Free energy from gravitation using Newtonian Physic

[Kator01]

Hello pequade and all,

may I draw your attention to the following Website :

http://www.unifiedtheory.org.uk/

Scroll down to Diagramm 13 :

INTRODUCTION TO STEEL INERTIA QUANTUM DRIVE

Hope you enjoy this.

Kator

[pequaide]

Sam; I think you are making this to complex. A rim mass wheel can be accelerated according to Newton?s Second Law of Motion (F = ma). You could wrap a vertically mounted wheel with a string and hang a mass from the string to accelerate the wheel. Or you could mount the wheel in a horizontal plane and drape a string, with a mass on the end, over a frictionless pulley to accelerate the wheel. Both methods should produce an F = ma relationship.

The cylinder and spheres experiment has given all the motion of from up to 8 units of mass to 1 unit of mass.  A large center disk is also stopped by two unwrapping pucks (on a frictionless plane).

[Eddy Currentz]

Hello pequade and all,

may I draw your attention to the following Website :

http://www.unifiedtheory.org.uk/

Scroll down to Diagramm 13 :

INTRODUCTION TO STEEL INERTIA QUANTUM DRIVE

Hope you enjoy this.

Kator

Thanks for the link! I like the way this guy thinks. I haven't laughed so hard in quite a while, he's got a good sense of humor. 

[pequaide]

P-Motion: I would make these alterations to your machine. Change length and mass to the metric system, the science world is much more familiar with metric. For example; you rarely see 32 ft/sec/sec as the acceleration of gravity, it is usually 9.81 m/sec/sec.

With this change you have 4.5 kg on each end of a one meter vertically mounted bar (the bar is very light weight) with a high quality center bearing.  Add an extra 4.5 kg to one of the end masses so that the overbalance 4.5 kg can rotates one half meter to the bottom. That will mean the extra mass will move from 90? to 180? (or from 3 O'Clock to 6 O'Clock). The acceleration rate should be one third that of gravitation 9.81* (4.5/13.5) = 3.27 m/sec/sec. This is from F = ma   44.145 newtons / 13.5 kilograms = 3.27 m/sec

At the end of the .5 meter drop (of the extra 4.5 kg) all the 13.5 kg will be moving 1.808 m/sec.  v = √ (2* .5m * 3.27 m/sec/sec) = 1.808 m/sec

As the extra (overbalanced) mass reaches the bottom release it into a light weight, .1 m radius, horizontally mounted wheel with a high quality bearing point. Now we have 9.0 balanced kg moving 1.808 m/sec in a vertically mounted wheel and 4.5 kg moving 1.808 m/sec in a horizontally mounted wheel.  Connect a string from the vertically mounted wheel to the horizontally mounted wheel so that it is winding up on the vertical wheel and unwinding from the horizontal wheel. Now release the overbalanced mass from the horizontal wheel but keep it attached to another string that has been wrapped around the horizontal wheel. While the overbalanced mass unwraps from the horizontal wheel it will absorb the momentum of the two 4.5 kg masses on the vertical wheel. This is the same phenomenon as the cylinder and spheres experiment. 

Newtonian Physics predicts that the overbalanced mass will now be moving 1.808 m/sec * 3 = 5.42 m/sec (from the Law of Conservation of Momentum) and it will rise 1.5 m. d = ? v?/a. It was dropped only .5 m.

Add the extra mass back to the vertical wheel at 90? after you have transfer 2/3 of its energy to another system. You can still start over because you have three times the original energy to work with. 

[Kator01]

Hello all,

I followed this with very much interst and I like to add a bit to this discussion :

I have understood that pequaide has seperated 1/3 of momentum and changed the orientation of
the momentum axis so that both axis are perpendicuar to each other. This is a key-feature in this system.
After this he transfers the momentom of the  bigger-momentum-system to the spun-off-system.

Regards

Kator