Science a GoGo's Home Page Forums General Discussion General Science Discussion Forum Harnessing the power of Hydrogen + Gravity

Me and Kallog have been discussing using the buoyancy
of HHO to recover the electricity that was used in converting water into HHO in the thread
"Harnessing the power of the sun " since the thread has
moved so far off course I have decided to move it to its own thread.

Kallog, since we were having such a hard time with the buoyancy I though I would make an image of a way to use gravity instead , perhaps we can agree on this one and dispose of all the complicated buoyancy and pressure stuff and get on with it.

in the below image the water is converted into HHO at the bottom and then the HHO rises to the top where it is converted back into water as it is used as a fuel.
the water then is put into the containers that were emptied at the bottom and gravity pulls them down.

click the below image for a larger one.
then click the larger one for an even larger one.



since it is the amount of weight pressing downwards that will determine the amount of electricity you can recover
from a HHO generating process , and since the height of this type of system does not require more electricity for a higher system to produce the HHO at the bottom.

and since the HHO is lighter than air and will rise to the top itself without requireing additional electricity
the amount of electricity you can recover or produce or both only depends on how high you want to build it.

or how much FREE ENERGY you can handle.

BTW

you can do the above alot cheaper by using a tall pipe , and let the water flow into the pipe at the top.

then at the bottom you can use a hydraulic motor , just let the pressurized water from the bottom of the pipe flow through the hydraulic motor at the bottom to turn a gear box (at whatever rpm your generator system requires) and then the generator.

Hi Paul. Good idea about a new thread.

Is the system sealed? If so then the HHO can't float up because there's no denser fluid (air) floating down to displace it.

If it's open to the air, and air sinks down the pipe to push the HHO up, then you have a higher pressure at the bottom than the top. The higher the tower, the more pressure, so the more energy needed to create the HHO. You told me yourself that HHO generation requires more energy at higher pressures.

I think we'll be running in circles until we can quantify it.

Or just build one! Look how cheap and simple it is!! Some bits of wire in a pipe, a water wheel, an old lawnmower motor and an alternator from the car wreckers.

Kallog


yes the system is sealed between the point where the HHO is generated and the point where the HHO is used as a fuel.

I forgot to include this in the image but the larger pipe
(B2) has air inside it , thats why its larger.

the HHO rises through the air due to its density.
air cannot flow through the pipe (B1) because of the
pressure inside (A1).

and because HHO is lighter than air the HHO will gather at the top of (B2).

the HHO that gathers at the top in (B2) also acts as a
storage area where more HHO can be drawn off when needed
for power surges in the power plant.

depending on the type of power plant you are using
ie.. if using a ICE engine or turbine then you would mix air with the HHO at the top after the HHO has passed through (B3).

unless you have a ICE or turbine that uses only HHO.

for a fuel cell you would not want a mixture.
but you would need a seperator to seperate the hydrogen from the oxygen.

but thats not really a part of this recovery system as it can vary greatly.

but if you were to need to build one that doesnt use a pipe that uses air to raise the HHO , and need to determine the difference in weight of the two
( air and HHO ) I have provided this below.

According to the CRC Handbook of Chemistry and Physics, the density of dry air at 20 degrees C at 760 mm of mercury (one atmosphere of pressure) is 1.204 milligrams per cubic centimeter.

1 cubic foot = 28,316.8467 cubic centimeters.

So, dry air weighs 34,093.48 mg per cu.ft.

Which is about 1.2 ounces per cu.ft.

if the pipe had a cross sectional area of 1 entire sq inch
this means that 144 feet of pipe could house 1 cu/ft of air or HHO at any given time.

so if the pipe is 144 feet tall then there would be a need to supply a force of 1.2 ounces , well not that much.

as HHO weights less than air.
1 cu inch of air weighs .00004 pounds
1 cu inch of HHO weighs .00002 pounds


1 cu ft of water weighs 62.4 lbs
1728 cu inches in 1 cu ft
62.4 / 1728 = .0361 lbs
1 cu inch of water weighs .0361 lbs

water expands 1800 times when converted into HHO
1 cu inch of water weighs .0361 lbs
so 1 cu inch of HHO weighs .0361 / 1800 = .00002 lbs

there are 12 inches x 144 ft of HHO in the pipe so it
weighs (12 x 144) x .00002 lbs = 0.03456 lbs

WOW

so for every 144 feet in height you would need to supply a
constant force of .03456 lbs in order to get the HHO to go into the pipe at the bottom.



well that has that pretty much quantified wouldnt you say




I think that building something should follow design work
after all you need to know how much electricity your system would use in producing HHO in order to design a proper recovery system to recover that electricity.

then theres the percentage that you want or need to recover
if you only want to recover the 22% lost durring conversion to HHO then the system could be much smaller.

if your providing hydrogen for the space shuttle or commercial air lines then your system would need to be much larger.

Cheers for working that out. Yea it isn't much, but the tiny weight has a downside -

After you burn the entire 0.035lb of HHO, you only get 0.035lb of liquid water. You've applied enough power to lift 0.035lb up 144ft, and at the same time the water wheel generates as much power as you can get from dropping 0.035lb down 144ft. It's the same!!

You'd need a 100% efficient water wheel just to provide enough power to pump that gas up pipe B2. So there's nothing gained at all.

If this doesn't sound right, remember the water wheel will be rotating extremely slowly. Its buckets are filling 1800 times slower than the HHO is going up the pipe.




It's half the story, but you completely omitted the energy generated. It has no value unless it generates more than it consumes.


So what? Mortgage your house and hire an engineer. No expense is too great. You're forgetting again that after you have it working, you'll be RICHER THAN GOD!! Not only that but you'll have solved all of humanity's biggest problems. Are you too selfish to spend some of your time and money to give billions of people better lives?

Or do you expect it might fail?

Kallog




I did include the calculations to determine the amount of force required to just lift the HHO up the 144 ft distance.

but I noted that this was if you needed to use it to determine
the amounts of force to do so , not that it should be done.

since the HHO would simply float up the 144 ft distance
and you have agreed that the water falling would supply the energy that would be used up if it were lifted up 144 ft , the only cost in HHO production would be the cost at (A1).

this cost is recovered as the fuel is burned at the top.

and since HHO floats on air due only to its density as you
have admitted to in the previous thread , there is no reason that the system could not be higher than 144 ft.

and the higher it is the more energy you can produce as the water falls.

even if the water falls slowly , the higher it is the more torque the weight of the water will present to a shaft.

and high torque low rpm can be converted to low torque high rpm.



why would I go to all the trouble to just have a incompetent engineer work all the efficientcies out of the system , no I wouldnt hire any engineers.

sorry !

.

OK. So we've established that pushing it up is not an option. That's progress, we've closed off one dead-end avenue.



You calculated the force per square inch required to hold up a 144ft column of HHO. This is the pressure that the HHO generator needs to work at. If we replace some or all of the HHO with air, its weight is greater, so the HHO generator has to work at a greater pressure. That's even worse. It doesn't matter that the HHO is floating up, we still have to push it into the pipe against the back-pressure of that column of air, which takes the same power as lifting the column of air itself, and in fact is what we are doing.

Don't believe me? Try floating helium into an inflated car tyre. Without a high pressure pump (using energy) it doesn't go in at all, despite being bouyant.


-----

I think an HHO generator would have maximum efficiency operating in a vacuum. You suggested something close to this yourself.

That means any pressure higher than absolute zero requires additional electrical power input. This column of air/HHO must necessarily have a pressure at the bottom of it which increases in proportion to the column's height. Every extra foot of height you give it increases the power consumption of the HHO generator (ignoring the atmosphere, which can be sealed out of the system anyway).

You could achieve equal or better efficiency by placing the HHO generator at the top of the tower. You would then have to remove the water wheel. But that destroys the whole idea.

That argument isn't convincing unless I show that the extra power consumed by the HHO generator is >= the power required to lift an extra column of gas at the same rate it's being generated.

I can't show that because I don't know the relationship between electric input power and pressure in an HHO generator. And you can't deny it because you don't know either. However by assuming conservation of energy we can conclude that the requirement in my previous paragraph holds.






Then what? Go to the grave taking the greatest discovery in human history? Something I noticed about perpetual motion is that nearly every proponent of it creates his own idea. I've never seen somebody take another person's concept and promote it, try to prove it or build it. You came up with this bouyancy/water wheel idea yourself instead of just repeating what others have said. So equally, nobody else will take your idea and develop it. Unless you finish it yourself it will be lost and ignored forever.

Are you happy for it to be lost?
Will you commit your life to this work?
Or do you accept that it may fail?

