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

[tinu]

If one looks carefully, the right side is shifted when the left side with the pendulum weight is in its highest position. That means, if the pendulum is stretched, the left side times distance is Ã,“heavierÃ,” than the right one, and so it is shifting the right side, whereas when the pendulum is moved to its right highest position, the right side wins the game with the rods. 

Cyberdust,

This is not correct. It?s precisely opposite: the right side is shifted when the left side with the pendulum weight is in its LOWER position. This happens due to the centrifugal force acting on the swinging pendulum. When the pendulum weight is in its highest position, the right side (the hammer) simply begins to fall under the gravity due to its own weight.

Static analysis of forces involved and their associated angular momentum is not enough to explain the device. Then analysis became soon very complex because the pendulum oscillates in a variable-accelerated reference system. One component of the acceleration is given by gravity (g) and it is well known but then there is the other one, adding to or subtracting from g, given by the lever movement. And the movement of that lever is not easily analyzable because the lever is stopped suddenly when the hammer reach its lower point. This involves a sudden change in acceleration which is felt by (is transmitted to) the pendulum; in theory this is a kind of discontinuity (infinite acceleration) but in practice the acceleration must have a finite value. Nonetheless, it may be muuuch larger than g, thus substantially affecting the pendulum.

Up to my understanding to this day, a significant part of the ?mystery? may be hidden in this very short collision between the hammer and its anvil. Otherwise, all we have there is a system composed of two classic oscillators.

For other questions, please have a look on my former post on the previous page. Here is an excerpt of it:

3. The whole problem of Milkovic is reducible imho to a basic question. It is known (from experience as well as from Newtonian physics) that a pendulum is in principle weightless at its upper point and it weights more that its own rest mass when crossing the vertical, due to centrifugal force. This is just basics, donÃ,’t shot yet. Now, if you give me a truly variable mass, I can easily build an OU machine, by simply lifting the said mass (and thus investing energy) when it is easy and then by letting it going down and perform work (and recovering energy) when it is heavier! If the mass would follow a variation like m=m0*sin(2*pi*f*t), like in a stationary pendulum, the OU device would work. ButÃ,…

4. If you followed the above, Milkovic is about a device that has a Ã,‘variable massÃ,’ on one side (on left side, in all of his practical devices but water pump). The problem with that variable mass is that it is no longer zero once you try to lift it. ThatÃ,’s due to inertia, of course. Then, the second problem is that it will no longer weight more than its rest mass once you let it drop free. I donÃ,’t know if most can follow me but to be short, the mass is not truly variable and in any case it does not follow a sinusoidal function. So, the analysis may be relevant in makings things clear for usÃ,… And thatÃ,’s why I started with 1 above. 

Restpectfully,
Tinu
"In the absence of light, dark prevails"

[i_ron]

Thanks for the explanations! My initial, brief observation of the video only focused on the pivot point, and my conclusion that there was no "apparent" downward force on that spot at the end of either swing, which I admit, I visualized at the horizontal.

One more question. Does the initial energy input derive from the potential energy in the ball before dropping it, the kinetic energy exerted to lift it, or a combination?

Dact

Hi Dact,

I am not sure as I follow your question. Treat the pendulum as the motor that drives the secondary arm. As the pendulum cycles it imparts a plus minus force on the secondary arm.This causes the secondary arm to oscillate... within the confines of it's stops. The secondary arm is just a teeter totter. The springs (or the striking of the hammer) return some energy to reverse the motion... the little feet that push off, if you wish.

The pendulum requires a source of input, the one flashlight for example, but the
input is least when the secondary arm is stationary. Increasing the arm travel
requires increasing the input to the pendulum.

http://www.sciences.univ-nantes.fr/physique/perso/gtulloue/Meca/Oscillateurs/botafumeiro.html

This is a fun site to play with and shows that the monks were actually raising the
pendulum at the bottom of the stroke. This will put the pendulum over the top!
Unfortunately we are allowing the pendulum to drop at the wrong time... thus we
should not over do the arm movement.... or we pay a penalty.

Ron

[Dingus Mungus]

I just want to add something to this discussion, but I've moved on from researching this device. During my months of virtual replication and emulation of the device I discovered several examples where a pendulum attached to this secondary oscillating arm would swing for longer and with greater kinetic energy then a duplicate pendulum. That is a rather interesting observation since the pendulum with more parts/friction/resistance was supplied the same starting kintic energy.

I theorized that the reason for this excess energy was improbability...

Let me explain:
The probability of the horizontal arm being at 0d with 0ke and the vertical pendulum arm being at 0d with 0ke is far less than a single vertical arm reaching 0d with 0ke. Small ammounts of gravitaional potential enery were taken from the horizontal arm when its <> 0d and the vertical arm == 0d, and visa versa.

I know thats a rather sloppy explanation, but one day I'll draw out some diagrams to further explain it. More importantly though... I was not able to find an example that provided anything close to 100% of efficiency, but I did find drastic efficiency improvements when using a duped pendulum to compare run times/input energy.

Remember though... This device is in no way even 100% efficient! If it was even 100% efficient the pendulum would never slow, and if it was 1200% efficient the pendulum would swing faster and higher the longer it ran. So clearly... The title of the thread is wrong. Altho it can make a pendulum 2x-3x More efficient. AKA runs 2x-3x longer...

[i_ron]

I was not able to find an example that provided anything close to 100% of efficiency[/b],

Dingus

There is no simulation program that models Veljko's pendulum.
Therefore your results are inconclusive... garbage in equals garbage out.


"*"This device is in no way even 100% efficient! If it was even 100% efficient the pendulum would never slow, and if it was 1200% efficient the pendulum would swing faster and higher the longer it ran."*"

But what is your method of extracting energy from the secondary beam and how are
you reapplying it to the pendulum? I posted pictures and video of my on going
experiment.... where are yours? How can you post conclusions when you have no
valid data?

What you are saying with that statement is you haven't a clue as to how this works. 

Ron

[Earl]

I believe this arguement does not apply to loadless mechanical feedback.

One can always insert between transmitter and antenna a magneto-ferrite electrical device called an isolator or circulator.  These are widely used for UHF, microwave and optical transmitters.

It should be obvious that an infrared transmitting diode is not affected by any infrared receiving diode.

Regards, Earl