Author Topic: 12 times more output than input, dual mechanical oscillation system ! (Read 2231711 times)

johnny874 · « **Reply #1830 on:** January 02, 2012, 06:21:10 PM »

Quote from: neptune on January 02, 2012, 04:56:37 PM

This seems to be my own private thread at the moment! No workshop time today . However I have been doing some serious thinking about the 2SO . I am a bit disappointed that I have been unable to prove OU up to now .There can only be two possible reasons for this . Either no one has built an OU 2SO , or my particular 2SO is not OU. Looking again at the work and videos of Raymond Head [rhead100 ] his demonstrations seem to show OU . However he , like Milkovic has no perfect way to measure input . The Milkovic fish scale method is inacurate , and his demo with the lever operated torch [hand lamp] is to me open to deception because one can continue to push the pendulum with the torch after its lever has completed its travel .
To measure the input continuously when the pendulum is pushed by hand every cycle . I suggest two methods .
1 . Tape a load cell to the palm of your hand , and monitor its output on
a computer.By logging pressure and duration we can compute input energy for each stroke .
2 . Instead of pushing it with the hand , push it with a lever . The lever is operated by a falling weight , and is reset by hand after each stroke . The distance the weight can fall is adjusted so it just maintains the pendulum amplitude . With this low tech method we know the exact input at each stroke .
For a mechanical device to be looped , it needs to be at least 2 times OU to compensate for losses .If it is only , say one and a half times , it could still have its uses , in muscle powered devices .
Accurate measurement is the key , gut feelings are unreliable . we can fall into what I call the bicycle syndrome . It is a proven fact , that if you walk a mile , you can cycle 3 miles for the same amount of energy . This does not mean the a bike is 3 times OU , it just means that walking is a relatively inefficient way to travel .
There is always the possibility that my method of measuring the input energy for just one swing of the pendulum is in some way flawed , but at this stage , I can not see how . I would appreciate opinions from everyone .

Neptune,
When I first tried something with OU in mind, I started with conservation of momentum.
When momentum is conserved, it can then be converted into what ever it is you want.
With a pendulum, one reason why they are so efficient is that their travel is mostly east-west. By this, I mean left to right, little lift or drop. This allows a little energy to allow for more motion.
One thing I have always felt is that a 2OS is avoiding the real issue. A single pendulum should be able to work. With what Tom thought of, the pendulums grates potential (when the bob is at it's highest point( can be used to lift another weight that would not have to fight inertia. And for any OU device, something obvious should be noticed. With scissors, it is obvious. A weight can travel a greater distance than the work put into it.
And if that doesn't work, then it would be known. And I think when it comes to swinging weights or even rotating ones for that matter, people have a tendency to ignore how much inertia is developed. It is equal to f=ma. This means any time a weight swings or drops to rotate, it's inertia is equal to it's potential. This is where a secondary system can break that cycle. it can convert f=ma (force equals mass times acceleration) to w=md (work equals mass times distance). And this would give it a chance.

Jim

edited to add; with what I am building, when a lever swings, it performs work. with a pendulum, some of the energy would need to be directed back into the system but is possible.

icanbeatbob · « **Reply #1831 on:** January 03, 2012, 01:32:37 AM »

Neptune,

Don't know if this has been though of before, but here is my idea. When the pendulum is at it's lowest point, it exerts it's greatest force downward against the pivot point, which is lost energy.

What if the weight pulling down was pulling down on a leverage point instead?

Brad

Cloxxki · « **Reply #1832 on:** January 03, 2012, 02:40:54 PM »

Allowing the weight to dip say 1cm lower will also take always from it's required next upswing.
And you can't take out extra CF by leveraging the extraction, as that would just slow down the weight even more, it will lift back to regular dead bottom and lay still there.

There is no energy without vertical work, and the weight can't to more work than its loss in height unless it's slowed down.
Laws of nature will not be rewritten to make the 2SO overunity.

fishman · « **Reply #1833 on:** January 03, 2012, 03:38:49 PM »

@Cloxxki,

"Laws of nature will not be rewritten to make the 2SO overunity"

The Laws of nature do not have to be rewritten to make the 2SO overunity. Your understanding of the laws of nature will.

icanbeatbob · « **Reply #1834 on:** January 03, 2012, 04:11:27 PM »

Understood. I am sure you guys know more about this than I do. It was just a thought.