Kallog



LOL

if the container at the bottom that is about to have the water in
it converted into HHO was at the top previously then it would have water and a small amount of air in it already and no HHO.

but its much simpler to use the pipe rather than the wheel , why you might ask?

because the pipe would cause less frictional loss than the wheel.

so I guess that removes your next straw.

using 2 pipes
(P1) to hold water, as it falls to the bottom.
and
(P2) to hold HHO as it rises to the top.

the water flows through a hydraulic motor at the bottom of (P1).
then enters the pipe (P2)that the HHO is in.

the water flows into the pipe (P2) via gravity.
water flows into pipe (P2) an into the HHO generator
located inside (P2)

pipe (P2) does not have an end on it at the bottom.

and the HHO is generated just above the bottom of the pipe
we have all seen that the HHO in all the videos floats upwards
in atmospheric pressure (1 atm) and there is no hardship placed
on the process due to atmospheric pressures or the buoyancy of HHO in air.

Im not sure that there wouldnt be some type of hardship placed
on it due to planetary alignment , or the coriolis effect

but I do know that HHO will float upwards in 1 atm pressure and that cannot be denied.

and I do know that water will flow down hill , that also cannot be denied.

now at the top of pipe (P2) the pipe is feeding HHO to
the power plant.

and the HHO at the top of (P2) will be slightly pressurized due to the buoyancy or density of HHO in air so it will
flow through to the power plant without undue strain or hardship.

dont believe me , try putting a glass held upside down over the top of a HHO generator , the HHO gasses will rise and push the air out of its way as it fills the upside down glass with HHO.

the power plant at the top will deposit the water from the power plant into the top of pipe (P1) which is open at the top to the atmospheric pressure.

now you could say that the water in pipe (P1) will evaporate
and that is true , some water will evaporate.

but if I add the amount of evaporated water at the bottom
of pipe (P2) then I dont need to lift it to the top , so no big deal as the company or home probably already has a 40 psi water
supply.

but you should also keep in mind that it rains these days
and a rain capture tank could be located at the top also !!

you should know by now that I dont believe in the energy conservation belief system that you believe in , as I know better.

I prefer to use the common sence method vs the theoretical
methods when determining if somethng will or will not work.

and just because you cant buy a FREE ENERGY system does not mean
that they cannot be built.

.

I think you ignored everything I just wrote. Generating HHO at 1atm uses more energy than generating it at lower pressure. You've said that yourself, and it sounds very reasonable. You find some evidence that it's wrong and you might have something.

I know you understand the idea that creating a bubble of gas lifts up the column of air above it. That consumes all the energy you might recover floating that bubble all the way to the top of the atmosphere. In other words dropping the column of air back to where it was.

Sure you can make the pipe higher and recover more of the wasted energy. At best you can approach recovering all of the energy by building a tower as high as the atmosphere.




If it depends on those things then it's not free energy, so who cares? Anyone can make a home hydro generator anyway. If it doesn't depend on those things then don't muddy the waters by mentioning them. They're a red herring.




Show me any evidence of violation of conservation of energy. You can't, so why believe in it?

Show me any free energy machine that's ever been built. You can't, because none exists to public knowledge. So why believe in it? Why not believe in magic carpets or little green men on the moon?

Common sense is the antithesis of science. By definition common sense can never lead to any new discoveries. You're always constrained by limited personal experience. Theoretical science can go much further than the feeble human imagination. That's why the Earth now goes round the Sun, instead of the common sense alternative that people once blindly believed.

But again, why aren't you building it? You still don't appreciate the spectacular value it will have? I can understand not building a rainwater electric generator - it's pretty mundane and only marginally useful. But free energy will SOLVE ALL THE WORLD'S BIGGEST PROBLEMS FOREVER!!

Kallog

the below system and its energy input and energy output (recovery) is opperating
durring a time frame of 1 minute.

if Im burning 1800 lpm at the top of the system.
and generating 1800 lpm at the bottom of the system
then I am putting 1 lpm of water in the pipe at the top each minute.


Michael Faraday has shown that 1 litre of HHO gas is produced
in one hour using 2.34 watts for a period of 1 hour.

2.34 watts X 1 hour = 2.34 watt hours each litre of HHO.

1800 x 2.34 watt hours = 4212 watt hours

if I use the 4212 watt hours in one minute this is my energy input



---------------------- ENERGY IN PER MINUTE --------------------------
252720 watts / 252720 joules / 4212 watt hours

---------------------- POWER IN PER MINUTE -------------------------

252720 watts / 252.7 kw
-------------------------------------------------------------------------


that part is now out of the way !!!!! we know our input energy.

---------- NOW FOR THE GRAVITY PART ------------------------------------


1 litre of water weighs 2.205 lbs.

the volume of 1 litre of water = 61.02374 cu/inches

if Im using a water pipe that has a inside cross sectional area of 61 sq/inches
then the water level in the water pipe will only drop 1 inch every minute because
I am only allowing 1 litre of water to pass through the hydraulic piston at the
bottom each minute , however I am putting back 1 litre of water in the top each minute.

to keep this simple as I like to do , we know we only have 1 litre a minute to work with

and I have to recover 4212 watt hours each minute to balance the recovery system out
(energy in = energy recovered) LOL.

so I will use a piston at the bottom and stroke it out once each minute.

if the piston stroke is 12 inches then the piston surface area will need to be
61.02374 / 12 = 5.002
5.002 sq/inches


this way every minute when it strokes it will remove 1 litre of water
from the water pipe at the bottom.

and when the piston strokes it will provide a force that can be applied over a 12 inch distance.

if the water pipe is 1000 ft tall , the water pressure at the bottom of the water pipe
will be 1000 X .433 psi = 433 psi

433 psi X 5.002 piston area = 2165 lbf applied to the piston
and applied for a distance of 12 inches or 1 ft in 1 minute = 2165 ft-lbf/minute

1 watt = 44.253 ft-lbf/minute.
2165 / 44.253 = 48.99

so if I apply the 2165 lbf over the 12 inches distance in one minute
I would recover a constant 48.99 watts during that one minute.

since my energy recovery system recovers 48.99 watts per 1000 ft in height
the ratio of energy recovery per foot in height would be
48.99 watts / 1000 ft = 0.0489 watts per ft in height.

I need to recover 252700 watts each minute as that is what it cost to produce the HHO.

so

252700 / 0.0489 = 5167689.16 ft high.

or 978.72 miles

since the energy out (recovery) is 252 kW/minute the energy that can be produced at the generator would be apx the same.

but the above is a way to produce more energy than you need , it would be much shorter if you were
focussing on just getting back the 22% you were loosing in normal water - HHO conversion.
and if its only 10% like you say , then thats even better.

and of course not everybody needs a 252 kW/minute power system.

the ISS is apx 199 miles high.

so that puts it right in the 22% range.

if I build a system this tall ((( ISS ))) ((( MARS ))) hint hint nanocarbons maybe !!!

I have recovered all of the energy I used to generate the 1800 lpm HHO

plus I still have all the energy that I produced at the top using the 1800 lpm HHO as a fuel in the power plant.

and if there are any inefficientcies found , just add a foot or so in height to compensate.


Im sure all of the stuff you write is important to you , and I do read it all , and you have found many
interesting obstructions that need to be considered , I just only reply to the stuff that you write that
applies to the discussion , I try not to focus on the energy conservation beliefs you have that you have
been taught and the what ifs that wont really accomplish anything.



the important thing about this is that it shows that the current usage of thermodynamics is a load of CR@P.


.

Excellent calculation! That really does cut through all the talk and get the point.

However! :P I spot a serious fault. HHO is only bouyant in air, not in vacuum. So it can't float up past the atmosphere.

Assume optimistically that the top of the atmosphere is at the ISS (200 miles, there are some air atoms at the ISS). Your machine can recover 51kW compared to 252kW used generating the HHO. That's 20% of the energy recovered.

Hmm. That's a curious coincidence. Do you suppose the commercial HHO plants have about 20% loss because they're operating under a column of air that could recover 20% using bouyancy? That suggests this machine is pushing us right close to 100% HHO generation efficiency.

To show that it goes over 100% I think you need to find an existing HHO generator with more than 80% efficiency at 1atm. And also find out how high HHO will really float. I imagine it can't reach the ISS without somehow losing its usefulness.


---

So you've said several times, but still haven't shown it.

I understand that you don't trust conservation of energy. So you'll notice that I only referred to it tentatively for things I wasn't sure of. None of my reasoning depends on it, because doing so would be circular. If I depended on it I would simply say "E_out > E_in, violates 1st law, goodbye".

