Just sharing some thoughts   
                                                                                               Jim

Hi dear Jim
 I want to inform you and all community of forum that all or most of all important videos was taken down by YT team and I am leaving this forum with our team because here we have published our videos and no where else.
 All what we have got as positive from here - we have got good , a few friends. All other our efforts were not evaluated in correct way by community and unfortunately it means this community has in its body not good sound interpersonal relationships.
 Mean time I am busy from my height position in Armenia Telecom to resolve the issue with YT, already connected with Google Irish guys to find the man who have done these idiotic things against of us. First of all we will find the country of frustrated man then we will find by his IP his home and address, then that guys will be called to police. I am anger as team leader that I have to call war to stupid persons, but unfortunately it must be done sometime as far as they are taking already more than half planet place...
Because of these incidents we are stopping our participation in this forum
Sorry guys. Not secure from idiots place here!
Truly
Rob

Show me the mathematical calculations that explain anything in detail and you will see the nature of the problem . Math is touted as some kind of foolproof logistical god of the universe when all it really does is help us keep track of our bank accounts . It cannot simulate, explain or otherwise exemplify reality , no matter what they taught on Star Trek . The best argument for perpetual motion is this: Bessler . He said it all . He was a career perpetual mobilist . ( How many of those in history ?) Only one . He said that the knowledge of these forces was known to children rather than adults . One  need only look at the state of our planet and system to realize that our children are superior to us in many ways .
Matt 11:25 "At that time Jesus said, "I praise you, Father, Lord of heaven and earth, because you have hidden these things from the wise and learned, and revealed them to little children." Food for thought .

This is an interesting example Parisd. If I understand you correctly, you are referring to a balloon in the context of the Archimedes principle, namely some sort of buoyant object moving up through a fluid and later falling back downwards due to gravity in the air whilst outside said fluid.


Note in passing that gases and liquids may each properly be characterised as working 'fluids' despite air being in a gaseous state.


Accordingly, we may as well take the example of water, as Archimedes himself originally did, and focus on any object or fluid with mass of less than approximately 1000kg/m3 because that is the density of water. Any object or fluid (such as oil which is less dense than water, or a balloon full of air) will rise up through our column of water.

I declined to talk about a Buoyancy motor too unpracticle to build, I was just talking of a montgolfier balloon that can transport persons in the air, the ballon does not require more energy to bring those persons to 20m elevation that it need to bring them to 10 meters elevation and guess what if one person jump at 10m his kinetic energy when reaching the floor will be much less than the one jumping out of the balloon at 20meters elevation.


We want the balloon to rise to the top of the column of water and then fall down due to gravity perhaps because it has a weight attached to it.


We know that when we release the balloon at the base of the water filled cylinder, that it will certainly rise to the top of our column of water because it is less dense than water and therefore has positive buoyancy.


So you are correct when you argue that less dense fluids or objects must always rise to the top of columns of water due to the greater density of water.


But this will be a one way trip for our balloon unless we can figure out a way of getting the balloon back to the bottom of the cylinder again, so it can perform useful work. So let us attach a small weight to the balloon, such that the balloon has positive buoyancy in the water filled cylinder (causing it to rise) but has enough weight to allow it to fall due to gravity through the air (heading back towards the base of the cylinder) when it reaches the top.


But it is here, at the base of the cylinder, (the balloon having travelled from bottom to top and back down again to the outside of the base of the cylinder) that we have to use the 'P' word. Pressure.


How do we get the balloon back into the base of the water filled cylinder?


The pressure at the base of a cylinder depends on the height of the cylinder, on the density of the fluid inside it and on acceleration due to gravity.


This is counter-intuitive because you might think that the diameter of a cylinder must be relevant to the pressure calculation, but it is not. Imagine a cylinder 500 metres in diameter, but only 10cm deep. By imagining such a cylinder, you will realise that the pressure at the base of it will be very low. But now try imagining a cylinder 500m tall and 10m in diameter. If you dived to the bottom of this cylinder, the pressure would be very high.


So in other words, the diameter of a cylinder is irrelevant to base pressure. The height of the cylinder is a key variable when calculating pressure, and I have provided below an example relating to a cylinder 25m high.


This pressure has been calculated solely with reference to the height of a column of seawater, and its density and gravity.

(height(25m) x density of seawater (1020kg/m3) x gravity (9.81 m/s/s)
= 250,155 Pa gauge pressure at the base of the cylinder
Adding atmospheric pressure of 101,350 Pa gives absolute pressure of 351505 Pa.

