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Free energy with Seesaw and kinetic movement


vikram_gupta11

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10 hours ago, vikram_gupta11 said:

Dear Sir,

I have tested the device only with tubes and with balls and I have already mentioned that I don't have piston generators to get electricity out from the system .you can try it with some balls and tubes and see the result.

Then show us your tests, please.  Recreate them and record them.  Otherwise, obey the laws of thermodynamics like all other humans must do and live a happy life.

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Ok, it doesn't sound like it makes any sense, but +5 for how polite you are. 

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Seriously, this has gone to 4 pages?

Perhaps it would serve the OP in corrective understanding by his posting his position to www.physicsforums.com

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On 11/3/2016 at 10:25 AM, vikram_gupta11 said:

<snip>

.you are talking about only one ball + tube system but ignoring one point that overall output will be greater than input.

<snip>

It is physically impossible to have an output larger than the input power. Physically impossible. I suggest you get a primer on physics and in the same breath *you* visit a renowned physics professor instead of asking us to do your work. Your are simply and, unequivocally, wrong. It is simply not possible anyway you slice it. 

You clearly do know very little, if anything of even the most basic of physics. Please do not waste our time.

Cheers,
Badeskov 

 

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7 hours ago, badeskov said:

It is physically impossible to have an output larger than the input power. Physically impossible. I suggest you get a primer on physics and in the same breath *you* visit a renowned physics professor instead of asking us to do your work. Your are simply and, unequivocally, wrong. It is simply not possible anyway you slice it. 

You clearly do know very little, if anything of even the most basic of physics. Please do not waste our time.

Cheers,
Badeskov 

 

Dear sir,

you are not understanding the concept .the applied force is equally distributing or each ball is getting equal amount of force  therefore overall output will be grater than input.

I'm recreating the model and will present a working model of this design.

 

Vikram

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9 minutes ago, vikram_gupta11 said:

Dear sir,

you are not understanding the concept .the applied force is equally distributing or each ball is getting equal amount of force  therefore overall output will be grater than input.

I'm recreating the model and will present a working model of this design.

 

Vikram

Oh, I understand the concept very well. From a physics perspective it is very simple and that is why posters (myself included) can tell you that you are simply wrong. You seriously need to learn some basic physics. Like very, very basic. And some really basic measurement skills, as you have obviously no idea of what you are actually looking at.

And, no, it is not up to us to do your work. Your claim, you need to prove it. It is that simple.

Cheers,
Badeskov

Added question: please explain to us in detail the amount of potential energy required to lift every ball in said tubes when one end of the seesaw goes up and then please explain to us where that energy originates.

Edited by badeskov
Added question
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10 hours ago, vikram_gupta11 said:

Dear sir,

you are not understanding the concept .the applied force is equally distributing or each ball is getting equal amount of force  therefore overall output will be grater than input.

I'm recreating the model and will present a working model of this design.

 

Vikram

With your design the force is not being equally distributed to each of the balls.  

The force equation for this design is Fe = Fl*dl/de where Fe is the effort force to move it, FL is the force of the load (weight), dl is the distance from the load to fulcrum and de is the distance from effort force to fulcrum.  But that is for only one load and one effort force while your design would have 5 effort forces and 5 load forces, or to put it mathematically it would be Σ Fe = Σ Fl*dl/de.  To simplify this I'm only going to focus on the right side of the equation and assume a constant arbitrary value for Fe cause if I don't it starts to get complicated since I would have to factor in different distance for each of the effort forces and potentially add in some time factor if all the balls wouldn't hit the lever at the same time. 

Using the greatly, if slightly unrealistic, simplified approach the equation becomes Fe = Σ Fl*dl/de which can also be written as Fe = Fl1*dl1/de1 + Fl2*dl2/de2 + Fl3*dl3/de3 + Fl4*dl4/de4 + Fl5*dl5/de5.  From this you should be able to see two very important things.  First that the force effort is a summation of force loads so no 1 force load can equal the force effort and second that distance from the fulcrum changes the distribution of the force so ones further away from the fulcrum get more force then those closer to the fulcrum.

Edited by DarkHunter
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If you really want free energy steal it from your neighbor.:whistle::lol: (serious don't do that)

The subject of free energy has always fascinated me. My problem was always with perpetual motion devices. If we really wanted to harness a lot of power, it would be best to invest in creating a dyson's sphere around the sun or earth. To harness all that solar energy.

I wonder if using a Dyson's sphere around Mars to generate a massive magnetic field would work?:hmm:

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7 hours ago, DarkHunter said:

With your design the force is not being equally distributed to each of the balls.  

The force equation for this design is Fe = Fl*dl/de where Fe is the effort force to move it, FL is the force of the load (weight), dl is the distance from the load to fulcrum and de is the distance from effort force to fulcrum.  But that is for only one load and one effort force while your design would have 5 effort forces and 5 load forces, or to put it mathematically it would be Σ Fe = Σ Fl*dl/de.  To simplify this I'm only going to focus on the right side of the equation and assume a constant arbitrary value for Fe cause if I don't it starts to get complicated since I would have to factor in different distance for each of the effort forces and potentially add in some time factor if all the balls wouldn't hit the lever at the same time. 

Using the greatly, if slightly unrealistic, simplified approach the equation becomes Fe = Σ Fl*dl/de which can also be written as Fe = Fl1*dl1/de1 + Fl2*dl2/de2 + Fl3*dl3/de3 + Fl4*dl4/de4 + Fl5*dl5/de5.  From this you should be able to see two very important things.  First that the force effort is a summation of force loads so no 1 force load can equal the force effort and second that distance from the fulcrum changes the distribution of the force so ones further away from the fulcrum get more force then those closer to the fulcrum.

 

Dear Sir,

I know about it but this is not important  as we can change the ball+tube distance by piling up each ball+tube on each side so that each  ball will get same amount of force from the fulcrum .

Let me recreate  the model  and see the  the model and make your conclusion after that.

 

Thank you Sir.

 

Vikram

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Are we there yet ? This reminds me of the pub with the Free Beer Tomorrow sign, all promises, no free beer. I do wonder if vikram's balls have dropped yet.

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Just let this entire issue fade away... It will, eventually.

Trust me.

The OP feels that 2+2=5

He is missing the +1 and he refuses to acknowledge that, which is why I referred him to www.physicsforums.com

Edited by pallidin
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