Well of course we all do assume the 1st law holds sometimes. But those cases are a bit too trivial and obvious to mention.



By the way, mixing imperial with metric, and miles with feet is the opposite of keeping things simple!!! Hehe that's why I don't check your calculations thoroughly. Too complicated :P

Additional data:

Below the turbopause, at 62 miles, the different gasses of the atmosphere are well mixed by turbulence, and not stratified by density.

If we assume that the relative density of HHO and air doesn't change with pressure, then we should expect that HHO will be bouyant all the way up to the turbopause, and somewhat beyond.

So I think 62 miles represents a _lower_ limit on the maximum height of the tower. At that height we can recover 16kW from the 1l/minute of water. That's only 6% of the power used to generate the HHO so this result isn't conclusive either way.

I suppose what we really need is to find the altitude at which the density of the atmosphere is equal to that of HHO at the same pressure. That would be the upper limit on tower height. Beyond that the HHO won't be bouyant anymore.

Kallog

what I said about the use of thermodynamics is true.

is physics becomming more like a religion these days?

for some reason I really think you know that the following
is true , yet you didnt seem to say anything about it.

if you reduce the scale of everything by say 100
then the 978 mile height becomes 9.78 miles
the 252 kW becomes 2.5 kW and so on.

you could even reduce the scale even further.


it would be entirely possible to make a floating platform at a 9.78 mile height , and using nanocarbon pipes that are supposed
to be stronger than steel , should make this entirely feasible.

now I know that the weather would present a problem but just the fact that this would work voids the first law as you use it.

but as I use it it will never be voided.

I could build something that could sit in your yard and it would be the size of a small swimming pool and it would produce HHO 24/7 and would power itself , and your home , and charge your electric cars batteries.

or you could just put one in you car.

you could put one in a airplane.

it would work in outer space , or on the moon , or on
mars... hint


but getting funding for anything that produces or saves any
large amount of energy is impossible because the people who supply the funding rely on people who have been taught wrongly for their opinions on the invention or process and they simply use the first law as their knowledge base and that pretty much dissallows any energy savings or free energy production.

I have tried many times , and am always shot down by thermodynamics.


even though thermodynamics is not used correctly its wrongfull usage is the correct way to ensure destruction.

.

Whatever other people might be doing isn't relevent. They're not holding you back, they're working on their own things. Your intention is to disprove part of the theory. So you don't have to depend on that theory. That's fine. Scientists are doing that all the time with lots of theories. Go for it. I'm also not depending on thermodynamics when I talk to you. Except of course the obvious cases that we both assume are true. For example, the density of a liter of water doesn't spontaneously double if you leave it sitting under the full moon and sing a special chant.



Correct. And the 1 liter of water becomes 0.01 liters.
then the water flow rate down the pipe reduces by 100.
And the reduced height causes the pressure at the piston to reduce by 100.
So the power generated by the piston reduces by 10,000 to 25.2W!!

You're welcome to repeat your calculation for any other set of numbers if you think it'll improve the result. You can even use an imaginary planet with a different atmosphere, different gravity, etc.



Go for it. But it's already a big step if you just show theoretically that it would work. That costs nothing except your time. But to persuade someone you have to counter every single critisism that anybody makes. It would also help if you explain how everybody else managed to be so wrong for so long. People will want to know that. I mean specifically how - what mistake was made in what experiment by who? Where did somebody forget to divide by 2? Where was a result extrapolated to an untested area? that sort of thing. Point your finger on the actual mistake so everybody can clearly see it. Maybe gravitational potential energy is excluded from the 1st law? Identify the actual problem. Then you can easily test your claim.

Kallog

yea I did forget to reduce the lenght of the pistons stroke by 100 , and I havent calculated the reduction in energy input
either so I might do that later but thats really not necessary.

so rather than putting up another example to use I think it
might be best to use your idea.



as thermodynamics should apply anywhere correct?

and since the 978.72 mile high example would work
if it were sitting in a hole here on earth.

we can still use the earth.
and the same atmosphere.
and the same gravity.

remember this is only an example just to prove my point
that the current use of thermodyamics is BULL$#!T.

not that its meant to actually build anything.

so if the 1 lpm 252 kW example was sitting in a hole here on earth.
and the hole was perfectly insulated from the heat of the magma.
and it had a roof over it to keep rain out.
and the great purple rinocerous was no where in the vicinity.

it would work if built in a hole on earth , correct?
correct!!!

your only reasons given previously to its inability to function was the fact that there is no buoyancy that high up.
does buoyancy work that far down?

if we venture out beyond the earth would we need to adhere to
the current usage of thermodynamics as it is supposed to apply anyplace , or should we throw it in the trash can where it belongs.


.

Yea of course thermodynamics works everywhere - as far as we know.

The hole in the ground suffers from a new problem. Your calculation depends on Faraday's relationship between electrical energy and volume of gas generated. That was done at 1atm pressure. At the bottom of the hole there's higher pressure, and you've said yourself that'll require more energy. Maybe it's not so much more that it breaks the idea, but unless you find out, you can't assume it's OK.

I know it sounds like I'm just making up a new silly excuse every time. But that's the nature of perpetual motion ideas - robbing from Peter to pay Paul, then robbing from John to pay Peter, then ... until you start to wonder if there's some fundamental roadblock that causes nature to always preempt your intentions and defeat them.

WOW

I guess it just wont work unless you can make it shorter
to remove the pressure differences at depths.

let me think about this today to see what I can come up with
to remove the pressue obstructions.

there is another system I would like to discuss with you
later that involves an added process in apx the same situation.

Kallog

I found a way to do this today but it does produce the HHO
under pressure only its at a 30 ft depth not 978 mile depth.

LOL

I either need to find the energy needed to produce HHO
under pressures or I need to find how much energy is required
to create a vacume.

but I will do that later , it shouldnt be hard to do as I
still have 57 kW left over from 309 kW recovered in the minute.

I still think this is a workable example because after the minute
I have 57 kW and 1800 litres of HHO have been used in the power plant.

Hmm using some of that 20% recovery rate is going to be tight. But I look forward to it.

Energy of a vacuum is easy. If you open up a 1m3 cube of vacuum on the ground, you're effectively lifting the column of air above it up 1m. So the force required is 1m^2 * 1atm pressure. Energy = force * distance:

E = 1m^2 * 100,000 N/m^2 * 1m
= 100,000 N m
= 100,000 J
= 27.8 W-hrs
= 1668 W-minutes to generate 1m^3 of vacuum or
1.67 W-minutes to generate 1 liter of vacuum or
3000 W-minutes to generate 1800 liters of vacuum.

A thought I had was to use an atmoshpere made from a gas that's just slightly denser than HHO. That way the atmosphere can be thicker while having the same pressure at sea level. The HHO would still float all the way up, especially if there's only one gas in the atmosphere so it doesn't stratify into lighter gasses near the top.

Kallog



well the 1800 litres HHO cost 252 kW / minute if I remember correctly , then I recovered 309 kW / minute.

50.4 kW would be 20% of 252kW.

100% would just be making up for the lost 20% normally seen in HHO production or if I only recovered 50.4 kW.

so I have 57 kW / minute to use for the vacume or the extra watts needeed to convert HHO in the higher pressure water before I go below 200% efficiency.

so for now I will wait until I find out how much more
energy will be required to produce HHO in pressures or
in a vacume.

but I already think that the vacume wont require much more
at all.

so right now Im at 200% efficiency plus 57 kW per minute.

btw.

thanks for the vacume info , how would you calculate
a 5 inch vacume?

vacume is measured on a guage by inches , at least thats
what my guage reads , I suppose I can just guess the force I apply to the vacume pump and measure its piston area x stroke , then measure the volume of air inside the flask I am throwing the vacume on since the guage will tell me what the vacume is , I should be able to make a close estimate this way.

it seems that the water wants to boil at apx 22 inches of vacume so certainly I wouldnt need that much and it takes several pumps with the hand pump to achieve 22 inches of vacume.

I say boil but its not boil like water and heat , its like
hydrogen comming out of the water using just a vacume.

anyway I think that the extra 57 kW / minute sholud be capable of doing this wouldnt you think?

Inches could be inches of mercury. 1 atmosphere is about 30 inches of mercury. But it would go backwards for vacuum so atmospheric pressure is 0in. and complete vacuum is 30in. What I worked out was for a complete vacuum, even tho we can't quite get there, and the pumps are probably horribly inefficient, but that's just a technological problem.

