The secret of extracting free energy from gravity ...

[pequaide]

I looked up the conservation laws, and I noticed that there was no Law of Conservation of Kinetic Energy. That should be a dead give away. Do you mean to believe that motion cannot be isolated from the other forms of energy so that only kinetic energy can be conserved in the system? Linear Newtonian momentum can be conserved in a closed system why can’t kinetic energy?

Nor do they bother to tell you that angular and linear momentum can not both be conserved in many laboratory experiments if not most experiments. NASA’s 3 kilograms stopping 1420 (yo-yo de-spin device) kilograms is one of them. If the tether length of the 3 kilograms is 24 times as great as the radius of the rocket (estimating that the rocket radius is a half meter; this is the early days of NASA); angular momentum conservation comes down to this proportion 1423 * .5 *.7070 (radius of gyration) to 3 * 12 (12 meters) or 503 to 36 this means that the 3 kilograms need only be moving 14 times faster (in m/sec) than the circumference velocity of the rocket. That would proportionally give it only 42 (14 * 3kg) units of linear Newtonian momentum, when it started with 503 units. Linear Newtonian momentum requires that the 3 kilograms be moving 167 times as fast the original circumference velocity of the rocket. At 14 times faster the tether, with the three kilograms on the end, would have less rotational speed (radian per second) than the original rotation of the rocket. This is not seen experimentally.

Place a thick immovable post in the center of a frictionless plane and start a puck rotating around it on the end of a string. As the string wraps around the post the puck maintains its same linear Newtonian momentum, while its angular momentum decreases. This is proof that angular momentum is not a conserved quantity.

[Michael]

http://en.wikipedia.org/wiki/Angular_momentum

[greendoor]

Have to admit ... the old ice-skater, spinning, bringing arms in towards body and speeding up trick sort of confirms the conservation of angular momentum principle ...

[greendoor]

Fletcher; you disengage the mass from the flywheel at 6 o’clock, and give all the momentum to the overbalanced mass. The flywheel does not need to be returned to 12 o’clock, it is balanced. The separated mass can spend its time or momentum lifting itself. It does not need to lift the flywheel. The same quantity of momentum that returns the flywheel and overbalanced mass to 12 o’clock can send the overbalanced mass alone far past 12 o’clock.

Thank you Pequaide - that makes a lot of sense to me. I had never considered that, but it seems so logical now.

That is a huge slice of angular momentum pie that simply isn't needed ...

Frankly - I am seeing a miracle here ...

The trick is obviously going to be in stopping the flywheel, without excessive loss of energy, and using all that momentum to lift the overbalanced weight.

Does it actually have to be transfered to that weight immediately? Could the energy be stored in something (spring, cross-bow, hydraulic pressure, whatever ...?)

It surprises me that impact is so lossy ... a Newtons cradle doesn't seem to display much loss ...

But for tonight, i'm absolutely stoked at this idea that the flywheel does not have to be returned from 6 o'clock ... brilliant!

[Fletcher]

So greendoor - perhaps use the idea of the newton's cradle & the ballistic pendulum to achieve your aim - the rim mass rides down to 6 o'cl & has a certain velocity - meantime the flywheel has some angular momentum gained from rotation - use the rim mass to impact a pendulum & bob with the same mass at the same radius - there should be kinetic energy from impact plus some carry thru from the angular momentum of the flywheel [like a golf club swinging thru after the shot] - the angular momentum should be severely depleted - even arrange to have the pendulum bob to have more mass than the rim weight so that the flywheel stops & reverses after impact [like a newton's cradle can do] - measure the height the pendulum bob swings to ?

P.S. you said it ...

The trick is obviously going to be in stopping the flywheel, without excessive loss of energy, and using all that momentum to lift the overbalanced weight.

There has got to be a way to physically test this hypothesis reliably to see if there is in reality excess energy available ?

