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NeilT

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Open to destruction of an idea
« on: February 12, 2015, 09:39:06 AM »
Nearly two years ago I posed a theoretical question to an idea of mine.  Namely that if I was able to harness the power of gravity to generate energy, would the government allow me to sell it?

OK so the discussion was vague and there were loads of "possibles".  Since then, offline through discussions, I've had to learn maths to explain my idea.  Apparently people who work with science can't read a diagram and say "yeah I see that", they say "you can't create energy, so prove it but you can't prove it because you can't create energy".

I never said I could create energy.   In a rather heated discussion with my brother, he told me that the way he would approach this is to try and destroy my idea every way possible and what is left, if it still works, is correct.

I recognise that this is good science.  Sadly I can't.  I've learned over the years that I don't see mechanical things they way other people do.  Just the way some people see languages as physical constructs and I see it as a total abstract created to convey ideas.  If it were a physical thing, there could only be one language.

OK so, after having to beat my brains out to learn some maths and physics I never learned at School, I can present my idea in one simple concept.  What I would like you to do is rip it to shreds for me.   That's why I have posted it.  A sort of crowd sourced scientific destruction session.  This is why it's in Walk the Walk.  After all this entire discussion is in the public domain....  I'm willing to take the ridicule to find out if what I see is right or if what I see is wrong.

If you'll bear with me for a few short paragraphs more, I'd like to explain the fundamental difference between my idea and many of the 0 net energy ideas out there.  It might help, it might not.

There are dozens of ideas out there which claim that they can use the power of gravity, whether in or out of water (other than something like a water mill), to create perpetual energy.  All of them, without fail, have one thing in common.  The energy they store in their wonderful contraptions is all moving in one direction.  So your energy balances out.

The energy I store in my idea works in the opposite direction to the initiating force.  Let me be a bit clearer here.  Gravity only works in one direction.  However the Effect of gravity can be felt in an opposite direction depending on the place and the medium.

Therefore allowing you to use the effect of the stored force to act with, not against the force creating the stored energy. 

OK that is one way of stating it but let's try it another way.

There is a place on this planet where the effect of gravity is  mirrored.  Just like any mirror, equilibrium is where the two meet.  Think of it as two reservoirs of gravity, both moving towards the middle.  If I take energy from the middle of one field, say the top one (falling) and transfer it to the middle of the lower field (rising), through a direct mechanical linkage, then I can use the energy of both fields to create more stored energy.  So long as neither forces (original and stored) reaches equilibrium, they continue to generate force and if I allow something to move (not necessarily both forces), then I can store more energy.  By allowing movement, I can stop the lower force acting on the upper force until I choose to do so.


OK that is the mechanics of it.

In physics it can simply be expressed as this.

I take the energy of mgd and store it as a force by generating pvg (fully submersed buoyant object).

After having created the pvg, I use simple mechanics to remove the force of mgd from the first buoyant object and apply it to another object which is non buoyant.  I also allow the force of the newly created buoyant object to act on the non buoyant object at the same time.  mgd continues to force the non buoyant object down onto a fixed point, pvd resists the downwards movement of the non buoyant object.  Both forces are applied to a piston inside the non buoyant object.  In short, the non buoyant object moves down and the water is ejected from it by the piston.  The size of the piston and the shape of the non buoyant object must be such that the pressure generated (f/a) is greater than the ambient outside pressure (1atmoshpere+additional pressure at depth of a liquid, water).

So long as the falling object mgd does not meet equilibrium, it will continue to create new buoyant objects.  So long as the buoyant objects do not meet equilibrium, they will continue to provide force additional to the force of the mgd falling object.

In reality the work is created by the fact that both the mgd falling object and the additional non buoyant objects continue to fall, using the force of mgd and the force of the static pvg to eject the water and create a buoyant object.

The relationship is this.

The falling object is 1000kg
Each non buoyant object is a chamber which contains 100kg of water
Inside the non buoyant object, The 100kg of water is ejected by a piston (integral to the object)
The weight of the buoyant object is considered to be 10kg depending on materials required to resist a maximum of 2 standard atmospheres
as the mgd object continues to fall, the force of the buoyant objects increases in 90kg increments.
The force of the pvg is only felt by the non buoyant objects until the force of pvg is greater than the original 1000kg force used for our mgd.  At that point, the force of pvd is connected directly to the original, upper, force of 1000kg mgd.

As the generated pvg is now greater than the original mgd, the whole assembly will rise.

Destroying the pvg is nothing more than flooding.  I'm sure I don't have to explain that mathematically???

I'll draw a diagram next, but not till tonight, UK time, because there is something else I want to have opened up to scrutiny and I'd like that to be in a diagram which shows the principle and I'd like to know why the numbers don't add up.

During trying to attack my idea I found one startling fact which I would like destroyed.  I don't think I'm Dominic Mpemba so I'd like the maths to work properly.  It has no real bearing on the fundamental workings of the idea, but it is rather disconcerting anyway.

After all you can't create energy.
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wehappyfew

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Re: Open to destruction of an idea
« Reply #1 on: February 12, 2015, 02:27:05 PM »
Quote
the water is ejected from it by the piston.

