Gents,
 
I think I should take distance from any dated comments I may have may with regards to the 2SO, as my current understand is less optimistic than when I first saw it.
 
I'd love the CW to free fall, and I know it will when unhooked. It's up there, it must come down. I am however unfortunately convinced that the pendulum will slow down greatly when not receiving the pull back from the CW.

Great point!, how did i not see that? I'll take some of that distance from dated comments also. Time for me look closer at the TSO.

@Ken, now i get where you are coming from better.
 

OK guys , just come from the workshop . I am now convinced that the lever CAN feed energy back to the pendulum given the right phasing . It is very difficult to get the pendulum to swing from a standing start . However if you start it swinging in about a 30 degree arc ,You can increase its swing by manipulating the beam by hand . Move the beam so the the pivot of the pendulum is at its lowest when pendulum is at bottom dead center , and slightly higher [0.5 cms in my case ] each time the pendulum is at its highest point . Try it . This may be significant , as this is exactly the phasing of a working 2SO .
       At Cloxxki . I have not yet 100% grasped your last post yet . However some tests I did today are connected with what I think you are saying . On my model , the up and down movement of the beam is limited by 2 adjustable stops . With no counterweight , the beam was in the UP position .To make it sit in the down position I have added 200 grams to the output end , which MUST NOT  be counted as part of the CW . The CW is limited by the stops to a vertical lift of 1.5 cms . The pendulum loses about 0.5 cm height in one swing . So input was 0.5 Kg cms [pend weighs 1 kg] .output is 350 grams x 1.5  cm = 0.425 Kg cms .or about 85 % efficient .Exact measurement is hard with poor eyesight , but that is my estimate . What I am saying is that to be fair , the beam must fall down WITHOUT the CW . Regards Ken .

OK guys , just come from the workshop . I am now convinced that the lever CAN feed energy back to the pendulum given the right phasing . It is very difficult to get the pendulum to swing from a standing start . However if you start it swinging in about a 30 degree arc ,You can increase its swing by manipulating the beam by hand . Move the beam so the the pivot of the pendulum is at its lowest when pendulum is at bottom dead center , and slightly higher [0.5 cms in my case ] each time the pendulum is at its highest point . Try it . This may be significant , as this is exactly the phasing of a working 2SO .
       At Cloxxki . I have not yet 100% grasped your last post yet . However some tests I did today are connected with what I think you are saying . On my model , the up and down movement of the beam is limited by 2 adjustable stops . With no counterweight , the beam was in the UP position .To make it sit in the down position I have added 200 grams to the output end , which MUST NOT  be counted as part of the CW . The CW is limited by the stops to a vertical lift of 1.5 cms . The pendulum loses about 0.5 cm height in one swing . So input was 0.5 Kg cms [pend weighs 1 kg] .output is 350 grams x 1.5  cm = 0.425 Kg cms .or about 85 % efficient .Exact measurement is hard with poor eyesight , but that is my estimate . What I am saying is that to be fair , the beam must fall down WITHOUT the CW . Regards Ken .

Thanks for testing.
With the crossbar level and the pendulum in rest, it's indeed hard to get it swinging. A non-level crossbar would surely get it going eventually, due to increased lateral movement of the pendulum pivot.
 
The way I currently see it (it's not gospel or fact by any means), the CW's main job is keep the crossbar down. The pendulum side with otherwise come down, crossbar up, and you've found.
Just by existing, the pendulum pulls on the CW via the central pivot and stiff bar, reducing the CW's load on its resting platform.
Indeed the 2nd stage side is perfectly timed to offer the pendulum feedback.
 
Another simple test would be with a CW that hammers its support. Make two clay balls, same size and shape. Get one impact of the CW in oscillation to squash it. A little cord gets it in place and away in time for the next strike.
Then do the same with a CW from the same high, in freefall. Is there is difference?
With a small device, the hand can be the guinea pig. I can tell the result. The oscillation's hammering action is greatly reduced, first from the weight of the pendulum itself, and possibly more from its CF pull and/or velocity change during the CW's downward movement.

