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### Author Topic: Rosch taking orders on OU Bouyancy device.  (Read 199933 times)

#### thngr

• Jr. Member
• Posts: 82
##### Re: Rosch taking orders on OU Bouyancy device.
« Reply #435 on: May 15, 2015, 05:48:32 AM »
1) A Striling or any other kind of heat engine needs both a hot and a cold reservoir.
[/font][/size]
yes you are right.

2) Rosch / Gaia claim that the system is self sustaining.  The compressor and each process that follows it is lossy.  The claim is false.

that system is lossy mine also but if you take out the heat of compressed air to stirling pump than there will be efficiency increase four times, this is what makes it going but they may lie about it.

• Hero Member
• Posts: 1842
##### Re: Rosch taking orders on OU Bouyancy device.
« Reply #436 on: May 15, 2015, 08:23:31 PM »
Why the calculation in http://revolution-green.com/rosch-smoking-guns-and-the-scientific-explanation/ or http://pesn.com/2015/05/13/9602617_scientific-explanation-for-Rosch-KPP-buoyancy-system/ is wrong:

The first power calculation in the article is:

== Formula 1 ====   Thus, we get: N = 9.81 x 2 x Q x 0.5 x H x 3 = 9.81 x Q x H x n ====
Further down one calculates:
== Formula 2 === N = 9.81 x 0.167 m3/sec x 2 m x 5 x 0,9 = 14.7 kW  =====
There is the error of putting "2 m" in again, because 2 was cancelled by 2 * 0.5 in Formula 1.
Therefore the corrected Formula 2 is:
== Formula 2 corrected === N = 9.81 x 0.167 m3/sec  x 5 x 0,9 = 7.35 kW  =====
But there is a further error in "Formula 2 corrected", which is the factor 5, because of  "5 working wheels above each other".
If there are 5 wheels above each other the water rises from one wheel to the other only 0.4 meters and not each time 2 meters. Therefore we can calculate with one wheel a height H of 2 meters, or for 5 wheels a height of H 0.4 meters each. We can calculate with one wheel with H = 2, or we can calculate with 5 wheels with a H = 0.4. So, either H = 2 or H = 0.4 x 5, which is the same.
Finally we end up with a correct "final Formula 2":
== final Formula 2 === N = 9.81 x 0.167 m3/sec  x 0,9 = 1.47 kW  =====
In addition we have to factor in a bigger loss than 0.9 because of the very optimistic height H = 2. If the water rises 2 meter it encounters friction at the wheel and some water is lost because it runs through gaps.
Further there will be losses in the gear connecting the wheel with a dynamo and the losses in the dynamo itself.
Therefore we will get much less than 1.47 KW.
Please note: I have not criticised the calculation itself (although one should do that), I only corrected obvious errors.

Criticism of "Formula 1":
The formula stems from water turbines http://en.wikipedia.org/wiki/Water_turbine#Power and assumes that water is falling down through the turbine. But one is not allowed to assume the same efficiency if "bubbly water (water air mixture with 50% air) rises through a turbine" rather than "smooth water is falling through a turbine". Much higher losses by friction of "bubbly water rising" in contrast to "smooth water falling" have to be assumed. And if things happen slowly (as is necessary in the "bubbly water rising case"), the formula is overly optimistic, because the formula only works with rather fast moving water.

#### d3x0r

• Hero Member
• Posts: 1433
##### Re: Rosch taking orders on OU Bouyancy device.
« Reply #437 on: May 23, 2015, 01:40:09 AM »

(for the TL;DR 'So ya, still can't happen.') [/size]

Light blue/cyan cells are primary input parameters - everything else is derived from them and constants.

white on Red cells are weight (given in pounds)
white on Orange cells are length (given in inches)
(first sheet is freely editable, a copy is protected readonly for original reference)

At the current state, the floats travel a greater distance than the pump is required to travel and the lift-weight of displacement is greater than the weight required on the pump head to pressurize the air...
so it's a 22% excess of power from the floats than required to pump 1 bottom float full of air.

- The excess gets greater by just adding separation between floats... since less air is required at greater depth, the stroke of the pump is less... but I guess even with a double action pump the excess distance is consumed to repressurize the second action from slightly less than 1atm(vacuum against valves to pull in new air) to the required pressure and then the additional distance to fill the float....

So I guess that's where the 'excess' would be going if accounted...

---
Note: There are 2 'required stroke to fill' the top one is to fill the float to capacity to full volume, the bottom one is required to fill a bottom float and have that be 100% full at the top; without losing air to expansion as pressure decreases on rise.

Hmmm... how to determine total stroke - oh just use the 1atm volume... so yes
any possible gain number turns into a loss

Still not satisfied that the final ratios don't turn out to be closer to 1; I guess it would if I assume that the double-action pump chamber will also self-balance a little... that is the compressed side will help the head move to pressurize the unpressurized side slightly...

So; again... really 1:1 best... unless you have a storage tank that gets depleted with time.
------
was listening to pesn video so decided to clean up the sheet and take another look to see if I missed something.  At least the input/output is in the same units now

I guess even on a continuous pump process the input at 1atm to X atm is a loss... cause really it's calculated from already having pressurized air... some sort of differential is required otherwise (integral over time?).

So ya, still can't happen.