That would mean 5" takes 5/30 = 1/6 of the energy needed to get complete vacuum on the same volume chamber.

Maybe the information on this page:

http://www.lhup.edu/~dsimanek/museum/themes/buoyant.htm

is useful to see why a buoyance-driven-perpetuummobile doesn't work.

the greatest amount of vacume that I can get using the hand pump
is 27 inches / -90 kPa , and at this vacume pressure the sides or surface of the flask is mostly covered by bubbles at this pressure.

if I agitate the flask then the larger bubbles float up and go out of solution , but new tiny bubbles form.

however if I strike the bottom of the flask against the bottom of a glass that the flask is sitting in , then thousands of tiny
bubbles stream up to the top !

if I continue striking the flask I keep getting the streams of thousands of tiny bubbles.

but the vacume pressure drops from -90 kPa to -80 kPa

at -80 kPa no more bubbles form even if I strike the flask.

ok I looked it up

my guage also reads kpa which at 30 inch guage
it is at -100 kPa guage.

the first stroke of the pump creates a -62.5 kPa vacume.
the second stroke of the pump creates a -82.5 kPa vacume.
the third stroke of the pump creates a -90 kPa vacume.

90 kPa is the highest vacume it can accomplish.

atmospheric pressure is 101.3 kPa

however 30 inch / -100 kpa can be exceded using a rotary vane pump and for higher vacume pressure a staged vacume pump
can be used.

.

the applications on that page do not require energy input
ie .. there is no force applied.

we are discussing proven methods here.

if you can use energy to displace water at the bottom
wthout exposing that energy to the water pressures.

then you can recover that energy as the buoyant force rises up through the water.

-1 + 1 = 0

now if the buoyancy is due to hydrogen gasses at the top of the water then you still have the energy in the hydrogen gasses to burn.

=1

Yea me and Paul are talking about something with an extra level of complexity, so it's harder to analyze than those examples.

The big difference is when the bouyant 'buckets' get to the top, they don't need energy to reduce their density and make them sink again, in fact we can extract energy while doing this by burning the gas and converting it to liquid water.

I suspect the problem is the energy required to convert this liquid water back into gas at the bottom, but havn't quite seen how that can use so much energy yet.

That's a pretty cool vacuum pump. I don't think it's necessary to worry about going beyond -90kPa. That leaves only 10% of the air in the chamber. Pretty close to none.

Instead of using a complex electro-chemical reaction to convert water to some gas, I would like to see an analysis of a simple mechanical equivalent.

For example: consider a box with a volume of 3 units (cm³, m³, whatever). Inside this Box, on each end, are two smaller boxes, each with a volume of 1 unit.

Both this smaller boxes are connected via a spring.

The weight of the object could be 1 unit (1g, 1kg, 1 metric ton, whatever)

By use of energy we are able to extend the inner boxes. This way we can reduce the density of this object from 1/3 to 1/5.
The energy to reduce the density is stored in the spring.

In my opinion this mechanical object features the same basic attributes as you are using from the H2O-molecules.


Lets assume this object is submerged in an environment with a density of 1/4 (in g/m³ or kg/m³).

In its extended state it will float up.
Since it displaces 5 units of volume, each with a weight of 1/4 It should experience an uplifting force equivalent to 5*1/4 - 1 = 1/4 units of weight.
Thats the weight difference between the displaced surrounding and the weight of our object.

On the way down our object will be in its retracted state.
Now the downward force is equivalent to 3*1/4 - 1 = -1/4 units of weight.

Lets assume our apparatus has a hight of 100 units of length (cm, m, whatever).

The energy gained on the way up should be force * distance = 1/4 * 100 = 25 units of energy.
On the way down we have a weight of 1/4 units.
E = m * h --> 1/4 units of weight * 100 units of distance = 25 units of energy.


So we have gained 50 units of energy, haven't we?

On the other hand: our object is ALWAYS submerged in the environment of 1/4 density.
How much energy do we need to switch our object from the retracted to the extended state?

As our apparatur is 100 units of length in height, the surrounding pressure on the bottom should be: p = rho * h.
p = 1/4 * 100 = 25 units of force.

To extend ONE of the two inner boxes we would have to spend 25 units of energy, to move the inner box outward against the surrounding pressure.
We are doing this for both boxes: voila 50 units of energy are used.

----

By using an electro-chemical reaction to change the density of your objects (H2O-molecules) you are obfuscating the need to use energy to change the density.

I guess you could complicate things furthermore by adding areas of different pressure.

------

The conservation of energy holds true for all experiments so far. That's why its called a law of nature - not because of some conspiracy.

Even if we observe some strange process where energy seems to vanish or appear: it is most likely we just don't understand the process.
For example: The neutrino was predicted because of an apparent loss of energy during radioactive decay.

Momos

the difference between the spring example you have and what we are zeroing in on is the extra energy that can be had durring the process.

ie.... the HHO

we are paying for the energy we use as we go , and generating extra energy durring the process.

Kallog

I've got it all figured out now but Im making some drawings so I will post them up when Im finished.

I have reverted back to buoyancy and kept the gravity also.

and the design is not obstructed by water pressure durring the HHO generation.

----------------
I have been trying to figure out something that seems odd to me.

when I pump up the vacume in the flask containing water the bubbles form , and the vacume bleeds down because of the bubbles forming.

so I pump it back up , it bleeds back down again because of bubbles forming but fewer and fewer , eventually when I pump up the vacume there are no more bubbles that form.

so something is happening to the water.

I just cant figure it out...

got any ideas?

It is the same principle.
In my mechanical equivalent you insert energy to extend the spring. The contracting spring can be used at any time to get SOME of this energy back.

That's the same as your idea of using energy for electrolysis and regaining this energy at a later time.

In both cases the problem persists: reducing the density of anything (Molecules, inflatable balloons, mechanical devices) in some medium will need exactly as much energy as you are gaining by letting it lift by buoyancy and fall down again.

My example uses just a mechanical device to see more clearly why it takes so much energy.

Just keep in mind: you will need more energy for splitting water to HHO then you can regain by recombination of HHO to water. (At least as long as the splitting is done under higher surrounding pressure then the recombination).

----

You water:
Couldn't this be dissolved gases, which are released in a low pressure environment? Sounds to me like the stuff happening to divers when they inhale gases at high pressure and then surface to lower pressure.

Momos
you compress the spring at the top and use 1 unit if energy
and the box sinks.
then if you decompress the spring under water so that it will expand and float up , then you are making the water level at the top increase , and you are lifting the water using the 1 unit of energy you put in compressing the spring.

you still have the energy from buoyancy as it rises to the top.

so you can compress the spring again.

but your problem is that you are working against the water pressures at the bottom.

if you didnt have to expend the 1 unit of energy at the bottom pushing against the water pressure
you could compress the spring at the top again.

and if you let the box rise more you could get more energy out from buoyancy.

this would be your extra energy over 100% efficicency.

Yea that'll just be water boiling into water vapour becuase the boiling point drops with low pressure. Then you have to suck that vapour out, bringing it back to low pressure, and allowing more to boil.

Maybe this is a serious issue, or maybe you can just do electrolysis on the water vapour?? Seems unlikely because it won't be very conductive.

Yes that agrees with what I was imagining. But how can you translate it into H and O molecules?

We're aware that when the 'box' is at the bottom, and you're reducing its density to make it float:
X joules of electrical energy is used, which goes towards
Y joules of chemical energy and
Z joules of work lifting the column of fluid above it.

Does X = Y + Z + losses? Conservation of energy says yes, but we're not allowed to rely on that. If we rely on that law then the whole system is automatically, and uninterestingly unworkable without any analysis.

From Paul's calculations which I checked too, it doesn't appear to add up right.

Not in Momos's system because it's already rising to the top of the column of water. If he'd had it rise half way, then you'd only need half as much energy to compress the spring at the top of its cycle. But also only get half as much energy out of the generator. So I think in Momos's system it doesn't matter how high you let it float, or how high the tank of water is. As long as the former is no more than the latter of course :P

I suspect this is the same with the HHO, but still can't see how the numbers add up.

Thats why I advise to concentrate on a full circle of one molecule of water (or its mechanical equivalent).
A guess paul wrote somewhere "the volume of HHO is 1800 times larger then water".
Thats where the energy is lost: you have to spend energy to do this expansion against the surrounding pressure.
Electrolysis under high pressure takes more energy then under low pressure.





Correct: the higher the tank: the more energy you gain from floating and falling. But the amount of energy necessary to change the density (do the electrolysis) increases accordingly.
Both processes are directly depending on the density of the surrounding and the force of gravity.



