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### Author Topic: Free energy from gravitation using Newtonian Physic  (Read 99746 times)

#### Kator01

• Hero Member
• Posts: 898
##### Re: Free energy from gravitation using Newtonian Physic
« Reply #195 on: January 22, 2009, 02:08:04 AM »
Hello pequaide,

I repost here what we have found some time ago concernig the Atwood-Simulation :

https://www.msu.edu/user/brechtjo/physics/atwood/atwood.html

Now I used 1kg for m1, 0 kg for m2 and 9 kg for m3 ( pulley ) which returns ( friction must be set, meaning pulley-mass influence is taken into consideration ) a accelleration of 1.7836363 m/ sec exp2 which is double the
value you have calculated in your above post.
I have to do the math again according to the correct formulas I had found in the net way back then.

There is a simple answer to tinselkoala concerning this angular-momentum-thingy which seems to get stuck im many peoples head :

Once the drive-wheel and the white-wheel has stopped there is no angular-momentum any more. This moment in time is a transition-point and the only thing one can use then is the velocity of the rim-mass which when suddenly brought to zero will mean a negative-accellation ( value is the velocity) which will fetch the Force = M x ( - a )

Regards

Kator01

#### pequaide

• Full Member
• Posts: 139
##### Re: Free energy from gravitation using Newtonian Physic
« Reply #196 on: January 22, 2009, 03:21:45 AM »
Kator01: a acceleration of 1.7836363 m/ sec exp2 which is double the value you have calculated in your above post.

That is correct: MSU assumes that the pulley wheel does not have all the mass in the rim (which without doubt is true) but is something more like a solid disk. They assume that the average rotational mass is half way from the center of rotation to the circumference. This would be true if the pulley were only spokes with a very light circumference mass.

#### pequaide

• Full Member
• Posts: 139
##### Re: Free energy from gravitation using Newtonian Physic
« Reply #197 on: January 22, 2009, 11:42:00 PM »
The MSU Atwoodâ€™s site accepts the acceleration of the pulley wheel as an F = ma relationship. If they thought there was a significantly different type of momentum in the pulley they would have set up a different category for the motion to fall into, but they didnâ€™t. If they thought there was a significantly different type of momentum in the pulley they would have probably mentioned the radius of the pulley for that would be necessary to evaluate the output angular momentum, but they didnâ€™t. If they thought there was a significantly different type of momentum in the pulley they would have had another window for angular momentum when the friction was turned on, and the quantity in the window would change as the mass of the pulley was changed, but they didnâ€™t. All of their motion falls into one category: Linear Newtonian momentum.

If you tripled the radius of the pulley (but kept the mass distribution of the pulley wheel the same) what effect would there be on the analysis?

I think there would be no effect. But with triple the radius angular momentum would be significantly different.

Take a 10 kg rim (or ring) that has a 2 meter diameter that has been placed vertically on dry ice. After accelerating the rim to 1 m/sec connect its spinning circumference to a string that is wrapped around the circumference of a 10 kg thin wall pipe, at rest, that has a .1 meter diameter. Place the pipe on its length horizontally so that it rotates in a vertically plane on dry ice. What kind of motion will the large rim share with the pipe, and what will be the final circumference velocity of the larger rim.

#### pequaide

• Full Member
• Posts: 139
##### Re: Free energy from gravitation using Newtonian Physic
« Reply #198 on: January 23, 2009, 02:37:02 AM »
Pictured: Four masses (brass bushings) were used to accelerate the wheel and a flag (hexagonal wrench) past two photo gates held at a uniform distance. This should be enough information to determine if F = ma. Do you agree?

I took four consecutive readings that were within 2/10,000th of a second from each other, amazing isnâ€™t it.

#### pequaide

• Full Member
• Posts: 139
##### Re: Free energy from gravitation using Newtonian Physic
« Reply #199 on: January 24, 2009, 12:52:35 AM »
F = ma is a mathematical relationship between force, mass, and acceleration. To prove that the statement is true you must show that a proportional change in force causes a proportional change acceleration. For example if force increases by 20% then acceleration increases by 20%, etc.

The experiment; four different descending masses were hung from the ribbon, The force they produce is proportional to their mass, this mass is only a small portion of the accelerated mass. The four descending masses were a brass bushing or a combination of several brass bushings. Bushing 1 was 37.4g, bushing two was 26.7g, three was 26.5g, and the fourth was 73.0g

Descending mass 1 was 37.4g, Descending mass 2 was 64.1g (37.4 + 26.7), descending mass 3 was 90.6g (37.4 + 26.7 + 26.5g), descending mass 4 was163.6g (37.4 + 26.7 + 26.5g + 73.0g),

The time interval for each force to make the flag cross the same distance between the photo gates (from the same starting point) was; F1 .6957 sec, F2 .5336 sec, F3 .4556 sec, and F4 .3420 sec.

