[The Spoon]

The spoon,Bored Chemist,
Both of you are not understanding is that you are taking it in wrong direction.
Let me clarify all things.
The scale length is 2 meter.
There is a 2 or 4 meter long tube.
Let's take tube length 2 meter.
This 2 meter long tube is mounted on left arm of seesaw in middle so it's 1 meter part is above fulcrum and 1 meter is below fulcrum.
A 10 kg.ball is resting at the bottom of the tube.
Now counterweight.
After calculating distance from fulcrum the counterweight will be 14.4 kg.to balance the seesaw as the distance of ball is 1.44 meter from fulcrum as the ball is in rest position in the tube.
Now the seesaw is in balanced position.
But I will take counterweight 20 kg.due to this excess counterweight the seesaw will be in tilting position at 90 degree angle.
Now when I lift up the seesaw then I will not have to calculate the input energy of  the 1 meter distance of tube which is mounted below fulcrum as I have added mass in counterweight.
Now calculate input using mgh formula
Mgh=10*10*1=100 Joule
But calculate output as ball will fall down from 2 meter height
Mgh= 10*10*2=200 Joule
Ball will again fall down from 2 meter height at the time of reversing
Mgh=10*10*2=200 Joule
So total output is 400 Joule but input is 100 Joule.
The very interesting point is that it is not important that the device must complete one cycle as ball has completed one cycle.
See the link.you can see momentum in the video.

It is obvious in that video you have to push the arm with the ball up every time. Therefore you are constantly inputting energy. You are not getting energy out.
Let's be clear, the energy released by the ball falling is no greater than the energy used to lift the ball. The formula you submitted is nonsense. You claim the ball fall 2 metres, but is only lifted 1 metre. The is logically impossible.

[vkrmvkrm 11 (OP)]

The spoon,
What I told you is very much correct.there will be no need of lifting the ball from 2 meter height as I have added the mass in counterweight.so the input will be required to lift the 1 meter only.
It is logically correct.if I will lift up the ball from 2 meter then why I will increase the counterweight?counterweight  must be 10 kg .but no ,counterweight will be increased to balance the seesaw .so there will be no need of lifting up the seesaw.

[The Spoon]

The spoon,
What I told you is very much correct.there will be no need of lifting the ball from 2 meter height as I have added the mass in counterweight.so the input will be required to lift the 1 meter only.
It is logically correct.if I will lift up the ball from 2 meter then why I will increase the counterweight?counterweight  must be 10 kg .but no ,counterweight will be increased to balance the seesaw .so there will be no need of lifting up the seesaw.

It makes no difference if you have a counter weight or not. If the ball falls 2m, then its start position is raised 2m. The counterweight is irrelevant. If a counterweight is present then part of the energy of the falling ball will be used to lift the counterweight. If the ball and counterwieght are perfectly balanced then the system will act like a fly wheel. There is no energy gain. 

[The Spoon]

And you keep stating that you are correct without providing any evidence.

[vkrmvkrm 11 (OP)]

The spoon,
See the sketch carefully .in the sketch the tube is mounted in middle with left arm so the ball is resting at the bottom of tube.the 1 part of tube which is above fulcrum will be not calculated as so the height will be only 1 meter but ball will fall down 2 meter.you can consult with any one on this point.the 1 meter part of tube mounted above fulcrum has nothing to do with input but output.

[vkrmvkrm 11 (OP)]

What I have not proven?these two video are sufficient to prove Overunity in this mechanism.
If there is no overunity then the formulas of mgh and 0.5 mv^2 are wrong.

[The Spoon]

The spoon,
See the sketch carefully .in the sketch the tube is mounted in middle with left arm so the ball is resting at the bottom of tube.the 1 part of tube which is above fulcrum will be not calculated as so the height will be only 1 meter but ball will fall down 2 meter.you can consult with any one on this point.the 1 meter part of tube mounted above fulcrum has nothing to do with input but output..

