Naked Science Forum
On the Lighter Side => New Theories => Topic started by: vkrmvkrm 11 on 17/01/2018 14:31:26
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As per sketch a 2 meter long tube is mounted on the left arm of a seesaw and this see saw is balanced due to counterweight.there is a 10 kg mass ball located in the in rest position in the tube.
See the link.
Now seesaw is balanced at 180 degree angle.when I tilt it this balanced seesaw then the ball fall down from 2 meter height and hit with upper part of tube .but interesting after hitting the seesaw will get back it's initial position without any external influence.so ball will again fall from 2 meter height.in this way the ball will fall down twicely but the main interesting point is that the input energy is almost zero as there is no torque due to counterweight.
There will be a lock mechanism to prevent the falling of counterweight at the time of tilting.
The ball will fall down after getting a certain angle as a pin will work to hold the ball to prevent it from sliding along tube at the time of tilting.
If there is no energy as a input as seesaw is balanced and torque is same then output is ,using mgh formula
Mgh= 10*10*2=200 joule at the time of tilting.
Mgh=10*10*2=200 joule at the time of reversing.
So total output is 400 joule but input will be almost free due to equilibrium position of seesaw.
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This is a challenge for everyone.i have consulted with many experts but all are confused.
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when I tilt it this balanced seesaw
There's your problem. Using your hand to tilt it is putting energy into the system. In the video, each time you want to make the ball move, you have to push the device with your hand. Make a version which doesn't require that step and then we can talk.
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I have proven what I want to prove(overunity).
If a seesaw is balanced then there will be need of almost 0 energy to tilt it at any degree.furthermore I have already mentioned that a pin will work to hold the ball and ball will fall down after getting a certain angle.
I will extract energy of falling ball ,using two piston generator mounted on each side of long tube an this energy will work to feed a lever system which will work to tilt it again and again.
Ok .tell me input and output.
But remember that seesaw is in equilibrium position if you are calculating input .
Tell me difference between input and output.
I'm ready to give up.
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The Box,you are considering yourself not less than Einstein but talking like in this way as if it is a very complex design.it is simple design.so I don't think there is a need of proving it more.
But I would like to tell you that I don't have resources to build it completely.if you could help me to build it then it will be a welcome step as there is a need of team work.
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I'm ready to give up.
That's good, because you are wasting your time (and ours).
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I don't think there is a need of proving it more.
You couldn't prove it less.
When you set it up and walk away, and it keeps going- forever- you will have proved it.
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These things always get me, ones where the motion is the show piece. Many things affect such bodies, magnetism grav8tational pull rotation of the earth, gravity of the sun. Have you considered these ?
https://www.scribd.com/doc/74784234/Gyro-Compass
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The Box,you are considering yourself not less than Einstein but talking like in this way as if it is a very complex design.it is simple design.so I don't think there is a need of proving it more.
But I would like to tell you that I don't have resources to build it completely.if you could help me to build it then it will be a welcome step as there is a need of team work.
You are mistaken if you think that the quote in my signature came from me.
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The Box,you are considering yourself not less than Einstein
Thebox has not posted in this thread.
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You must have stood on the shoulders of giants. Then fell off and bashed your head.
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I would like to know if a seesaw system is in equilibrium position then the torque will be same on both arm of seesaw.all of you are ignoring this valid fact.it doesn't matter that the ball is below fulcrum as system is balanced due to counterweight.that's why there is no need to adding energy in the system.so all gravitational forces,magnetism,and sun rotation will work same on both side of seesaw and required energy ( input)will be almost free to tilt it .
There is nothing beyond the formula of kinetic energy and this formula is proving overunity in it.
So no one can challenge it.
I have asked all of you that tell me input and output.
I am not wasting anyone's time.i am presenting it on the basis of pure physics..
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If there is no overunity in this mechanism then all of you are proving the formula of kinetic energy wrong yourself.
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I again insist that there is no need of adding energy in this mechanism as it is balanced due to counterweight and if there is doubt then all of you test it .all doubts will be cleared immediately regarding input and output.
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So no one can challenge it.
So finish your device and go get your Nobel Prize already. Take out a loan to build it if you have to. If the device is as fool-proof as you claim, then you'll quickly and easily make all of your money back many times over for providing an unlimited, nonpolluting energy source to the world.
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I have added counterweight more so if ball is 2 meter below from fulcrum then there will be no need of adding I put energy or lifting up the device due to counterweight.
This one point is working to get overunity in this mechanism.
Take it very seriously and make efforts to build it completely as a team work and collaboration.
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Take it very seriously and make efforts to build it completely as a team work and collaboration.