I was thinking that the string would be taught, so when at the bottom, it would not travel down any substantial distance to lose energy for the upswing. Just goes to show you how much I know.

Anyway, I should have tried it myself first. Thanks for the information guys.

Brad

johnny874 · « **Reply #1835 on:** January 03, 2012, 05:34:36 PM »

@All,
Here is a link to one thought he had, and so far it hasn't worked.
Stefan asked when he started this thread if it was what Bessler had found.
I believe so. What I posted in the other thread has a good chance of working.
And it would be what Vjelko considers a feedback loop. It seems the math
has eluded everybody but me.
As things are, my rotary tool bit the dust last night so I will be taking a break
until I can get myself in a better build position.
As for Bessler and the 2OS, Good Luck.

Jim
edited to add; it's about 2/3rds down the page, this is the heading
MATHEMATICAL & THEORETICAL ANALYSES

http://www.veljkomilkovic.com/OscilacijeEng.html#analyses

fishman · « **Reply #1836 on:** January 03, 2012, 06:58:24 PM »

Tom's innovation migth be worth checking out. And as I mentioned, it is something that could support Vjelko's work.
It would be a big plus for what we are trying to do. Something to think about anyways. And Merry Christmas everyone.
Jim

@ Jim, what is "TOM's innovation"?
Any links...

neptune · « **Reply #1837 on:** January 03, 2012, 06:59:06 PM »

Hi guys , thanks for replies and a happy new year to you all . @Cloxxki .So from what you say , are we to infer that you believe the 2SO is not overunity ? If this is the case do you believe any mechanical device is OU ? I am not willing to say ay#t this stage if I think 2SO is OU or not . Obviously I think it could be , or I would not waste time on it .My take on how it might work are as follows . Friction and air drag are real forces and my experiments have confirmed that if the pendulum is released from the horizontal , the highest point it will reach is , say 2 Cm below horizontal . So if we allow the pendulum pivot to drop 1 cm at bottom dead centre , then the highest point it will rise to is 3 cms below horizontal . But mathematics show that the pivot moves down with a force of 3 times the pendulum weight due to the addition of centrifugal force .If we ignore or eliminate wind drag and friction , the pendulum will rise to 1 cm below horizontal , for a pivot fall of 1cm. So to restore the system , we need to raise the pendulum bob 1cm at a time when its weight is normal , near its highest point . So if the bob weighs 1 Kg ,input is 1Kg -centimeter , and output is 3 Kg- cm , or an OU of 3 minus losses .

johnny874 · « **Reply #1838 on:** January 04, 2012, 12:21:59 AM »

Quote from: fishman on January 03, 2012, 06:58:24 PM

Tom's innovation migth be worth checking out. And as I mentioned, it is something that could support Vjelko's work.
It would be a big plus for what we are trying to do. Something to think about anyways. And Merry Christmas everyone.
Jim

@ Jim, what is "TOM's innovation"?
Any links...

fishman,
Here's the basic concept. As the pendulum swings downward, it lifts a weight at the bottom of a scissor. If the scissors (2 sections) hang down, there length is 4x (x= length of one lever). When the top scissor is pulled from both sides to open to a spread of 90 degrees (2 45 degree angles), the lateral movement is x*70% and the lift is 4x*30%.
The second section I don't think would require any extra energy to work. This would happen because as the top section opens, the lower section is mirroring it's motion while the load remains the same. Thus more movement with the same energy expenditure.
And when the fulcrum releases the weight that was lifted, it's energy can be put back into the system to keep the pendul swinging and still have a surplus which could rotate a wheel that would act as a shell to cover the mechanics much like a clock has a face.

Jim

edited to correct math. if lateral movement is equal to 70%, then lift is 30%.
And 4*.30= 1.2 which of course is more than 1*.70= 0.70
Anyway, it looks good on paper

Cloxxki · « **Reply #1839 on:** January 04, 2012, 01:01:40 AM »

Quote from: neptune on January 03, 2012, 06:59:06 PM

Hi guys , thanks for replies and a happy new year to you all . @Cloxxki .So from what you say , are we to infer that you believe the 2SO is not overunity ? If this is the case do you believe any mechanical device is OU ? I am not willing to say ay#t this stage if I think 2SO is OU or not . Obviously I think it could be , or I would not waste time on it .My take on how it might work are as follows . Friction and air drag are real forces and my experiments have confirmed that if the pendulum is released from the horizontal , the highest point it will reach is , say 2 Cm below horizontal . So if we allow the pendulum pivot to drop 1 cm at bottom dead centre , then the highest point it will rise to is 3 cms below horizontal . But mathematics show that the pivot moves down with a force of 3 times the pendulum weight due to the addition of centrifugal force .If we ignore or eliminate wind drag and friction , the pendulum will rise to 1 cm below horizontal , for a pivot fall of 1cm. So to restore the system , we need to raise the pendulum bob 1cm at a time when its weight is normal , near its highest point . So if the bob weighs 1 Kg ,input is 1Kg -centimeter , and output is 3 Kg- cm , or an OU of 3 minus losses .