All forces worked over the same distance but they did not take the same amount of time to do it. To find acceleration from time we can us the formula d = Â½ a * t * t because all the distance are the same the only variable are time and acceleration.  Rearranging the formula to solve for a we get 2 * d / t * t = a, again distances are all the same so the proportional  acceleration of F1 is 1 / .6957 *.6957 = 2.078, F2 is 1 / .5336 * .5336 = 3.512, F3 is 1 / .4556 *.4556 = 4.818, F4 is 1 / .3420 * .3420 = 8.55.

The proportion of force to the proportion of acceleration is close to as follows.
F1   37.4g      2.078

F2   64.1g     3.512

F3   90.6 g    4.818

F4   163.6g     8.55

But there is one more correction to make. The fourth force F4 is accelerating a larger mass than the first force. This is in a proportion to its total mass and the total mass accelerated by the smaller force F1, that proportion is the rotational mass (inertia of the wheel, 2600g) of the wheel plus each mass that caused the force. F4 accelerates the larger mass slower in a proportion of 2637.4g / 2763.6g = .954; this means that F4 would have accelerated the smaller mass 2637.4 faster giving it a higher acceleration by 1 / .954 = 1.048 * 8.55 = 8.96.

The proportion of force to the proportion of acceleration is as follows.
F1   37.4g      2.078

F2   64.1g     3.512 corrected for the greater mass being accelerated 2664.1g / 2637.4g to 3.55

F3   90.6 g    4.818 corrected for the greater mass being accelerated 2690.6g / 2637.4 to 4.92

F4   163.6g     8.55 corrected for the greater mass being accelerated 2763.6 / 2637.4 to 8.96

So does a proportional change in force (which equals a proportional change in descending mass) equal a proportional change in acceleration? Or does: 37.4g / 163.6g = 2.078 / 8.96

37.4g / 163.6g = .2286,    2.078 / 8.96 = .2319    .2319 / .2286 = 1.0145

37.4g / 64.1g = .5834,   2.078/3.55 = .5853    .5853/.5834 = 1.0033

37.4g / 90.6g = .4128,   2.078/4.92 = .4223     .4223/.4128 = 1.023

These are within two percent of a perfect F = ma relationship.  So yes; F = ma is true for tangent forces working on the circumference of a circle. This is true even though the wheel is not a rim mass wheel, and this is Linear Newtonian Momentum.

#### pequaide

• Full Member
• Posts: 139
##### Re: Free energy from gravitation using Newtonian Physic
« Reply #200 on: January 25, 2009, 04:33:04 AM »
I have been using the ribbon to drop different masses from the edge of the wheel. By using the photo gates to record their acceleration rate I can determine the rotational inertia of the wheel. The rotational inertia of the wheel is about 2500g. That means the dropping mass accelerates as if it were accelerating a sled with a mass of 2500g.

For example; I dropped a 137g mass .5715m. This mass and the average mass of the wheel had a velocity of about .7692 m/sec. This is an acceleration rate of .5176m/secÂ². 9.81 /.5176 tells us the relative mass of the wheel to dropped mass.

So after a 137g mass has dropped only .5715m we have 2.02 (2630g * .7692 m/sec) units of momentum.

By using the cylinder and spheres principal we can give all the motion to the 137g. For .137 kilograms to have 2.03 unit of momentum it will have to be moving 14.8 m/sec

The 137g object can rise 11.1 meters with a velocity of 14.8m. d = 1/2vÂ²/a.

The 137g mass was only dropped .5715m. This is an energy increase of 19.4 times the original energy. 11.1m / .5715m.

#### Fred Flintstone

• Jr. Member
• Posts: 73
##### Re: Free energy from gravitation using Newtonian Physic
« Reply #201 on: January 25, 2009, 06:28:37 AM »
@pequaid

Why dont you build a setup so that will cause the 137g object to rise 11.1 meters ?

#### pequaide

• Full Member
• Posts: 139
##### Re: Free energy from gravitation using Newtonian Physic
« Reply #202 on: January 27, 2009, 03:54:30 AM »
The acceleration of the dropped mass divided by the standard acceleration of gravity is proportional to the mass dropped divided by the total mass being accelerated. The total mass being accelerated is equal to the dropped mass plus the rotational inertial of the wheel.

