Does it really work? [Answered - No]

[ABV]

Hey.

 

This site contains unclear theory with experiments. But Idea is briliant.

 

What do you think about it?

 

Paradox of classical mechanics

[Kyrisch]

Too many spelling and grammatical errors to decipher...

[ABV]

Are these mistakes corrupt the main idea?

[Kyrisch]

Are these mistakes corrupt the main idea?

 

Yes they are present so much that I'm having trouble figuring out what you're trying to say. It is very important to be especially clear in math and science because of the general imprecision of language to begin with.

[ABV]

The idea is very simple.

Centripetal force doesn’t care about moment of inertia. It means, body with shape causes same centripetal force on curve trajectory. Centripetal force just depends on mass, velocity and curve radius.

But bodies with different shapes but same velocity and mass have different kinetic energy. From the other words we have to spend DIFFERENT amount of energy to reach same velocity for ring and disk. But bodies will cause SAME centripetal force on same curve trajectory.

So, in U-track case acceleration and break energy will have same direction, but amount of them will be different for disk and ring. Otherwise centripetal forces for disk and ring are equal. If use only rolling friction for break bodies, vector of momentum will be different. (Not all break energy will be spent to internal works)

This is “two words” explanation. Sorry about my English. I just try to bring a brilliant idea to world

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Full kinetic E = m*V*V/2+J*W*W/2

The Disk moment of inertia J= m*R*R/2. The Ring moment of inertia J = m*(R1*R1+R2*R2)/2.

If internal and external radius of ring close to each other R2-R1 ->0, then Ring radius is infinite R1->oo, R2->oo. And the ring full kinetic energy E->oo.

Other words, the ring with same velocity and mass will get a huge kinetic energy. But this won’t affect curve part of track, because centripetal force doesn’t depend on moment of inertia.

Both disk and ring with different kinetic energy will cause same centripetal force on curve part of track F=m*V*V/R. But need complete different chunk of energy to accelerate them to same velocity V. Or reduce their velocity V to 0.

[swansont]

The idea is very simple.

Centripetal force doesn’t care about moment of inertia. It means, body with shape causes same centripetal force on curve trajectory. Centripetal force just depends on mass, velocity and curve radius.

But bodies with different shapes but same velocity and mass have different kinetic energy. From the other words we have to spend DIFFERENT amount of energy to reach same velocity for ring and disk. But bodies will cause SAME centripetal force on same curve trajectory.

So, in U-track case acceleration and break energy will have same direction, but amount of them will be different for disk and ring. Otherwise centripetal forces for disk and ring are equal. If use only rolling friction for break bodies, vector of momentum will be different. (Not all break energy will be spent to internal works)

This is “two words” explanation. Sorry about my English. I just try to bring a brilliant idea to world

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Full kinetic E = m*V*V/2+J*W*W/2

The Disk moment of inertia J= m*R*R/2. The Ring moment of inertia J = m*(R1*R1+R2*R2)/2.

If internal and external radius of ring close to each other R2-R1 ->0, then Ring radius is infinite R1->oo, R2->oo. And the ring full kinetic energy E->oo.

Other words, the ring with same velocity and mass will get a huge kinetic energy. But this won’t affect curve part of track, because centripetal force doesn’t depend on moment of inertia.

Both disk and ring with different kinetic energy will cause same centripetal force on curve part of track F=m*V*V/R. But need complete different chunk of energy to accelerate them to same velocity V. Or reduce their velocity V to 0.

 

 

Centripetal forces are perpendicular to displacement on a circular path, so they do no work and are not responsible for changing kinetic energy.

[ABV]

Centripetal forces are perpendicular to displacement on a circular path, so they do no work and are not responsible for changing kinetic energy.

 

This is correct. This is explaining why centripetal force doesn’t depend on bodies shape.

For 2 cases, rolling disk and ring mass m – const, velocity for curve part of track V – const kinetic energy E depends on shape.

The thin ring has a big radius R. Even if ring and disk has same mass and velocity, the ring kinetic energy will be higher than disk kinetic energy. Centripetal force for curve part of track won’t change on both cases.

m - const, for curve part of track V – const, R(ring)>>R(disk), E(ring)>>E(disk), F(centripetal) = F(ring) = F(disk) - const

Even if ring rolling friction equal to disk rolling friction.

