which means one thousand. The M stands for mega, which means one million. So 10K ohms = 10 kilohms = 10,000 ohms, and 1M ohm = 1 megohm = 1,000,000 ohms. You need a resistor that is between 5 and 7.5 ohms, so for instance something like 6.8K ohms is way too much.

The wattage of a resistor tells you how much power can pass through it safely. Most resistors are rated at a quarter of a watt, which is 250 milliwatts. If your turbine is going to produce more than that, you'll need a resistor with a higher power rating.

illusionist wrote:The rules state that the resistor should be between 5-7.5 ohms. At radioshack, they have resistors ranging from 10-1m ohms. Which one is right for wind power? Also, they say that there are 1-watt to .25 watt resitors. Which one should I get?

The wattage rating of a resistor is the amount of power it can handle (dissipate) continuously. Resistors can handle much higher loads for brief periods (a few seconds, at most, depending on the load)

P = I x E
P (Power in Watts) = I (Current) x E (Electromotive force in Volts)
and
E = I x R
E (electromotive force in Volts) = I (Current in Amps) x R (Resistance in Ohms)

P = E^2 / R
P (Power in Watts) = E^2 (Electromotive force in volts squared*) / R(Resistance in Ohms)

If your turbine generates 2 volts across a 7.5 Ohm resistor, it would dissipate 2^2/R watts or .53 watts. Any resistor capable of dissipating .53 watts or more would be sufficient.

Look for the square "sandblock" resistors. They generally come with low resistance in 5 or 10 watt capacities. Another solution might be the 10-ohm 1W 5% Metal-Oxide Film Resistor (2-Pack) from RadioShack. If you connect them in parallel, you will have a 5 ohm, 2 watt resistor. You can connect other values in series/parallel configurations to get any resistance/power capacity tou want.

Example:

10, 75 Ohm, 1/4 watt resistors connected in parallel will produce a 7.5 Ohm, 2.5 watt resistor
10, 75 Ohm, 1/2 watt resistors connected in the same configuration would make a 7.5 Ohm. 5 watt resistor.

OK, thank you. radioshack didn't have any of the 7.5 ohm resistors, so we're considering another store called electric connection. I don't understand how two 10 ohm resistors in parallel would make 5 ohm resistance?

1/Req=1/R1+1/R2, so here 1/10 +1/10 = 1/5. See http://www.physics.uoguelph.ca/tutorial ... allel.html for derivation.

Ohh, that makes sense. I feel stupid now:P