Free energy from gravitation using Newtonian Physic

[Kator01]

Hello pequaide,

assume the same setup as you have tested and presented here - but the only difference is that the tethers of both pucs are fastened to a ball-bearing on a center-axis. The disc also can rotate around the
center-axis with another ball-bearing. So the difference wound be that the tethers do not wind up on the axis thus becoming shorter at each turn but can rotate free araound the axis and keep their length.
I have no idea of the effects of this additional degree of freedom. There are some key-variations to be tested to fully understand the principle which make the full-stop of the disc or cylinder possible.

Anyway your last description ( disc rotates backwards at 90 degree-position of the pucs = bouncing back )  indicates to the elastic-impact-case  if the mass of the pucs is too big and the half-elastic-case if the masses are in the right proportion.

I have to build it. There is no other way to find out.

Regards
Kator

[pequaide]

I will send a diagram.

Point D is where the string enters the cylinder or drapes over a pin on the disc.

Line C-A is tangent to the disc or cylinder at point D.

Line D-B is perpendicular to line C-A

If the disc was moving counterclockwise, a puck on the end of a string in quadrant ADB would force the disc to accelerate clockwise. A puck on the end of a string in quadrant CDB would force the disc to accelerate counterclockwise.

So you might see the puck in quadrant ADB slow the counterclockwise rotation of the disc; stop it; and restart it in a clockwise direction. As long as the puck is in quadrant ADB all acceleration is in the clockwise direction. If the disc is too massive the puck will not stop the counterclockwise rotation of the disc. I have stopped cylinders nine times heavier than the spheres.

While the puck and string are on line D-B the puck places no rotational force upon the disc.

While the puck, on the end of a string, is in quadrant CDB (and still attached to the disc) it will accelerate the disc counterclockwise. If the disc is stopped the puck will accelerate it counterclockwise. If the disc is still moving counterclockwise it will accelerate counterclockwise. If the disc is moving clockwise it will accelerate counterclockwise. If the disc is to light there will be two stops of the disc; the motion will be counterclockwise (original motion); clockwise (stopped in ADB and restarted); counterclockwise (puck in CDB), I have seen it many many times.

The challenge is to add just enough mass to the disc so that the disc stops while the string and puck are on line D-B. If the string is then released from the disc the disc will remain stopped. Or if the string enters a slit in the cylinder it will remain stopped until the string reaches the other end of the slit and accelerates it counterclockwise.

Fireline is extremely inelastic; especially with very light pucks on 10 lb test. I envision no elastic activity. 

Okay; lets say your two bearing (one axis) design is placed on an air table. The bearing for the string is just above the bearing for the disc. The bearing for the disc is below the surface of the disc. The string of the puck is fastened to its bearing just above the disc and the string proceeds out to the circumference of the disc; the string then drapes (horizontally) over a pin and proceeds along the circumference clockwise about one disc radius.

You will probably need to hold the pucks in place until just at release. How are you going to accelerate the disc?

It should work. If the disc is about three or four time the mass of the puck it should get you in range of adding just enough mass to the disc to make the disc stop while the puck is on line D-B. I can?t give you exact mass numbers because I don?t know the length of the string that wraps around the disc. If the string of the puck is clear of the disc (just above) it should work great.

[pequaide]

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

On the Atwood site the Green S in the upper left corner of the page identifies the site as originating from Michigan State University, I would imagine the site was made by students but checked over by their professor before release. Therefore the concepts presented by the site represent those of the University. It should not be surprising that one of the United States? major universities thinks F = ma.

This gives us a source of extra momentum. The spinning wheel (pulley) has a much larger amount of momentum than the momentum developed from the freefall for the same distance of the same mass that gave the momentum to the wheel.

A five kilogram rim mass pulley accelerated by dropping a one kilogram mass one meter (as in an only one suspended mass Atwood) has 9.04 units of momentum; and the one kilogram that could have freefell the same distance has only 4.43 units. When the 9.04 units of momentum are given to one kilogram it will rise 4.167 meters, 4.167 times higher than the dropped kilogram.

We know from ballistic pendulums that linear motion and circular motion are completely interchangeable and experience no loss of motion in either direction. So the rotating five kg rim (above) has 9.04 units of momentum. v = Sqrt (1 m * 2 * 1kg/6kg * 9.81m/sec; mv = 1.808 m/sec * 5 kg = 9.04 units

We know that a one kilogram mass moving 5 m/sec will combine with a four kilogram mass at rest and the combined 5 kilograms will be moving 1 m/sec., for momentum conservation. Can going from 5 kg moving 1 m/sec to one kilogram in motion do anything but also conserve momentum?

[Kator01]

Hello pequaide,

thank your for your detailled explanations, I think I have all information necessary for building the setup I have in mind.
I will use spheres held in place by two electromagnets powered by lithium-polymer-batteries. The have enough amperage ( 1.5 Ampere/h ) to feet the electromagnets. Electromagents and batteries are mounted on the bottom-side of the disc and thus rotate with it. There are 2 vertical holes in the disc.  Two  iron pins in these holes are magentically powered from the electromagnet below thus holding the spheres in place on top at the rim of the disc. Via remote-control ( high-frequency-sender ) and the receiver-board ( also rotating on the disc ) the electromagnets are shut off at definite speed thus releasing the spheres.

By chance I lately was getting a ceiling-mounted horizontal-working fan. The rotating disc ( with the fan-blades ) is powered by an electromotor ( AC) and has quite a good amount of mass. Motor and the shaft ( central axis)  are mounted together. So the shaft is fixed on the ceiling. Motor and shaft do not move, just the disc with the blade is rotated by an inner gear. This is exactly what is needed for this experimental setup I have in mind.

You  also can use such a fan for your experiments. The only thing to change here is a ball-bearing to be mounted on the upper shaft with the tethers and spheres attached to it.
I will post a picture of this fan. You will understand it better then.

Regards

Kator

[Kator01]

Hello pequaide,,

concerning the atwood-calculations can you please give more details as to the basic assumptions of the rim-mass-pulley ?

The mass-inertia of a rim-mass depends of the mass-distribution  and the radius.

How did you calculate the spin-momentum of the pulley ?

Using 1 kg for mass m1 and 5 kg for m3 ( pulley ) I get 2.8028572 m/sec exp^2 in the simulation.

Now you have to assume a radius first before you can calculate the angular momentum.

Thank you

Kator