Your design looks pretty simple. Why would you need a team of people to build it? You could probably find everything you needed at a hardware store.
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You are correct but I think two is better than one.but its ok.i try it.
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This is a challenge for everyone.i have consulted with many experts but all are confused.
What 'experts'?
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I have proven what I want to prove(overunity).
If a seesaw is balanced then there will be need of almost 0 energy to tilt it at any degree.furthermore I have already mentioned that a pin will work to hold the ball and ball will fall down after getting a certain angle.
I will extract energy of falling ball ,using two piston generator mounted on each side of long tube an this energy will work to feed a lever system which will work to tilt it again and again.
Ok .tell me input and output.
But remember that seesaw is in equilibrium position if you are calculating input .
Tell me difference between input and output.
I'm ready to give up.
You may have 'proven what you ant to prove', but you have failed to show convincingly to anybody else that this is valid. It is for you to tell the difference between input and output as it is your idea. Why should we? We get perpetual energy loonies on here every week.
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The Spoon,what I have not proven.
If a seesaw is balanced then the torque will be same on both side of seesaw.is it not valid?
If I tilt it from rest position to 180 degree more then the input will be very minimal as it is balanced.is it not valid?
If a 10 kg weight ball fall down from 2 meter height then kinetic energy will be ,200 joule.is it not valid? The system will get back it's initial position without any external influence due to counterweight,impact energy and gravity.is it not valid?The 10 kg ball will fall down again at the time of reversing from 2 meter height then the kinetic energy will be again 200 joule.is it not valid.
Friction is not an issue.P.E.is being increased,input is very minimal due to counterweight,then what should I prove?
Even surprisingly physics formulas are supporting overunity in it and you better that physics laws are not against Overunity but some properties like friction,air resistance works against to achieve overunity.but in this mechanism friction and air resistance are not an issues.is it not valid?
If you are Confused on input then try it as I tried .you will get convinced.
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If you have a iron ball and a plastic tube then test it.but remember that also use a pin to hold the ball or prevent sliding of ball .mounted it on a beam and mounted counterweight and balanced the seesaw.now tilt it and post your comments.
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I promise you that you will found that input is very minimal.but when ball will fall down then you will feel some energy.
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is this another thread on the so-called perpetual motion machine?
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Yes ,it is an another thread.
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This is a challenge for everyone.i have consulted with many experts but all are confused.
What 'experts'?
Again, what 'experts' are you referring to? Physicists? Engineers? Or some local bloke who thinks he knows a bit?
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I consult with some physics professor s in my area and they are confused only on one point that input is minimal and ball is completing one cycle after falling twicely then
There must be overunity but says they are confused and can't tell exactly but also says that I can win the race.
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I consult with some physics professor s in my area and they are confused only on one point that input is minimal and ball is completing one cycle after falling twicely then why there is no overunity.
Are you sure he is not confused by your explanation?
You say 'If a seesaw is balanced then the torque will be same on both side of seesaw.is it not valid?'
If it is balanced where is the torque? There is no rotational movement if in a state of balance.
You also say 'Friction is not an issue.P.E.is being increased,input is very minimal due to counterweight,then what should I prove?' Of course there is friction - unless you have invented frictionless bearings and this is no air....
You keep saying that input will be minimal. That means there is input. Therefore, you have to put energy into the system.
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Off course I say that input is minimal but why are you not considering output?why don't you tell me about output?
If output is more than input then it is not considerable to calculate input.
I say friction is not an issue.it means that friction will not create any hurdle in the device to get back it's initial position.
I have shown it in the video.there is lot of friction but ,however,seesaw is reversing.so friction is not an issue.
The 99.99% overunity devices couldn't survive due to friction but in this mechanism gravity will work to overcome the friction.
Remember the ball will work to generate energy not arm .the arm is working only to increase to he potential energy of the ball.
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Maybe you have forgotten when you posted this stuff before and Colin pointed out:
However, the item is question is not the input energy, but the output energy for which your calculation is incorrect.
Your potential energy calculation is correct for a non-counterbalanced weight, but the falling weight also has to lift the counterbalance weight and the energy required to do this should be subtracted from your calculation in order to show the minimal amount of output energy available from the seesaw.
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The spoon,the minimal input is only 20 Joule if I take friction loss 10% but output is 200 Joule from a 10 kg falling ball from 2 meter height.so when ball will hit with upper part of tube at the time of tilting then kinetic energy is 200 joule.now minus the input 20 Joule from 200 Joule
So overall output is 180 Joule.
But you are forgetting that the seesaw is getting back it's initial position without any extra energy so ball is again falling from 2 meter height then output is 200 Joule again.
