Mathematically proven Overunity mechanism

[Rlyeh]

1 hour ago, vikram_gupta11 said:

At first calculate the torque.

A10 kg ball is resting at the bottom of a  2 meter long tube so it's distance will be 2.44 meter from fulcrum.

Now torque=r*f=2.44*10*10=240 nm

The counterweight will be 24 kg to balance the seesaw

Now torque will be same as the seesaw is balanced so calculate just 10 % of 240 nm torque as a friction loss.that will be input .

Input=24 Joule

 

This bull**** doesn't answer the question.  

How much energy are you using to push this device against the force of gravity and how are you measuring it? If you can't answer it then you haven't measured it.

Edited January 19, 2018 by Rlyeh

[Tatetopa]

12 hours ago, vikram_gupta11 said:

No friction is not an issue and it will not slow down and stop as load connection is not a problem in this mechanism as work is being  taken out  with kinetic energy of ball not with system.

How about the energy loss of the bearing, the resistance of the ball rolling in the tube, the ball moving through the air.  Those things convert gravitational potential energy to heat, not a lot but always a little.  The heat dissipates, the energy is lost for doing useful work. Third law of thermodynamics is hard to escape.  You impart energy to get  it moving, when that energy has been used up, your system stops.

[vikram_gupta11]

21 hours ago, Tatetopa said:

How about the energy loss of the bearing, the resistance of the ball rolling in the tube, the ball moving through the air.  Those things convert gravitational potential energy to heat, not a lot but always a little.  The heat dissipates, the energy is lost for doing useful work. Third law of thermodynamics is hard to escape.  You impart energy to get  it moving, when that energy has been used up, your system stops.

See the image .everything is same but I have increased the length of tube.now tube length is 4 meter and it is mounted in middle with left arm of seesaw.

Now what is input and output.

Input is mgh

So in previous design input was 200 Joule at the time of booking and ball was falling down from 2 meter height so obviously Output was 200 Joule .so input is totally free.

Ball was again falling down from 2 meter height at the time of reversing so output is 200 Joule .but you say this output will not work due to friction and losses.

Ok.but now what will you say after viewing the sketch. 

As per sketch input will be same as it was in previous design but output?

Output will be 400 Joule at the time of tilting and again 400 Joule at the time of reversing.

So total output will be 800 Joule now without increasing input .

One more important point all of you are forgetting the energy of counterweight.

When ball will hit with upper part of tube then it will exert force on counterweight equally.so output from counterweight will be also include.

If ball energy is 400 Joule then counterweight energy will be also 400 Joule.

So now total output

400 Joule at the time of tilting

400 Joule at the time of reversing

400 Joule of counterweight

Total 1200  Joule

Now. Subtract the input of previous design output which was 400 Joule

1200-400=800 Joule

Now what will you say?

 

 

 

 

 

[vikram_gupta11]

On 1/19/2018 at 7:43 PM, Rlyeh said:

This bull**** doesn't answer the question.  

How much energy are you using to push this device against the force of gravity and how are you measuring it? If you can't answer it then you haven't measured it.

See 53 no.post

[Rlyeh]

On ‎21‎/‎01‎/‎2018 at 3:19 AM, vikram_gupta11 said:

So in previous design input was 200 Joule at the time of booking and ball was falling down from 2 meter height so obviously Output was 200 Joule .so input is totally free.

Is that why in the video you had to put in more energy? Totally free. You don't seem to understand your video contradicts your math.

 

On ‎21‎/‎01‎/‎2018 at 3:19 AM, vikram_gupta11 said:

Now what will you say?

I say do the experiment, show this device running on it's own output energy.

[vikram_gupta11]

On 1/27/2018 at 0:32 PM, Rlyeh said:

Is that why in the video you had to put in more energy? Totally free. You don't seem to understand your video contradicts your math.

 

I say do the experiment, show this device running on it's own output energy.

Mr,Ryleh,

See the sketch.it is being proven mathematically and is there any need to prove it more.

As per sketch if the long tube is mounted very near to fulcrum with 10 kg weight and counterweight(13kg) is mounted 1 meter from fulcrum then I think input energy will be less.
There is a spring based brake system to store the energy of counterweight as well as to provide it momentum.
So input energy will be only 30 Joule if counterweight is 13 kg.

You just calculate input and output but remember you please calculate output energy twicely as ball will fall down twice.

You also calculate output energy of counterweight ball.

[Rlyeh]

43 minutes ago, vikram_gupta11 said:

Mr,Ryleh,

See the sketch.it is being proven mathematically and is there any need to prove it more.

The fact remains the experiment disproved your math, no amount of sketches will change this.

