The Paradox Engine

[gravityblock]

Yes, good point!  It is shown by experiment the conservation of mass doesn't hold either!  So, how can the kinetic energy equation be internally self-consistent when we have a conservation of mass violation, and mass itself is inside the kinetic energy equation?

Gravock

Theoretically, if we can instantly transfer the whole momentum from a heavier object to a lighter object, then we can create as much excess energy as we like according to the kinetic energy equation of KE = ½mv².

A 5 kg mass moving 1 m/s has 5 units of momentum and has a kinetic energy of 2.5J.  A 1 kg mass moving 5 m/s has 5 units of momentum and has a kinetic energy of 12.5J

12.5J > 2.5J !!!  Is this internally self-consistent?

Thanks,

Gravock

[telecom]

Theoretically, if we can instantly transfer the whole momentum from a heavier object to a lighter object, then we can create as much excess energy as we like according to the kinetic energy equation of KE = ½mv².

A 5 kg mass moving 1 m/s has 5 units of momentum and has a kinetic energy of 2.5J.  A 1 kg mass moving 5 m/s has 5 units of momentum and has a kinetic energy of 12.5J

12.5J > 2.5J !!!  Is this internally self-consistent?

Thanks,

Gravock

from this link:
http://www.besslerwheel.com/forum/viewtopic.php?t=2580&postdays=0&postorder=asc&start=1440&sid=124f83a7597ad50ff3c6977aba01a0b2











PostPosted: Tue Jan 14, 2014 9:47 am    Post subject: re: energy producing experiments   Reply with quote  Report Post to Admin
http://video.mit.edu/watch/mit-physics-demo-low-friction-atwood-machine-3138/

Yes: the 10 gram is causing all the motion; it is the only unbalanced force. This is 10 grams accelerating 1110 grams
.
I did not say their experiment made energy. But add to this the experimental concept proven by the double and triple Atwood's and you have a means of making energy. This concept is that the Laws of Levers applies to Atwood's pulleys connected over the same axis but with different radii. As described below.

Let's say that the radius of the pulley is 8cm.

550 grams on the left side at 8 cm places a certain torque on the point of rotation.  This torque is perfectly countered with an equal amount of torque from 550 grams on the right side at 8 cm.

The extra 10 grams on the right side is the accelerating force of .01 kg * 9.81 N/kg = .0981N which is expressed as torque upon the center point of rotation. These 10 grams of measured force remains constant throughout the acceleration. There is not one force for the static position and then another force for the acceleration. Acceleration can be determined from F = ma; F/m = a; .0981 N / 1.110 kg = .08838 m/sec²

The individual torque of each 550 grams was caused by the gravitational pull on the mass. The one side's torque canceled the others side's torque but the inertia of the 550 (1100 grams) grams is not canceled.  This inertia is proportional to the previously existing torque.

In Newtonian Physics inertia is measured by the change in momentum (linear Newtonian momentum) caused by an applied force. The change in linear Newtonian momentum of both 550 gram sides is equal. The bobs are viewed as going in the same positive direction.

55 grams at 80 cm has an equal amount of torque as 550 grams at 8 cm. If you try to move the 55 grams at 80 cm you will find that it acts like it has an equal amount of inertia from the point of rotation as 550 grams at 8 cm.  When moved; it also has an equal amount of linear Newtonian momentum, and an equal amount of centrifugal force. The system is balanced and it acts like it.

The 10 gram will accelerate the 55 g at 80 cm just as easily as 550 g at 8 cm; in fact it will accelerate two 55 grams on each side at 80 cm just as easily as it accelerates two 550 grams at 8 cm on both sides. The final velocity for the 10 grams (left at the 8 cm position) with two 55 grams at 80 cm will be exactly the same as when you had two 550 grams at 8 cm; with the additional 10 grams at the 8 cm position. And the velocity of the 10 grams will be .42 m/sec. Velocity can be determined by a rearrangement of the distance formula: v = the square root of (2 * d * a) = the square root of (2 * 1 m * .08837m/sec²)

