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You are here: > News > April 24, 2015; 23:00 GMT

Part IV: Math for expanding gas as floats rise shows Rosch KPP buoyancy device is overunity

Based on a comment noting that the gas will expand as the floats rise, Ryan Ross redid his math and showed that, indeed, by including this factor in the math actually predicts, scientifically, an overunity system. While the numbers are encouraging, they don't come close to accounting for what is being achieved.

by Sterling D. Allan
Pure Energy Systems News

Earlier today, I posted the following story regarding some reasonable skepticism regarding the two stories (Part I | Part II) we posted yesterday about the Rosch / GAIA technology that harnesses power of rising floats using pressurized air, which then descend after the air escapes above the water chamber.

We received a response from the CEO of GAIA, saying that he doesn't have a scientific explanation, but that the bottom line is that it works. Their self-looped system produces a net 4.8 kilowatts with zero input from external sources.

Ryan Ross, who is friendly to free energy, but wanting to understand the physics behind it, was going back and forth with me by email, and I reminded him that these new phenomenon are likely to involved overlooked or not-yet-understood principles of science, and he should be thinking about considerations that he had not included in his original math.

Here is the dialogue I had with Ryan:


Adding expanding gas to math

From: Ryan Ross
To: Sterling Allan 
Sent: Friday, April 24, 2015 10:54 AM
Subject: Re: Fw: need help with KPP buoyancy device skepticism

Hi Sterling

Quite excited at the moment. I was reading through some of the comments of part 2 and somebody pointed out that the bottom float can only be partially filled with compressed air as it will expand on its way up the column. In my maths derivation I assumed each float was closed - but they aren't. The gas expands as it goes up. Per mole of gas, more buoyant work is done the further up the column it goes - I assumed it was constant.

There may actually be something to this system...

I'll be home in about 3 hours and I can redo the maths and see if OU is possible



From: Sterling Allan 
To: Ryan Ross 
Sent: Friday, April 24, 2015 11:15 AM
Subject: Re: Fw: need help with KPP buoyancy device skepticism

Hi Ryan,

There has to be a second principle at play that we're overlooking. Something to do with the nature of buoyancy itself. 

I'm quite sure your math is correct with the assumptions you're making.

The problem is that you're not taking something crucial into consideration.

Comparison: magnet holding stuff up on a fridge. No "work" is being done, but there is a lot of energy being manifest.


From: Ryan Ross 
To: Sterling Allan 
Sent: Friday, April 24, 2015 1:33 PM
Subject: Re: Fw: need help with KPP buoyancy device skepticism

Holy crap I think I have it

Let me just check the maths one more time

From: Ryan Ross 
To: Sterling Allan 
Sent: Friday, April 24, 2015 2:32 PM
Subject: Re: Fw: need help with KPP buoyancy device skepticism

Well... that person in the comments might have been right. The reason this thing could possibly work is due to the fact that the gas in each float expands as it rises, displacing more water within the float thereby increasing the buoyancy of the float allowing that float to do more work to the generator.

Here is my maths <<ATTACHMENT>> redone taking the expansion of the rising gas into account - and dare I say it there might be something to these buoyancy generators.

Well… is this it? Is it really this simple?

You're welcome to post my investigation and let other people have a look at it. I don't think I've made any mistakes but it is entirely possible that I have so I need other people to check it.

But if the maths is right, we have a fundamental proof of this thing working.... That is worth far more than just a demonstration - it's an explanation, which in turn will allow others to test and replicate it (if it works)

It's late here now and I have an early start tomorrow - don't know how I'm going to sleep being so excited.

Kind regards


Still Missing Something

by Sterling D. Allan

I will comment that while this is encouraging, I don't think it's complete. There is at least one other factor we're not including yet. The earlier numbers Rosch measured showed an output:input ratio of nearly 3:1 (12 kW out with 4.2 kW input), including losses, with a tower of only about 3 meters, so to overcome losses, we should expect that the math will predict a ~4:1 ratio, without losses, when we have all the factors properly tabulated. At 3 meters, Ryan's math would only account for maybe 1.1x efficiency.

What is nice about this information is that it gets us past the hump of "impossible," using presently-agreed-upon scientific formulas.

Corrected Math; Heat Pump Factor

From: Ryan Ross 
To: Sterling Allan 
Sent: Saturday, April 25, 2015 1:08 AM
Subject: Re: Fw: need help with KPP buoyancy device skepticism

Hi Sterling

I had an idea: we've been looking at this thing wrong the whole time.

