Muddypause wrote: If gravity is used to change the volume of the float when it is at the top of it's cycle, then the same force will be insufficient to change it back again at the bottom of the cycle. You have to do more work expanding the float than contracting it because there is greater pressure on it. It doesn't matter how you arrange this mechanism, you cannot get gravity to make a net gain in the work it can do.
On the site
http://perpetuum.monsite.wanadoo.fr there are 3 engine:
01 10913 - at summer tested on a prototype in 1994.
01 11357 - is easily verifiable by arithmetic calculations.
02 00723 - this engine was not checked by calculations nor tested on a prototype.
Method of checking by calculations of the engine 01 11357:
It is considered, that the wheel axle sup. is on the level of overrates water, and one selected the weight of mass.
For ex. one selected the weight of the mass = 100 kg.
For ex. the distance enters the axes of the wheels = 5 m.
From here, the pressure has the depth 5 m = 0.5 kg/cm². (to understand, to see the point "D" fig. 1).
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/cm², consequently:
400 kgf/0.5 kg/cm² = 800 cm² (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/cm² ("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 cm² (the surface of piston) X 0.3 kg/cm² (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/cm².
The piston of 800 cm² operates the force, causes by the pressure of water (0.8 kg/cm²). 800 (piston overrates it) X 0.8 (pressure of water) =
640 kgf.
This force operates 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 =
640 kgf.
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 "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 to 5-10-20 kg. To calculate these weights account should be held, primarily, of the losses for frictions, which comes from the springs with gas.
The operation of the springs has gas is explained here:
http://www.stabilus.com/frameset.asp?sp ... =menue.asp
To choose
Gas Springs.
Explanation of operation of only one float.
We start by manufacturing a float in such a way that, folded up it "weighs" 10 kg under water and that once deployed it "weighs" -10 kg. By weighing we understand that the push that it undergoes from the Archimedes' principle and its actual weight gives the weight indicated. In other words, 10 kg: it runs, -10 kg it floats.
We place this float at a 3 m depth. The pressure associated with the weight with the concrete mass will cause the répliage float, displacing air through flexible tube towards the opposite float and decreasing the density of the float. We have a mass of 10 kg which run towards the bottom of the basin, until a 8 m depth, that is to say 5 m of way with a sufficiently slow speed that to avoid to the maximum the losses of energy due to friction of water.
In the same time, the expelled air of the first float will cause the deployment of the opposed float, in partnership with the weight of the concrete mass which, once the turned over float, brings a deployment of the unit. This increase in volume (20 liters) involves a reduction in the density of the float which "will weigh" -10 kg then and thus will float towards surface, producing a work identical to that produced at the time of the phase of descent. Once reached the 3 m of depth, the opposite float will be him to 8 m and both carry out a rotation of 180°, thus starting again the cycle.
We spend of energy for the phase of rotation, but let us produce we at the time of the floats run and float.
The difference between energy spent and that produced gives the useful work of a float. It is possible that the float is turned over all alone with the proviso of using a system of rails and of butted in such way that once traversed env. 90% of the 5 m displacement the float ridge against the thrust placed slightly side compared to the centre of gravity of the float.
The float will thus circumvent the rail and will be turned over. The float having changed direction, the mass of concrete will cause the expansion or the reduction volume. What avoids us having to even spend of energy by us and to extract it directly from the system.
The system thus produces energy in "consuming" forces: the force of gravitation at the time of its changes of volume, the force of Archimedes for its displacements. The actual weight does not have any importance since what imports is the density (variable) floats. It goes of course from oneself that several sets of floats (at least two plays thus 4 floats) are necessary along the chain in order to allow the reversal of the floats in phase of reversal (the only moment when they do not produce any work, they consume energy for their rotation has 180°).
P.S. Computer translation.
