Milkovic Two Stage Oscillator - Dec 2008

[graham]

Hi Ralph, here is a quote from your post :

Starting with a balanced beam... on the up swing the pendulum 'appears' to lose weight, thus allowing the output end to fall...

So right here off the bat there's a problem.

If the output end falls then the pendulum end will rise and in doing so will rob the pendulum of some of its angular velocity resulting in a lower starting point for its downswing.(we're already in trouble). A little push by hand helps here.

Do I have to proceed ?

Graham

[ruggerodk]

Hi Graham,

I disagree on your conclusion.

and believe my point is clearly demonstrated by Nebrojsa Simin: ( http://www.veljkomilkovic.com/Oscilacij ... escription ) under "Excess" Energy

Quote: "It practically makes no difference whether the lever is doing something or not. That work does not reflect the oscillation of the pendulum, and the pendulum in return, as it oscillates, makes the lever to oscillate too."

A good read is "Mechanical Feedback Loop Problems..." or the "How To Pump a Swing" by Tareq Ahmed Mokheimer...

For elevation and continous pemdulum movements I find this link to the "Cyclostyle Ball Clock" extremely interesting:
http://www.betrisey.ch/ecyclo.htm

regards
Ruggero ;-)

[greendoor]

Quote: "It practically makes no difference whether the lever is doing something or not. That work does not reflect the oscillation of the pendulum, and the pendulum in return, as it oscillates, makes the lever to oscillate too."

I think that might be more of a wish than a statement of fact.

If the output end falls then the pendulum end will rise and in doing so will rob the pendulum of some of its angular velocity resulting in a lower starting point for its downswing.(

graham has phrased what I was trying to say more succinctly.

However - the beam bounces back up, because the pendulum weight is cycling between weightless and double weight. As long as the beam returns to it's starting position with each pulse, I think this counterbalances the argument that angular velocity is being lost.

I may be wrong in my assumption that a motorised 360 degrees rotation can emulate a pendulum effect. With a pendulum, the time when the mass is weightless is obviously the point where the mass changes direction at the highest point. I'm thinking a rotary motion that doesn't change direction may or may not cause a weightless peak - maybe depending on actual speed .. ?? Probably opens up a whole nuther can of works.

I think maybe it's better to stick with a pendulum, and power it with a solenoid to supply timed impulses. An optical sensor could provide the feedback for timing purposes to synchronise the impact. The power input could then be measured accurately.

Somehow, I think the Milkovic water pump might be similar to an experience of somebody I work with who recently rode a modern racing push bike. He couldn't believe how easy it was to ride the thing - he was really surprised. Simply by providing optimum gearing/leverage, one human can supply an amazing amount of energy if it is made easy and fun. I think this device can make pumping a lot of water very easy and fun, which is a good thing anyway.

I have both doubt and hope - i'm not satisfied yet that all the power is provided by the hand. The maths are just a bit beyond me. The leverage effect is easy to see and calculate. But the total power input and output needs a lot of calculus of areas under sine wave graphs, and i'm a bit lost.

It's probably fair comment that this device has been talked about for years and nothing has come of it, so it's probably a lemon.

[ruggerodk]

Dear Greendoor,

I really like your 'lemon'...;-) It reminds me of a 'mouse-wheel'...

Anyway - Though I do agree with you that the pendulum loose both velocity and height after each cycle, I was just making the point of fact, that the loss of velocity after each swing-cycle is not caused by the mainlever's fall (of the hammer side) and not by the opposing rise of the pendulum side, but is due to the natural physics of pendulum.

The pendulum will loose the same amount of velocity whether or not it's mounted to the lever or not.

Regards
ruggero ;-)

[Gravitronic]

Well I did a query and here is the response I received. This response is from a friend and contact involved with the Milkovic research.

Keep in mind that there are two power impulses to the beam.

