Free energy from gravitation using Newtonian Physic

[Kator01]

Hi P-Motion,

more explanations -> new questions. Please bear with me.

I have to apologize but my english misses the meaning of the term ramp. I like to be more precise : what does this ramp look like ?
Is this a device fixed against ground which catches the overbalanced mass pushing it back upwards in the guiding socket ?
I understood the process but not the technical detail of this ramp

Very intersting idea

Kator

[pequaide]

P-Motion: You must transfer the momentum of the entire wheel to a subset of itself, and then release it.  Since your ramps and lever arms don?t do this; I predict that your wheel will not work.

A very light one meter bar with a quality bearing point at .5 meters has six kilograms on one end and five kilograms on the other end. The six kilogram side (the six kilogram lever arm of .5 m length) will rotate to the bottom with an acceleration in accordance with F = ma. If all eleven kilograms remains attached to the bar the six kilogram side will rise to the same level from which it was dropped. That leaves you with no change in energy.

If the momentum of the system (11 kg at .9444 m/sec velocity) is transferred to the one extra kilogram, when it is at the bottom of a .5 meter drop, it will have 10.38 units of momentum and 10.38 m/sec velocity. When released this velocity will allow it to rise to 5.499 meters and it was only dropped .5 meters. 

Please don?t ignore and avoid this fundamental difference between the laws of levers and The Law of Conservation of Momentum.

What the lengthening of the lever arm gains in leverage it loses in distance. The side that has the longer lever arm but equal mass will indeed rotate to the bottom, but when it gets there you are stuck with an attempt to rotate it back to the top. If you leave it in one piece and don't consolidate the momentum, it will cost you the same going up as that which you gained going down.     

[Kator01]

Hello P-Motion,

did you calculate in a simulation the kinetic energy of the overbalance-mass at the point where it is beginning to move up  ( to be pushed up ) by the ramp and compare this with m x g x h ( 3 cm ) ?

As far as I understand this process cannot  be calculated with the impuls-formulas because the time for the upward-movement of the overbalanced mass must be measured first. As your mass is in a continous movement all along one cycle, I do not see a impuls-transfer so that the wheel-mass is fully stopped. According to the formulas that exactly is necessary in order to get excess-energy. In your desing finally the mass consumes m x g x h ( 3 cm ) in a slow-motion-fashion compared to full impuls-transfer. If there is Ekin >> Epot you will win.

I concentrate on getting rid of the standard-tunnel-view for finding  unconventional ( new ) physical/technical means to achieve what the formula promises.Some iddeas have come to my mind which I will discuss here soon.

I wish you the best and hope you will proove me wrong.

@peqaide : what is the mechanism you use to keep the spheres in the openings while accellerating the spin of the system and then release it ? As I understand it you have to first reach a certain speed so a certain amount of centrifugal force can develop which then can be released.
What setup did you use doing this with gliding masses on a almost frictionles spinning CD-Disk ?


Kator

[pequaide]

Kator01 question: What is the mechanism you use to keep the spheres in the openings while accellerating the spin of the system and then release it ? As I understand it you have to first reach a certain speed so a certain amount of centrifugal force can develop which then can be released.
What setup did you use doing this with gliding masses on a almost frictionless spinning CD-Disk ?

Pequaide answer: In the mechanically dropped and released models the spheres are held up against the cylinder with arms made of plywood (the arms were painted black for photographic reasons). The seats in the plywood are padded with hard rubber. Strong springs are triggered that jerk the arms away from the spheres. This arrangement releases both the spheres and the cylinder at the same time. The cylinder is held square with a piece of plywood mounted on the spinning rod that holds the arms and springs.

The rod spins at 3.25 rotations per second. The diameter used for the 5 in.  O.D. PVC pipe was 4.81 in. (distance to the center of mass of the cylinder). The diameter used for the spheres was 5.00 in. (their center of mass seats on the outside diameter of the cylinder.  This gives you an initial velocity of 1.297 m/sec for the spheres and 1.247 m/sec for the cylinder.

In the most recently posted pictures the spheres are held in place with the side of the thumb and the side of the second finger. In the hand held models the RPS is unknown; the mass of the inserted pipe is altered to get the cylinder to stop just as the strings on the spheres are entering the slit in the cylinder. These, as all models, are video taped and clearly show the cylinder (or center disk in the air table model) stopping.

In the air table models the pucks were held up against the large center disk with a string that is pinched up against a raise wall in the center of the large disk. The raised wall is used to accelerate the large disk, while the fingers (or spring loaded padded wooden dowels) are pinching the strings that hold the pucks in position at 180?. In this way the pucks and disk are released at the same time. The strings are then dragged by the pucks across the air table, causing a small amount of friction. The strings that allow the pucks to swing out are of course above the table.

The RPS of these models is also unknown, if held by hand. Some calculations were made using the mechanically released models (the spring loaded padded wooden dowels).  My measurements and calculations showed energy increases in all models, as the spheres reached maximum velocity. I can post a few pictures of the arms and center disk, but I will have to do it tomarrow.   

[pequaide]

Kator01 one 1M picture