Correction - sorry..


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Posted by Øystein Rustad (130.67.81.1) on May 08, 2003 at 12:50:24:

In Reply to: Re: Another quick clarification request - Darren posted by Øystein Rustad on May 08, 2003 at 12:47:57:

Ehhh... mixed negative and positive work there... :-) sorry..

I Just put in a comment on the question, that was directed to Darren, if I may ?

: %OU for a gravitywheel would be Input related to output where
: Input = negative work, and output is positive work acting as torque * distance acting on the axle or rim..
: OU = Wdown (negative work) / Wup (positive work) * 100 - 100

: So input is not the starting "hand-push" but the lifting of weights that the wheel is responsible for..and the work the falling weights have to perform and hopefulle have some work left to drive some external machinery :-)

: Øystein


:
: : Thanks Darren for the earlier clarification...

: : In that previous post, you said that a thoeretical design you had worked on "also produced OU (about 20%) "

: : I`d be grateful if you could explain in general terms how you arrive at a figure for "O/U" in the context of a gravity wheel.

: : To say something produces 20% more than unity implies IMHO that you have a notional value for "input energy" (or rather work, because in such a wheel gravity is constantly putting energy in)

: : I can see how one quantifies output energy (and given a rotational speed) work, but how do you quantify the 'input energy' bit?

: : If you start from "energy needed to start the wheel" then you end up with an absurdity, because if the output power is thereafter 'always available' the ratio of output/input power keeps rising - the aggregate value for "output power" grows with each revolution.....(even if the machine limits its own speed due to friction etc, there is always an increasing value for output power)

: : Or do you mean that the available turning force is 20% greater than that for a unity machine...? But that can't be right either because in the latter case the aggregate turning force is zero (a unity machine does not turn by itself). Therefore the ratio

: : torque of O/U design
: : --------------------
: : torque of unity machine

: : will always be infinite (because the lower term is always zero)...

: : How are you quantifying the 'input bit' to arrive at a notional 20 % O/U in this context?

: : Thanks

: : Nick




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