Well, I'm unable to calculate the amount of energy (Z)to seperate the atoms of the molecule against the surrounding pressure. Thats why I used my mechanical boxes as an equivalent. Do my figures add up?

Sorry, probably my english is not good enough to understand everything. But let me try:



Thats not my initial model.
In my model I use the energy at the bottom to stretch the spring. And at the top I get some of this energy back by letting it recontract.
Same as you using energy for electrolysis at the bottom and regaining some of this energy at the top.


But, ok. Lets assume we are compressing the spring at the top, using 1 Unit of energy. The Box will now fall downwards. At the Bottom we have the let the spring stretch again.
The problem is: at the top we have less surrounding pressure then at the bottom. So the energy used at the top to compress the spring is NOT ENOUGH to let the spring stretch on the Bottom.

Lets quantify this: We build an offshore-box-driven-power-plant in some part where the oceans depth is 10km.


Tower height: 10.000 m.
Density of Water: 1000 kg / m³

Weight of Box: 6000 kg.
Volume of Box at the top: 3 m³
Density of Box at the top: 2000 kg/m³ (will sink)
downward force while sinking: [6t - 3m³ * 1t/m³ ] * 9.81m/s² = 29.43 t*m/s² ~ 30 kN

Volume of Box at the bottom: 12 m³
Density of Box at the bottom: 500 kg/m³ (will float up)
upward force while floating up: [12m³ * 1t/m³ - 6t] * 9.81m/s² = 58.86 t*m/s² ~ 60 kN.

Energy gained by floating up and falling down:
10.000m * 60kN + 10.000m * 30kN
= 900.000.000 N*m = 900 MJ.
(or 882.9 MJ using the exact numbers)


How much energy do we need to convert the box from the sinking to the floating state?

At the bottom we have to extend the volume of our box from 3m³ to 12m³. So we have to push 9m³ of water away.
The surrounding pressure at the bottom would be: 10.000 m * 1000 kg/m³ * 9.81 m/s² = 9.81 * 10^7 N/m² = 9.81*10^7 Pa = 981 bar.

The energy needed to gain 9m³ of volume at this pressure would be: 9m³ * 9.81*10^7 N/m² = 88.29*10^7 N*m = 882.9 MJ.


All generated Energy is lost if you try to complete a full cycle.



If we try your latest approach and compress the spring at the top we have the same problem.
Either we use a light spring, easy to compress.
But then the stored energy in the spring won't be able to overcome the surrounding pressure. The box won't float up again. Or we use a heavy spring, but then we have to apply the needed energy at the beginning for compressing it.








Now, this is a different model!
Now you are complicating things by changing the water level during the floating phase.
I guess now the problem is:
a) inserting the box at the bottom of the water-tower at the start of the second run. (pushing against the now elevated water level means pushing against more pressure loosing all energy gained by increasing the falling and rising distance.
b) You could try to sink the water level again, before inserting the box at the bottom. But now you have the problem of re-elevating the water level. You are loosing again exactly the same amount of energy.
c) You could use water from a separate source (like rain), but then you have just an open system, gaining nothing more then the potential energy stored in raindrops.

Hi Momos. I think it's getting confusing with all the different systems. I know you won't want to read all the history, but me and Paul have discussed various configurations, including doing it underwater, and come to one he described clearly in message #34053.

I noted that it required more height than the atmosphere had. But this is where my thoughts are at this stage:

a) Convert a liter of water to HHO, let it float up (throw away that energy), burn it back to water (throw away that energy), then drop the water back to the ground (recover that energy).

b) We know how much energy you can get from a liter of water falling through a height of 978 miles (4212 W-hr)

c) We know how much energy it takes to convert a liter of water into HHO at atmospheric pressure (4212 W-hr)

d) If you had an atmosphere that was >978 miles high, but had a pressure at ground level the same as our atmosphere, then it looks like you come out better than even. This atmosphere would have to be made of a gas less dense than air, and a tiny bit denser than HHO.

e) You don't even need such an extreme because you can recover more from the "throw away that energy" stages in a).

f) There's another option of pumping out a vacuum to generate the HHO in.

Maybe my requirement in d) is impossible. I have to work that out, but I'd rather wait for Paul to present a new complete system.


A big stumbling block is not knowing how much extra energy is needed to produce HHO at higher pressures.

Thanks for your explanations!

> A big stumbling block is not knowing how much extra energy is
> needed to produce HHO at higher pressures.

Thats the point: you need at least as much energy as needed to overcome to surrounding pressure and to enhance the volume.

Thats the whole point of my mechanical box: to make it easier to calculate this needed energy.

The surrounding pressure at the bottom, the buoyance and the falling distance are directly related to each other.



> d) If you had an atmosphere that was >978 miles high, but had
> a pressure at ground level the same as our atmosphere, then it
> looks like you come out better than even.
> This atmosphere would have to be made of a gas less dense than
> air, and a tiny bit denser than HH

The pressure at groundlevel is directly depending on the density of the atmosphere and the gravitional pull.

pressure = height * density_ath * g
energy_gain = energy_gain_up + energy_gain_down

energy_gain_down = height * g * volume_sinking * density_ath

energy_gain_up = height * g * volume_floating * density_ath

energy_loss = pressure * (volume_floating - volume_sinking)

total_energy = energy_gain - energy_loss

As you can see the total amount of energy can be calculated by this simple formula which is only referring to the basic parameters of your system:

= (height * g * volume_sinking * density_ath) +
(height * g * volume_floating * density_ath) -
(height * density_ath * g * [volume_floating - volume_sinking])

since "height * g * density_ath" is always the same constant factor k, we can write:

= ( k * volume_sinking ) + ( k * volume_floating ) - ( k * [volume_floating - volume_sinking])

I hope it is clear this formula will always result in 0 !
It doesn't matter how big or small this factor of k is!

You can cange the density of the atmosphere, the gravitational pull and the height of your tower in ANY COMBINATION, it will only result in some different constant factor "k".

We only know that because we assume conservation of energy. We're not allowed to make that assumption, because then the argument is circular - "the machine won't violate conservation of energy because you can't violate conservation of energy".

Kallog and Momos

I have finished and would like to ask you one question.

if I have a glass that is half filled with water.
and I put a straw in the glass , then put my finger over the top of the straw and lift the straw almost all the way out of the water.

1) would the water pressure at the bottom of the glass increase?

although the water height has increased , the water pressure at the bottom of the glass has not increased because the water in the straw is supported by the vacuum in the straw , but the water height from the bottom of the glass to the top of the water in the straw has increased.

But I'm not assuming conservation of energy.
I just calculated the amount of energy needed to extend the box/molecule at the bottom.
It turns out it's the same amount of energy we gained before.
I'm not using some abstract principle but a simple calculation of the necessary forces.

> would the water pressure at the bottom of the glass increase?

I think the water level in the rest of the glass is sinking, according to the amount of water you are removing/lifting.
The pressure at the bottom would decrease.

Of course you need energy to lift the straw with the water, but so far everything seems fine.

well yea but the height of the water in the glass would only drop by the amount of water that the straw displaced while it was submerged in the water in the glass.

which would not be much , when compared to the new height
of the water in the straw and glass.

the height has almost doubled , but the pressure has dropped at the bottom of the glass.

do you agree to this?

I'm not sure

On the surface of the water the weight of the atmosphere is applying pressure. This pressure is also driving the water into the straw.

Inside the straw the elevated water has weight, and is trying to fall downwards.


The height of water in the straw marks a state of equilibrium: the force of the atmospheric pressure and the force of the weight of the elevated water + the pressure of the enclosed air at the top of the straw are equal.

The water level in the glass (apart from the straw) has dropped due to the now missing elevated water in the straw.
I still expect the pressure in the glass to drop.

I'm not entirely sure...

The pressure at the bottom of the glass will be: (weight of dropped water level + atmospheric pressure).
The pressure at the bottom of the glass underneath the straw will be: (weight of dropped water level + weight of the elevated water + reduced air pressure in the straw).

Momos

the water in the lifted straw is not putting any pressure on the water in the glass.

it cant fall because of the vacuum at the top of the straw.

this same scenario occurs in larger pipes.

take a 2" pipe 8 ft long and put a cap on one end.

then submerge it in your swimming pool , then lift it straight up with the capped end pointing upwards.

the water will stay in the pipe until you completely pull the pipe out of the water.

ie...when the air can rise into the pipe.

if you hold your hand directly under the lifted pipe you will not feel any pressure from the water in the pipe.

untill air can rise up and allow the water to fall by removing the vacume in the pipe.

or use a 2 liter coke bottle , fill it with water then
turn it upside down in a coffee cup , as soon as the water in the cup fills up and touches the top of the coke bottle
the water will stop comming out.


just like a pet waterer , the pet drinks water , the water level falls , air moves up into the tank , water falls down again.

the water is held up by the vacume.