[georgexbailey]

The trick is obviously going to be in stopping the flywheel, without excessive loss of energy, and using all that momentum to lift the overbalanced weight.

If a physical experiment cannot be easily achieved, then can someone with math skills and access to WM2d build a model that may help answer some questions. For example, in WM2d, build a flywheel, place a weight on it and then determine the momentum of the flywheel after the weight has dropped to 6:00. Next, apply this momentum to a stationary weight of the same mass as was applied to the flywheel. (this part I have no idea how to do in software but maybe it can be done with math). At least this will give us another way to verify pequaid's math.

I played around with a test flywheel last night by placing a weight on it at 1:00 and stopping the flywheel with my hand when the weight was at 6:00. The flywheel could be stopped so easily that it is hard to imagine that transfering the momentum to the weight could propel it back to the top in any way. Obviously this was not a very good test but it did cause me to question whether any of this is true.

GB

[Oystein]

I have tested this in WM2D and momentum (m*v) is not conserved, or else energy would not be conserved! (big mass - slow speed....v^2) This seems to be the case with "pendulums" in general.
The masses involved can easily end up with alot of more momentum totally than one weight achieves from falling alone. (aproaching "infinit" numbers) (example; a weight falling fra 12 o`clock to 6 o`clock driving a ....heavy rim)

The problem is when talking about angular momentum vs. linear momentum, because angular momentum has a totally different notation.

Best
Oystein

[pequaide]

Well at least WM2D has shown us that we have to choose one from the other. Is Newtonian physics false (The Three Laws of Motion and the Law of Conservation of Momentum) or is the Law of Conservation of Energy false. I hope I have made it clear which side I stand on, such as sarcastically mentioning about the fact that no one can find the alleged heat in a ballistics pendulum. And I don’t pretend that angular momentum is conserved. I also mentioned that there is no Law of Conservation of Kinetic Energy; it always needs an imaginary friend like heat. The Law of Conservation of Energy is false.

With 252g of added mass on one spoke my balanced wheel currently has a mass of 3,680g. Its radius of gyration is apparently at .64, because the wheel responded to applied mass as if it were a rim with a mass of 2,360g. There is an extra 252.4g mass taped inside the rim that can drop a distance of 39.7 mm. These two 252 gram masses are different masses and are similar only by coincidence. Acceleration is 252.4 / (2360 + 252.4) * 9.81 or .9476 m/sec/sec, for a final velocity of: the square root of (.397 * 2 *.9476) or .867 m/sec for a momentum of .867m/sec * (2.360 kg + .2524 kg) or 2.26 units of momentum. So if you apply 9.04 newtons of force for ¼ sec you can stop the wheel. 9.04 is 92% of the force needed to lift one kilogram. No: that is not much force; it would not be too hard to stop the wheel.

But if 252.4g of mass had 2.26 units of momentum it would have to be moving; p = mv or 2.26/.2524 = 8.95 m/sec. At 8.95 m/sec the 252.4g overbalanced mass will rise (d = ½ v²/a) 4.08 meters and it was only dropped .397 meters.

[georgexbailey]

Acceleration is 252.4 / (2360 + 252.4) * 9.81 or .9476 m/sec/sec, for a final velocity of: the square root of (.397 * 2 *.9476) or .867 m/sec for a momentum of .867m/sec * (2.360 kg + .2524 kg) or 2.26 units of momentum. So if you apply 9.04 newtons of force for ¼ sec you can stop the wheel. 9.04 is 92% of the force needed to lift one kilogram. No: that is not much force; it would not be too hard to stop the wheel.

But if 252.4g of mass had 2.26 units of momentum it would have to be moving; p = mv or 2.26/.2524 = 8.95 m/sec. At 8.95 m/sec the 252.4g overbalanced mass will rise (d = ½ v²/a) 4.08 meters and it was only dropped .397 meters.

pequaid - Thank for your continued work. I see that you are convinced beyond doubt of your discovery. You have done your best explaining this. Your math skills are very impressive. I definitely have an open mind. BUT...you (us) must find a way to verify your math so this is undisputable. There is no reason to continue with the math. Either you are right and we are sitting on one of the greatest discoveries in the history of the world... OR, you are making some sort of mistake.

