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

[norman6538]

Neptune you are absolutely right about the wrong numbers from Raymond Head.


Neptune how did you get the 5:1 ratio in this statement?
"With a fully rotating pendulum , as opposed to an oscillating one ,
          The bob has a down force at bottom dead centre of 5 times its static
          weight ."


I did my suggested tests and they do not look promising but will continue to persue that.
I was very disappointed to see a short pendulum swing seriously dampen the pendulum cycles. A longer swing
was more consistant with a regular pendulum.
Norman

[johnny874]

Hi Johnny and thanks for that input . I am not so hot on mathematics , but I found a simple formula that will help me . Aparrently , centrifugal force = mass x velocity squared , and divide the result by Radius .I am hunting parts to build an off balance wheel model , but need to get some local help with welding . I have reached the reluctant conclusion that in its basic form , the 2SO is not overunity . I think Milkovic`s pump just makes the pumping easier because it is a more convenient way to do the job .I also think at this stage that my hero , Raymond Head has got his calculations wrong . In a video he shows a 140 pound pendulum lifting  a claimed 70 pounds , with an estimated hand push of about 10 pounds . His lever is 3 to 1 . So in reality , about 50 pounds of the "load "is used to balance the static weight of the pendulum , and the real load is 70 minus 50 = 20 pounds .
         With a fully rotating pendulum , as opposed to an oscillating one , The bob has a down force at bottom dead centre of 5 times its static weight . The big question is of course , what price has to be paid for that increase . My next quest is to answer that question .

   Neptune,
What the math ignores is a velocity less than a (acceleration of gravity) and not being perpendicular to it.
A weight moving towards top center could have no cf.

[neptune]

@Cloxxki . If you read exactly what I said that I did not claim that a heavier pendulum was more efficient , merely that it makes more swings before stopping .Indeed I was at pains to point out why I felt that a heavier pendulum would not give more efficiency .
@Norman . You asked where I got the fact That in a fully rotating pendulum ,the bob weighs 5 times normal . This was a direct quote from the paper by Jovan Marjanovic . It would make sense to me because , in a pendulum released from horizontal , the figure is 3 . Pendulum frequency is [almost] independent of amplitude . The time taken for a high amplitude swing is the same as for a low amplitude swing . So , with a maximum swing , the speed will be faster at bottom dead centre , and as centrifugal force = M Vsquared over radius , The centrifugal force will be proportional to the velocity squared . So if the velocity were doubled [which it is not ] C force is multiplied by 8 . Anyone agree/disagree ? As I said last time , what matters is what price we have to pay .
  @Johnny874 . You say that a weight moving towards top dead centre can have no CF . That may be no bad thing . After the bob rises above horizontal we have already harvested the CF at max velocity , and are going into the beam reset phase . The main gain above the horizontal is when the bob is falling , and gaining extra acceleration towards bottom dead centre . Laters , Ken .

[Cloxxki]

@neptune
I was not attacking you in person, only anyone who'd come up or believe such a notice without making remarks.

Both the inventor and prime replicator practice fuzzy math, and are left manually pushing a device with no output work, which in theory should be swinging at least as well as a simple pendulum (indefinitely), yet doesn't.

I see the @SO extracting energy from a pendulum by moving it pivot up and down, along a slight curve. But there doesn't seem to be any useful output. If the CW hits a surface, I suppose that's the "work" done, create vibrations. If a frictionless 2SO (magnetic pivots, inside vacume chamber) would have the CW swinging freely, I actually wuld not know why the 2SO would slow down. If it did, I would probably learn stuff about efficiency of energy transfer.

Think of an engine's crank shaft. Input from the cylinder is only temporarily available, less than perfect leverage (only 90º is) will result in energy loss through heat. A dropping CW has energy to transfer, but will it always go 100% somewhere useful within the system? I fear there are going to be losses unless elaborate escapements and added.

[neptune]

@Cloxxki . No worries , no offence taken . Raymond Head admits that he is not the worlds best mathematician , and indeed has asked for help on at least one of his videos . But the great thing is that he gets stuck in and builds things . I have come to the conclusion that even those who claim to be professional mathematicians  are still confused about this device . Mathematical proofs of the OU of the 2SO are a dime a dozen .But where is the self runner . At least Jovan Marjanovic , the mathematician has built small wooden models .At this time , I am fairly sure that the 2so is not overunity . I am not sure if it might be using a rotating pendulum , and so, rather than try to calculate , I intend to do practical tests . I had a sudden burst of energy today , and made a start on my new rotary model . I work on a zero budget using 100% recycled parts . I found a nice piece of timber yesterday whilst out cycling , and that inspired me to get started. I also did an experiment today . I inverted my bicycle and added a 1 KG weight to the front wheel . I put the bathroom scales under the handlebar . At bottom dead centre , the scales showed a weight gain of about 4 Kg . Watch this space .
Regards to all , Ken .