I'm no-one's expert, but have been reading up on such systems for a few year, looking for the same holy grail we all are.
Of course I want to believe mechanical overunity to exist.
As I've been reading and pondering on the subject, the idea is forming that it will be quite advanced. I unfortunately lack both the technical English (2nd language) and the physics education to put that into words.
There's speed, acceleration, increased rate of acceleration, etc.
The 2SO I would put on the acceleration level.
I am suspecting that for OU to exist, it will be at least one level higher up, and locked really well.

If you're able to put into formulae what you explained above, I'd be interested to study that.

Someone who above claims 3.25x overunity from math, but can't loop is, IMO is full of it.
Make me a simulated 2SO (hydralics multiplying input) that offers just 2x OU, just as claimed, in the fashion that it's claimed, I'll find a child in the street under the age of 10 to loop it, using only Lego and similar toys. In the span of an hour. I will be supporting of course to make sure the kid doesn't rush off to the game computer prematurely.

A loose thought.
The additional tension on the pendulum can be calculated, right? Not that hard.
If you move the pivot downward at the bottom of the swing though, would that not reduce tension right there and then?

All that you extract from the pendulum and invest into the second stage, the second stage will need to give right back, or the pendulum will hang still shortly.
I do not see a free lunch. Just a lot off misinfo, empty claims, and very bright brains being stuck figuring out a puzzle possibly designed to not fit, but intrigue the best of puzzlers for an eternity, keeping them away from the next puzzle, from progress.

fishman · « **Reply #1840 on:** January 04, 2012, 10:00:28 AM »

Quote from: neptune on January 03, 2012, 06:59:06 PM

My take on how it might work are as follows . Friction and air drag are real forces and my experiments have confirmed that if the pendulum is released from the horizontal , the highest point it will reach is , say 2 Cm below horizontal . So if we allow the pendulum pivot to drop 1 cm at bottom dead centre , then the highest point it will rise to is 3 cms below horizontal . But mathematics show that the pivot moves down with a force of 3 times the pendulum weight due to the addition of centrifugal force .If we ignore or eliminate wind drag and friction , the pendulum will rise to 1 cm below horizontal , for a pivot fall of 1cm. So to restore the system , we need to raise the pendulum bob 1cm at a time when its weight is normal , near its highest point . So if the bob weighs 1 Kg ,input is 1Kg -centimeter , and output is 3 Kg- cm , or an OU of 3 minus losses .

You're kind of correct, but not completely. The key point you left out is.
1. To raise the bob 1 cm, to restore the potential energy, you can,
(A) Lift the bob cm1 which means the work is 1KG x 1 CM.
OR
(B) Push the bob downward to restore it's potential energy. This achieves the same thing at a fraction of the cost. Maybe 1/10th the cost or less.

Just ask yourself who in there right mind lifts the bob-weight to raise a pendulum? Answer, know one.
The reason we push a child on a swing set is because there is an great advantage to push with the influence/acceleration of gravity over lifting against gravity. Try it you'll quickly dsicover this to be true.

Pushing w/gravity & your "3:1" goes MANY times higher than that. (note your centrifugal force number is too high for the standard small TSO, but your point is right on.)

If you really want to give you OU test a fair shot, you need a longer swing arm, 1 meter at least, 2 meters is much better. More weight also helps a lot.
At 2 meters even if you make mistakes with the build, you should see what you are looking for.