We know the quantity of mass being dropped (tenth column) and we know standard gravitational acceleration: and the acceleration of the dropped mass can be determined from the final velocity given by the photo gates.

Knowing the distance dropped and the final velocity given by the photo gates we can use the rearranged distance formula (d = 1/2vÂ²/a) to determine the acceleration of the dropped mass (seventh column). We now have all the components necessary to determine the rotational inertia of the wheel; which is in the twelfth column and is called total mass minus dropped mass.

#### pequaide

• Full Member
• Posts: 139
##### Re: Free energy from gravitation using Newtonian Physic
« Reply #203 on: January 28, 2009, 01:40:42 AM »
Fred: Note that an increase in the dropped mass gives you a proportional increase in the acceleration of the wheel and dropped mass. The ribbon is of course tangent to the surface of the wheel. So this means that a given force F pulling tangent to the surface of the wheel gives you an F = ma relationship for the velocity changes in the wheel.

So if you apply a tangent force for a certain period of time you will get a certain velocity for the particles in the wheel. And if you are to get that velocity back out of the wheel and return the wheel to rest you are going to have to apply equal force (in the opposite direction) for an equal period of time, or twice the force for half the time, or half the force for twice the time etc..

When the steel puck is unwinding from the white disk it is a line tangent to the wheel with a force F that is sufficient to stop both the white disk and the rotating wheel. The force times time relationship in the fishing line must be equal to the force times time relationship in the ribbon that started the motion. But the force in the fishing line is not just an F = ma relationship that is working on the wheel, it is also working on the puck.

Newtonâ€™s Third Law of Motion tells us that the momentum change in the wheel must be equal to the momentum change in the puck. The above experiment proves that the momentum change in the wheel is linear Newtonian momentum, and the momentum change in the puck must be linear Newtonian momentum. If the wheel and white disk and the puck have six times as much inertia as the puck then the puck must be moving six times as fast when the puck has all the motion. This is an energy increase. All data collected thus far has confirmed that this energy increase does occur and that Newtonian physics is correct.

So the answer to your question â€œwhy donâ€™t you build a setupâ€œ is that I already have. I already have such a machine. We know how high objects will rise given a certain horizontal velocity. Catching an object on the end of a cable and allowing it to rise as in a pendulum is something for an engineering department not for a research Lab.

#### pequaide

• Full Member
• Posts: 139
##### Re: Free energy from gravitation using Newtonian Physic
« Reply #204 on: February 03, 2009, 03:05:24 AM »
The only visual equipment that works well, for now, is my camera.

I used a suspended dropping mass off of the big wheel for a uniform velocity; the suspended mass hits the floor about the time that the ribbon releases the tab on the gray puck.

In picture one the white disk and gray puck are achieving maximum velocity. It appears that the tab on the puck has already pulled out from underneath the ribbon. I can see I need to improve this release system.  The suspended mass on the big wheel will soon hit the floor and the disk and puck and wheel will gain no more velocity.

The second picture is a different run of the same set up. I just take a lot of pictures and then try to arrange them in the appropriate order. But this is a different event and not near as good as a frame but frame evaluation that can be achieved with a video, I will have to replace my DVD recorder.

The second picture shows the puck (with glaring white tab) coming clear of its seat but it has not yet passed under the ribbon. The ribbon is not loose because the momentum of the big wheel is pulling on it. The white disk is slowing down because the gray puck is pulling on it. As the white disk slows down the momentum of the wheel pulls on the disk through the ribbon. There is a dark line between the two layers of plastic; this is a carpenterâ€™s mark not the fishing line. The fishing line is harder to see.

In picture three the puck is headed under the ribbon, the ribbon is still taut. Note the black square on the white disk, there is a small angular increase of this square between picture three and picture four. The tightness of the ribbon and the angular advancement of the square means the big wheel is still turning.

Shortly after picture four the big wheel and the ribbon and the white disk are stopped. The fishing line enters an open area and is attached nearer to the bearing. This closer attachment of the line causes the momentum exchanges to occur more slowly. Since the gray puck is not released it will reaccelerate the white disk in the same direction.

By picture five there is a large angular change in the black square as the puck reaccelerates the white disk. It will not restart the wheel however because the ribbon is limp, it can not push the wheel. There is not a uniform period of time between photos.