So base pressure in the cylinder = 352kPa (+/- 1kPa)

Pressure at other points in the cylinder will be lower as one moves up (as the height of the column of fluid decreases relative to the position of the balloon).

We know from Pascal's law that pressure applied to a confined fluid is transmitted undiminished with equal force on equal areas at 90 degrees to the container wall. In other words, the pressure at the side of the cylinder (by the base) will be the same pressure as we find at the bottom of the base so to speak. Depth or distance from the top of the cylinder is the only relevant consideration.

The point is that if we have a 25m high cylinder, and we want to introduce an air filled balloon into the base of it (or indeed introduce any object or fluid with lower density than seawater (1020kg/m3) or lower density than regular water (1000 kg/m3), we will need to expend energy.

But how much energy will we need to spend?

The answer to this question is much more difficult. To answer it, I would try to phrase a parametric equation to provide an accurate answer, but we can use less complicated mathematics to get an idea of the energy involved.


The first thing to realise is that the means by which we get the balloon back to the bottom of the cylinder and the means by which we get the balloon INTO the fluid at the bottom of the cylinder are path dependent.

In other words, we can use inefficient methods of getting the balloon to the base (hamster wheels) or more efficient methods (e.g. gravity), and obviously we are going to pick an efficient method of doing it because we want to generate electricity or mechanical power from our buoyancy machine.

Whatever system we devise, we will always need to use sufficient force (Pressure = Force per unit area) to cram the balloon back into the base of the cylinder once the pulley or weight attached to the balloon has allowed it to fall to the outside of the cylinder base through gravity.


But again, how much power must we spend in watts to cram the balloon back inside the cylinder base?


We know the pressure is 352,000 Pascals at the base of the water filled cylinder. As a general rule of thumb, you will need between twice as much and four times as much power to cram a balloon back into the base of a cylinder as may be generated by the balloon when attached to an alternator motor.


The shaft of our alternator motor is attached to the rising balloon on the way up and to the falling balloon on the way down (we have assumed it falls due to an attached weight due to gravity back to the base of the cylinder), so that it is generating electricity as it moves up and down. But it is the act of getting the balloon back inside the base of the water filled cylinder that represents the practical problem.


By way of example, imagine trying to force a balloon full of air to the bottom of a swimming pool. A lifeguard (or some such other powerful swimmer) would find it difficult to force the balloon to the bottom of the pool.


Even if we assume that a powerful swimmer is capable of generating 300 watts of power for short intervals (which is very unlikely...175 watts for a professional cyclist would only be possible for a few moments), there is still no way the rising and falling balloon can generate 300 watts by rising and falling when attached to an alternator motor. The balloon motor would most likely generate no more than 5 to 10 watts (and this is an upper limit). But why is this?


The reason for the low power output is precisely the same reason that led to positive buoyancy in the first place. If an object (such as a balloon) has low mass, then yes it will have positive buoyancy and rise to the top of a column of fluid. But for precisely the same reason, it will not generate very much force when used to power an alternator motor even with a weight attached to it. Why not?


This is because Force = mass x acceleration, and therefore a slowly rising and rapidly falling balloon with low mass will generate very little force (perhaps 3 or 4 Newtons of force if you are lucky).


3 or 4 Newtons of force will be enough to generate a fraction of a watt (if you can find an alternator motor with sufficiently low torque (N.m) to get it to work in the first place.


By way of conclusion, the device will 'work', but you will get considerably less electrical or mechanical power out of it (in Watts) than you have to spend to get the balloon to move up and down when attached to an alternator motor. Most of the losses will stem from trying to cram the balloon into the high pressure environment at the base of the cylinder.


A more interesting idea would be to use glycerol or castor oil as a working fluid via two connected cylinders, and have an air compressor force the working fluid through a denser substrate such as seawater. However, the maths I did on this were inconclusive and a PhD in electrical engineering from a top ten university (as well as the OU member Fletcher) told me it would not work and provided reasoned mathematical arguments demonstrating why it would not work. This was because the air compressor is tantamount (using Lump matter discipline) to a battery, and the one cannot get more energy out of a system than is put in. Gravity is not a source of energy. It merely converts potential energy (water at altitude) to kinetic energy as it falls, and no energy is added in this process. in fact, at least 50% of the force of the falling water is immediately lost due to delta mom (the change in momentum of the blades of a turbine) when the falling water hits the cups of the turbine.


I hope this has been helpful.