I am not giving up and I really want to get to the bottom of this. There has to be a way to modify your test setup so that it will raise the weight.

George

[georgexbailey]

The problem is when talking about angular momentum vs. linear momentum, because angular momentum has a totally different notation.

Pequaid - you determined that the angular momentum of the wheel is 2.26 units...then you said that if the 252.4g mass had 2.26 units of momentum it would rise 4.08 meters. Is it a mistake to apply the 2.26 angular momentum directly to the mass? I assume the mass would have a linear momentum. Does 2.26 units of Angular Momentum = 2.26 units of Linear Momentum? (sorry if this has already been covered or is a dumb question).

From wikipedia..
"Linear momentum is a vector quantity, since it has a direction as well as a magnitude. Angular momentum is a pseudovector quantity because it gains an additional sign flip under an improper rotation. "

Linear Momentum p= mv
Angular Momentum L = r x p

GB

[pequaide]

2.26 is in linear units of momentum. The quantity of motion around the circumference of the circle times radius of gyration.

[greendoor]

EDIT - post deleted as apparantly my input is unwelcome in this forum.

[Oystein]

The same happens with linear momentum, when accelerating a big horisontal moving mass with a small falling/vertical mass.
(If mass is placed only on the rim of a wheel, most of the momentum is close enough to linear anyways, for rough calculation purposes)

But if you try to convert all that "gained" momentum back to the small mass, the momentum is back to its "original" value.
(the value the smaller weight would have if it just fell the same distance)

Best
Oystein

[pequaide]

Oystein statement: But if you try to convert all that "gained" momentum back to the small mass, the momentum is back to its "original" value.
(the value the smaller weight would have if it just fell the same distance)

You have no experiments to back up this statement. The statement is a violation of all Three of Newton’s Laws of Motion. It is a violation of the Three Law's equivalent, The Law of Conservation of Momentum. You can’t give one example of such a tremendous loss of Newtonian momentum. Please try.

The horizontal wheel that concentrates all the motion into the over balanced mass must be moving in a circular path. But the wheel need not have any mass other than that which is needed for structure. The over balanced mass must also be moving in a circle. But the great portion of the mass can be linear. I have used linear motion to accelerate the wheel. I used a sled. You could use a sled the size of a locomotive with a horizontal wheel on each end of the sled’s back and forth motion.

The pulley of an Atwood’s machine can be placed horizontally, with the introduction of two more bearing points. The pulley then could be used to toss the overbalanced mass. If you are concerned about angular momentum conservation being true then think of this. A 20 meter radius rim mass wheel that has a mass of 1000 kilogram will accelerate at the same rate as a .1 radius rim mass pipe of the same mass if you hang the same mass from each. . Use ten kilograms to obtain 1 m/sec velocity on the circumference of the big rim and then transfer all that alleged angular momentum to a small horizontal wheel that is used to toss the overbalanced mass of the ten kilograms. I have used a big wheel to accelerate small horizontal wheels. In reality the 20 meter wheel has no more momentum than the .1 meter pipe. But you need not worry about angular momentum being conserved it simple isn’t. If I am wrong you have the big rim with absolutely huge angular momentum, but don’t break your piggy bank on hopes of angular momentum conservation.

[Fletcher]

For those interested - this morning I made a simple wm2d sim of a flywheel & rim riding weight - in many ways similar to both what Oystein & Pequaide have said - I am taking screen shots of various configurations & will also post the results in table form as soon as I can - I have only used Ke as that is the measure of the capacity to do work & left Momentum right out of it - I'll post a sim if anyone else wants to doctor it & measure the momentum etc.

Will edit this post with results.