After sending you the email last night something kept bugging me: one would expect that as the gas expands inside of the float doing PdV work against the surrounding water that the gas would cool to get the energy to "balance", and I didn't include it in my calculations as I simply assumed constant temperature, which was a mistake on my part.

However, the key point of this system is that the floats are immersed in a column of water which as we know has a high heat capacity and heat conductance. So if the floats are moving slow enough, there is enough time for heat to transfer BETWEEN the surrounding water and the expanding gas to keep it at constant temperature.

And that's it.

The buoyancy generator is NOT a Free Energy device. It does not break any conventional laws of thermodynamics. It is simply a type of Energy Pump. It converts ambient heat energy (that is in the air) into buoyant work energy which is then converted into electricity. It operates almost exactly the same as an air-conditioner or "heat-pump". The energy you put into the system is used to create an entropy difference to get energy to flow in the direction you want. Like an air-con can do 3kW of heating with a 1kW electric motor input, these buoyant generators can output more electric energy than it takes to drive the compressor.

I'd really like to build my own one now.

So assuming the floats move slow enough to allow for heat transfer to keep the expanding gas at constant temperature, then my maths above holds.

Assuming my maths is correct, I'm satisfied that buoyant generators work.

Last thing, I cannot account for the 3x efficiency using a column of 3 meters - but I did make simplifying assumptions such as ideal gas to see if the maths works. Rosch might be genuine with their claims...

Kind regards


From: Ryan Ross 
To: Sterling Allan 
Sent: Saturday, April 25, 2015 6:30 AM
Subject: Re: Fw: need help with KPP buoyancy device skepticism

Hi Stirling, fixed my maths Mark E. spotted. Same result! <<ATTACHMENT>>

I also explain here how it works as an energy pump, and we aren't breaking any laws of physics.

Can you imagine the possibilities if this actually works? I've got a friend who said he is willing to help me try make one of these things, so I think we'll start that on Tuesday. I'll keep you in the loop.



"Free Energy" = harvesting wheelwork of nature, which is what a heat pump does

by Sterling D. Allan

Nice work, Ryan.

I would different terminology. You say this isn't free energy because it's basically a heat pump, but you then describe how it harvests ambient energy. That, in my vernacular, is free energy. You don't have to pay for it. It comes from the wheelwork of nature. 

Remember, in my vernacular, solar, wind, tide, geothermal, waste-to-energy, boimass, etc, are all "free energy" for the same reason, and none of them violate any known/accepted laws of physics; but they all harvest the wheelwork of nature (or of human consumption and waste). See 25 Exotic FE and 25 Conventional page at PESWiki.

My question would be if the system is going to be heating up in the process, and therefore have limits, or if the harvesting and dissipation of heat is in equilibrium. Maybe you could give us an energy balance flow chart, showing where the energy is coming from and being transferred to. 

I understand the principle of a heat pump, as I've heard engineers talk about it many times, but I don't have my head wrapped around just how that works.

Livestream Update

From Roberto Reuter, CEO, GAIA; 9:47 am

Don’t wait on that.. a lot technical issues go along with that.. open ports etc.. all that stuff is hard to make it secure..

but beside that - we have had a very good first day !!!

Parameters of Heat Pump Principles

From: Simon DERRICUTT 
To: Sterling Allan ; Ryan Ross 
Cc: Mark Euthanasius 
Sent: Saturday, April 25, 2015 1:01 PM
Subject: Re: Fw: need help with KPP buoyancy device skepticism

Hi Sterling,

Ryan has a clever idea. The expansion of the air will indeed draw energy from the water and I'd missed that. If you take the compressed air input at the same temperature as the water and ignore the losses in compressing the air and cooling it down to the water temperature then the actual chain of floats will be over unity. It's also possible that the whole system might be over unity if the compressor is designed for peak efficiency at the delivered pressure.

The water will cool down, though, and take energy from the ambient. As the system stands, there isn't a good heat-transfer between the environment and the water so there will be a limit to the energy you can pull before the water freezes. That could be fixed if you tell them why it works and that they should be measuring the water temperature. 

Despite the possibility of this working, practically I'm not sure if the end to end efficiency will be good enough to take advantage of it. I thus don't think Ryan's maths tells enough of the story, since a compressor is generally somewhere on the adiabatic scale and you'd want to change the design a bit to get it to a more isothermal version with the excess heat from compression being cycled into the water for recovery by the system. If it's hotter than the water when it enters the float then it will start by shrinking and you lose some efficiency. 