Your failure to perform the experiment according to your calculations only confirms you know it's wrong. These sketches are your only defence.

[Krater]

Over-unity (AKA perpetual motion) is pure fiction and physically impossible in this universe. If you don't understand this, then no one can help you (because even a 10 year child student knows this). 

Your "math" and "reasoning" are utter nonsense and you're just embarrassing yourself. You should find a simpler, more productive hobby (maybe jigsaw puzzles) because this pursuit is a complete waste of time.

 

There is no such thing as free energy. Period.

[vikram_gupta11]

1 hour ago, Rlyeh said:

The fact remains the experiment disproved your math, no amount of sketches will change this.

Your failure to perform the experiment according to your calculations only confirms you know it's wrong. These sketches are your only defence.

But tell me where I am wrong.if the mass of ball in the long tube is 10 kg and it is very near to fulcrum then counterweight will be also 10 kg.

But I have taken it 13 kg.so the device will be in vertical position.

Now there is 30 nm torque difference between long tube and counterweight.

You are forgetting that the 2 meter long tube is mounted in middle so why are you calculating 1 meter part,which is above fulcrum,in your calculations?

The above 1 meter part of tube has nothing to do with input energy.

It is very simple to prove it using mgh formula.

Use these figures and mgh formula.

[Duke Wellington]

On ‎18‎/‎01‎/‎2018 at 5:40 AM, Rlyeh said:

Why does the video show someone holding it if this seesaw is meant to be balanced?

The video contradicts your statement of no external influences and input energy.

Correct and dont forget gravity is doing work here too.

[Rlyeh]

52 minutes ago, vikram_gupta11 said:

It is very simple to prove it using mgh formula.

Use these figures and mgh formula.

The experiment disproves it, that shows your figures are wrong. 

Your video shows the device isn't generating the same amount of energy it is using, no figures you throw out is going to change this.

[+joc]

On 1/18/2018 at 8:03 AM, vikram_gupta11 said:

You will get surprised that these are not  issues in this mechanism.there is friction ,air resistance,heat but gravity will work to overcome these issues.

Friction, air resistance, heat...

I think you are missing one little thing....It is because of gravity...that these things exist at all.  Gravity cannot overcome what it is the prime factor of.

[vikram_gupta11]

1 hour ago, Rlyeh said:

The experiment disproves it, that shows your figures are wrong. 

Your video shows the device isn't generating the same amount of energy it is using, no figures you throw out is going to change this.

The purpose of video is only to understand the concept.there are several things in it.

The big mistake,what you are doing is that you are calculating the rest 1 meter part of tube above fulcrum in your calculations at the time of input energy.

[vikram_gupta11]

Just now, joc said:

Friction, air resistance, heat...

I think you are missing one little thing....It is because of gravity...that these things exist at all.  Gravity cannot overcome what it is the prime factor of.

But you are missing one point that the device is not creating energy but only ball will work to create energy so the friction ,heat,air resistance are not an issues in it.

[Rlyeh]

26 minutes ago, vikram_gupta11 said:

The purpose of video is only to understand the concept.there are several things in it.

The big mistake,what you are doing is that you are calculating the rest 1 meter part of tube above fulcrum in your calculations at the time of input energy.

The experiment doesn't support your math, there is nothing more to say.

[+joc]

59 minutes ago, vikram_gupta11 said:

But you are missing one point that the device is not creating energy but only ball will work to create energy so the friction ,heat,air resistance are not an issues in it.

Gravity will eventually stop the ball from moving.  Nice try...

...the large mistake in your work is in believing the premise that it can be done.  It cannot.  Not in an environment of gravity.  

Consider the Solar System itself.  The effects of gravity are still present, but held in check....by a mysterious thing called 'space' or dark energy.  You cannot overcome the affects of gravity.  Nice try though.  

[vikram_gupta11]

1 hour ago, Rlyeh said:

The experiment doesn't support your math, there is nothing more to say.

Ok.then you calculate input and output  and post your calculations.

[vikram_gupta11]

1 hour ago, joc said:

Gravity will eventually stop the ball from moving.  Nice try...

...the large mistake in your work is in believing the premise that it can be done.  It cannot.  Not in an environment of gravity.  

Consider the So nolar System itself.  The effects of gravity are still present, but held in check....by a mysterious thing called 'space' or dark energy.  You cannot overcome the affects of gravity.  Nice try though.  

You are forgetting one point if there is overunity then it will not stop as after each cycle the energy will be created more and more and more and more so ...there is no question that these issues will create hurdle in it as this excess energy will work to overcome these issues after each cycle as input is very  less compare  than output.