This requires doing work against the pressure of the surrounding water.

That amount of work should be exactly equal to the amount work extracted by the overall device. Add some friction, and you are at net negative.

"If we’ve been bamboozled long enough, we tend to reject any evidence of the bamboozle. We’re no longer interested in finding out the truth. The bamboozle has captured us. It’s simply too painful to acknowledge, even to ourselves, that we’ve been taken" - Carl Sagan

NeilT

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Re: Open to destruction of an idea
« Reply #2 on: February 13, 2015, 12:18:45 AM »
OK so I'll take that at face value.

Right.  So let me explain this.

I have a chamber with a piston in it which has a rod protruding out of the bottom.  It is sized so that it is 0.5m tall and it contains 100l of water.  I use the 1,000kg to force the water out, at 7m of depth pushing the rod against a fixed surface to create pressure, and create 90kg of buoyant force by driving the water out of the chamber.  OK I have to have an air tube connected to the bottom of the chamber to avoid a vacuum and it has a 10kg effect on the whole thing.  But, in the interest of brevity.

I have moved 0.5m and I create 90kg of buoyant force.

My second chamber is the same as the first, but the shaft holding the piston goes right through it and comes out the top.  Sealed naturally.

What happens when I, mechanically, disconnect that 90kg of force from the weight above the water and connect it to the top of the piston on the second chamber?

Then continue the same action.  New fixed surface, new piston being driven down by 1,000 kg of force, new 0.5 meters of movement.  Except, that the 90kg of buoyant force pulling upwards on the top of the piston shaft.

If we continue this, assuming that the weight started this action 7 meters above the water, after 13 iterations I will have 1080kg of buoyant force.  At a fall from 7m to 13.5m depth, the pressure has risen from 1.7bar to 2.3 bar. But my pressure in this chamber created by the piston is always 5kg/sq cm.  Or, in other words, sufficient to expel the water at each depth.

So now I have 1080kg of buoyant force, connected to a weight which is 1000kg.  the top of the buoyant force is 7m below the surface.  The bottom of it is 6.5m below that.  Which means the entire 6.5m long buoyancy chamber can rise 7 meters, because we started this exercise 7 meters down.

Now I'm totally confused here and, remember, I haven't even begun to calculate what each cumulative 90kg force does to the equation when acting on the top of the piston.

So, first, are my calculations correct.  Do I have enough energy to expel the 100l of water from the small buoyancy chambers for each 0.5m of fall of the weight?

Second.  Will the 90kg of buoyant force work with the falling weight (1000kg in the air above the water), work against it or do nothing because it is not moving

Third, why do I calculate

Energy used by falling of the 1,000kg weight as 68,670n (1000 * g * 6.5)
Potential energy of the flotation chamber as 74,163n ((pV-mg)gd or 1080 * g * 7) Even at 6.5 it would be more.

Fourth. Because I started 7m down and I have only fallen an additional 6.5m, I can rise a further 0.5m before the first chamber reaches equilibrium or the surface of the water

Fifth.  If I lock the entire frame at that point where the first chamber reaches equilibrium, I can then use the buoyant force of the 6.5m long chamber to collapse and flood itself [edit] and sink back to 7m depth.

[Edit] ignore this bit.Sixth.  I can now fall 7m with a force of 1000kg and a series of chambers which are 0kg buoyant force in order to start again.
That is for later when I don’t use 5kg/sq cm

Let’s keep it as  I gaind 0.5m.

OK so having put all that in, how does this match with 

Quote
This requires doing work against the pressure of the surrounding water.

That amount of work should be exactly equal to the amount work extracted by the overall device. Add some friction, and you are at net negative.

Because I can tell you, for me that does not add up.

Did I get the calculation wrong.  Is it not possible for the 100kg weight to expel that 100l of water at 7m depth?  I calculate the pressure created as 5kg/sqcm

1000 (weight)/200 (sq cm piston)

I calculate the ambient pressure at 7m to be 1.7bar or 1.733 kg / sq cm.  Which means my 5kg will always force the water out.

I calculate the final ambient pressure at 13.5m to be 2.4 kg/sq cm

I'd like someone to tell me why it does not add up. Because it's not in line with the statement above.

Also remember that in a balanced system each 90kg of buoyant force would act against the 1000kg which is pressing down.  In this system it does not although, at some point, fluid dynamics would act to provide pressure against the downward force until the fluid was vented.  However the net effect would still be that all force goes to creating more buoyancy, both buoyant force and the 1,000kg force above the water.  Not in counteracting the original force.

So from beginning to end, the 1,000 kg force always exerts itself fully against the chamber being vented.

Please now take it apart without generic statements of a balanced system.  This is not a balanced system.  I'm looking for some help here to describe this.
« Last Edit: February 13, 2015, 12:40:42 AM by NeilT »
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wehappyfew

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Re: Open to destruction of an idea
« Reply #3 on: February 13, 2015, 02:07:30 AM »
I'm a more visual person. I can't visualize your device from your description (not your fault), so I will wait for the diagram.