Putting it into perspective, all the excess power you get out of it has to come from the water, and if you're running it continuously then that heat has to come from the ambient air (and cools the air). The water must thus be running (steady state) below air temperature, and if the air-temperature is below zero (Celcius) then it can't produce power. If you're thinking of running it indoors to get around this, then the power it produces comes from cooling the room until that gets too cold. 

If you use the water to cool both the compressor and the generator, then those normal "lost energy" sources will be utilised. Antifreeze in the water would be somewhat useful though also somewhat expensive. It needs a better heat-transfer between the water and the environmental air. At a certain air-temperature, the maximum power out would happen when the water is near freezing. 

I still think their measurements have been over-optimistic, and that the efficiency won't be as good as the ideal case Ryan has calculated. I'd need to look at each section to get a good feel as to the actual power-in versus power out. 

I'll put up a somewhat shorter version of this email as a comment. Well done Ryan - why it's useful to have people thinking about these things. I didn't think deep enough on this one. There is a slight problem in that it won't give you power in winter when you want it, but in summer it could provide air-conditioning and free power.

Best regards, Simon

From: Ryan Ross 
Cc: Sterling Allan ; markeuthanasius
Sent: Saturday, April 25, 2015 2:57 PM
Subject: Re: Fw: need help with KPP buoyancy device skepticism

Hi Sterling and Simon

*Just a note here, if everyone is willing can this conversation be public? If we have some ideas its almost a certainty someone else will have / has had them too. Public domain and someone cant go and patent it and profit off it

Last check is everyone happy with V3 of my maths? I'm prone to silly mistakes as Mark E. pointed out (haha thanks) and I'd appreciate if you guys can take a last look at it. That way I will be happy with the efficiency curve it gives.

I think this device will operate best if we maximise the heat transfer between the floats and the surroundings. So something like a copper float with an internal radiator (also copper) should speed up the heat transfer allowing for a greater power density of the device as a whole.

In terms of obtaining heat for the column of water isn't a problem. You can circulate the water outside to a solar collector. Even in freezing weather if you concentrate the sunlight we should be able to get the water heated sufficiently. A couple of square meters for a couple of kW power isn't an unreasonable thing to ask for in the winter. Also, provided enough air circulation in summer months it may remove the necessity for an outside heat collector (there's an awful lot of enthalpy flow through a room with two open windows and a few m/s wind)

In terms of the compressor I'm torn between running it adiabatically or isentropically. The argument for adiabatic is that if the gas is the same or lower temp than the water, it wont suffer from contraction when first injected into the float. That being said I'm leaning towards running it isentropically; There is no reason you can't put the compressor's air reservoir inside of the water column itself. That way you can ensure recycling of heat losses. Maybe liquid cool the motor casings and the compression mechanism with column water? Even if you didn't allow for direct heat transfer between the compressor components and the column, you are still heating the surrounding air which is where the energy is being pulled from anyway.

I agree that the system will operate best at Steady State when the water is near freezing. I'm not sure how familiar with heat transfer you are Sterling, but the short of it is that the greater the temperature difference between the column water temperature and ambient, the greater the heat transfer between the two (and therefore the more available to do buoyant work)

I'm going to try make one starting Tuesday. So far my end objective is a 10m model with 22 cubic floats total (10 on each side, one on top and one at the bottom at any point in time). I think I will make the column out of sheet metal (high heat conductivity), but make one side have view ports with Plexiglas every couple meters. I'm expecting at most 37% excess, but hey who knows.

While making it I'll stop at about 3 or 4 meters to replicate that first video. If I get 3x output I'll be sure to let you guys know (although I expect about 13% to be generous)

Last thing, does anyone understand what Alastair was saying about multiplying efficiencies per float? I was with him until he said that. To me per float the work comes from the gas expansion. The more floats on the chain the more gas you need to have compressed to get it to run.

It's late here and I have another early start tomorrow. I'll be online in the next 7 hours or so

Kind regards


- - - -

QuestionEverything wrote in the comments below:

This apparatus is certainly not a heat pump. Thermodynamic effects are far too small to yield any relevant contribution to the output energy.

- - - -

Over at Fess wrote: (translated by Google)

Hello everybody,

I am new to this forum but have been following since
Summer 2014, the OU posts AuKW about.