Yes, but when the ball falls, it has to travel back to its start position. At its lowest, it will be 1m below the fulcrum point, and at it highest 1m above the fulcrum point. Therefore, after it has been lifted initially and is set in motion, it has to gain 2m height.

Lets ask anybody on here shall we?

[The Spoon]

What I have not proven?these two video are sufficient to prove Overunity in this mechanism.
If there is no overunity then the formulas of mgh and 0.5 mv^2 are wrong.

No they are not. They show you lifting a tube and letting it go. It then stops moving until you lift it again. If you think that is proof then you are more deluded than I thought in the first place.

[Bored chemist]

The main purpose is to get more output than input.

Then don't come back until you have measured both.

[Bored chemist]

What I have not proven?

What you have not proven is that the energy out is more than the energy you use in lifting it.
And that's the only thing that would make this thread worthwhile.

Without proving that,  your video is no more than bad "performance art".

[Kryptid]

Bored chemist,
The main purpose is to get more output than input.
But tell me one thing if it move continuously then can I get more output than input.No.the output will be same in both condition if it moves continuously or it stop and will need to push it up again and again.

Here's another way to do this without a generator: build two of these machines and connect them to each other such that the energy released by the downswing on one device is used to power the upswing on the other device. If the energy released by the ball falling is truly more than the energy needed to lift it back up again, then these two devices will be able to power each other endlessly.

[vkrmvkrm 11 (OP)]

As per video at first see the initial position of device.the Device is positioned at some angle.so I will have to tilt it only 5 cm and ball will fall down.

I am using my finger as a lock system so it will work to prevent the device from counter rotating. The latch pin in the tube will work to hold the ball and ball will fall down only after the device get 180 degree angle.

The ball(10 kg) is falling down from 2 meter height so it's kinetic energy will be 200 Joule. The counterweight is some heavy then ball+ tube.so the device will be in tilting position initially.

This impact energy will work to provide momentum as the counter weight will be lifted up.and device will rotate.

It is very simple.

Even the input energy can be reduced. The input is 13 Joule as per mgh formula even I entire mass of device is 26 kg. But output is 200 Joule. The impact energy of the ball can be extract using piston generator and this energy will be used again to tilt the device.

But I would like to tell that the device will not rotate in a circle but oscillate but it doesn't matter as output will be same in both cases if it complete one cycle or doesn't.

Hope this time all of you will understand the whole mechanism.

You can see the length of tube which is just 20 cm.

But in kinetic energy formula time is in square.

You can also see momentum in this second video.the video is just for understanding the mechanism.

[vkrmvkrm 11 (OP)]

I would like to tell that I am interested in impact energy of ball as all of you will say that it will not oscillate forever due to friction and other losses but a piston generator mounted on each side of long tube will work to generate energy and this energy can be used to feed the lever again and again to tilt the device. –
I have tested it.when I released the ball then the ball did two work at a time.it compressed a spring and provide the momentum so mounting generator will work to solve the all problems regarding LOSSES.
It will run at constant speed so I don't think that Noether theorem will create any hurdle in it. –

[Bored chemist]

t will run at constant speed so I don't think that Noether theorem will create any hurdle in it. –

Then you haven't fully understood the theorem.

As Kryptid explained, it's  perfectly simple to convince us that you are right. Build two and have each drive the other. If it's over-unity then the pair will keep going indefinitely.
If not, well, at least you learned something.

[vkrmvkrm 11 (OP)]

Bored Chemist,
I will build it definitely.but you don't think that is our physics not so mature that we could predict the feasibility if it is being proven mathematically.
Math is universal truth and there is nothing beyond mgh formula.

[vkrmvkrm 11 (OP)]

Bored Chemist, I have understood it very correctly but I think this time you should consult with some Physicist .

[Bored chemist]

Bored Chemist, I have understood it very correctly but I think this time you should consult with some Physicist .

Prove it.
Set up the machine + show it working.

[vkrmvkrm 11 (OP)]

Bored Chemist,
Very soon

[Bored chemist]

Bored Chemist,
Very soon

I will buy some popcorn.