Originally Posted by
James_
I did some math and I realized one issue that I've had with the metric system and that is calculating work. What I did was to calculate work in N-m's of work.
Everyone knows W = M * D (work = mass x distance) so what I did was to multiply N-m's of force by distance. Force is just newtons (N). If you multiply Newton-metres by meters you end up with Newton-meters^2. By definition, a N-m is a Joule (J), so the unit you are describing is a Joule-metre (Jm). I may be wrong, but I don't think a Joule-metre is a meaningful measure of anything.
And with what I showed in Mt 20 it does show a net force.
While Bessler showed his weights rotating and hitting stops, they might have. Witnesses said that they heard 8 knocking sounds per revolution. An easy way to account for this is to have the weights rotate at about (45° ATC). And with 8 weights, one would be placed every 45°.
With the math, hopefully someone will go over this;
This is what I keep coming back to;
(5.635 N-mf - 4.9 N-mf) * (1.15m * Pi) = 2.65 overbalanced weight's net work
4.9 N-mf * (0.15m * Pi) = 2.31 out of balance weight on a long ever
The 2 figures are N-m's of work, this is why Pi is a part of the calculation. And 0.34 N-mf does show potential but offers no guarantee.
What this does take into consideration is the much longer path the overbalanced weight takes.
4.9 N-mf is a 1/2 Kg weight. What I'm having trouble considering is that a weight rotating 15 cm's from the axle (on the long levers) has the same force as a weight that is 15 cm's further from the axle than it's opposing weight (overbalanced weight) yet will travel 1.57 meters more. That could be why I find Mt 20 intriguing. It's not exactly what someone would consider thinking rationally.
Jim