Below I'll explain my theory why
the AuKW might work, because when energy
Source comes to my mind the
Thermal energy of the ambient air in question.

In thought-experiment it comes to compression and cooling
of air in an ordinary bicycle pump, the I
Represent opinion that using the compression work
the earth-atmosphere extracts heat energy and
is thus stored in the compressed air!

When adiabatic compressed air as an ideal diatomic gas, then
goes according to school physics, the entire compression work into heat energy through,
and if one removes this heat by a Isochoric process,
then still remains unchanged compacted volume overpressure
exist, can perform which work against the ambient pressure, whereby
the compressed gas of course always cool below the ambient temperature.

This process can be also very good in a pV diagram for
Ideal gas law (PV = nRT) represent (s. Annex).

With each adiabatic compression of the Earth atmosphere is so energy
deprived of what at presentation of the Earth atmosphere as a closed system
is also evident. This advantage should be exploited, and I still ponder,
if you ever need to recover such AuKW this free energy.

I find it awkward when technician was doing compression work over
Calculate Integral take the easy way out on the internal thermal gas power
to go (see Fig. 2c, for example, solution to the task of Dr. Patommel to the air pump, link 1 below).
If he has pressure and volume of the compression already, then he really needs
only the difference between the inside thermal gas energy to the state before compression
calculate (W = 2.5 * p2 * V2 - 2.5 * p1 * V1 because interior air thermal energy U = 2.5 * p * V).
Instead of air when using an ideal gas atomiges-1, would then U = 1.5 * p * V, since f = 3rd

On the theory I came when I read in the isothermal state change
did that even there the heat extracted directly corresponds to the work performed.

It's interesting that this thermal energy from the earth atmosphere for
Expansion remains, and this consideration was confirmed to me by a university.

For real equations of state and atmospheric conditions is the invoice
Although somewhat more complicated, but the described effect disappears in no case.

I have checked the theory by some textbook examples with solutions (. S link 2 below)
and div. Wikipedia entries such as Adiabatic / Isochore / Isothermal change of state.

It is, as already mentioned, also logical, since the earth atmosphere in the compression with
Assists 101 kPa and expanded this / cools. The compression work brings the piston
from the pressure equilibrium and compacted "along with the Earth atmosphere," the
Air in the cylinder. The work of the Earth atmosphere remains after deducting
Thermal energy U = V = W at const. exist in the cylinder.

I hope that in Spich are not fraudsters at work, because in the current
Earth warming would be a power-generation of heat air "rescue".

Thanks to Harti for this informative forum!










To reduce drag

by Sterling D. Allan

Question. It seems to me that one of the major impediments here is the friction of the floats through the water. If the center of the apparatus where nominally sealed, could you get a flow of water that nominally follows the flow of the floats, over the top, down one side, under the bottom, up the other side -- kind of like a stream? I would think that would improve the system efficiency. I don't see a reason to have the floats go above the water at the top of their cycle, to improve the continued flow of water by having it go above the floats at the top of the cycle. And maybe the bottom of the tank could be rounded in such a way to encourage the smooth flow of water through it.

Use large-piston compressor that runs from the chain speed

Simon Derricutt writes:

What would be nice to see would be a large-piston compressor that runs from the chain speed (very slow, therefore) that compresses the air to put into the floats. This would save a lot of gearing and other losses and of course the bore/stroke could be set to pump the right amount of air into each float. This should then run itself and you'd need to attach a (geared) generator to provide power out. No need to even plug this into the wall except maybe to start it up, and that could be done with a small tank of compressed air.

That would be a demo we couldn't ignore.

April 26: Dialogue Continues

From: markeuthanasius
To: Sterling Allan ; Ryan Ross 
Cc: Simon Derricutt 
Sent: Saturday, April 25, 2015 11:06 PM
Subject: Re: Fw: need help with KPP buoyancy device skepticism

It is then the same wrong result. If you do the integral math correctly you will find that the theoretical efficiency ignoring all loss mechanisms is 1.00. One of the Germans did the integral math that looks correct to my inspection and got that exact result. Ryan should look through that analysis and see if he disagrees with it.

If the working fluid (gas) were incompressible, then at the bottom water in the tank is lifted, and that energy is what you have available to perform work later. You do not get anything extra. 