Why are you not considering the falling of ball at the time of reversing.
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The spoon,the minimal input is only 20 Joule if I take friction loss 10% but output is 200 Joule from a 10 kg falling ball from 2 meter height.so when ball will hit with upper part of tube at the time of tilting then kinetic energy is 200 joule.now minus the input 20 Joule from 200 Joule
So overall output is 180 Joule.
But you are forgetting that the seesaw is getting back it's initial position without any extra energy so ball is again falling from 2 meter height then output is 200 Joule again.
Why are you not considering the falling of ball at the time of reversing.
See above.
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The spoon,answer is that the energy will be generated by the kinetic energy of ball not by system.the ball will work to press a piston generator not system.so when it fall down then obviously the counterweight will be lift up but the output will be not subtracted .
In this mechanism ball is working to generate energy not device as device is working to increase the potential energy.
The counterweight has already been lifted up otherwise how ball will fall down?
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The spoon,answer is that the energy will be generated by the kinetic energy of ball not by system.the ball will work to press a piston generator not system.so when it fall down then obviously the counterweight will be lift up but the output will be not subtracted .
In this mechanism ball is working to generate energy not device as device is working to increase the potential energy.
The counterweight has already been lifted up otherwise how ball will fall down?
Te ball is not 'generating' electricity it is releasing energy from being lifted.
As I pointed out, we get loonies claiming to have invented perpetual motion every week. It is therefore pointless engaging with you further.
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The spoon,I am very much surprised that how are you taking it.
The ball is being lifted up but you are forgetting that the seesaw is balanced so input is no matter.
The ball is not generating energy ,it very strange but is releasing 200 Joule energy or 400 Joule energy twicely and you say that it is not perfect all motion.will you please explain then what is perpetual motion or overunity?
You yourself has contradictions in your post.
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The spoon,I am very much surprised that how are you taking it.
The ball is being lifted up but you are forgetting that the seesaw is balanced so input is no matter.
The ball is not generating energy ,it very strange but is releasing 200 Joule energy or 400 Joule energy twicely and you say that it is not perfect all motion.will you please explain then what is perpetual motion or overunity?
You yourself has contradictions in your post.
If you are lifting the ball of course there is energy input - it doesnt matter if the see saw is balanced - to lift the ball from a starting position you require an energy input. That will be the equivalent of the energy output when it falls. Or are you proposing the use of magic to levitate it? Perhaps you are thinking of unicorns or wizards or some other nutty mechanism.
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Ok but what about falling down of ball at the time of reversing.as the seesaw is getting back its original position without any extra energy.
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Ok but what about falling down of ball at the time of reversing.as the seesaw is getting back its original position without any extra energy.
That may happen if there is enough energy to overcome friction from the bearings, air etc. But that will not go indefinitely due to losses from friction - same as with any perpetual motion device. It will come to a stop when the energy is dissipated.
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The spoon,
See the sketch carefully.
I have just changed the length of long tube .now it is 4 meter and it is mounted in middle with the Left arm of seesaw.
Now calculate input and output.
You please also calculate output from counterweight as I had forgotten to mention about it.
Everything is same but I have just increased the length of tube so input will be same like previous design.
But output???????
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Yes ,it is an another thread.
Why have you started a second thread?
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So that everyone could understand it correctly without any confusion.
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Why not carry on in the first thread?
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As I mentioned that I have done some changes in the mechanism so that everyone could understand it completely to avoid confusion.
It will be also easy to a new registered member to read the entire thread separately.
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The spoon,
See the sketch carefully.
I have just changed the length of long tube .now it is 4 meter and it is mounted in middle with the Left arm of seesaw.
Now calculate input and output.
You please also calculate output from counterweight as I had forgotten to mention about it.
Everything is same but I have just increased the length of tube so input will be same like previous design.
But output???????
Why are you asking me to calculate input and output? It is your idea and therefore up to you to show that it works and do the calculations.
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The spoon,
I have already proved it mathematically.you are not convinced that's why I asked you.otherwise you don't think that I don't know.everone know what will be input and output?
Out put will be always greater than input.
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What part of this device are you unable to construct on your own?
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Kryptid,
The part is ,to mounted generator on each side of tube.
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I have already proved it mathematically.
What do you think you have proved?
Do you mean this
Mgh= 10*10*2=200 joule at the time of tilting.
Mgh=10*10*2=200 joule at the time of reversing.
So total output is 400 joule but input will be almost free due to equilibrium position of seesaw.
Because that's not the correct arithmetic.
The net output is not 400J
What you need to do is subtraction, rather than addition.
The 200J released in one stage of the cycle is used to provide the 200J needed to lift the weight again.
