12 times more output than input, dual mechanical oscillation system !

[hansvonlieven]

Hans you should know better than that, guessing is not scientific... do you want me to have omni have a word with you?  lol

Ron

Sorry Ron,

Since the method of measuring was not supplied (In the best of unscientific tradition by the way) I took it the documented approach until now applied ( Bathroom scales, springscales, fishscales and other accurate scientific equipment ) as the method of measurement used in the example. If you have a better guess, be my guest.

Maybe some of Omni's rigourous scientific testing is more to the point, maybe he will even tell us what it is. 

Until then, I am still guessing LOL

Hans von Lieven

[exnihiloest]

...
After all, you only have to lift a weight once, and then it manages, after falling, to rise again, and after falling, to rise again, and again, and again.  With a low friction pendulum, we have what, a 1000:1 overunity ratio?
...

You're right. We don't need any energy to maintain a pendulum swinging except to compensate for the frictions (air and mechanics).
In this paper announced in the last sunday news at peswiki.com (http://www.veljkomilkovic.com/Images/Merenja/Ronald_Pugh_Input-Ouput_Measurement_Mk5.pdf) there is a Milkovic Output/Input measurement at COP 1.46.

First we must admit we are far from the "12 times more output than input".

Then when we see the calculation, we understand it is completely wrong.
The author argues that the work done on the counter weight (used instead of the hammer) is the product of its weight by the height at which it rises (he's right).

Then by an obscur reasoning not only involving the time but also confusing the forces measured in the referential of the constraint jauge on the counter weight and of an observer at rest, he calculates the "work" and finds watts instead of joules!
He finally calculates the COP using the mean electric power (which seems to be that for maintaining the movement, not that just needed to rise the height of the counter weight).

At least we can say that the height at which the counter weight rises then oscillates is 0.044 m (1.75 inches). The weight is 40 pounds = 18.18 Kg. Thus the work done is 0.044*18.18=0.8 Joule. If the time needed by the counter weight to attain 0.044 m is 1 mn, then the required power is 0.8/60=0,013 W.
9W of electric power were used, so COP=0,013/9=0,0015. What a fantastic over unity :-)))

[tagor]

Hans you should know better than that, guessing is not scientific... do you want me to have omni have a word with you?  lol

Ron

I am sorry I_ron but the maths are erroneous and we need the protocol used

the moving weight up and down is not a real usefull output work !

all the moving parts are depending each other and in equilibrium ... so the output work is egal : 0
the cop = 0

you need to put , with the moving weight up and down , a load

this load has to do a real usefull work ... and you can look at this work ...
and you see that the real output work is less than the input one !!

[Cloxxki]

If the pendulum could drive a car jack, moving the car only upward, that would be interesting.

But, I'm pretty sure it can't.

[tagor]

If the pendulum could drive a car jack, moving the car only upward, that would be interesting.

But, I'm pretty sure it can't.

moving a car or water ( like in the water pump )

I am sure this pendulum can pump water , but ...

with bad efficiency !!