After a drop of one meter for the 10 grams; the 110 (55 + 55) grams on a 80 cm wheel will be moving 4.2 m/sec; this (1/2mv²) is .971 joules of energy. The input energy was (10 grams dropped one meter.01 kg * 9.81 N/kg *1 m =) .0981 joules. Roughly 10 times the energy
For a logical proof place 550 grams on one side at 8 cm and 54 grams on the other side at 80 cm. The 550 grams will rotate the 54 grams.  Place 550 grams on one side at 8cm and 56 grams on the other side at 80 cm. The 550 grams will be rotated by the 56 grams. With 55 gram on the other side at 80 cm there will be no rotation.

Now place 550 grams on one side at 8cm and 549 grams on the other side at 8 cm. The 550 grams will rotate the 549 grams.  Place 550 on one side at 8 cm and 551 grams on the other side at 8 cm. The 550 grams will be rotated by the 551 grams.  With 550 grams on both sides there will be no rotation.

Conclusion: the Difficulty of rotating 55 grams at 80 cm is somewhere between the difficulty of rotating 549 grams at 8 cm and 551 grams at 8 cm.

When an outside force is asked to accelerate 55 grams at 80 cm it will find it as difficult (somewhere between 549 grams at 8 cm and 551 grams at 8 cm) as rotating 550 grams at 8 cm.  Don't buy into the myth that a different force takes over as soon as motion begins. Their experiment proves that the 10 grams of force starts the motion and it stays with the system throughout the one meter drop. The 4.80 seconds is predicted for the 10 grams of force (.0981 N) by F = ma. d = 1/2at² = 4.757 seconds.
The 110 grams moving 4.2 m/sec could rise .889 meters for .110 kg * 9.81 N/kg  * .889 m = .9702 joules

Detractors will try to angular momentum you; but Atwood's are linear and Angular Momentum is for space.

[Tusk]

I get the impression from this:

When they make a balance of the energy before and after, they find a huge gap - they conclude that the collision was unelastic,
since they can't find any other explanation for the gap...

and this:

Theoretically, if we can instantly transfer the whole momentum from a heavier object to a lighter object, then we can create as much excess energy as we like according to the kinetic energy equation of KE = ½mv2.

......that you are both standing at the edge, willing yet not quite ready to take a leap of faith. Very interesting points made, but that OU device still not quite revealing itself. Whilst I am uncomfortable presenting work for which I have not yet formulated a robust explanation perhaps in this instance there should be an exception. Not knowing if you have accepted the secondary reaction at the cm/axis, there is the risk of firing blanks but here goes.

Allow me to repeat an earlier statement re the secondary reaction:

a force equal to the applied force - which has the unique ability to accelerate a mass while maintaining frame of reference with that mass; or more simply, much like a rocket engine, velocity has no impact on either the work done or the acceleration - although the rocket must eject mass to function whereas the secondary reaction does not. The rocket engine (or reaction engine if you prefer) always applies force from the frame of reference of the body on which it acts so that acceleration from say 10m/sec to 20m/sec costs no more than acceleration from 110m/sec to 120m/sec.

In case you missed it, a mass accelerated from 110m/sec to 120m/sec gains substantially more KE than the same mass accelerated from 10m/sec to 20m/sec (thanks v²). Which typically suggests more work is done to achieve that gain (since the distance covered by the point of force is far greater, and this typically costs us dearly). But with the secondary reaction, which manifests in the FoR of the cm regardless of it's velocity, we now have the means to realise this gain. While not actually missing it the first time around, I am culpable in terms of distracted focus; perhaps beguiled by the dynamics of the PE with it's curious motions.

Anyway, since we can easily bring our drive disk/s back to a manageable rate of rotation ready for another acceleration cycle without losing E in the main rotor arm - using regenerative braking btw, therefore much of the disk E is recovered - it is inevitable that at some point our main rotor arm will achieve OU.

