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vikram_gupta11

Mathematically proven Overunity mechanism

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Rlyeh
1 minute ago, vikram_gupta11 said:

It doesn't matter that seesaw is balanced or not as it will get reversed .if it is getting it's initial position due to gravity then where do you see external influence? Why are you not considering that at the time of reversing it is not using energy.so what is this?

If it doesn't matter why did you state it?

You are tipping or pushing the device with your free hand, gravity then causes it to fall back to it's original position. Is that really so hard to understand?

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Rlyeh
4 minutes ago, vikram_gupta11 said:

This is using chemical energy.

It's using thermal and gravitational energy.

https://science.howstuffworks.com/innovation/science-questions/question608.htm

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I'mConvinced

Let us for a moment imagine this works. Where do you think the 'free' energy is coming from? Gravity can't impart more energy than is needed to overcome itself, by itself.

Please bare in mind that I'm an average joe without any background or qualification in physics so please keep the maths simple.

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vikram_gupta11
1 hour ago, Rlyeh said:

If it doesn't matter why did you state it?

You are tipping or pushing the device with your free hand, gravity then causes it to fall back to it's original position. Is that really so hard to understand?

I am tilting it because it is necessary but why are you forgetting that to the input is very minimal to tilt it.why are you not understanding it? The seesaw is balanced so input is very minimal but output?

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vikram_gupta11
1 hour ago, I'mConvinced said:

Let us for a moment imagine this works. Where do you think the 'free' energy is coming from? Gravity can't impart more energy than is needed to overcome itself, by itself.

Please bare in mind that I'm an average joe without any background or qualification in physics so please keep the maths simple.

This free energy will come from gravitational field as potential energy has been increased .Effort will be very less due to counterweight as I have already added energy in counterweight.

The math is simple.

Use mgh formula

Suppose the weight of ball is 10 kg.and it is falling down from 2 meter height

So mgh= 10*10*2=200 joule

But device will be reversed due to counterweight and gravity so ball will fall down again from 2 meter height

So mgh=10*10*2=200 joule

P.E.=K.E

So total output will be 400 joule

Now input

Though input is not zero but it will be almost zero so there is nothing to think about input.

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Rlyeh
14 minutes ago, vikram_gupta11 said:

I am tilting it because it is necessary but why are you forgetting that to the input is very minimal to tilt it.why are you not understanding it? The seesaw is balanced so input is very minimal but output?

Clearly not that minimal if I can see it as plain as day.

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vikram_gupta11
9 minutes ago, Rlyeh said:

Clearly not that minimal if I can see it as plain as day.

So tell me how much energy will be required to tilt down a balanced seesaw?but remember balanced seesaw.

Consult with someone who has knowledge of physics.

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I'mConvinced
1 hour ago, vikram_gupta11 said:

The math is simple

I think this is where you're going wrong. The reason you can't get it to work is because you are oversimplifying the maths. Where are your equations that take into account the energy losses in the system and how do you account for those?

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vikram_gupta11
21 minutes ago, I'mConvinced said:

I think this is where you're going wrong. The reason you can't get it to work is because you are oversimplifying the maths. Where are your equations that take into account the energy losses in the system and how do you account for those?

I think this where you are not understanding it.if input is minimal and output is 400 Joule then just minus 15% or 20% though it will be not more than10% as a mechanical losses.now calculate output.i didn't show it as I thought you will understand it.

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I'mConvinced
Just now, vikram_gupta11 said:

I think this where you are not understanding it.if input is minimal and output is 400 Joule then just minus 15% or 20% though it will be not more than10% as a mechanical losses.now calculate output.i didn't show it as I thought you will understand it.

Input is not minimal and output will not be 400 joules. The ball might produce this energy when falling but there is no way to capture that energy without loss. 

If you are so convinced of this minimal input then do the math for us here, including accounting for losses in the system. Then produce a working model that gets faster as it goes without any external input. At this point we will get excited.

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vikram_gupta11
6 minutes ago, I'mConvinced said:

Input is not minimal and output will not be 400 joules. The ball might produce this energy when falling but there is no way to capture that energy without loss. 

If you are so convinced of this minimal input then do the math for us here, including accounting for losses in the system. Then produce a working model that gets faster as it goes without any external input. At this point we will get excited.

Maths is very simple.how many times I will tell you that if a seesaw is balanced the there will be need of minimal energy due to same torque.you can add 10% frictional losses in input so it will be 20 Joule but this will be depend on ball bearing.if I use magnetic bearing then it will be more less.