The strange thing is: if there would be no pressure in the straw, nothing could float up inside it. Buoyance is nothing more then the resulting force of pressure differences.

AT the top of the water in the straw we most likely won't have vacuum, just low pressure. And exactly speaking the vacuum doesn't hold the water up (e.g. is not applying some upward force). It is the air pressure of the surrounding atmosphere pressing the water into the straw, because the low pressure in the straw can't compensate.

Thats why the water level in the straw rises, as long as the weight of the water column equals the pressure difference between the surrounding air pressure and the lower pressure at the top of the straw.

At least that's what I guess

This leads to the following conclusion:
If you have a perfect vacuum in the straw (no air pressure) there is a maximum height to wich we can lift the water along with the straw. This height is determined by the air pressure.

Since air pressure is 1 bar = 100,000 pa = 100.000 N/m².
A "straw" can hold at most a ~10m column of water, after that lifting the straw wont lift the water anymore.

Actually, for the same reason this is also the maximum lifting height of any liquid achievable by a suction pump.

I think Paul has it spot on.

Yea things would still float in the straw. Because there's still a pressure gradient from 1atm where it enters the glass, to >0 at the top of the straw. The pressure gradient is the same as in the glass or under the sea, so bouyancy would be identical.

I have no doubt things would still float in the straw.
So far I'm agreeing with both of you.

the pressure inside the straw is not the reason that something would float up inside the straw.

it is the density differential that would cause a object to float up inside the straw.

the water in the straw has no pressure in any direction.

at the top of the straw the plug is supporting all the weight of the air and water in the straw.

the weight of the water in the straw is causing a vacume in the air between the plug and the water.

the water in the straw is trying to fall , but cannot because water will not stretch , so its just hanging
there exerting no pressures at all.

picture a board that is nailed to the ceiling.

the board would fall if it were not supported by the nails.

the nails in the straw example is the vacuum in the straw.
the plug is the ceiling.
the water is the board.

Vacuum = nothing. Nothing cant hold anything.
Try holding a board by NOT attaching it to the wall

And water is an incompressible fluid, there is virtually no difference in the density of the liquid.
Anyway: a minimal density difference in the water would be THE RESULT of pressure compressing the water.
If you were right the water in the straw would have no density gradient. Nothing could float there.

The water is hold by the air pressure, hence the maximum lifting height of suction pumps.

And buoyancy is the result of pressure differences between the top and the bottom side of the object.
This picture might give you an idea what I mean:
http://de.wikipedia.org/w/index.php?title=Datei:Auftrieb_Archimedes_1.svg&filetimestamp=20080324085043

red dot at the top: pressure of the liquid at top, force downward.
red dots at the sides: equal pressures, canceling each other out.
red dot at the bottom: pressure of all the liquid above it, this pressure is more then on the upside of the object.
Resulting in an upward force.

There has to be no density difference whatsoever.

Ah, found it: http://en.wikipedia.org/wiki/Suction
"Suction is the flow of a fluid into a partial vacuum, or region of low pressure. The pressure gradient between this region and the ambient pressure will propel matter toward the low pressure area. Suction is popularly thought of as an attractive effect, which is incorrect since vacuums do not innately attract matter. Dust being "sucked" into a vacuum cleaner is actually being pushed in by the higher pressure air on the outside of the cleaner.

The higher pressure of the surrounding fluid can push matter into a vacuum but a vacuum cannot attract matter."


This should be obvious. If vacuum had some magical attractive force, what would happen to the ISS? The realy BIG vacuum of outer space sucking in one direction, the tiny vacuum in direction of earth sucking with less force in the other direction....

momos



any pressure below 0 guage pressure is refered to as a vacuum.

http://en.wikipedia.org/wiki/Vacuum_gauge

there are low vacuums , medium vacuums , high vacuums
and ultra high vacuums and extremely high vacuums.

http://en.wikipedia.org/wiki/Vacuum



and the pressure is the result of density.
so density is the true reason for buoyancy.



http://en.wikipedia.org/wiki/Buoyancy



you are refering to a open system that is exposed to the atmospheres pressures , not a system that is exposed to a vacuum.

in a open system pressures are distributed in layers
and the only reason that water in a open system ever exerts pressure outward or upward is due to the weight of water
above it.

in a closed system pressure is distributed equally in all directions.

you continue with outer space which is also an open system
there is a difference between open and closed systems.

1) open systems are able to interact with atmospheric pressures

2) closed systems are not.

the straw system is a closed system and its bottom is the water pressure that is in a open system.

it should be clear to you that this straw in a glass system is a combination of both open and closed systems.

something you should try.

get a glass bottle fill it with water.
fill a pot with water , now turn the bottle upside down into the water in the pot.

now lift the bottle slowly out of the water observing the top of the bottle very closely.

the bottle can come completely above the surface of the
water in the pot , yet the water in the pot rises up to the bottle as you lift it.

still no water comes out of the bottle.

even though it is clear that there is not enough water pressure along the mouth of the bottle that could be supplied by the water pressures in the water in the pot
to support the weight of the water in the bottle.

so wheres the pressure that is holding the water in
the bottle?
remember in order to have water pressure you must also
have water at a elevation.
but theres no water higher than the mouth of the bottle.

so is it just atmospheric pressure that holds the water
in the bottle?

but if I continue to lift the bottle the water comes out.

so it cant be just atmospheric pressures holding the water in the bottle.

its the vacuum at the top of the bottle that holds the
water in the bottle , not the atmospheric pressure pressing down on the water in the pot.


I used a 1 gallon glass bottle when I tried this.

.

The pressure holding the water in the bottle is the athmospheric pressure. You can actually see it! If a high-pressure weather system coming along the water in the bottel will rise even more. You can use it as a barometer.

You can also calculate it: The weight of the elevated water in the bottle matches exactly the athmospheric pressure.




Yes.



Sorry, but you are wrong.
The reason the water comes out when you rise the bottle above the surface is: water is a liquid. The atmosphere pressing against the water in your bottle will form bubbles. Each bubble displaces its own volume of water.

Try this: hold a cardboard over the opening of your bottle, just before rising above the surface.
Now the atmospheric pressure has some steady/inflexible board to press against. You will see, the atmosphere is pressing against the board with enough pressure to hold the water inside.



You are wrong. Please read again the section about the vacuum-cleaner and suction. Or ask any manufacturer of suction pumps. Or you could try it yourself:
You would need a tower of 11 m height, a hose of 11m length and a tub with water.
Even with a perfect vacuum at the top of the hose you can elevate water higher then the atmospheric pressure compensates (around 10.3 meters).




Lets try a thought experiment: Assuming you are correct and "nothing" (perfect vacuum) is somehow inducing a "suction force".

If you have 1 Liter of vacuum this should be some amount X of "suction force". 2 Liter of vacuum should induce twice as much suction force.

Lets consider a closed system: A box, completely sealed from surrounding pressures. Inside this box we have a vacuum. In this box is also a movable separator, dividing the box into two compartments. Like a piston?

If you were right about this "suction force" the piston would always tend to move into the middle. If the piston is somewhere else one of the two compartments will have more volume of vacuum this producing more suction.

Reality behaves differently: in reality the piston doesn't care on which side more vacuum exists.
Because vacuum (nothing) doesn't exert any force.

Pressure is the result of a force applied. That's the very definition.

Force (in our case) is not the result of density.
You can have a stone (hight density) side by side to some piece of wood (low density) there won't by any force pressing the wood against the stone.

The force (in our case) is the result of mass and gravity.
In a Zero-G-Environment you won't have this force, so you won't have pressure and you won't have buoyancy.
That's independent of any density differences.

-----

http://theory.uwinnipeg.ca/physics/fluids/node10.html


But still there is a connection between density and buoyancy.
It just isn't "the true reason for buoyancy".


This is an applet simulation a submarine, showing the air-pressure, water-pressure and weight along with the resulting force.
http://www.didaktik.physik.uni-muenchen.de/materialien/inhalt_materialien/auftrieb/auftrieb.zip

Momos

no , in our case it IS THE DENSITY.

it doesnt matter how you twist it around to fit your world.

the ONLY REASON THAT BUOYANCY OCCURS is the differences in



DENSITY

in our case were not in a zero g environment
so im not sure why you are refering to zero g

but since you mentioned zero g

if you pour out a gallon of water inside the ISS
a sphere of water will form.

if you push a ping pong ball into the sphere of water in the near zero g environment the pingpong ball will be pushed back out , because of the differences in the density of the water and the density of the ping pong ball.