I thought using monkeys as inventors of OU would be enough to make clear that I am ridiculing the OU claim of OU. The little monkey is after all really tired, else he'd kept pestering his dad until he'd come off the see-saw and chase him down.
With Dad's side being supported by packed dirt, he won't feel his kid playing on the other side until there is a momentary overbalance there. Most seesaws have a car tire to dampen the impact. This offers some spring action, and you do feel load of the other side as you're being lifted tire-assisted for the first couple of inches.
Even if Dad's side is down to 1kg of load to the ground from 200kg without kid, the big Gorilla is not to know. The ground knows, but doesn't care.
The kid's intial leap to reach the seat does a signifant part of the work towards reaching balance. But he can swing all he wants, he can only get Dad to be lifted for a limited height, a short time. And he need to keep swinging, because Dad going back down isn't translated 100% in kid swinging back up again. This is why swinging is more tiresome than just hanging in a tree, enjoying the freedom. Kid can do that for minutes. Een swinging on a stiff branch is easier, no output, no limited feedback.
 
A 2SO where the CW rests fully on the ground, I think, might be one that has the pendulum come up 90º both sides. CW reaches 100% of load on ground for 2 infinitely short moments, each highest pendulum swing point. All the rest of the time, the pendulum has a force on the crossbar, which relieves the CW. In a 180º swing, it goes from doing nothing initially, slowly building up, and then after 45º start to ramp up the strain. After dead bottom, the other way around, quick diminishment, and in the last 45º there's hardly any support for the CW to speak off. On average, the pendulum does the same hanging still as swinging. I've seen experiments aimed at contesting that, but not convincingly.

I have worked out the basic physics of the Milkovic two stage oscillator as below.

the two stage components involved in its operation

x. leverage advantage of small pendulum weight to lift the heavy weight
x. lifted/stored work of the heavy weight
x. since the pendulum swings less and less each time
 ie. it does not swing as high as its dropped point.
 extra lift is required to get it back to the starting point
x. use the lifted/stored work to raise the pendulum high enough to get
back to its starting point



proposed test
 1. lift pendulum to a know stopping point and let it swing and get
 caught/latched at its furtherest swing by a oneway stopper/limiter.
 and mark that point.
 2. Then find out how much more additional lift is required to get the
 pendulum back where it started.
 3. Then catch the lifted weight in its up position with a stopper/limiter
 and see if the stored weight can lift the pendulum up to the required
  point measured in 2 above to get it back to where it started.

Then If the force in 3 is adequate then it will self run if the switching forces are not too high.

Norman

@ Fishman . would it be possible to show a diagram of your version of the 2SO please . I am finding it hard to visualise .
     In one his videos rhead 100 says that if we make a pendulum heavier without altering its length or aerodynamic profile , it obviously takes more energy to raise it to its start position . This is true . but than he says that it takes no more energy to maintain its amplitude .  . If you look through his videos , you will find one entitled something like " Weight of pendulum does matter" , and he shows some experiments with a bike wheel . The video is a bit muddled and needs editing because he has a problem with the bearings . Watch the second half . He has two weights of identical size but one is twice the weight of the other . He makes a pendulum by attaching each weight in turn to the wheel rim . He shows that with the heavier weight , the pendulum makes twice as many strokes before stopping . Look carefully and you will see that the heavier pendulum loses less height per stroke than the lighter one . So to maintain its amplitude , we would only need to lift the weight half as far .But it weighs TWICE AS MUCH so therefore the energy input is the same for both pendulums . So Raymond looks to be right on this . The centrifugal force on the larger pendulum will be double , but the 2SO will need twice the "dead weight" on the output arm to balance the system BEFORE we add the counterweight to balance the system .So , I think that there is no real gain from a heavier pendulum, if as I suspect , centrifugal force is directly proportional to weight .Her are the results of a crude experiment I did today . I turned my bike upside down , and made the front wheel into a pendulum by attching a number of microwave oven magnets to the rim in a cluster . In each test , I started with the weight at the top , and let it do almost a full turn until it stopped .The wheel has 36 spokes , and so , 36 gaps between spokes . By counting the gaps we can see by how much the wheel stopped short of top dead centre . each gap is 10 degrees .
       Number of Magnets                          Degrees away from top dead centre at end of swing
     1                                                      35
     2                                                      35
     3                                                        33
     4                                                        25


    6                                                         20
    8                                                         15
 From this crude data we can see that as the weight increases , the height loss per swing decreases  . In a perfect test the relation ship is probably a directly proportional one , but here we are testing not a pure pendulum , but one with an attached flywheel I just realised a made a mistake , I should have measure the VERTICAL HEIGHT short of TDC rather than the angle . Doing this would bring the data more in line with the theory ..Can anyone tell me for sure , is centrifugal force directly proportional to weight ?
 ADDED LATER .  I just repeated the experiment with vertical measurements . As near as I can measure, if we double the weight , the vertical distance short of TDC is halved , as I suspected .
 

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 .