I have already mentioned that a piston generator works with 50 % efficiency,however,output will be more.

I am considering a spring based generator as spring can restore 99% energy and can increase the velocity of ball.

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I'mConvinced
11 minutes ago, vikram_gupta11 said:

Maths is very simple.how many times I will tell you that if a seesaw is balanced the there will be need of minimal energy due to same torque.you can add 10% frictional losses in input so it will be 20 Joule but this will be depend on ball bearing.if I use magnetic bearing then it will be more less.

I have already mentioned that a piston generator works with 50 % efficiency,however,output will be more.

I am considering a spring based generator as spring can restore 99% energy and can increase the velocity of ball.

Cool, then all we are missing is your working machine. I look forward to seeing more of you in the future once you've unveiled it to the world and become a superstar.

I'll start a book pulping company in the meantime, I'll make a killing when they have to rewrite all those textbooks!

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vikram_gupta11
Just now, I'mConvinced said:

Cool, then all we are missing is your working machine. I look forward to seeing more of you in the future once you've unveiled it to the world and become a superstar.

I'll start a book pulping company in the meantime, I'll make a killing when they have to rewrite all those textbooks!

Ok.Thanks

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ChrLzs

Note that I am simply posting this for any other 'free-energy' or perpetual-motion or over-unity devices.

Maths can be simple or COMPLEX.  It obviously depends on the situation, and to just say that it always simple is a very, very stupid thing to say.  If you do not account for the entire process, losses (eg friction) and all external and internal outputs, then your maths is not only completely wrong, it also shows that you are way out of your depth.

In this case, the biggest problem is that the OP has no idea how to describe the entire process or to map the mathematics properly to that process, and in fact you can see from what was posted, he has completely left out a hugely important step in the process.

There are MORE problems in the 'maths', but I'll just focus on the biggest howler...

4 hours ago, vikram_gupta11 said:

This free energy will come from gravitational field as potential energy has been increased.

No, it won't.  You will lose exactly what you gain from gravity

4 hours ago, vikram_gupta11 said:

Effort will be very less due to counterweight as I have already added energy in counterweight.

And if do the maths properly, which clearly Vikram will never do.., you will see that stuff like "effort will be very less" is just meaningless word salad

4 hours ago, vikram_gupta11 said:

The math is simple.

NO, it isn't.

4 hours ago, vikram_gupta11 said:

Use mgh formula

Vikram's 'mgh formula' is the one for gravitational potential energy.  It's not completely correctly used here, if you want to estimate the energy output of the device, but no matter, let's assume it was...

4 hours ago, vikram_gupta11 said:

Suppose the weight of ball is 10 kg.and it is falling down from 2 meter height

So mgh= 10*10*2=200 joule

So Vikram is calculating the potential energy that is available from letting the ball drop AFTER raising it up.  It's near enough, but does he think the ball got up there by ... magic?  If you read on, clearly that's exactly what he thinks.

4 hours ago, vikram_gupta11 said:

But device will be reversed due to counterweight and gravity so ball will fall down again from 2 meter height

So mgh=10*10*2=200 joule

P.E.=K.E

So total output will be 400 joule

WOAH.  Stop right there.  Why didn't Vikram discuss how the ball got back up?  Let alone how the ball got up there before the first 'cycle'? Where did that energy come from, and why didn't he include it in his 'calculations'?

He completely ignores any energy inputs he provides.   Could that omission possibly be from lack of knowledge?  If so, that is just breathtakingly ignorant.  If it is an attempt at a scam, then it has received the reception it deserves.

Though input is not zero but it will be almost zero so there is nothing to think about input.

Seriously, that is face-palmingly wrong.  Anyway, it is clear that Vikram is not making any serious attempt at anything even remotely credible.  These sort of threads are really a complete waste of time, unless someone ELSE learns from them....

 

If anyone wants to delve further into the 'wrongness', let me know.  Vikram, don't bother asking for help - you don't listen to anyone anyway.

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Rlyeh
10 hours ago, vikram_gupta11 said:

So tell me how much energy will be required to tilt down a balanced seesaw?but remember balanced seesaw.

Consult with someone who has knowledge of physics.

That's your job, you're doing this experiment.

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vikram_gupta11
5 hours ago, ChrLzs said:

Note that I am simply posting this for any other 'free-energy' or perpetual-motion or over-unity devices.