Since the working fluid is compressible, that means that work is performed both lifting the water by purging it initially, and compressing the gas. Assuming that the bucket is only purged to an extent that the air pressure inside never exceeds the surrounding pressure outside, including at the top of the travel, (otherwise we would see bubbles coming out particularly near the top), then energy stored in the compressed gas lifts additional water on the way up. It would in the best case be zero sum gain. 

At the end of the day we are still lifting and dropping mass. We can add all kinds of nuances but what we will invariably find is that none of those nuances contribute towards an energy gain. Each in fact contribute to additional loss.


From: Ryan Ross 
To: markeuthanasius
Cc: Simon DERRICUTT ; Sterling Allan 
Sent: Sunday, April 26, 2015 1:23 AM
Subject: Re: Fw: need help with KPP buoyancy device skepticism

Hi Mark

I don't argue with any of the reasons you have for mechanical losses for the system. I agree with you whole heartedly that the system is far more complicated than the model I used when performing my calculations. I aggree that the German model is correct too as it finds unity (as my first set of calculations did)

Specifically, in my model I neglected to calculate the work done by the gas to expand the water as it rises, which is why my maths seemingly leads to a COP greater than one. In another way my key assumption was that temperature remains constant inside of the float.

For argument sake, this is what I believe is happening in the floats:

Say we have a volume of compressed gas inside of the float at a height of x that is rising, where the temperature of the gas at that height is T which is in equilibrium with the column water temperature. At the heigh x-dx, the pressure of column liquid has changed by rho.g.dx . At this point from an entropic point of view, the gas wants to expand by dV. However, it must do work against the volume of water dV in order to take up that space. The only energy available to the gas at this point is it's enthalpy, therefore it expands by dV by decreasing enthalpy by dH. i.e. cooling down to T - dT.

The key point here is that there is now a temperature difference between the gas and the column of water. Therefore there is heat transfer between the gas and the water as by Fourier's law q = -kAdT/dx. The gas is brought again to temperature T but now at a pressure P - dP and volume V + dV. BUT at the same time the buoyancy of the float has increased by rho.g.dV.

Now of course this assumes the float is moving slow enough for heat transfer to occur at the same rate as the expansion of the gas - which if the float is metal should not be a problem.

In conclusion, the energy comes from the column of water helping to heat the gas as it expands on its way up. In turn the column of water is heated by surroundings.

I'm saying the the losses are accounted for in this. The heat energy transferred to the gas is at least equal to the PdV work against the water integrated along the height of the column plus the energy losses.

What is interesting to me is that any work that needs to be done against friction will in turn come from the enthalpy of the trapped gas, which is immediately replaced by the enthalpy of the column of water. The energy required to overcome these losses comes from the heat energy of the water, because the gas still wants to expand and the float still wants to rise due to the difference in density.

That being said, I wont know unless I build one for myself and take all my own measurements. I'll be sure to include lots of photos as well as all the raw data I can obtain.

Kind regards


In this email, Mark inserted his comments into Ryan's email above. Rather than repeat Ryan's text, I'll just give the last couple of sentences from Ryan, followed by Mark's comments in blue.

From: markeuthanasius
To: Ryan Ross 
Cc: Simon DERRICUTT ; Sterling Allan 
Sent: Sunday, April 26, 2015 1:49 AM
Subject: Re: Fw: need help with KPP buoyancy device skepticism


The key point here is that there is now a temperature difference between the gas and the column of water. Therefore there is heat transfer between the gas and the water as by Fourier's law q = -kAdT/dx. The gas is brought again to temperature T but now at a pressure P - dP and volume V + dV. BUT at the same time the buoyancy of the float has increased by rho.g.dV.

The reduction in pressure caused by heat transfer results in a negative feedback of the form G/(1+GH), where the coefficients are partially determined by the rate at which the surrounding water passes ambient heat to the gas.  A very high rate of exchange would result in expansion that almost but not quite matches the simpler model.  Lower rates all leave the expansion at a deficit against the simpler model, reducing the total amount of water lifted. 

It may seem weird that if the environment is supplying energy that we don't have a gain.  But that is because we haven't accounted for the energy deficit we created when we rejected heat to the environment via heat when we pumped the gas into the bucket down at the bottom in the first place.  If we want to count heat transfer during expansion, in order to be honest we have to count it during compression.  So, when all the calculus gets done, what we find is that this heat exchange mechanism on the way up is trying to pay back a debt we did not previously acknowledge.  At best we could reduce the heat exchange loss to zero only in theory, but in all practice it burns us, no pun intended.

