Wind harmonic generator?

I had this idea a while ago. There's a nylon strap strung between two trees. It's let's say 2 cm wide and 5 meters long. A strong wind is blowing across it so that the strap is resonating, displacing through, say, 20 cm at 10 Hz. What is the maximum power involved? If you had one or more wire coils running the length generating current, what kind of efficiency over all could you expect? I'm thinking of this as an ultra cheap and easy means of wind generation.

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I took a break from yard work, using your numbers for dimensions the area swept is 1.33333 square meters so the max power using say 7 m/s wind speed is 288 watts.
Wow, really? That's more than I expected. It occurs to me also that a climbing tape or similar would actually be about 5 cm wide. Does that make it 720 watts? What kind of efficiency could I reasonably expect from this setup, if say the tape could orient to always face into the wind, and would tune itself to the resonant frequency? And what kind of efficiency does a flexing wire coil generally produce?
he width of the tape isn't in the calculations, I just calculated the power of the wind for the area defined. There would be some happy medium where the width of the tape maximized the energy capture of the wind, but I don't know what it is. That's funny your thinking of that, I just finished a feedback class and have been thinking of building a windbelt that is always in resonance across a range of wind speeds.
Right, gotcha. It should be fairly simple to make the thing autotuning, some kind of vane that catches the wind and puts more or less tension on the strap depending.
I'm not sure about that- it seems like when the thing is resonating it would be under more tension so would work against your tuner to detune itself enough to reach the balance point between the two forces. Also consider that IIRC, when you load the tape (as an electrical generator would do), the resonant frequency will move. I'm not sure how you intend to generate the electricity but a reciprocating linear motion is usually best rather than twisting motion.
SolarFlower_org (author)  PKM8 years ago
Good point on the detuning, tho it should be possible to have a worm gear or similar mechanism to isolate the direction of force.

I'm not sure what you mean about loading the tape and linear vrs twisting?

It turns out this has been already done (which could save me a lot of hassle)

tho I'm thinking of something larger, simpler, and tuneable.
Someone told me that if you bend or otherwise deform a coil it generates current, but I can't find anything on it and am strarting to question it.
What the windbelt guy has going in the link above is probabaly a pretty good option.
I suspect that bending/twisting the coil would have to happen in a magnetic field, so the current would be generated in the usual way (a wire moving relative to a magnetic field). I'm not sure if this method is used in many practical applications, I believe rotary or back-and-forth motion tends to be more practical. The diagram in your link shows what I was thinking of- as the belt "flaps" you capture that motion and use it with a conventional moving magnet/coil arrangement to generate power. I'm always surprised at the efficiency you can get from this sort of generator, though I'm not sure the windbelt is the most efficient way of capturing wind energy (though it has the obvious advantage of mechanical simplicity). You seemed to do pretty well with the solar tracker/collector, why not rig up a test model or two and see how much movement you can get from the windbelt before adding the generator?
I also wanted to add that since you have a length and a max displacement you have three points from which to determine the equation of a parabola (which this can be modeled as) than simply integrate from 0 to L and multiply by 2 to get the actual area. Or you could model it as a sine wave, whatever.
You need to calculate the area swept by your ribbon, then Max power is calculated by

.5*rho*area*velocity of the wind3 Where rho is the density of air (1.23 kg/m3)

Your actual power will be some small fraction of that number