Roland

neptune · « **Reply #1841 on:** January 04, 2012, 03:31:38 PM »

Hi guys and thanks for the intelligent replies . @ Cloxxi . Your English is excellent . Moving the pendulum pivot down does of course take energy from the pendulum . According to Marjonovic [see later] , this is minimised if we prevent the pendulum pivot from falling until just before bottom dead centre , and prevent it rising untill near top dead centre . If you want to see my last post expressed as a formula , read Marjonovic .
@ Fishman . Your theory about pushing the bob down instead of lifting it may be right . Or we may be falling into a trap , and it seems easier because it is more convenient . It is easier to lift a car with a jack , rather than by hand . And yet in theory using the jack uses more energy due to losses in the jack . On the other hand cycling uses less energy then walking , in spite of mechanical losses in the bike . Experiments are definitely needed here . When I have done all I can with my small model , I plan to make a large 2 metre model , but it will take time . Incidentally , if anyone has a large model already built and has a few minutes to spare , perhaps someone could use my method to measure efficiency . It needs only a piece of plastic or cardboard . Full details in recent posts . Any questions please ask .
Now back to Marjanovic . These papers are essential reading for serious 2SO fans . I do not know how to create links . Search on google for Jovan Marjanovic Secrets of free energy from the pendulum . His other papers are ,
1 Keys of gravity machines
2 Theory of gravity machines
3 Mechanical feedback loop .
Please do not miss these . Working together , we can lay this thing to rest , either positively or negatively . Even if we can prove it does NOT work , we can help to stop people wasting time on it .

Cloxxki · « **Reply #1842 on:** January 04, 2012, 03:37:55 PM »

Quote from: fishman on January 04, 2012, 10:00:28 AM

You're kind of correct, but not completely. The key point you left out is.
1. To raise the bob 1 cm, to restore the potential energy, you can,
(A) Lift the bob cm1 which means the work is 1KG x 1 CM.
OR
(B) Push the bob downward to restore it's potential energy. This achieves the same thing at a fraction of the cost. Maybe 1/10th the cost or less.

Just ask yourself who in there right mind lifts the bob-weight to raise a pendulum? Answer, know one.
The reason we push a child on a swing set is because there is an great advantage to push with the influence/acceleration of gravity over lifting against gravity. Try it you'll quickly dsicover this to be true.

Pushing w/gravity & your "3:1" goes MANY times higher than that. (note your centrifugal force number is too high for the standard small TSO, but your point is right on.)

If you really want to give you OU test a fair shot, you need a longer swing arm, 1 meter at least, 2 meters is much better. More weight also helps a lot.
At 2 meters even if you make mistakes with the build, you should see what you are looking for.

Roland

Sounds great, although too vague (or complicated) for me to understand.
Could you please explain when and why pushing with gravity is advantageous, and in which frame? Time, Kinetic Energy, height, etc?
Formulae for pendulums, KE and CF are quite straightforward. If you see a gain, please offer the math?

If that were to be true, the 2SO is needlessly complicated. We could extract the excess on the left side of a simple pendulum, and use only part of it to input on the right, contact gain. I've seen other inventions on that premise, but never seen measured gains.

johnny874 · « **Reply #1843 on:** January 04, 2012, 04:27:55 PM »

Unfortunately, one pendulum can't accelerate another pendulum
when it's motion is parallel to the lever transfering the force. Basically
why a 2SO will never work.

Cloxxki · « **Reply #1844 on:** January 04, 2012, 06:43:43 PM »

Quote from: neptune on January 04, 2012, 03:31:38 PM

Hi guys and thanks for the intelligent replies . @ Cloxxi . Your English is excellent . Moving the pendulum pivot down does of course take energy from the pendulum . According to Marjonovic [see later] , this is minimised if we prevent the pendulum pivot from falling until just before bottom dead centre , and prevent it rising untill near top dead centre . If you want to see my last post expressed as a formula , read Marjonovic .

I once opened a paper the group published, and even with my highschool psysics I could tell within seconds they were making up incorrect formulae with incorrect units of measurement as outcome. A waste of time. It's not up to us to school them, they are to school us.

If they know it so well, they'd build it. They sure have the building skills.
Making wild claims, and letting the world figure out a way to make it work, after which they can claim invention.

If they could point out the underlying math without (intentional or unintensional) errors, this forum would build a working device, 2SO or otherwise, instantly.
Present me a precise pendulum path and the amount of energy to be extracted from CF, and I'll tell you how to build the device to do so.
Same for the second stage (weren't we suppose to extract all those buckets of FE there?). Tell us how much to extract, and we will. Don't blame us when the result is a stalled 2SO.
This is no hint of FE, not even anomalous behavior that I'm aware of.

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Author Topic: 12 times more output than input, dual mechanical oscillation system ! (Read 2231711 times)

johnny874

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