The net output is not
200J +200J = 400 J
the correct equation is
200J - 200 J = 0
The net output is zero which is exactly as expected from the conservation of energy.
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The spoon,
I have already proved it mathematically.you are not convinced that's why I asked you.otherwise you don't think that I don't know.everone know what will be input and output?
Out put will be always greater than input.
Where have you proved it mathematically? Do you mean this?
The spoon,the minimal input is only 20 Joule if I take friction loss 10% but output is 200 Joule from a 10 kg falling ball from 2 meter height.so when ball will hit with upper part of tube at the time of tilting then kinetic energy is 200 joule.now minus the input 20 Joule from 200 Joule
So overall output is 180 Joule.
But you are forgetting that the seesaw is getting back it's initial position without any extra energy so ball is again falling from 2 meter height then output is 200 Joule again.
Why are you not considering the falling of ball at the time of reversing.
That is not mathematical proof. Where do you get your figures from input and output? If the output is 200 joules from a ball falling from 2 metre height, how can your input be only 20 joules when you have to lift it to 2 metres height? lso what about the loss for lifting after it has fallen and needs to use momentum to lift it back up to 2 metres height? Where did you get the 10% friction loss figure?
Also see Bored Chemists comments on your so called calculation.
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Hello Bored Chemist,
See this sketch carefully.
In the sketch the tube length is 4 meter.
So input is 200 Joule correct but output.
Output is 400 Joule as ball is falling down from 4 meter height at the time of tilting.
Ball is again falling down from again 4 meter height at the time of reversing.
So output is 400 joule.
So total output is 800 Joule.
Now subtract 800-200=600
So overall output is 600 Joule.
If there is 10 % friction loss in it then the output is 540 joule.but I reduce it more 140 joule.still it is 400 Joule.
But you are calculating only one time output.
Calculate input and output after viewing the sketch.your all doubts will be cleared immediately.
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The spoon see the sketch.and read my post to Bored Chemist.
I have just increased the length of tube from 2 meter to 4 meter.
Now input will be same as it was in previous design when the tube was 2 meter long but output will be increased very interestingly.
The Bored is forgetting the reversing of device as ball is again falling at the time of reversing.
I took 10 % friction loss or 20 Joule due to reversing of arm.
All of you are correct that input will be same 200 Joule and output is also same 200 Joule but it will be only for one side falling.
All of you forgetting the reverse falling of ball again from 2 meter height.
But no problem see the sketch to clear your doubts.
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Please show your calculations for inputs and outputs with workings. All we have is your claim for inputs and outputs. That is not a calculation.
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If you have viewed the sketch carefully.hope you will understand it.
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The spoon.
All of you at first input at the time of tilting ,then falling of ball as a output,then again falling of ball at the time of reversing but at the time of reversing don't calculate input as the device is getting back its original position without any external influence.
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If you have viewed the sketch carefully.hope you will understand it.
Yes I have seen the sketch and I understand it perfectly. It is a seesaw with a tube containing a ball at one end with a counterbalanced weight. In position 1, the ball is at the bottom. In position 2 the ball has been lifted up. The ball then falls to the bottom of the tube causing the 'seesaw' to swing around. You claim it will swing back to position 2 because of the kinetic energy resulting from the ball dropping. Correct?
That is fine, in as much as it is a conceptual model demonstrating what you think will happen.
However, I did not ask this. I asked how you calculated the inputs and outputs showing your workings which you seem reluctant to do. You are claiming output is greater than input. Without workings and evidence for these inputs and outputs, you are just another nutter who claims to have invented a perpetual motion machine.
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your all doubts will be cleared immediately.
I have no doubts.
it will not work, because it can not work.
If you don't understand that fact then it is you who needs to do the calculations.
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Kryptid,
The part is ,to mounted generator on each side of tube.
Then don't worry about using it to generate power just yet. First just get a working model constructed that can move on its own indefinitely despite friction. If you can do that, then finding someone to help you connect it to a generator should be easy.
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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.
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it will not work, because it can not work.
If you don't understand that fact then it is you who needs to do the calculations.
If it was in any useful way "overunity" you would not need to keep pushing it.
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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.
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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.
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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.
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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.
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And you keep stating that you are correct without providing any evidence.
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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.
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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.
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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?
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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.
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The main purpose is to get more output than input.
Then don't come back until you have measured both.
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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".
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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.
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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.
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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. –
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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.
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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.
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Bored Chemist, I have understood it very correctly but I think this time you should consult with some Physicist .
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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.
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Bored Chemist,
Very soon
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Bored Chemist,
Very soon
I will buy some popcorn.