Kallog

you agree that it is the density differences that cause
buoyancy , so I suppose that we can continue our discussion and if momos wants to continue
thinking it is the pressures then that is his problem.

what do you think , do we have this covered already?

I didnt know you had posted another reply momos so
I will respond to it.



so all I would need to do to increase atmospheric pressure outside the bottle is to increase the water inside the bottle.

but what if I have 2000 lbs of water in a bottle?

are you saying that the atmospheric pressure outside the bottle would become 2000 lbs or are you saying that the 2000 lb of water looses its weight and becomes 14.7 lbs?



yes I tried this and then I poked a hole in the top of the bottle and the cardboard still held up all the weight of the water inside.

there is something magical about the atmospheric pressures.

then I woke up , but there was no water floating around because of the MASSIVE atmospheric pressures.

I went to get a cup of coffee , but stumbled over a block of atmospheric pressure along the way.

No it won't. No gravity = no weight = no pressure = no pressure gradient = no buoyancy.
This air bubble has less density then the surrounding water: still in zero g it stays submerged:
http://www.youtube.com/watch?v=ZyTwLAW-Z8c&feature=related

The density difference between the ping pong ball and the water doesn't "do" anything. Exactly like the density difference of a ping pong ball and a brick wall at its side doesn't exert any force.

Also there wouldn't be any mentionable density difference in the water (even with gravity). Water is basically incompressible. You can read this in any boom about physics.


You donÄt have to believe me, just take a take a look here (same source you got your "density" picture from:
http://en.wikipedia.org/wiki/Buoyancy#Forces_and_equilibrium
Or:
http://www.answers.com/topic/buoyancy
Or:
http://answers.yahoo.com/question/index?qid=20100401145652AAtLQew

Or if you don't like reading watch this:
http://video.answers.com/Q/learn_about_fluids_-_part_5_99162287 (actually these videos of kahn academy are quite good).




Now you are playing stupid?
Of course not.
It means you can't elevate the water more then the air pressure allowes you to.
If you elevate the straw further then 10 meter the water would still stay at 10m.
You could lift up a straw of any length, kilometers if you like, the water column in the straw would still be at a height of 10 meters.




I'm saying: You can't support 2000 lbs of water in the bottle. At least not by using the "suction of a vacuum".
The pressure of the weight of the water in the straw will always match the surrounding air pressure.
If you lift the bottle/straw further, the water will stay where it is, you will just have a bigger empty place at the top of the straw.

Strictly speaking you can support 2000 lbs of water in the bottle: AS LONG as the height of the water column is less then 10.3m. You could just make your straw wider.
It's a question of pressure: Weight of the water / Area of the straw = downward pressure <= atmospheric pressure.

momos

I was a little quick with my ping pong example.
I did not take into consideration surface tension in microgravity.

however even surface tension forms spheres in microgravity and even small spheres have gravity , pressures , and buoyancy.

you carried the discussion into outer space for some reason
and I made a speedy incorrect reply.

however if the water sphere were larger , or if the sphere were made of some other material such as tiny beads that would not have surface tension as water does , then the bead sphere would push the ping pong ball out if you tried to submerge it.

but only if the beads that surround the ping pong ball have a greater density than the ping pong ball.

even if you begin building a sphere starting with the ping pong ball then adding the tiny beads to the ping pong ball the mass that will form in the shape of a sphere in a microgravity
will eventually push the ping pong ball outward.

the problem with the water example is that the surface tension of the water attracts objects that come in contact with it ,
but the beads would only be attracted to each other by gravity

now in the bead and ping pong ball example there is gravity , pressure and buoyancy.

just like the earth has.

now you say that the reason that buoyancy occurs is only the pressure differential between the top and the bottom of an object.

and you say that Im wrong when I say the true reason for
buoyancy is only density.

the only reason that you have any pressure is because of gravity pulling a weight/density of a mass toward the earth.

without density all you have left is nothing.

no gravity no pressure no buoyancy.

I dont really think you believe that it is only pressures
that cause buoyancy , I think you are just refusing to admit
that the true reason that buoyancy occurs is density.

or maybe you truly do believe that.

if so , how would you propose that you could have any pressures without a object that has density acting against another object that has density?

as I said , I really dont think you believe that , I just think
you are holding out for some reason.

so why dont you show me an example where there is pressure without density and then I might believe you.


heres something to ponder while I wait.

take two identical sized spheres.

1 sphere has a weight of .0000001 lbs

the other sphere has a weight of 5000 lbs

they are both submerged and held at the same exact depth 10 feet below the waters surface level.

so the same pressure is being placed on the two spheres
on the top and on the bottom.

if you release the two spheres at the same exact time
what would happen?

I say that the 5000 lb sphere will sink because of its density.
even though there is a pressure differential between the top and bottom of the sphere.

and the .0000001 lb sphere will rise because of its density.
even though there is a pressure differential between the top and bottom of the sphere.

according to you they will both rise or they will both sink.
because they have the same exact pressure differential acting downwards and upwards on the two spheres.

and according to you density has nothing to do with buoyancy , it is ONLY the pressure differential between the top and bottom of a submerged object.

ie...post #34142

Haha wow you guys have really gone off on a tangent of miscommunication! I have a feeling you both know exactly how it works :P

Let me lay the smack down:

*** BUOYANCY IS CAUSED BY A PRESSURE GRADIENT APPLIED TO AN OBJECT ***

A pressure gradient is usually caused by density of the surrounding fluid and a gravitational field. On Earth, the pressure gradient is only a function of the fluid's density. So it's good enough to say the buoyancy force on an object is only a function of the surrounding fluid's density (and the volume of the object).

In the above I'm ignoring the self weight of the object, which of course needs to be added to convert buoyancy force into total, measurable force. Here we have a semantic issue. Paul, your two-spheres-underwater example assumes it's the total force, but I think it's more correct to ignore the object's weight (and density) when defining buoyancy force. Either way, it doesn't change reality when we apply whatever definitions we each use.



-----


In the case of a ball of water on the ISS, you raise an interesting point Paul. The water would have its own local gravity field, which would cause a pressure gradient inside it. That would cause boyancy of a ping-pong ball. Even tho this force would be amazingly tiny, what would stop it? The viscous friction of water approaches zero as speed goes to zero. So I'd think the ping-pong ball would actually move to the surface. But it could be extremely slow, and maybe other tiny forces like photons hitting it, or the microgravity of Earth would overwhelm buoyancy.

Kallog

it doesnt matter how you word it.

when its time to turn the cards its density that wins.

you can say its the great purple rinoceros if you like.

you can call it what ever you choose , but density by any other name
or deffinition is still density.

you can call it weight if you like.

but in my world what causes buoyancy or the reason a object floats
or sinks in a fluid is the density dfferential.

and I dont care how far under water an object is , even if the waters pressure is
at 50,000 psi.

the surrounding water will have basically the same density per unit volume
as it has at the surface.

even if the water is inside a container that has a large vacuum thrown on it
where the water pressure at the bottom of the container is at -100 kpa

in this situation the pressure gradient you speak of is negligible or
non existant.

and the only two forces that apply in that type of situation are the force of gravity and the resultant force of buoyancy of an object due only to the density differential between the water and the object.

if I take a 10 ft tall pipe and attach a pressure guage and a vacuum guage at the bottom of the pipe and at the top of the pipe and seal it at the bottom and fill it with water and seal it at the top in 1 atm

then when I look at the vacuum guage at the bottom it reads zero
if I look at the pressure guage at the bottom it reads 4.33 psi.

if I add 1 lb of air pressure at the top

the vacuum guage at the bottom of the pipe still reads zero.
the pressure guage at the bottom of the pipe reads 5.33 psi

if I remove the 1 psi air pressure and throw a - 50 kPa vacuum on the air at the top

the vacuum guage at the bottom of the pipe reads -50 kPa
the pressure guage at the bottom of the pipe reads zero

but in each situation either in 1 atm , pressurized , or a vacuum
the only reason a object will have buoyancy is the differential between the
density of the water and the object.

and to try to continue our discussion if a HHO cell is at the bottom of the 10 ft pipe
more energy will be required to convert water into HHO at 4.33 psi guage pressure and even more
at 5.33 psi guage pressure

and less at -50 kPa guage pressure.

.