In simple terms we have a constant force (although possibly periodic as we cycle the disk/s) applied to a mass (the main rotor arm and all parts thereof) accelerating (therefore) at a constant rate due to a (nearly) constant input. I say nearly because as the rotor arm rotation rate increases the apparent disk rotation rate over the central axis of the unit increases, which will eventually create an upper limit. But I firmly believe this is just a matter of optimal engineering. 

[CANGAS]

Hmmmm. You have gone beyond the boundaries of the proposed experimental parameters CANGAS. If I may borrow a relevant term from the sporting realm, 'foul!' (blows whistle).

Since your question was aimed specifically at gravock there seems little point in engaging with it. With some relief, bearing in mind the obvious challenge it represents, a clear 'calling out' much like the ornery gunfighter in the saloon scene of an old western. So I guess there's no suprise you went straight for that 44 stuffed down your breeches. And I thought you were just pleased to see me.

But staying with the metaphor, I suppose the job of barkeep in this little drama is down to me. I'm simply trying to keep the peace and help maintain some focus on the main topic. We don't want no trouble around here stranger. This is a quiet town. So quiet there's days you couldn't snuff your cigar in the spittoon.

The raison d'etre for this thread is the PE device and associated concepts, which may well call for a discussion of v² and the "1/2" in the energy equation but would probably be best served by an absence of gunfights. I can either pour you another glass of rotgut or fetch the big double out from under the bar.

So; you skin that smoke wagon and we'll see what happens.   

LOL! Better than Groucho Marx!

Now, to help your short memory span break out of your hallucinatory phase and get back to reality.....I asked somebody if he REALLY did not understand, as he had claimed he did not understand, how the v SQUARED belonged in the Newtonian physics  based Kinetic Energy Equation, and likewise the 1/2. An alternative being that he could be leading up to try to fool some of us.

It is extremely interesting that he has avoided making a direct statement verifying or denying his previous assertions that the SQUARED and the 1/2 don't really belong there in the framework of Newtonian physics logic.
And it is extremely interesting that you are jumping to try to be his bodyguard.

Maybe YOU don't understand how the SQUARED or the 1/2 got there.

Tell you what, forget about about your imaginary hallucinatory shotgun and explain to us the Newtonian physics derivation of the Kinetic Energy Equation especially pointing out the conventional rationale for the SQUARED and the 1/2. 

I'll know if you are lying because I have derived the derivation and I know exactly why everybody (pretty much) accepts the presence of the SQUARED and the 1/2. I'll even give you a strong hint.....it has something to do with ARBITRARY DEFINITIONS.

Start explaining, bad boy.


CANGAS 75

[gravityblock]

LOL! Better than Groucho Marx!

Now, to help your short memory span break out of your hallucinatory phase and get back to reality.....I asked somebody if he REALLY did not understand, as he had claimed he did not understand, how the v SQUARED belonged in the Newtonian physics  based Kinetic Energy Equation, and likewise the 1/2. An alternative being that he could be leading up to try to fool some of us.

It is extremely interesting that he has avoided making a direct statement verifying or denying his previous assertions that the SQUARED and the 1/2 don't really belong there in the framework of Newtonian physics logic.
And it is extremely interesting that you are jumping to try to be his bodyguard.

Maybe YOU don't understand how the SQUARED or the 1/2 got there.

Tell you what, forget about about your imaginary hallucinatory shotgun and explain to us the Newtonian physics derivation of the Kinetic Energy Equation especially pointing out the conventional rationale for the SQUARED and the 1/2. 

I'll know if you are lying because I have derived the derivation and I know exactly why everybody (pretty much) accepts the presence of the SQUARED and the 1/2. I'll even give you a strong hint.....it has something to do with ARBITRARY DEFINITIONS.

Start explaining, bad boy.


CANGAS 75

Where did I make the claim I do not understand how the v² and the 1/2 belonged in the kinetic energy equation, as you wrongly assert?  In addition to this, where did I make a previous assertion that the squared and the 1/2 don't really belong in the framework of Newtonian physics logic, as you once again wrongly assert? 

Gravock