I removed 448g from the big wheel because the white disk and wheel could not be stopped when the gray puck was fully extended. The length of line is a determining factor in how much mass the puck can stop. Longer lines can stop greater quantities of mass because the time over which the force acts has increase. This is not because of angular momentum conservation; remember you could use a twenty meter big wheel and then the puck would have to be moving a bullet speed to conserve angular momentum.

A few obvious improvements should be made in the experiment. I hope to build two electronic releases; one for a pinpoint release position of the puck from the seat of the white disk. And then I would like to release the puck from the rest of the system when the white disk and big wheel are stopped. This would leave the puck moving in a straight line (apparently) headed for the back wall.  I also think the dropping mass can be incorporated in the mass of the wheel itself. Then the challenge would be to release the gray puck toward the back wall at precisely the same time that the white disk is stopped and the wheel is stopped with the extra mass at six oâ€™clock. And then of course I should place the photo gates between the release point and the back wall.

I would be delighted if people would repeat any of the experiments. I endeavor to give enough detail to make replication possible.  I would not see replication as a form of distrust, but only as good science. And the simpler experiments of the â€˜cylinder and spheresâ€™ or the â€˜disk and pucks on the air tableâ€™ should not be passed up. Their lack of bearings makes them deadly accurate, and they are inexpensive. Their low mass makes them susceptible to air resistance, but you design the experiments different ways: slow and small with no bearings, faster and bigger while using bearing, etc.

#### pequaide

• Full Member
• Posts: 139
##### Re: Free energy from gravitation using Newtonian Physic
« Reply #205 on: February 07, 2009, 03:37:42 PM »
I donâ€™t think making energy in the lab is an insignificant advancement. It would be one the most valuable scientific achievement in history; it would have to rank up there with the invention of the wheel and the control of fire. Of course it must be followed by functional machines, but you could not prevent that if you tried. You will have to bolt the doors to keep from getting trampled to death.

A one kilogram mass that is dropped straight down in freefall (9.81m/secÂ²) for one meter will have a velocity of 4.429 m/sec and it will have a momentum of 4.429. A one kilogram pendulum bob that has dropped down one meter has a 4.429 m/sec velocity and 4.429 units of momentum.

Near 6 oâ€™clock the force applied to the bob (in the direction of travel) in a pendulum is small but the time over which the force acts is proportionally greater. So in a simple pendulum you end up with the same velocity as that of the freefall mass. Much like suspending the mass near the axis, the time over which the force acts is longer but the force itself is proportionally small, and you will be getting the same overall velocity change.

The angles over which the force acts in a pendulum are identical to the working angles of the force in the wheel.

A one kilogram mass on a string that is suspended over a frictionless pulley (just off the edge of the table) with the string attached to a nine kilogram block on a frictionless plane will accelerate at a rate of .981m/secÂ² (1/10th of 9.81). This is because only 1 out of 10 kilograms is under gravitational acceleration.

The same is true for the balanced nine kilogram wheel with one kilogram of overbalance. The same quantity of force that was available to you in the simple pendulum is now available to you in the wheel. And the force is going to accelerate the other nine kilogram just like the suspended mass over the pulley accelerated the block. Now we have 1/10 the acceleration rate that got us 4.429 meters per second which will now give use 1.4007 m/sec. This is accepted Newtonian physics. Even though the acceleration is 1/10 the final velocity is greater than 1/10 and the final momentum is greater than 4.429. The velocity is determined by this formula d = 1/2vÂ²/a, where v is the square root of (2 * d * a) which in this case is (2 * 1m * .981m/secÂ²) = 1.4007m/sec. That puts momentum at 10kg (all the overbalance wheel is moving 1.4007 m/sec) times 1.4007 m/sec for 14.007 units of momentum.

Now you put the 14.007 units of momentum into a yo-yo de-spin device and transfer all of the motion to the small overbalance mass of 1 kilogram. If (a very small if) Newtonâ€™s Three Laws of Motion apply to this event then the one kilogram must be moving 14.007 m/sec. At 14.007 m/sec the one kilogram will rise (d = 1/2vÂ²/a) 10 meters, and it was only dropped one meter. All measurements that I have taken confirm Newtonian physics, and that the Laws apply to this event.

I intend to do an experiment of mass attached directly to the wheel, but I am quite sure I will be checking bearing resistance not Newtonian physics. Well; but indeed, I will be checking both.

I placed a mass on the end of the red ribbon which was wrapped around the big wheel

I taped a photo gate flag (hex wrench) to the big wheel. I placed the two photo gates at a fixed distance from each other and in a position that the flag (after acceleration) interrupted the photo gate beams.