F(fr)(ring)==F(fr)(disk)

Ring momentum from rolling friction will be higher than disk momentum. Because ring kinetic energy higher than disk kinetic energy.

E(ring)>>E(disk)

P(fr)(ring)>>P(fr)(disk)

This means U-track with thin ring will move into one direction, but the law of momentum conservation will work locally.

Edited May 25, 2009 by ABV

[swansont]

There's no paradox, though.

[ABV]

No any new physic laws. This is correct.

But the paradox is isolated system moves without any external forces. And all law of conservation is working. All momentums are present.

The Ring just passes some “extra” kinetic energy thought curve part of track.

Acceleration force charges high energy to thin ring. Isolated System causes high momentum. And Isolated System will cause same high momentum on friction force acting.

Forces have a same direction on acceleration and friction acting time.

But System will cause low momentum on centripetal force action time. Just this momentum will opposite to others 2 momentums.

P(centripetal) << P(acceleration)+P(friction)

Edited May 26, 2009 by ABV

[J.C.MacSwell]

Hey.

 

This site contains unclear theory with experiments. But Idea is briliant.

 

What do you think about it?

 

Paradox of classical mechanics

 

Assuming an idealized system where the oval track doesn't move:

 

It looks like the differences in the centripetal forces at either end of the oval are offset by the forces on the track that are not shown in the drawing during the acceleration/deceleration stages halfway between.

 

Free to move unconstrained:

 

All this system would do is wobble around without it's center of gravity changing velocity, with no net change in momentum, not even temporarily.

[ABV]

Assuming an idealized system where the oval track doesn't move:

 

It looks like the differences in the centripetal forces at either end of the oval are offset by the forces on the track that are not shown in the drawing during the acceleration/deceleration stages halfway between.

 

Free to move unconstrained:

 

All this system would do is wobble around without it's center of gravity changing velocity, with no net change in momentum, not even temporarily.

 

4 forces are present on the oval: 2 centripetal acceleration and deceleration.

For solid disk the oval won’t move to anywhere and net momentum will be zero.

P(s_disk)(centripetal_1)= P(s_disk) (acceleration) + P(s_disk) (deceleration) + P(s_disk) (centripetal_2)

But the thin ring has more energy than solid disk. Even if acceleration/deceleration forces are equal to same solid disk forces, to reach same velocity, acceleration/deceleration acting time for thin ring will higher, than the same solid disk action.

P(t_ring)(centripetal_1) <> P(t_ring)(acceleration) + P(t_ring)(deceleration) + P(t_ring)(centripetal_2)

 

The law of momentum conservation works locally. During acceleration action extra momentum passes to extra momentum deceleration action. This extra momentum won’t affect momentum from centripetal force. Because curve part of track centripetal force depends on mass only. And do nothing to the rolling body moment of inertia.

 

P(t_ring)(centripetal_1) = P(s_disk) (centripetal_1)

P(t_ring)(centripetal_2) = P(s_disk) (centripetal_2)

 

P(t_ring)(acceleration) = P(s_disk) (acceleration) + P(extra)

P(t_ring)(deceleration) = P(s_disk) (deceleration) + P(extra)

 

These two extra momentums have opposite direction on strait track. And sum of these momentums will be a zero.

But these momentums have same direction on U-track. And sum of these extra momentums not equal zero.

Edited May 26, 2009 by ABV

[J.C.MacSwell]

Centripetal forces are perpendicular to displacement on a circular path, so they do no work and are not responsible for changing kinetic energy.

 

It might be hard for you to convince him of that.

 

What is labeled "centripetal force" in the diagrams (correctly in the case of a static track) can do work if the track is free to move.

 

Of course when that arises it changes the magnitude and direction of the centripetal force, (which as you correctly point out, does no work, but less strictly you could allow that it does), and the work is done by a component tangential to the path of the disc/ring/flywheel, as the track is displaced.