#### MarkE

• Hero Member
• Posts: 6830
##### Re: Rosch taking orders on OU Bouyancy device.
« Reply #438 on: May 23, 2015, 04:17:48 AM »
buoyancy is not a source of energy but with one addition to compressor it going to be; if one know how use that heat with stirling pump.(stirling engine but without a work piston insted check valves) but please for give me; I do not have to teach you to design a perfect stirling pump.[/size]
Your statement self contradicts itself.  All the energy that if any that reaches the output shaft of te buoyancy machine comes from energy supplied by the compressor.  The compressor that Rosch/Gaia used had a capacity of ~75W puping air to 5 meters depth.

#### MarkE

• Hero Member
• Posts: 6830
##### Re: Rosch taking orders on OU Bouyancy device.
« Reply #439 on: May 23, 2015, 04:18:57 AM »
[/font][/size]
yes you are right.

2) Rosch / Gaia claim that the system is self sustaining.  The compressor and each process that follows it is lossy.  The claim is false.

that system is lossy mine also but if you take out the heat of compressed air to stirling pump than there will be efficiency increase four times, this is what makes it going but they may lie about it.
The efficiency is less than unity, far less.

#### RomanEmpire

• Newbie
• Posts: 18
##### Re: Rosch taking orders on OU Bouyancy device.
« Reply #440 on: May 24, 2015, 12:37:52 PM »
It may also work with a system that exploits an effect similar but opposite like this?

#### markdansie

• Hero Member
• Posts: 1471
##### Re: Rosch taking orders on OU Bouyancy device.
« Reply #441 on: May 25, 2015, 03:15:10 AM »
It may also work with a system that exploits an effect similar but opposite like this?
Please send your diagram to PESWIKI where it will be warmly welcome.

#### memoryman

• Hero Member
• Posts: 758
##### Re: Rosch taking orders on OU Bouyancy device.
« Reply #442 on: May 25, 2015, 03:19:42 AM »
Please make sure you give a percentage of the sales to Sterling.

#### LibreEnergia

• Sr. Member
• Posts: 332
##### Re: Rosch taking orders on OU Bouyancy device.
« Reply #443 on: May 25, 2015, 03:29:21 AM »
It may also work with a system that exploits an effect similar but opposite like this?

This system cannot work. At best it will move until the fluid comes to an equilibrium condition and then it will stop.

#### Pirate88179

• elite_member
• Hero Member
• Posts: 8366
##### Re: Rosch taking orders on OU Bouyancy device.
« Reply #444 on: May 25, 2015, 03:33:24 AM »
Please make sure you give a percentage of the sales to Sterling.

Also, he needs to donate a working device so Sterling can power his house with it.  I believe this will make like 15 O.U devices Sterling will be using to power his house.  My guess is, that he still pays an electric bill every month.

Bill

#### Spilled Fluids

• Full Member
• Posts: 138
##### Re: Rosch taking orders on OU Bouyancy device.
« Reply #445 on: May 25, 2015, 03:33:54 AM »
This system cannot work. At best it will move until the fluid comes to an equilibrium condition and then it will stop.

It only works as long as the circulator(water pump) is running.

#### Pirate88179

• elite_member
• Hero Member
• Posts: 8366
##### Re: Rosch taking orders on OU Bouyancy device.
« Reply #446 on: May 25, 2015, 03:35:49 AM »
It only works as long as the circulator(water pump) is running.

...and is plugged into the mains through a hidden power cord.

Bill

#### memoryman

• Hero Member
• Posts: 758
##### Re: Rosch taking orders on OU Bouyancy device.
« Reply #447 on: May 25, 2015, 03:42:16 AM »
"...and is plugged into the mains through a hidden power cord." do you have a problem with that?

#### MarkE

• Hero Member
• Posts: 6830
##### Re: Rosch taking orders on OU Bouyancy device.
« Reply #448 on: May 25, 2015, 09:44:50 AM »
This system cannot work. At best it will move until the fluid comes to an equilibrium condition and then it will stop.
Thank goodness there is a check valve to keep the thing from going into reverse and flooding from below.  Perhaps the circulator is supposed to be a generator driven by the rising water.

#### RomanEmpire

• Newbie
• Posts: 18
##### Re: Rosch taking orders on OU Bouyancy device.
« Reply #449 on: May 25, 2015, 10:08:44 AM »
I seem to see a bit of irony in your answers............. I admit that my diagram is elementary and my thinking is simple, but in my defense I say that I posted it on a site called "Overunity - search for free energy and discussing free energy" and your answer seems rather a group that writes on a hypothetical site called "Never Overunity - free energy does not exist and if you believe it you are a fool", I mistakenly thought that this was the right place.. Or maybe you.
However, I think you have me mistaken for someone else. To answer some allusions:
Yes, the circulator* is connected to a source of energy and the hope is, which is probably unfounded,  that using the weight of the water the generator can produce more energy than the circulator can consume. Even the container where there is the mechanism and the generator is part of the closed circuit.

I view of the fact that a small circulator can move up to 4 cubic meters per hour of water, which means 4000 kg of water in an hour, or about 66 kg in a minute, and it consumes a maximum of 75 watts/hour. I wondered if it was possible, but apparently it is not (this is the link of the circulator of which I speak http://productfinder.wilo.com/en/IT/product/0000002e00037bbd0001003a/fc_product_tendertext )

Thank you for your valuable attention and sorry for the loss of time, enjoy your stay on Overunity.com

*Yes -Circulator: http://p.globalsources.com/IMAGES/PDT/B1055392017/Hot-Water-Circulator-Pump.jpg