 

Edited January 28, 2018 by vikram_gupta11

[vikram_gupta11]

1 hour ago, joc said:

Gravity will eventually stop the ball from moving.  Nice try...

...the large mistake in your work is in believing the premise that it can be done.  It cannot.  Not in an environment of gravity.  

Consider the Solar System itself.  The effects of gravity are still present, but held in check....by a mysterious thing called 'space' or dark energy.  You cannot overcome the affects of gravity.  Nice try though.  

If the formula of potential energy mgh is correct then no one can debunk it as this formula is clearly showing Overunity in this mechanism. 

There is nothing beyond this formula.

I will win in both condition s.if it works or doesn't . if it doesn't work then formula of potential energy or kinetic energy will be proven wrong.

 

 

Edited January 28, 2018 by vikram_gupta11

[Rlyeh]

46 minutes ago, vikram_gupta11 said:

Ok.then you calculate input and output  and post your calculations.

Why should I? It's your experiment that doesn't match your calculations. 

The experiment failed and rather than changing anything you continue to drum on about your calculations. 

Edited January 28, 2018 by Rlyeh

[toast]

 

[vikram_gupta11]

40 minutes ago, Rlyeh said:

Why should I? It's your experiment that doesn't match your calculations. 

The experiment failed and rather than changing anything you continue to drum on about your calculations. 

I have already mentioned that the input is one time but output is three time in this mechanism.

My calculations are very much correct but you are not trying to understand it.

Why don't you post calculations if you are not agree with me.

The experiment is to understand the mechanism.

Post your calculations as I want to know your mathematical ability.

[Rlyeh]

1 minute ago, vikram_gupta11 said:

I have already mentioned that the input is one time but output is three time in this mechanism.

The experiment doesn't reflect this.

 

1 minute ago, vikram_gupta11 said:

My calculations are very much correct but you are not trying to understand it.

Yet the experiment contradicts your calculations.

 

1 minute ago, vikram_gupta11 said:

Why don't you post calculations if you are not agree with me.

The experiment is to understand the mechanism.

Post your calculations as I want to know your mathematical ability.

That doesn't make sense. You built the device and performed the experiment that disproves your calculations. What does that say about your mathematical ability?

[DarkHunter]

Since no one else has seem to done this yet I'll do it.  Using simple dynamics and physics it's really easy to disprove this whole system even though the rather poor drawings do make it a bit challenging as the counter weight, either a mass or second ball in a tube, seems to magically rotate.

To start off might as well begin with the first diagram on this page, assuming my post is on page 3.  

You said the tube was 4 meters in length and I believe the mass of the ball was mentioned to be 10 kg, even though it won't matter as the counter weight is the same.  When the ball is dropped it will fall 4 meters and in that 4 meters it will hit the bottom at some velocity.  Using the kinematic equation of Vf^2=Vi^2+2*A*dy you can get the velocity of the ball as it hits to be 8.85 m/s.  From knowing the velocity and mass you can get the momentum of this ball which is p=m*v which in this case is 88.5 kg*m/s.  Momentum is conserved so using the relation m1v1 + m2v2 = m1'v1' + m2'v2' you can get the velocity that the mass and ball will be moving at after impact which would be 88.5 kg*m/s + 0 = 10*v1' + 10*v2' which since they are connected by a rigid object simplify down to 88.5 kg*m/s = 20*v' as both objects need to be traveling at the same velocity, which will be 4.425 m/s.  The significance of finding the new velocity of both is that the mass has to rise some height to reset this mechanism.  Not having the length from the masses to the fulcrum makes this next part more of an estimation, but to reset this mechanism the counter weight will have to rise up in height the distance from the mass to the fulcrum.  The energy required to do this is found using the gravitional potential energy equation so you would need 9.8 m/s^2 * 10 kg * h plus a tiny bit more to get it to fall.  The energy the counter weight has on hand though we do know since it's the kinematic energy equation which is 1/2*m*v^2 which ends up being 97.9 J.  Only having 97.9 J of energy to move the counter weight is a problem as energy required to reset it will be 98*h J of energy.

This took longer then I thought it would, and being busy, I'll have to do the seco d drawing later tonight or tomorrow.

[ChrLzs]

Darkhunter, kudos to you for taking the time to show the PROPER way to go about calculating momentum/energy.  Sadly, I suspect Vikram will completely ignore this and go back to his patented mgh equation, which he is misapplying horrendously.

Vikram has been here before - never listens, never learns.  Even though he himself proves his devices don't work - if they did he'd be showing them running without input....

 

But hopefully other aspiring toy makers will read your stuff and start taking the whole thing a little more seriously than Vikram does.