 

Still nothing free, kinetic energy is lost by the disc as it is gained by the track, Conservation of energy is maintained at all times, Conservation of momentum is maintained at all times etc. (the thing doesn't work as described)

[ABV]

It might be hard for you to convince him of that.

 

What is labeled "centripetal force" in the diagrams (correctly in the case of a static track) can do work if the track is free to move.

 

Of course when that arises it changes the magnitude and direction of the centripetal force, (which as you correctly point out, does no work, but less strictly you could allow that it does), and the work is done by a component tangential to the path of the disc/ring/flywheel, as the track is displaced.

 

Yes, I agree. It changes the magnitude and direction of the centripetal force. I’ll create a new picture where 2 bodies running synchronously in opposite direction. This will compensate all other centripetal force directions.

 

Still nothing free, kinetic energy is lost by the disc as it is gained by the track, Conservation of energy is maintained at all times, Conservation of momentum is maintained at all times etc. (the thing doesn't work as described)

 

Yes. And all laws work. But if you closely look into the Isolated System, you’ll see acceleration and deceleration part have same direction (not opposite). So, if thing ring gain more energy from the track on acceleration part. This kinetic energy will push back to track from thin ring on deceleration part. Both these parts have same direction. Just centripetal force will opposite to them. And this centripetal force won’t change, even if rolling body has a different moment of inertia.

Other words, following by law of momentum conservation, Isolated System will move into one direction

Edited May 27, 2009 by ABV

[J.C.MacSwell]

Yes, I agree. It changes the magnitude and direction of the centripetal force. I’ll create a new picture where 2 bodies running synchronously in opposite direction. This will compensate all other centripetal force directions.

 

 

 

Yes. And all laws work. But if you closely look into the Isolated System, you’ll see acceleration and deceleration part have same direction (not opposite). So, if thing ring gain more energy from the track on acceleration part. This kinetic energy will pull back to track from thin ring on deceleration part. Both these parts have same direction. Just centripetal force will opposite to them. And this centripetal force won’t change, even if rolling body has a different moment of inertia.

Other words, following by law of momentum conservation, Isolated System will move into one direction

 

OK. Only the variable moment of inertia flywheel can accelerate or decelerate without external inputs. The ring gains nothing by comparing it to a disc of equal mass any more than the disc would gain by comparing it to another disc of equal mass with an even smaller moment of inertia. (I won't get into the errors of the equal centripetal force claims despite the rings larger diameter as they are largely irrelevant)

 

Now look at the acceleration and deceleration of the flywheel. Both in the same direction as you say, correct? Toward the faster end.

(a momentum gain each time from the traction on the track)

 

Now look at the change in velocity at the faster end. Assuming no track movement there is a full 180 degree shift in velocity at constant speed, from an acceleration, toward the slower end (a momentum gain each time from a force perpendicular to the track)

 

This is opposite but greater than what happens at the other end, and exactly counteracts the momentum changes from the changes in speed at the midpoints.

 

The flywheel "borrows" momentum from the track, all in the same overall direction at the slow end and both midpoints, and returns it all at the faster end.

 

(it also 'borrows" angular momentum at one midpoint, and returns it at the other)

[ABV]

OK. Only the variable moment of inertia flywheel can accelerate or decelerate without external inputs.

This is the next step of implementation part. You would probably saw that on a bottom of document. Even if the flywheel can change velocity without external forces, during acceleration/deceleration time the track causes acceleration/deceleration force. This is result of difference between a new angular velocity of flywheel and old flywheel linear velocity. This will equate through these forces.

The ring gains nothing by comparing it to a disc of equal mass any more than the disc would gain by comparing it to another disc of equal mass with an even smaller moment of inertia. (I won't get into the errors of the equal centripetal force claims despite the rings larger diameter as they are largely irrelevant)

 

Now look at the acceleration and deceleration of the flywheel. Both in the same direction as you say, correct? Toward the faster end.

(a momentum gain each time from the traction on the track)

 

Now look at the change in velocity at the faster end. Assuming no track movement there is a full 180 degree shift in velocity at constant speed, from an acceleration, toward the slower end (a momentum gain each time from a force perpendicular to the track)

 

This is opposite but greater than what happens at the other end, and exactly counteracts the momentum changes from the changes in speed at the midpoints.

