Free energy from gravitation using Newtonian Physic

[pequaide]

 I don’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/sec²) 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 o’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/sec² (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 = 1/2v²/a, where v is the square root of (2 * d * a) which in this case is (2 * 1m * .981m/sec²) = 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) Newton’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 = 1/2v²/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 don’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 = 1/2at². 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 = 1/2at², v = at or t = v/a and t² = v²/a², substituting v²/a² for t² the distance formula is now d = ½ v²/a. So lets check and see if it is correct. After one second in free fall the distance dropped is 4.9 meters (s = 1/2at²) and the velocity is 9.81m/sec (v = at) so lets plug in 9.81m/sec velocity in (d = ½ v²/a) instead of time in (s = 1/2at²): d = ½ * 9.81m/sec * 9.81m/sec / 9.81m/sec² = 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 = 1/2at²) and the velocity is 19.62m/sec (v = at) so lets plug in 19.62m/sec velocity in (d = ½ v²/a) instead of time in (s = 1/2at²): d = ½ * 19.62m/sec * 19.62m/sec / 9.81m/sec² = 19.62m, yep it works.

If you leave the one kilogram attached to the wheel ‘no’ the 14 units of momentum will not be enough to return the one kilogram to12 o’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?

Pequaide answer: Yes.

If a suspended mass of 1 kg accelerates a 0 kg rim mass balanced wheel the acceleration will be 1/1 * 9.81 m/sec² or 9.81m/sec². 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 (1/2mv²) 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/sec² or 1.962m/sec². 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 (1/2mv²) 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/sec² or .239m/sec². 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 (1/2mv²) 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/sec² or 4.9m/sec². 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 (1/2mv²) 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/sec² or .00980m/sec². 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 (1/2mv²) 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 v²/a, and (1/2mv²)

[sushimoto]

Hi,
sorry for bumping in in such a dumb manner,
but may i ask you, where you have got this flywheel from?

I really would like to follow your experiments hands-on,
but sometimes reality bites hardly.
Getting the stuff together is sometimes harder than to figure out theories... :/

Thanks,
sushimoto

[pequaide]

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]

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]

http://www.youtube.com/watch?v=YzwC6Oa8-X0

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