Starting with a balanced beam... on the up swing the pendulum 'appears' to lose weight, thus allowing the output end to fall...
but when the pendulum swings back down, the fall under gravity makes the pendulum heavier than when it is at rest, driving the
output end of the arm up.

To illustrate... take a bucket of water and swing it by your side, when the bucket is straight out there is a pull straight out on your arm,
for sure, but how much does the bucket weigh? nothing.

Now stand on the mall roof and have your buddy drop the bucket of water over the side on a rope... how much does the bucket weigh
at the end of a 10 foot drop? enough to pull you off the roof, right?

So these are the two input forces to the arm. I am not familiar with the 74 pound weight lift experiment, so hesitate to comment on it but it depends on whether this is including part of the counter balance weight? If it does not... then there is no weight transfer between sides. But if this is part of the counter balance then there will be... The example problem needs to be more clearly defined.

Here's what I see happening and it is in regard to the quote above regarding the bucket. I used a ball instead but I think the idea is the same.

Quite simply a lever with equal length sides will exert a force on the output side equal to that applied on the input side. Everybody knows that! Now imagine this for a moment... attach a steel ball of ten pounds to one side of the lever with, lets say a piece of aircraft cable with a ten thousand pound rating. Drop that weight from a height that will allow it to reach terminal velocity. When the weight brings the cable tight it pulls the pendulum down. The force due to acceleration is greater than simply what would be ten pounds of force. Now since the Milkovic device is using a rigid member and not a cable. Some of the energy from the falling mass is lost because it's rate of decent diminishes due to the the fact that it begins to rotate on it's axis.

The energy from the impact is there, we just don't observe it as a collision. It is what is known as an "inelastic collision". Ramond Head shows in a couple of videos

http://www.youtube.com/watch?v=EQv1Z6Tbyus,
http://www.youtube.com/watch?v=EQv1Z6Tbyus,

that with a lever in a 3.4 to one ratio and the long side being the work side, that the Milkovic device is still able to lift more on the long side than what is inputted on the short side. This is in contrast to the way a lever should work. It should actually output less.

The reason as I see it is because the force of ten pounds acting on a lever is additive with the force of the impact collision that we never really notice but is there. Without getting too crazy with a bunch of numbers, Imagine all the forces as static just for a moment.

Ten pounds of static force plus the force of the inelastic collision, minus the losses of the inelastic collision due to diminishment from the rotation of the pendulum's column is greater than ten pounds of static force.

This is why this device behaves the way it does as Raymond Head has showed. This device will self run as I see it with the proper mechanical configuration. Someone just needs to figure out how and I think I have it figured out as you will all soon see.

[Besslers Assistant]

To Greendoor

Thank you for highlighting the Two stage oscillation, I like the comedy of the beginning statements form graham and Michael. I am very slow to understand Mechanical mechanisms, but when I do I can manipulate it very easily. The two stage oscillation has two environments and that is inside the pendulum and outside the pendulum. Milkovic is working outside the pendulum, but Bessler worked inside the pendulum. Think of a weight held by a spring that is inside a hollow pendulum. When the pendulum is swung the weight inside will bob up and down. At rest the spring experiences 'mg', when swung at 60 degrees the spring experiences '2mg', at 90 degrees '3mg' and so on. The trick is to get this internal force to keep the pendulum going. Bessler figured this out through his views of mechanics and clockwork mechanisms, and I think if we apply the modern mechanisms to this two stage oscillation there will be many different design of PM devices.

[Grimer]

Sounds good to me. Resonates with what I see going on in the Rubber Band Motor.

[nicbordeaux]

The reason that Mr Milkovic insists on hand input is that the power input moment (and intensity) varies if minimal input for a given result is required. Can't tell you why really.

To make that simpler, the timing is not constant, or to be more precise, exactly predictable. Or, one could say the that the system is mildly chaotic. Chaotic and unpredictable in the way that a double arm pendulum is, albeit to a lesser extent.