[are] my findings

My Problem with Buoyancy Generators
Dear S. Allan. I have a problem with all proposed buoyancy generators in that I do not see how it is
that they are able to operate in a state of over-unity. I don’t think I am wrong and I welcome criticism,
but at this point in time I cannot fathom how it is possible for them to operate. Allow me to explain:
In order for the buoyancy generator to work, the amount of energy used to compress the air must be
less than the energy extracted through buoyancy. In order to pump air into containers (I like to call
them balloons) at the bottom of the swing, the pressure of air must be greater than that of the water
at that depth. For instance, say the depth of water used in the generator is 𝑥 meters, and that there
are an arbitrary number of steel balloons that hold the compressed air to become buoyant on both
sides of the swing. Here are a few equations. (Note g = 9.8 m/s2 and 𝜌𝑊 = 1000 kg/m3)
𝑉2
𝑊𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑖𝑠𝑒 = − ∫ 𝑃𝑑𝑉
𝑉1
𝑥2
𝑊𝑒𝑥𝑡𝑟𝑎𝑐𝑡𝑒𝑑 = ∫ 𝐹𝑛𝑒𝑡 𝑑𝑥
𝑥1
𝐹𝐵𝑢𝑜𝑦𝑎𝑛𝑐𝑦 = 𝜌𝑊 𝑉𝑔
𝐹𝑛𝑒𝑡 = 𝐹𝐵𝑢𝑜𝑦𝑎𝑛𝑐𝑦 − 𝐹𝑤𝑒𝑖𝑔ℎ𝑡
𝑃1 𝑉2
𝑉2 𝑃2
=
𝑜𝑟 𝑉1 =
𝑎𝑡 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑇
𝑃2 𝑉1
𝑃1
𝑃𝑉 = 𝑛𝑅𝑇 𝑜𝑟
𝑊𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑖𝑠𝑒 is the isentropic work required to compress a gas (that is reversible work – minimum work
required to do so). At sea level (101.325 kPa or P1) and constant pressure this equation simplifies to
𝑊𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑖𝑠𝑒 = −𝑃1 (𝑉2 − 𝑉1 )
Where 𝑉1 𝑎𝑛𝑑 𝑉2 is the volume of gas to be placed in each balloon before compression and after
compression respectively.
For the work extracted by the generator, the buoyancy of the balloons forces a chain upwards which
turns a cog connected to an electric generator. Assuming there are an equal number of balloons on
each side of the chain (the chain side on its way up and the chain side on its way down) we can neglect
the weight of the metal balloon casings. Therefore the net force experienced by the buoyant balloon
on its way up is used to turn the generator.
𝐹𝑛𝑒𝑡 = 𝐹𝐵𝑢𝑜𝑦𝑎𝑛𝑐𝑦 − 𝐹𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑒𝑑 𝑔𝑎𝑠 ≅ 𝐹𝐵𝑢𝑜𝑦𝑎𝑛𝑐𝑦
So per balloon the maximum work extracted is (assuming a chain height of 𝑥 meters)
𝑥2
𝑊𝑒𝑥𝑡𝑟𝑎𝑐𝑡𝑒𝑑 = ∫ 𝐹𝑛𝑒𝑡 𝑑𝑥
𝑥1
𝑥
= 𝐹𝐵𝑢𝑜𝑦𝑎𝑛𝑐𝑦 ∫ 𝑑𝑥 = 𝜌𝑊 𝑉1 𝑔(𝑥) = 9800𝑉2 𝑥 [𝐽𝑜𝑢𝑙𝑒𝑠]
0
In order for the compressed gas to displace water in each balloon, the pressure of the gas must be
greater than the pressure of the water at the bottom of the chain. That is to say that 𝑃2 , the pressure
of gas entering the bottom balloon must be at least:
𝑃2 = 𝑃1 + 𝜌𝑊 𝑔𝑥 = 𝑃1 + 9800𝑥
Substituting for 𝑉1 =
𝑉2 𝑃2
𝑃1
and 𝑊𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑖𝑠𝑒 = −𝑃1 (𝑉2 − 𝑉1 )
𝑊𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑖𝑠𝑒 = −𝑃1 (𝑉2 −
𝑉2 (𝑃1 + 9800𝑥)
𝑃1 + 9800𝑥
) = −𝑃1 𝑉2 (1 −
)
𝑃1
𝑃1
Bearing in mind
𝑊𝑒𝑥𝑡𝑟𝑎𝑐𝑡𝑒𝑑 = 9800𝑉2 𝑥
Define system efficiency as
𝜂=
∴𝜂=
𝑊𝑒𝑥𝑡𝑟𝑎𝑐𝑡𝑒𝑑
𝑊𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑖𝑠𝑒
9800𝑉2 𝑥
9800𝑥
=
=1
𝑃1 + 9800𝑥
𝑃1 𝑉2 (
− 1) (𝑃1 + 9800𝑥 − 𝑃1 )
𝑃1
Therefore it is not possible to run a buoyancy generator over-unity. Under absolutely ideal conditions,
the best you’ll get is as much energy out as you are pumping in using the compressor.
While I’m completely open to correction if someone can show me otherwise, I’m fairly confident in
the maths here. As of this moment, I do not believe it is possible for any buoyancy generator to operate
in an over-unity fashion.
Hope you’re having a wonderful week
Ryan