Free energy.

The operation of a float of the engine 01 11357.

The operation of a float. Ex.:

We manufacture the float (variable volume) in such way, that in state of minimal volume it weighs 10 kg UNDER WATER, and in a maximum state of volume it weighs less than water (-)10 kg, so that it emerges worm surface. Ex., if its weight is = 110 kg, its volume is = 100 liters in a its state of min. volume and 120 liters in the state of max. volume.

The float is hermetic, inside is the air and its mechanism, by ex.:

The mechanism includes/understands the mass (any concrete mass e.g. or a volume filled with sand), which is fixed at the end of an arm of lever. Inside are them (or it) springs with gas, and the piston. The spring with gas it is a cylinder with the piston, filled by nitrogen under the pressure. The springs with gas, e.g. are used in the cars for the maintenance of the back door (or of cap), which opens upwards.

The float is under water, e.g. with the depth 3 m, the piston turned worm the top, as shown on the drawing:

We analyze that happens: The mass (e.g. of a weight = 100 kg) will move worm bottom, it will move (will attract inside) the piston, by decreasing the volume of the float and by compressing the springs with gas (to which we store the potential energy).

With the depth 3 m the pressure of water = 0,3 kg /cmÂ². This pressure operate the piston with the force, proportional A overrates it piston. If piston overrates it = 800 cmÂ², water with the 3 m depth will operate the piston by the force of: 800 * 0,3 = 240 kg. At the end of arm of the lever one will have 240/2 = 120 kg. If the springs with gas are at the end of the lever, on the springs one will have the weight of the mass (100 kg) and forces it of 120 kg, coming from the piston. At the end of arm of the lever one will have: 100 + 120 = 220 kg. We choose the springs with gas of a force of thorough = 220 kg.

The float decreases its volume, its weight under water becomes = 10 kg and it cule worm the bottom. Let us admit that the depth of the pissine is = 8 m. The float passes from the 3 m depth until the depth of 8 Mr. Chemin traversed = 5 m.

Let us admit that the height of the float makes it possible the mass to move at the distance = 50 cm. The course of the mass = 50 cm. Since the piston is pushed by the medium of the lever, it will pass the distance 50/2 = 25 cm. Piston = 800 cmÂ² overrates * 25 cm = 20 000 cm.cube = 20 liters. The displacement of the piston with decreased the volume of the float to 20 liters. For a weight of 110 kg and the volume (of the float under water) of 100 liters, its weight (under water) is = 10 kg. The float of 10 kg "falls" from a 5 m height (under the effect of the gravitation). The float "fell" (run) to the depth 8 m. The mass inside the float is in bottom, the piston is moved (is attracted) inside.

Now we turn over the float to 180 Â°. For that one must raise a weight of 10 kg to the 50 cm height (the float weighs under water 10 kg, the mass is in bottom) so that the mass passes to the top. Mass = 50 cm traverse. The float will be found in this position:

The gravitation and the springs with gas (they will return energy accumulated) will move the mass worm bottom, will push the piston worm bottom (interior worm outside), having increased the volume of the float by 20 liters. Maintaining the float is lighter than water, it weighs under water (-)10 kg and will assemble worm the top, worm depth the 3 m, there one stops it.

The float, "while falling" from the 5 m height produces energy and to turn over it, it is also necessary to spend energy. C.t.d., the float (10 kg)"fell" from a 5 m height and to turn over it, it should be raised (its weight of 10 kg while turning over it) to a 50 cm height.

From here:5 m - 0,5 m = 4,5 m- the distance, on the which float produces energy. Energy does not depend on the trajectory, but only on the height. For the top (with the depth 3 m) it is necessary turned over again, while spending the same quantity of energy that in bottom, and it runs, etc.

So that the floats are turned over themselves and to increase the power of system, we fix the floats at a chain (or chains), which circumvent the wheels and the floats are turned over, by circumventing the wheels.

Ex. like that: (the diagram)

The floats are turned over to 180 Â° in top and to 180 Â° in bottom, by circumventing the wheels and they produces useful work (energy) while going down (while running) and while going up (while floating worm top).

Calculation of engine 01 11357(exemple).

It is considered, that the wheel axle sup. is on the level of overrates water, and one selected the weight of mass.

Ex. one selected the weight of the mass = 100 kg.

Ex. the distance enters the axes of the wheels = 5 m.

From here, the pressure has the depth 5 m = 0.5 kg/cmÂ². (to include/understand, to see the point "D", diagram).

From (for this depth) we calculate the surface of the piston, in not do not forget that one with the springs has gases, which are compress (the spring has gas were compressed by the mass, when the float this found in top, of dimension right, on the level of the wheel axle sup.).

To counter the pressure of water the 5 m depth there is lays out:

100 kgf (weight of the mass) + 100 kgf (the force of pushed springs has gas) = 200 kgf (this force is at the end of the lever).

Holding account that the piston is thorough (worm outside) by the medium of the lever, the force is multiplied by 2: 200 kgf X 2 = 400 kgf.

Since the distance enters the axes = 5 m and to the pressure of water to the 5 m depth = 0.5 kg/cmÂ², consequently:

- 400 kgf/0.5 kg/cmÂ² = 800 cmÂ² (piston overrates it).

From here one calculates coefficient 800/100 = 8. (it is for the convenience, it is easier to make calculate them more detailed with him).

Now, that one found the surface of the piston, one makes the correction for the depth, to which this finds the engine.

The engine is under water, the higher axis this finds with the 3 m depth by ex.

From here: - the pressure has the 3 m depth = of 0.3 kg/cmÂ² (not "B" fig. 1).

We lay out of this pressure on the piston with the point "B" and it will produce the force of thorough on the piston of 800 cmÂ² (the surface of piston) X 0.3 kg/cmÂ² (pressure of water) = 240 kgf. Since this force is applied to the medium of the lever, then, at its end we will obtain: 240/2 = 120 kgf, which is added to the force, which comes from the mass, weighing 100 kilogrammes (fig. 1 point "B").

On the springs with gas there will be (to compress them) the force of 100 kgf (weight of the mass) + 120 kgf (coming from the pressure) = 220 kgf, which one will store in the springs with gas (the force of propulsion, that one will use, when the float moves at the point "D").

The distance between the axes = 5 m, that means that the lower axis this finds with the depth: 3 m + 5 m = 8 m. With this depth (not "D" fig. 1) pressure = 0.8 kg/cmÂ². The piston of 800 cmÂ² operates the force, causes by the pressure of water (0.8 kg/cmÂ²).

800 (piston overrates it) X 0.8 (pressure of water) =640kgf This force operate the piston of outside worm the interior.

Now let us look at the force on the side opposite of piston:

- 100 kg - the weight of the mass.

- 220 kg - the force of propulsion of the springs.

Total: 100 kgf + 220 kgf = 320 kgf. This force acts on the end of the lever. In the medium of the lever (and on the piston) one will have 320 X 2 =640kgf.

The forces of propulsion are identical on two sides of the piston (outside and interior).

One little to leave like that. In this case in high A right (not "B") the piston will move worm the lower interior (increases the pressure of water in connection with the increase depth), and in bottom left it has will move worm outside above point "D"; (the pressure of water will decrease).

But one little also to increase the weight of the masses with 5-10-20... kg to calculate these weights it is necessary to hold account, primarily, of the losses for frictions, which comes from the springs with gas.

It is all.