I raised the mass 296mm above the point where the flag was half way between the photo gates. After being dropped the mass accelerated the wheel and the flag which crossed the photo gates in .0272, and .0272 seconds, in two different runs.

I then taped the same mass to the inside of the rim of the wheel and raised its position 296 mm above the same photo gate position. After release the flag crossed the distance between the photo gates in .0269 and .0269 seconds.

So if the mass is suspended from the circumference with a ribbon or fixed to the wheel, it is the distance the mass drops that determines the final velocity of the wheel. The velocity is also dependent upon the wheel inertia and the dropped mass of course.

If a mass is dropped the same distance off the circumference or fixed inside the rim the final velocity will be the same.

I donâ€™t see anything that is left to speculation, NASA and RCA both performed the de-spin stops. And I have accomplished stops of disks, cylinders, and wheels. I have timed the spheres and pucks in these stops and have confirmed Newtonian physics. The wheel wrapped with a ribbon will give Newtonian accelerations and an inside (the circumference) wheel attachment will also give Newtonian results. So we know that the original input motion is achievable.

The distance (or displacement formula; s) formula is s = 1/2atÂ². I think I have seen d used instead of s so I went with that, who would think of s being distance. So now we have d = 1/2atÂ², v = at or t = v/a and tÂ² = vÂ²/aÂ², substituting vÂ²/aÂ² for tÂ² the distance formula is now d = Â½ vÂ²/a. So lets check and see if it is correct. After one second in free fall the distance dropped is 4.9 meters (s = 1/2atÂ²) and the velocity is 9.81m/sec (v = at) so lets plug in 9.81m/sec velocity in (d = Â½ vÂ²/a) instead of time in (s = 1/2atÂ²): d = Â½ * 9.81m/sec * 9.81m/sec / 9.81m/secÂ² = 4.9m, yep it works.

Lets double check it at two seconds. After two second in free fall the distance dropped is 19.62 meters (s = 1/2atÂ²) and the velocity is 19.62m/sec (v = at) so lets plug in 19.62m/sec velocity in (d = Â½ vÂ²/a) instead of time in (s = 1/2atÂ²): d = Â½ * 19.62m/sec * 19.62m/sec / 9.81m/secÂ² = 19.62m, yep it works.

If you leave the one kilogram attached to the wheel â€˜noâ€™ the 14 units of momentum will not be enough to return the one kilogram to12 oâ€™clock, but if you separate the one kilogram from the wheel and give it all the momentum then it will rise 10 meters. Separate the overbalanced mass from the wheel and transfer all the momentum to it. The cylinder and spheres (or the yo-yo de-spin device) separates the mass and transfers all the motion to the small mass, and energy is made.

A question:  If I graphed time on the X axis, and velocity on the Y axis - is distance traveled the area under the graph?

If a suspended mass of 1 kg accelerates a 0 kg rim mass balanced wheel the acceleration will be 1/1 * 9.81 m/secÂ² or 9.81m/secÂ². Just brainstorming here, this is free fall. At the end of a one meter drop the velocity will be 4.429 m/sec and its momentum will be 4.429 units of momentum, and its energy will be (1/2mvÂ²) 9.81 joules, and the mass will rise 1 meter.  This is all the energy that is needed to reload the system, less a little friction.

If a suspended mass (on the end of a ribbon wrapped around the circumference of a wheel) of 1 kg accelerates a 4 kg rim mass (nearly all the mass is in the rim) balanced wheel the acceleration will be 1/5 * 9.81 m/secÂ² or 1.962m/secÂ². At the end of a one meter drop the velocity will be 1.98 m/sec and its momentum will be 9.90 units of momentum. If all that momentum is given to the one kilogram of overbalance its velocity will be 9.90m/sec its energy will be (1/2mvÂ²) 49.05 joules, and the mass will rise 5 meters.  9.81 joules is all the energy that is needed to reload the system, less a little friction.

If a suspended mass of 1 kg accelerates a 40 kg rim mass wheel the acceleration will be 1/41 * 9.81 m/secÂ² or .239m/secÂ². At the end of a one meter drop the velocity will be .6917 m/sec and its momentum will be 20.05 units of momentum. If all that momentum is given to the one kilogram of overbalance its velocity will be 28.36m/sec its energy will be (1/2mvÂ²) 402.1 joules, and the mass will rise 41 meters.  9.81 joules is all the energy that is needed to reload the system, less a little friction.