This sort of thing happens with each bit of sophistication that we add to the model.  When we are done, we find the system is far below unity efficient.  We would have been much better off never building the thing and having connected wires from the mains directly to whatever electrical load we have.


Kind regards

One of the most useful things that one can learn by spending time perusing the Museum of Unworkable Devices is how easy it is to overlook important effects.  Many of the puzzlers presented take more than a few seconds to work out what exactly is wrong.


From: Simon DERRICUTT 
To: markeuthanasius; Sterling Allan ; kannan c.d. 
Cc: Ryan Ross 
Sent: Sunday, April 26, 2015 1:52 AM
Subject: Re: One point on Rosch KPP analysis

Hi all,

I need to apologise for jumping in without thinking enough. Ryan's idea is still a good one, but of course the work needed to compress the gas, even in an idealised situation, will stop this thing running when the environmental air temperature is the same as the water. The work needed to compress the air even isothermally will only be got back by the isothermal expansion of the gas in the float.

In order to self-run, the water needs to be hotter than the environmental air. We might get a bit of extra lift from the water evaporation too (input air will be normal humidity, output air will be 100% humid) but this will be a small factor anyway until the water gets noticeably hot and you'd see it steaming at the top of the column.

If the compressor input air is colder than the water, then the work needed to produce the air bubble in the water will be less than you can get from it as that float rises. The system should be able to exploit a fairly small temperature differential to produce work.

Ryan - you've worked out the expansion side, but you missed the input work for the compression which also rises as the height of the machine changes. In ideal conditions and everything at the same temperatures this gives a zero sum, but in real life there are losses. I think this can be made to work with solar heating of the water, but on a financial basis I think you'd be better off with standard solar panels. Don't go jumping into a big build before you've considered all the energy transactions.

Again I'm sorry for the too-fast enthusiasm. I've been somewhat short of sleep for a bit since my mum has lost track of time of day (and what day it is) and that had been a bad night before. Despite that, the idea does give us the possibility that they will demonstrate self-running with power out, but I don't think they'll measure the water temperature (or tell us). The question really is whether they know this and are trying to keep it quiet. If they've got measurements from a few minutes/hour runs, then wait for the 2-week runs and see if it still works. For short runs, the energy stored in the hot water can make it work, but on a longer run it will cool and thus stop. 

Best regards, Simon

From: markeuthanasius
To: Simon DERRICUTT ; Sterling Allan ; kannan c.d. 
Cc: Ryan Ross 
Sent: Sunday, April 26, 2015 2:13 AM
Subject: Re: One point on Rosch KPP analysis

Simon, I am sorry you are dealing with an elderly parent suffering dementia.

No, there is no possibility that there is hidden heat engine operating on temperature differentials within this device.

The real elephant in the room is the pathetic burp of bubbles seen from each bucket as it approaches the top on the rising side, and essentially no bubbles on the falling side. As a reasonably close approximation: The average flow of those bubbles at the density of water approximates the water mass being lifted and dropped per unit time in the machine. We are talking a small fraction of a liter per burp, and a burp about once every two seconds. We can be generous and call a burp 250ml: 0.25kg. That's 0.125kg/s raised 4.5m every second or: ~5.5W. That's what the machine has to work with before losses.


From: Simon DERRICUTT 
To: markeuthanasius; Sterling Allan ; kannan c.d. 
Cc: Ryan Ross 
Sent: Sunday, April 26, 2015 7:31 AM
Subject: Re: One point on Rosch KPP analysis

Hi Mark,

Thanks - it severely impacts the time I'm able to put to other things, and is getting worse. 

Given the situation, I rarely watch videos and on this one I certainly have not analysed it to the extent you have. I thus based my view on the size of the floats and didn't expect them to be massively oversized. 

If the air-fill is as small as you think, then your calculations imply that they intended the perspex demo system to only show people the idea and not to either self-run or produce power. I still think there is a possibility of a hidden heat-engine in there for the larger versions, but I don't see much hope of a lot of power delivered since there isn't that much available and the efficiency of their system isn't that great. It looks to be expensive power if it does work.

Since the demo should be running now, maybe we'll get some figures to work on. At the moment it's looking like either the output will be pretty small or that they'll be economical with the truth when publishing the data. 

