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

[tagor]

Milkovic Output/Input Measured at 1.46 - Ronald Pugh, a Canadian machinist, reports on the pendulum-lever system efficiency measurement he conducted.

ok , ok
if this penduleum is so efficient ...
it has to be easy to loop this stuff ? and so the proof of OU !!

I have my own replication of the dual pendulum but ... I am sceptical

Once again, he confirms the super efficiency of Milkovic's Two-stage Mechanical Oscillator.

no it is the first time that I see measurement of this pendulum

we are waiting for confirmation from other replication !!

[fritz]

============================================================
Work done: (1)
It lifts 40 pounds 1.75 inches 100 times a minute
that is 40 pounds 175 inches in one minute
divide that by 60 to get distance in one second = 2.9166 inches or .24 feet
per second
40 times .24 = 9.722 foot pounds per second divided by 550 = .01767 HP
746 watts per HP divided by .01767 = 13.18 watts
Therefore, 9 watts input, for 13.18 watts output
= COP, 1.46
========================================================
I think this calculation is BS.

The 40 pounds are lifted 100 times a minute 1.75inch (against gravity)- but -
this excess energy(!!) is reflected back into the other oscillator during go down.
so (==0).
This statement would only be true for 40 pounds weight going up and down - and an extra weight of 40 pounds  which loads extra during going up.
Then there would be energy "comming out".
With this initial assumption wrong - there is no go.

This 2nd order oscillations are quite nice when it comes to transform _POWER_.
You can achieve strongest momentum pumping energy from the first oscillator into the 2nd one...

Its amazing how a tiny oscillating weight can cause the big move - but its just conversion of kinetic energy into potential energy back and forth. The movement of the big mass has to be seen as an oscillation with high Q, less losses.

If using as hammer - tiny part of the kinetic energy deforms the material - the remaining part is reflected back - recycled into the first oscillator.

Even if this is a very nice technique - I see no excess _ENERGY_ here - even if there is lots of _POWER_ accessible (for a certain time).

For machines with stop-move-stop operation - things like that can save a lot of power.

[Nabo00o]

============================================================
Work done: (1)
It lifts 40 pounds 1.75 inches 100 times a minute
that is 40 pounds 175 inches in one minute
divide that by 60 to get distance in one second = 2.9166 inches or .24 feet
per second
40 times .24 = 9.722 foot pounds per second divided by 550 = .01767 HP
746 watts per HP divided by .01767 = 13.18 watts
Therefore, 9 watts input, for 13.18 watts output
= COP, 1.46
========================================================
I think this calculation is BS.

The 40 pounds are lifted 100 times a minute 1.75inch (against gravity)- but -
this excess energy(!!) is reflected back into the other oscillator during go down.
so (==0).
This statement would only be true for 40 pounds weight going up and down - and an extra weight of 40 pounds  which loads extra during going up.
Then there would be energy "comming out".
With this initial assumption wrong - there is no go.

This 2nd order oscillations are quite nice when it comes to transform _POWER_.
You can achieve strongest momentum pumping energy from the first oscillator into the 2nd one...

Its amazing how a tiny oscillating weight can cause the big move - but its just conversion of kinetic energy into potential energy back and forth. The movement of the big mass has to be seen as an oscillation with high Q, less losses.

If using as hammer - tiny part of the kinetic energy deforms the material - the remaining part is reflected back - recycled into the first oscillator.

Even if this is a very nice technique - I see no excess _ENERGY_ here - even if there is lots of _POWER_ accessible (for a certain time).

For machines with stop-move-stop operation - things like that can save a lot of power.

Your are completely wrong in your assumption that energy is reflected back from each turn.
If there was a spring which was compressed when the hammer went down then yes, it would have saved energy in each turn. But smashing metal down in the table will not make the metal bounce back up, it will transfer most of the momentum in the metal into heat, sounds and other vibrations.
Of course using a hammer to prove more work out is a very crude way to convince someone that it can actually do more useful work, so it should be connected to something, like a wheel maybe, or perhaps a pickup coil. The evidence put forward however is clear, and I cannot see how you can dispute it. You are of course free to prove me wrong.

[fritz]

Your are completely wrong in your assumption that energy is reflected back from each turn.
If there was a spring which was compressed when the hammer went down then yes, it would have saved energy in each turn. But smashing metal down in the table will not make the metal bounce back up, it will transfer most of the momentum in the metal into heat, sounds and other vibrations.
Of course using a hammer to prove more work out is a very crude way to convince someone that it can actually do more useful work, so it should be connected to something, like a wheel maybe, or perhaps a pickup coil. The evidence put forward however is clear, and I cannot see how you can dispute it. You are of course free to prove me wrong.

If you hammer manually - you "charge" the hammer with potential energy on going up - accelerate the mass of the hammer on going down - and if the hammer hits the target - the kinetic energy of the hammer is released into deformation, vibration, bounce back.
By choosing the right hammer with the proper weight and optimum length you can reduce the effect for a special purpose.
If your hands feel broken at the end of the hammering day - you obviously used the wrong hammer - and your hands had to absorb too much energy.

This case is different. The hammer is mounted on a solid lever which allows only movement in one axis. So any energy which cannot be absorbed of the target is reflected back into the lever - feeding energy to the other weight at the other side.

Depending on the absorption of the hammered target - part of this energy is reflected back - adds immediate to the kinetic energy at the other side.
In the case of a non compressible, absolutly solid target - the entire energy is transformed back minus the losses by friction in the bearing of the lever.
Think about pool billiard, think about softer and harder balls,....
depending on the involved material - same story.

Part of the deformation is reversible (thats the spring you are missing), other part of the deformation exceeds the material constant where deformation gets inreversible.

The hammer itself doesn´t need extra energy to go up and down (without touching). The only energy needed is to compensate friction in the bearing.
If you limit the movement of this oscillation - you can harvest the kinetic energy at that point. Depending on how you do that - you can have the entire energy - or only part of it.
Hammering into a sand bed will absorb very much energy. If you assume perfect non-deformable lever, perfect rocksolid mounting which stops the weight on that point - well where should the energy go ? If this mounting is more solid than the hammer - the hammer will bounce.

This setup works only well with perfect solid lever - otherwise part of the energy would be lost due to dynamic deformation of the lever. (Well, part of this energy would be come back but the "leverage" would decrease

[hansvonlieven]

It would be interesting to see just how the "output" was measured.

My guess is it was done with a spring scale, in which case most of the energy would have been fed back into the system. Let it hit a solid surface and then see how it performs. If my experiments and simulations are an indicator, very poorly !

Hans von Lieven