If a suspended mass of 1 kg accelerates a 1 kg rim mass balanced wheel the acceleration will be 1/2 * 9.81 m/secÂ² or 4.9m/secÂ². At the end of a one meter drop the velocity will be 3.13 m/sec and its momentum will be 6.26 units of momentum. If all that momentum is given to the one kilogram of overbalance its velocity will be 6.26m/sec its energy will be (1/2mvÂ²) 19.62 joules, and the mass will rise 2 meters.  9.81 joules is all the energy that is needed to reload the system, less a little friction.

If a suspended mass of 1 kg accelerates a 1000 kg rim mass (nearly all the mass is in the rim) balanced wheel the acceleration will be 1/1001 * 9.81 m/secÂ² or .00980m/secÂ². At the end of a one meter drop the velocity will be .14000 m/sec and its momentum will be 140.14 units of momentum. If all that momentum is given to the one kilogram of overbalance its velocity will be 140m/sec its energy will be (1/2mvÂ²) 9819 joules, and the mass will rise 1001 meters.  9.81 joules is all the energy that is needed to reload the system, less a little friction.

This math is F = ma. And d =1/2 vÂ²/a, and (1/2mvÂ²)

#### sushimoto

• Full Member
• Posts: 202
##### Re: Free energy from gravitation using Newtonian Physic
« Reply #206 on: February 12, 2009, 10:01:54 PM »
Hi,
sorry for bumping in in such a dumb manner,
but may i ask you, where you have got this flywheel from?

but sometimes reality bites hardly.
Getting the stuff together is sometimes harder than to figure out theories... :/

Thanks,
sushimoto

#### pequaide

• Full Member
• Posts: 139
##### Re: Free energy from gravitation using Newtonian Physic
« Reply #207 on: February 13, 2009, 01:08:15 AM »
I found the wheel in a trash container at work. One of its two bearing was missing so I assume that it had smoked a bearing and instead of maintenance buying a new bearing they bought a new wheel. That may have been their best choice but I bought a bearing for \$6 or so and the wheel is very accurate. It was one of two wheels from an upright band saw; I would hate to think what it would cost new. It would probably be cheaper to find a used band saw and strip the wheels out of it. You could go to a dealer and ask him what he wants for a replacement band saw wheel. I have been wishing I had two, and then I could attach ribbons to the white disk from both sides.

#### pequaide

• Full Member
• Posts: 139
##### Re: Free energy from gravitation using Newtonian Physic
« Reply #208 on: February 14, 2009, 09:50:41 PM »
Someone once said that ice skaters appear to conserve angular momentum. Well if they do it is because they use their muscles to increase their angular velocity. The increased linear velocity of the parts as they are pulled in is transferred to angular velocity.

Here is an experiment that proves that inanimate objects can not conserve angular momentum in the lab.

I was in a warehouse once that the builder stated that the entire floor surface did not vary by one millimeter. The floor was hardened to support fork lift trucks that would stack skids in racks six high.  So the 200 meter by 200 meter floor had to be free of vibration.

On such a floor you could take a one ton cart and place it on a 100 meter horizontal cable fixed on the other end. After accelerating the cart to one meter per second you could have the cable come in contact with an immovable post that was 3.33 meters from the center of mass of the cart. The Law of Conservation of Angular Momentum requires that the cart must increase in linear velocity by 30 times the original linear motion. To maintain the original angular momentum the radians per second must increase by 30 times; which can only be accomplished by the linear velocity increasing by 30 times.  A radian per second is the travel distanced of one radius around the circumference in one seconds.

1000 kg * 1/100 radian per second * 100 meters * 100 meters = 100,000

1000kg * 1/3.33 radians per second * 3.33 meters * 3.33 meters = 3,333

3,333.4 * 30 = 100,000    the linear velocity must increase by 30 times to maintain angular momentum conservation.

If angular momentum conservation is true then now we have a 1000 kg cart that is moving 30 m/sec. At 30 m/sec we can catch the cart on the end of a pendulum cable and it will rise 45.87 meters. An object need only drop .051 m to achieve a velocity of one meter per second, so after we patch the hole in the roof we can start making energy.

Of course the truth is that angular momentum conservation in the lab is false. It may appear that a skater or a person on a spinning stool conserve angular momentum, but appearances can be deceiving or they do it with their muscles. Probably a little of both.

#### pequaide

• Full Member
• Posts: 139
##### Re: Free energy from gravitation using Newtonian Physic
« Reply #209 on: February 19, 2009, 04:38:27 AM »