 

The flywheel "borrows" momentum from the track, all in the same overall direction at the slow end and both midpoints, and returns it all at the faster end.

 

(it also 'borrows" angular momentum at one midpoint, and returns it at the other)

 

Assuming the ellipse track with solid disk won’t move anywhere, because all momentums from centripetal, acceleration/deceleration forces give us zero on result.

But the thin ring has more kinetic energy comparing to the solid disk. And this extra energy will work between thin ring and track during acceleration/deceleration time only. The thin ring centripetal forces will be the same as for disks forces. Both Isolated Systems should have same velocities everywhere. The difference only is acting time on acceleration/deceleration part of track. This difference will bring different acceleration/deceleration momentums, which will move Isolated System.

Edited May 27, 2009 by ABV

[J.C.MacSwell]

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This is the next step of implementation part. You would probably saw that on a bottom of document. Even if the flywheel can change velocity without external forces, during acceleration/deceleration time the track causes acceleration/deceleration force. This is result of difference between a new angular velocity of flywheel and old flywheel linear velocity. This will equate through these forces.

 

 

Assuming the ellipse track with solid disk won’t move anywhere, because all momentums from centripetal, acceleration/deceleration forces give us zero on result.

But the thin ring has more kinetic energy comparing to the solid disk. And this extra energy will work between thin ring and track during acceleration/deceleration time only. The thin ring centripetal forces and “extra” linear momentums from the track ends will be the same as for disks forces. Both Isolated Systems should have same velocities everywhere. The difference only is acting time on acceleration/deceleration part of track. This difference will bring different acceleration/deceleration momentums, which will move Isolated System.

 

Same speed and more kinetic energy. So what? That had to be put in place as an input, as an initial condition. It's not a source of continuous energy and certainly doesn't empower it to break any laws of physics.

 

Where are you adding the additional energy that you think you will get out of it?

 

It's like claiming something heavier is a continuous source of mass, and how is that possible?

Edited May 27, 2009 by J.C.MacSwell
Consecutive posts merged.

[ABV]

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Same speed and more kinetic energy. So what? That had to be put in place as an input, as an initial condition. It's not a source of continuous energy and certainly doesn't empower it to break any laws of physics.

Just keep in mind, from same momentum the thin ring won’t reach same velocity as the solid disk. It’s rolling body which has movement on 2 directions, linear path and circular path.

If thin ring and solid disk will get same acceleration/decoration momentums, velocity on curve part of track will be a different for them. Otherwise, to keep same velocity on curve part of track, the thin ring and solid disk should have different acceleration/deceleration momentums.

 

 

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Where are you adding the additional energy that you think you will get out of it?

 

It's like claiming something heavier is a continuous source of mass, and how is that possible?

 

In thin ring case, additional energy comes from the angular part of energy. You may create very thin ring with huge radius and energy comparing to solid disk wish small radius and energy. This will work just for rolling bodies. And it doesn’t work for linear moves

[J.C.MacSwell]

Just keep in mind, from same momentum the thin ring won’t reach same velocity as the solid disk. It’s rolling body which has movement on 2 directions, linear path and circular path.

If thin ring and solid disk will get same acceleration/decoration momentums, velocity on curve part of track will be a different for them. Otherwise, to keep same velocity on curve part of track, the thin ring and solid disk should have different acceleration/deceleration momentums.

 

Which momentum, angular or translational? You cannot add them together and each is conserved in an isolated system. Do not confuse them with kinetic energy.

 

 

In thin ring case, additional energy comes from the angular part of energy. You may create very thin ring with huge radius and energy comparing to solid disk wish small radius and energy. This will work just for rolling bodies. And it doesn’t work for linear moves

 

This is true. It has more kinetic energy. But no amount of energy can change the momentum (angular or translational) of an isolated system, unless it comes from outside the system, in which case it wouldn't be an isolated system.

[ABV]

Which momentum, angular or translational? You cannot add them together and each is conserved in an isolated system. Do not confuse them with kinetic energy.

