| daestrom 2005-10-10, 10:21 am |
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"Scott A Crosby" <scrosby@cs.rice.edu> wrote in message
news:oyd1x2uqn58.fsf_-_@bert.cs.rice.edu...
> On Sun, 09 Oct 2005 12:50:28 GMT, abelshapiro@pinex.com writes:
>
>
> FYI: 300MW is about .1% of what the US uses.
>
>
> Here's a counterproposal for 'free' energy from a gyroscope. Well, not
> free, I believe the force it is exploiting is coming from earth's
> rotation, but I suspect there's a mistake somewhere because I don't
> think I'm conserving angular momentum.
>
> Put a partially gimballed gyroscope on the equator with its axis of
> rotation aimed straight up/down. As earth rotates, the gyroscope will
> continue to rotate in the same plane, but the earth will rotate
> underneath it flipping it end over end once a day. Now, we only allow
> the gyroscope's axis of rotation to rotate through a vertical plane
> --- the equator. Now we have a gyroscope whose axis of rotation will
> flip end-over-end 360 degrees a day.
>
> Attach a generator, generate electricity.
>
> Ah, it does purporteldy work!
> http://www.lhup.edu/~dsimanek/museum/advanced.htm
> 'gyrogenerator'
>
> Now how do I work out the math for how much energy it would produce?
The speed of rotation is fixed (1 rev/day). So the power is simply
proportional to the force it can exert on some mechanism. Of course, too
large a force and it stops rotating, or rotates slower than the planet (i.e.
the drag of the apparatus causes it to gradually 'rotate' around). Too
little drag and the unit doesn't generate useful energy. I suspect the math
would show that enough force to cause it to rotate in space once a day would
be the optimum.
How much force it would take to cause it to rotate in space once a day would
be a function of its moment of inertia and speed of rotation (i.e. its
angular momentum).
daestrom
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