Posted by Nick Hall (195.74.122.243) on May 08, 2003 at 10:32:51:
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