 

This is the rolling body. This body has 2 movements. Yes, this 2 momentums are perpendicular to each other and impossible add them together except rolling friction for deceleration (let just talk about this part). On rolling friction this 2 momentums work together. Linear part momentum will gains momentum to track, but extra angular energy gives more rolling capacity. This rolling capacity increase net momentum, which rolling body gains to the track.

This is true. It has more kinetic energy. But no amount of energy can change the momentum (angular or translational) of an isolated system, unless it comes from outside the system, in which case it wouldn't be an isolated system.

 

I absolutely agree. Isolated System should not move without any external forces.

But. I proposed whole document about it. In this particular combination, Isolated System has unusual behavior.

Edited May 27, 2009 by ABV

[J.C.MacSwell]

This is the rolling body. This body has 2 movements. Yes, this 2 momentums are perpendicular to each other and impossible add them together except rolling friction for deceleration (let just talk about this part). On rolling friction this 2 momentums work together. Linear part momentum will gains momentum to track, but extra angular energy gives more rolling capacity. This rolling capacity increase net momentum, which rolling body gains to the track.

 

 

I absolutely agree. Isolated System should not move without any external forces.

But. I proposed whole document about it. In this particular combination, Isolated System has unusual behavior.

 

Which comes back to:

 

No, it doesn't work as described. If it did then congratulations, you are breaking known laws of physics, as what is described is impossible.

[ABV]

Which comes back to:

 

No, it doesn't work as described. If it did then congratulations, you are breaking known laws of physics, as what is described is impossible.

 

It does not break any laws of conservations. It just makes condition, which moves Isolated System without any external forces and law of conservation works internally.

 

Anyway, thank you for you posts.

[J.C.MacSwell]

It does not break any laws of conservations. It just makes condition, which moves Isolated System without any external forces and law of conservation works internally.

 

Anyway, thank you for you posts.

 

That is not possible.

[ABV]

That is not possible.

 

Yes. I know it should not.

But in this particular case, I see where law of conservation works internally and allow Isolated System moves.

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I would agree about, this Isolated System won’t move anywhere, if the thin ring rolling resistance acting time has same value for solid disk too. But the rolling thin ring kinetic energy value much bigger than rolling solid disk kinetic energy. This thin ring “extra” energy part with same value of rolling friction force gains “extra” momentum to track. This is normal behavior of rolling bodies, which easy to see on straight line track (Isolated System also). Why rules should change to ellipse track? Inside ellipse acceleration/deceleration parts of track use straight line also.

[J.C.MacSwell]

Yes. I know it should not.

But in this particular case, I see where law of conservation works internally and allow Isolated System moves.

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I would agree about, this Isolated System won’t move anywhere, if the thin ring rolling resistance acting time has same value for solid disk too. But the rolling thin ring kinetic energy value much bigger than rolling solid disk kinetic energy. This thin ring “extra” energy part with same value of rolling friction force gains “extra” momentum to track. This is normal behavior of rolling bodies, which easy to see on straight line track (Isolated System also). Why rules should change to ellipse track? Inside ellipse acceleration/deceleration parts of track use straight line also.

 

 

With no outside forces, what accelerates the ring on the straight part of the track?

Edited May 28, 2009 by J.C.MacSwell

[ABV]

With no outside forces, what accelerates the ring on the straight part of the track?

 

Let's accelerate thin ring inside u-track without external forces. In this case thin ring gains momentum to track. Correct?

Thin ring will cause centripetal force on curve part of track. Track gains another momentum with opposite on this period of time. Correct?

During deceleration time the thin ring gains momentum to the track again. Direction for acceleration/deceleration momentums on u-track will be identical. Correct?

For thin ring Net momentums curve part of track will be less than sum of acceleration/deceleration momentums. For solid disk they will be equal to each other. Why? Because the thin ring has more energy than solid disk has. And during deceleration time (let's talk about that part. I didn't describe how to accelerate thin ring) thin ring gains bigger momentum to track than solid disk can do.

Inside Isolated system law of conservation will work. But for external viewers Isolated System will oscillate and move into one direction on each circle.

 

BTW.

I updated site. I did a lot of changes. Please take a look.