I agree with kallog.
Maybe I expressed myself unclear: The measured buoyancy is the result of the (downward) force due to the weight of the object and the the resulting force of pressure differences.

And of course I don't deny the connection to the density of the liquid and the object. Since this density is the result of the volume (causing the pressure differences) and the mass in a g-field (causing the downward force).

But it's not the density difference as such causing the buoyancy! That's why I'm always referring to a zero-g environment.
Without gravity the density differences between liquid and object won't change. But there won't be any pressure difference in the liquid or downward force of the object anymore, resulting in a buoyancy force of zero. (I admit: I'm ignoring microgravity, thermal diffusion, surface tension, quantum mechanics and a lot of other aspects since they are not part of the principle in discussion).

Why am I always nagging about this?
Paul designed a device with an elevated water column, which is fine. I guess he will try to use some object (possibly hydrogen) to float along this column and sink again. That's ok.
When we are discussing the energy gained or lost during this process we will have to calculate the work necessary to achieve any change of density, movement and so on at different points.

That's why it is important for us tho have the same understanding which pressure and forces are applying to an object somewhere in the water column.

In my opinion the vacuum (or low-pressure air) at the top of the straw is not exerting any negative force (suction) at the water column below it. In case of a vacuum it just doesn't exert anything (how could it since "it" isn't there).
The water in the column has still its weight and would fall downwards. The air pressure on the open part of the system is not matched by the air pressure on top of the water column.
That's why this pressure is pressing the water into the straw, until the weight of the water in the straw balances the pressure difference between air pressure and the vacuum.

In the water column we have the usual pressure gradient (from 0 to 1 atmosphere) resulting in just the same buoyancy of any object as usual.

If Paul agrees to this explanation, I have to apologies since any misunderstanding could very well be on my part and a possible misuse of the word buoyancy


----

on a side note:
Since water is dielectric we could apply an electric field from the outside on the water column which exerts an upward force on any water molecule. Thus we could compensates the weight of the water (letting it really elevating it). The water column would have no internal pressure gradient, any object unaffected by the electric field would sink. Regardless of its density.

[Edit: found a video of levitating frogs and water in a magnetic field]

momos

I dont agree , because you still think that the vacuum is not
supporting the weight of the water in the straw.

water will not expand or stretch inside the straw.

the partial vacuum at the top of the straw is generated by the falling water in the straw.

when I lift the straw the water in the straw pulls down just as if a piston were pulling downwards on the air above the water in the straw.

the vacuum is exerting a upward force on every molecule
in the straw , why should I apply a electric field to accomplish the same result?

but if you want to spend a few billion dollars to lift the water
go ahead.

then you can float a few frogs when your finished.

I'll just use the $30.00 vacume pump.

"the vacuum is exerting a upward force on every molecule
in the straw"

That's just plain wrong. I give up, you don't believe me, you don't believe internet sources, apparently you don't believe in scientific textbooks.

http://www.tpub.com/content/construction/14265/css/14265_206.htm: "Assuming the liquid is water and there is a perfect vacuum below the piston, atmospheric pressure pushes water up into the cylinder to a height of 34 feet, even though the piston may be raised higher than 34 feet"

Go ahead, try to build a PM without basic knowledge of pysics. Good luck.

--

You shouldn't apply the electric field. It was just a thought experiment and actually that's part of a scientific process: You have an Idea (like pressure gradient is responsible for floating) and then you try do think of an experiment to falsify your claims (e.g. water columns without a pressure gradient shouldn't let things float).

In this video you can see a drop of water, levitating in a magnetic field. Notice the air bubbles: they are not floating upwards, despite the still working gravity and the density difference. Instead they are behaving like they would in zero-g. (actually they should sink downwards, I guess the thermic motions are to strong or the air itself is diamagnetic to).

http://www.ru.nl/aspx/download.aspx?File=/contents/pages/534727/drop2a.mpg&structuur=hfml

its easy to see that they are not levitating anything in the
video , they have built a area that is not affected by gravity and that is all they have done.

which is quite a feat none the less.

the vacuum is pulling on the top of the water in the straw.
water cannot stretch so the remainder of the water below the surface is pulling on the surface.

etc etc etc...

use your brain for thinking.

not just for reading.

Yep. As long as we all understand each other. Which I think we surely do because this is such a simple phenomenon that we've all done with a rubber duck in the bath when we were a kid.


Not on the ISS it doesn't. No gravity = no bouyancy.


No, the pressure gradient is constant and non-negligible. That's where the force comes from in the first place. The downward force on top is less than the upward force at the bottom, so the thing get a net upward force. Without that difference there's no buoyancy. Equally of course without a density difference in the two things, there's no buoyancy either, when you define buoyancy to include the object's own weight. Different colored rhinoseroses.




Yep. Can you describe this whole new system?

I think for our purposes it doesn't matter if you say the vacuum is pulling or the atmosphere is pushing. The net force adds up to the same. Just have to remember that vacuum's pulling force is limited, as Momos has explained.

So can we get onto the actual device? Or are you still working it out?

Whoa! When did they invent antigravity?

Well , I havent finished completely I still have to do a
little more drawing and wording so I suppose you can have
a look at the below images from a water engine Im designing.

I didnt include the HHO cells in the images yet , but
they are located inside the water in the tanks at the bottom of the images.

also I havent shown the piston rods at angles as they
push the crankshafts around in a circle.

just use your imagination.


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I was waiting for your explanation.

I can see basically how it works. But how's it different from a normal engine?

Im really not sure !!!

but I was thinking that the vacuum generated in the one side will require less energy to generate HHO.

also as the piston strokes into this vacuum the piston will
have an extra force applied by the vacuum.

so what we have here is HHO being generated in a vacuum.
and HHO being burned in a higher pressure.

plus the extra umph from the vacuum as the piston strokes towards the vacuum.


3 plusses.

it should be capable of providing enought power to run a generator , that should provide the electricity to generate
the HHO.

and if anythings left over then that is free energy.

so the real difference is that it contains its own fuel
supply that never dimminishes.

because the water is recycled from water to HHO then back to water then back to HHO , etc etc etc.

you just need a battery to start it , just like a car
engine.

it probably wont work as fast as a car engine , because there is a time factor involved in the HHO generation
but by using a heavy flywheel the force from the explosions can be transfered into the flywheel that will hold the force until it is used.

although the flywheel could be turning very fast , the
engine rpm could be very low , by using a geared ratio
between the engine and flywheel.

this way the engine can have a slower rpm that will allow
more time between explosions and allowing more time for the build up of HHO gasses.

and you could have several of these inline to achieve more power.

and they could be timed so you would have a more balanced
fireing order and power input to the flywheel.

I think it really would work !!!

can you think of anything that would stop it?

???

Sorry, but this must be the most inefficient engine ever!
Just use an electric engine, and you will be much much better off.

"that it contains its own fuel "

No, it doesn't. It contains a very inefficient indirect way to convert electrical energy to mechanical motion.

--

Just think of a conventional combustion engine with hydrogen.

maybe you could explain your reason for thinking its
inefficient.

or should we just take your word for it.

just give it a try , your so good at it , and you know everythng
about energy , it shouldnt take much effort.

Making the piston pull a vacuum consumes energy. Then you fill that vacuum with HHO, so you can't recover the lost energy by having it pull the piston back again.

You've said that free energy is already possible. Why not investigate one of these "possible" methods that others have talked about? There's no need to improve on them if they already work.

kallog

when the spark fires the HHO ignights , all of the HHO in the chamber explodes
sending the piston forward.

forcing the piston to move.

transfering the force from the explosion to the heavy flywheel.

immediately after the explosion the HHO converts back into water.

causing the vacuum.

the vacuum is caused by the imploding HHO not the piston.

then as you say the chamber fills with HHO again but the HHO
is being generated in a vacume that is being held steady as the
piston moves further out.

producing more HHO for less energy.

we have already discussed the fact that HHO cost less to produce
in a vacuum , so if you explode the HHO in a higher pressure
then you naturally get more energy from the explosion than you put into making the HHO that caused the explosion.

this is all rudimentary , currently used processes.

this is a highly possible and applicable method as everything points to its viability.


dont you agree that HHO can be generated at a lower cost
in a vacuum?

dont you agree that HHO would supply more force if exploded
in a compressed atmosphere.

you have previously stated each was true , has something changed.

we know already that hydrogen has 140 mj of energy compared to gasolines 40 mj

we know that a gasoline engine has a
power stroke
exhaust stroke
intake stroke
compression stroke

and this engine only has a
power stroke !!!

and it has 3 times the amount of energy in the fuel that powers these strokes.