Maths can be simple or COMPLEX.  It obviously depends on the situation, and to just say that it always simple is a very, very stupid thing to say.  If you do not account for the entire process, losses (eg friction) and all external and internal outputs, then your maths is not only completely wrong, it also shows that you are way out of your depth.

In this case, the biggest problem is that the OP has no idea how to describe the entire process or to map the mathematics properly to that process, and in fact you can see from what was posted, he has completely left out a hugely important step in the process.

There are MORE problems in the 'maths', but I'll just focus on the biggest howler...

No, it won't.  You will lose exactly what you gain from gravity

And if do the maths properly, which clearly Vikram will never do.., you will see that stuff like "effort will be very less" is just meaningless word salad

NO, it isn't.

Vikram's 'mgh formula' is the one for gravitational potential energy.  It's not completely correctly used here, if you want to estimate the energy output of the device, but no matter, let's assume it was...

So Vikram is calculating the potential energy that is available from letting the ball drop AFTER raising it up.  It's near enough, but does he think the ball got up there by ... magic?  If you read on, clearly that's exactly what he thinks.

WOAH.  Stop right there.  Why didn't Vikram discuss how the ball got back up?  Let alone how the ball got up there before the first 'cycle'? Where did that energy come from, and why didn't he include it in his 'calculations'?

He completely ignores any energy inputs he provides.   Could that omission possibly be from lack of knowledge?  If so, that is just breathtakingly ignorant.  If it is an attempt at a scam, then it has received the reception it deserves.

Seriously, that is face-palmingly wrong.  Anyway, it is clear that Vikram is not making any serious attempt at anything even remotely credible.  These sort of threads are really a complete waste of time, unless someone ELSE learns from them....

 

If anyone wants to delve further into the 'wrongness', let me know.  Vikram, don't bother asking for help - you don't listen to anyone anyway.

Hello,I think you have not taken a ride on a seesaw in your childhood that's why you are talking like a child.

Have you ever seen a conventional balanced system to weight stuffs?

If not then try.

Now I come to the point.

The input is 20 Joule after calculating 10% frictional loss.

Now you say that the ball is getting the height magically.

Yes there is magic and I have mentioned this magic several times that the seesaw is balanced and a pin is working to hold the ball .

Once the tube get certain angle the ball will fall down from 2 meter height 

So output is 200 Joule 

But you are again forgetting one thing that ball is again falling from 2 meter height at the time of reversing and this time there is no need of extra energy as gravity is working.

So again output is 200 Joule.

So total output is 400 Joule but 

Total input is 20 Joule.

It doesn't matter that ball is mounted 2 meter below from fulcrum as I have balanced the seesaw by already adding the counterweight.if you are thinking that counterweight will be also 10 kg .then you are wrong as it will be more than 10 kg.to balance the seesaw.

Once ball hit the upper part of tube after tilting then the counterweight will become heavy due to distance from fulcrum so it will work to reverse the device .

I am not scammer.iam presenting it on the basis of pure physics.

You please consult with any Physicist as it will be better for you to understand what I want to say.as a physicist will understand it quickly.

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vikram_gupta11
Just now, Rlyeh said:

That's your job, you're doing this experiment.

I have mentioned the answer and if you think that Im asking you then it means not that I don't know but I want to convince all of you.

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Rlyeh
Just now, vikram_gupta11 said:

I have mentioned the answer and if you think that Im asking you then it means not that I don't know but I want to convince all of you.

Nearly zero isn't an answer.

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vikram_gupta11
9 minutes ago, Rlyeh said:

Nearly zero isn't an answer.

See  no.41 post

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Rlyeh
1 hour ago, vikram_gupta11 said:

See  no.41 post

How did you arrive at 10 joules?

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Tatetopa

You can't take work out of this system.  Over time, it will slow due to friction.  If you ty to take work out, it will slow down and stop sooner I believe.

 

Best regards.

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vikram_gupta11
54 minutes ago, Rlyeh said:

How did you arrive at 10 joules?

It is 10%not 10 Joule .this 10% is frictional losses and it is valid in any kind of system.

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vikram_gupta11
32 minutes ago, Tatetopa said:

You can't take work out of this system.  Over time, it will slow due to friction.  If you ty to take work out, it will slow down and stop sooner I believe.

 

Best regards.

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.

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Rlyeh
46 minutes ago, vikram_gupta11 said:

It is 10%not 10 Joule .this 10% is frictional losses and it is valid in any kind of system.

How are you measuring the input?

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vikram_gupta11
4 hours ago, Rlyeh said:

How are you measuring the input?

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

 

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