Like Sterling, I find it hard to believe that people would spend so much time and money on making something without having better measurements than they have published. On the other hand, I've seen a lot of evidence that people do do this. Apaprt from the infamous QEG there was also a magnetic oscillation device a while back that had some high-powered backing (ex-government people and a ton of money) that also can't work and also seems to have been quietly dropped. That 30-ton gravity machine had a second copy built, too. It could well be that this is the same and people have fooled themselves. 

I really can't raise a lot of enthusiasm for this device. If it does work at all it's going to be massive, unusable for part of the year and the power will be more expensive than getting it from the wall socket. We're better off getting a solar panel. 

If the design does get published in detail then maybe I'll work through the details and produce a predicted output level. Or maybe not.... If they don't publish the working temperatures then I'll assume they are doing that deliberately. Although I do try hard (maybe too much) to not assume either bad measurements or deliberate falsification of results, it's sad how often it happens.

Best regards, Simon

From: Stefan Hartmann 
To: Sterling Allan 
Sent: Sunday, April 26, 2015 7:40 AM
Subject: Re: you going to GAIA demo in Cologne, Germany?

Hi Sterling, you can publish this:

Well have a look at this diagram: 
There is definitely less output than input calculated by normal physics.

Here is an Excel Spreadsheet to calculate it all: 

Here is the calculation of engineer Brugmueller: 

Hope this helps.

There is now a first witness report an it says, that Rosch-Gaia says:

Believe it and buy or not...

No technical measurements allowed in this moment... maybe only at the last day and then only on certain measurement points they provide...

They even did not publish a complete picture of their system !

Sounds all very fishy...

Regards, Stefan.

From: markeuthanasius
To: Sterling Allan ; Ryan Ross 
Sent: Sunday, April 26, 2015 9:25 AM
Subject: Re: Fw: need help with KPP buoyancy device skepticism

The heat pump concept was a stream of consciousness idea from Simon that is like many of Simon's ideas very creative. I do not see any way that the idea could hold up to even the point that it could turn the mechanism without any load. Consider that the small unit supposedly contains 750kg of water. If perfectly insulated, that requires almost 1kWh to increase its temperature by 1 degree C. That's a damned big kettle of tea to try and warm up. Question Everything is spot on.


From: Simon DERRICUTT 
To: markeuthanasius; Sterling Allan ; Ryan Ross 
Sent: Sunday, April 26, 2015 9:50 AM
Subject: Re: Fw: need help with KPP buoyancy device skepticism

Hi All,

The heat-pump idea was Ryan's, I think, but it was certainly creative. I do however agree with Mark that there just isn't enough power in there to do anything useful. Various other calculations have shown similar problems where the real output power can at best be over an order of manitude less than the ~5kW quoted. There's not enough mass of water being displaced and it's not moving fast enough to produce 5kW even if the air was free.

Sorry, but the obvious conclusion at this point is that they are messing up the measurements and it's hard to avoid the conclusion that it's deliberate. That means that the kWh meter on view is probably nobbled in some way (heat the magnet inside it and it loses the drag and misreads) and I'd use my own kit rather than trust theirs. 

If that meter has been nobbled that way (or another way) then this ends up as a scam rather than something really amazing. OK, the word is out now.....

Best regards, Simon

From: Ryan Ross 
To: markeuthanasius
Cc: Sterling Allan ; Simon DERRICUTT 
Sent: Sunday, April 26, 2015 10:34 AM
Subject: Re: Fw: need help with KPP buoyancy device skepticism

Hi all

Mark, I don't disagree with any of your reasons for there being losses in the system that were unaccounted for in the simplified expansion model. Yes there are many factors unaccounted for, it was just an exercise to see if there is possibly anything to this system. However, even if you introduce efficiency fudge factors to both the Wpressurize and Wexpansion equations, they factorise out as prefactors in the COP / efficiency equation. Given enough column height, eventually the overall COP will be greater than one. I don't know if that is 10 meters or 1000 meters or if you cant do it with water and need something like liquid mercury. I don't even know if it is possible to ever reach a COP greater than one, but I'd still like to try and see how close you can get, especially with more than a few floats.

I agree that the injecting gas needs to be at the same or lower temp than the column of water. That isn't a problem. You can place the compressor reservoir within the column itself, and have an approximately equal temperature of inlet gas to the water. The excess heat energy of the compressed gas can go straight into the water reservoir. I live in a particularly sunny part of the world so a few 10s of kWh of solar water heating per day is not an issue for me, especially with the column height it is convenient to have a thermosiphon in place.

