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

[neptune]

@Cloxxki . I feel you may have mqade a mistake . You state "Cross beam is three tines as long on PENDULUM side". I think you meant three times as long on counterweight side . Regards, Ken

[Cloxxki]

@Cloxxki . I feel you may have mqade a mistake . You state "Cross beam is three tines as long on PENDULUM side". I think you meant three times as long on counterweight side . Regards, Ken

The distance from pendulum's pivot to the central pivot in my example is 3x that of the central pivot to the CW's pivot.

I don't know what's currently claimed the most OU relationship.
I chose the weights, 100kg CW and 20kg pendulum as with this crossbar placement relationship, the CW would rest on the hammerplate (to some extent), but the pendulum could lift it for a part of it's downswing.
Now it could look like the 20kg pendulum lifts the 100kg counterweight. Truth, in my example, is that the 20kg already pulls on the CW by 60kg when just handing there. Any CF from the pendulum is also trippled, in negative, to the other side.

I'll investigate my mistake.
A few pages back I used some rough numbers and the formula I found for centrifugal force.
I remember earth is around 6,000km radius (12,000km diameter).
Maximum equatorial circumference (dia * pi) is travelled by an object placed there, evere 24hour.
This makes for a rim speed of 436m/s

FC = (mv²)/r, I believe.
I'll take a 100kg weight
FC = (100* 190385)/6,000,000= 3,17N
3.17N~0.3% of 100kg
Seems I was off by a facor of 100, my sincere apologies.
This would render my concern to zero.

[neptune]

Hi Cloxxki . The mistake was mine not yours , sorry . I was confused by the fact that your cross bar proportions are different to what Milkovic normally uses . He normally has the counterweight end three or three and a half times as long as the pendulum end . Marjanovic says that there is advantage in not letting the pendulum pivot drop untill just before bottom dead centre , thus allowing the pendulum to gather max speed before extracting energy .This can be achieved by using a heavier counterweight or by some method of locking the beam pivot at the right time . Regards , Ken .

[Cloxxki]

Which weights and pendulum length are typical at this inverted crossbar proportion to mine? I would expect a relatively small CW then...

[johnny874]

Which weights and pendulum length are typical at this inverted crossbar proportion to mine? I would expect a relatively small CW then...

  Cloxxki,
I kind of wish everyone understood torque like Tom/Webby1 does. It can generate a lot of force.
The primary reason a TSMO won't work is because when a fulcrum is lifted, it's weight is not
lifted with it. This moves the weight closer to center lowering it's drop.
One way it might work is modifying what I suggested to Tom and that is when one pendulum starts
it's swing, it's torque can increase the velocity of the other pendulum. An example of this is if a pendulum
is 1 meter from center of fulcrum to center of it's weight, 1kg and @ an angle of 30 degrees the weight has
1/2 kg of force @ 1 meter. How much potential does the weight have as far as torque goes ?
This is what everyone has been over looking. Missing a huge energy potential because it's at the beginning
of the stroke. And most likely it would only take a burst of energy transfer. If so, then the pendulum
powering it's sister would lose minimal height. Of course, like a car's engine, timing would be important.
 
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