Another thing I was thinking about is the mechanical friction / losses of the drive chain / output shaft mechanism. The friction / losses are a function of the speed of the drive chain - not the number of floats present. If you can get each to float to rise with a COP of say 1.1, with enough floats you can overcome the friction because at a certain drive chain speed there will be an amount of friction per float. If the sum of the 0.1 excess COP of each float along the chain all exerting buoyant force on the chain eventually overcomes the friction - the chain will spin and you will have useful output work. Whether you can ever get this, I don't know - but we wont know unless someone tries and documents all of the raw data and makes it available for everyone to analyze.

I'm going to go see a thermos expert at my university on Tuesday to get a better understanding of the compression / expansion thermodynamic interactions. I will share his opinion on the device and I will take a copy of everyone's arguments to him. Is everyone okay with that?

Kind regards


From: markeuthanasius
To: Ryan Ross 
Cc: Sterling Allan ; Simon DERRICUTT 
Sent: Sunday, April 26, 2015 1:38 PM
Subject: Re: Fw: need help with KPP buoyancy device skepticism

The basic analysis to me comes down to this: If one is going to try and create a heat engine running from the ambient, it is a non-starter. No matter how many hoops one jumps through, at the end of the day Sadi's immutable equation dictates that the amount of work that can be had from (T1-T1)/T1 is identically zero. There are people around who think they have found escape clauses but no proposed macro level escape scheme has ever been shown to actually work. So that leaves heating the water and trying to build an engine that operates from that. Now, Sadi tells us that our efficiency will be less than 1.0 even if we could expand the working fluid infinitely so as to drop the temperature to 0K. I do not see a starter in any of this. [...]

From: Simon DERRICUTT 
To: markeuthanasius; Ryan Ross 
Cc: Sterling Allan 
Sent: Sunday, April 26, 2015 3:36 PM
Subject: Re: Fw: need help with KPP buoyancy device skepticism

Hi Ryan,

I have an ambition to prove Mark wrong - so far it's unachieved. Although Sterling describes this as a heat pump, as a straight hydrostatic problem it won't work so the only way to get anything out is a heat engine configuration. 

If you take the ideal air pump for this, it will be a cylinder of the same capacity as the float at the top of the system, and will discharge that quantity of air inside into the float at the bottom. The compressor starts with air at atmospheric pressure, and will start to transfer air to the float when the pressure is sufficient at (with the 5M tower) around 1/3 of the stroke. The work needed here is the isothermal compression followed by the other 2/3 of the stroke at constant pressure. This compression will cause heat to be produced, and you want to do it slowly enough to be able to disperse that heat into the environment. An isothermal compression, in other words. In practice you can't actually achieve that but can get fairly close. If it is held at the temperature of the water, then you're back to the hydrostatic problem and a zero-sum. If the compressed air is at a lower temperature than the water, then you get some free expansion of the air from the heat of the water and there can be a gain. It won't be that much - say the air goes in at 300K and the water is at 330K (water at around 57°C) then you can get 10% of the lifting power for free. If you look at PA32R's estimate of the most power available from the whole paternoster, he says 174W. It's a near-enough estimate for the 5M tower. That means you end up with less than 20W gain. The compressor will need that 174W put into the compressed air, and if it's 90% efficient then that's just used up all the power you got from the hot water. I doubt if a motor and compressor will actually be 90% efficient (I'd expect around 80% tops) so you'd need the water another 30° hotter to just break even, and that gets to 87°C. Almost the right temperature to brew coffee, in fact. To get some energy to light a lamp, you'd need the water at boiling-point. Make tea.... Here though we still haven't looked at the mechanical losses and generator losses. To allow for those you'd need to chill the input air.

Although I can see the attraction in building something that works, even if it doesn't produce much, I'd suggest you don't spend money you can't afford. Even if you can get it self-running (which would be a great achievement) the available power will be very low. Get all the calculations done before spending any money, unless you want a very large paperweight. 

That's about it. You have my respect for thinking of the loophole, but the loophole isn't big enough to do anything useful. You don't however find out until the numbers are put in. Mark just told me that the prospective customers at the demo are walking away, too. 

Best regards, Simon

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Page composed by Sterling D. Allan
Last updated May 19, 2015




"It is harder to crack a prejudice than an atom." // "I'd rather be an optimist and a fool than a pessimist and right." -- Albert Einstein

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