Common mistakes for underwater Perpetual Motion Machines would be:

 - forgetting to add in air pressure acting on the top of the water column
 - pressure differences across the height of a submerged vessel (top vs bottom)
 - changing pressures as a chamber fills or empties
 - expansion of air (or changing pressure) as a chamber rises

...

If you can't identify a net loss of energy somewhere in your device (like falling water losing potential energy in a hydroelectric generator), then you can't extract energy from it without violating one of those important Laws.



"If we’ve been bamboozled long enough, we tend to reject any evidence of the bamboozle. We’re no longer interested in finding out the truth. The bamboozle has captured us. It’s simply too painful to acknowledge, even to ourselves, that we’ve been taken" - Carl Sagan

NeilT

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Re: Open to destruction of an idea
« Reply #4 on: February 13, 2015, 07:54:53 AM »
Thanks this is good. I'll post the picture after the weekend.

OK air pressure.  I didn't add that as soon as I raise the piston, I draw in air pressure at atmospheric pressure.  So, now, I'm not working against 1.7bar, I'm working against .7bar.  correct?  As I have the atmospheric pressure of air working on the "up" side of the piston.

Also I did add atmospheric pressure.  The initial pressure at sea level is 1 bar.  This increases with depth at 0.1bar per meter.  Hence 1.7 bar at 7 meters.

Height from top to bottom.  It is a fully enclosed pressure vessel.  The pressure externally is 0.5 bar higher at the bottom than at the top, however I am venting the water at the top. So I only have to deal with that pressure plus the pressure of 0.5m water depth in addition to the atmospheric pressure water that is in the chamber.  Gravity being constant in an enclosed vessel, it is not affected by the 7m depth because it is isolated from it.

I think I dealt with the pressure change as it empties, I fill it with 1bar of pressure from the bottom.  Although there is some resistance to the air being drawn in.

As the vessel is rigid, there is no only a temporary change in pressure.  It starts at atmospheric, the water inside experiences 5kg/ sqcm of pressure and vents.  Atmpsoheric pressure replaces it on the other side of a fixed barrier.  this remains constant until it floods again.

Oh and there was one point I got wrong.  The first chamber falls 0.5m and then is static.  so there is 7.5m of d available not 6.5 or 7.

I have a few diagrams available.  But I won't be able to post until after the weekend or maybe Sunday night.



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Neven

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Re: Open to destruction of an idea
« Reply #5 on: February 13, 2015, 08:54:31 AM »
Neil, I'm really bad at this, but how do you get those 1000 kg up? Is the energy needed to do that included in your calculation?
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Re: Open to destruction of an idea
« Reply #6 on: February 16, 2015, 03:12:26 AM »
Quote
There is a place on this planet where the effect of gravity is  mirrored.

no, gravity works the same anywhere on the globe.
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NeilT

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Re: Open to destruction of an idea
« Reply #7 on: February 16, 2015, 08:34:16 AM »
Neil, I'm really bad at this, but how do you get those 1000 kg up? Is the energy needed to do that included in your calculation?

So am I.  Which is why I'm opening this up to some help here.  It is a one time effort in the construction and is the start point.  After the machine has completed one cycle, it is returned to the start point.  Consider the start point as a created equilibrium, the energy is part of the infrastructure, the job of the engine is to return itself to that start point without consuming more energy than was given by Gravity in order to move from the start point, back to the start point again.  Fall and rise in this case.
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NeilT

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Re: Open to destruction of an idea
« Reply #8 on: February 16, 2015, 08:45:05 AM »
Quote
There is a place on this planet where the effect of gravity is  mirrored.

no, gravity works the same anywhere on the globe.

Gravity does. Absolutely and I rely on that.

However the effect on an object in a particular space and time, can be reversed.  That is why I used the word effect.

Consider the case in point here, it's used all the time.  If I take an empty flotation bag down 20m in the sea and inflate it with air, what direction is the Effect of the force of gravity?

The Effect is to rise to the surface of the water.  Gravity, through the action of the column of water above the volume of space occupied by the flotation bag, forces the bag with it's air, to the surface.

This is exactly the same with a hot air balloon.  Gravity acts on the atmosphere, compressing the atmosphere, when anything lighter than air is introduced, the effect of gravity is to force it upwards.  My engine would work in the air too, but, given that water is 784 times more dense than air, the energies produced are so insignificant that friction would overwhelm it long before it ever got anywhere.

I'll post the diagram tonight, travel shock is not playing well with me and my wife thinks that I am a house building resource at the weekends.  She thinks I spend all week "playing" with a computer so I should not touch one at the weekend.

I need to re-draw it because I need to change the way it is represented to make it clearer.

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NeilT

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Re: Open to destruction of an idea
« Reply #9 on: February 16, 2015, 08:59:24 AM »
Whilst we are at it, I'd like one thing explained.  It was never part of my design, but working with the maths, I don't understand it, although I do have a few ideas as to why it is.

if I use my 1Tonne weight to vent one 100l chamber at 7m depth, I force the chamber down 0.5m and create a buoyant force of 90kg at 7.5m depth.

The energy used is mgd or if we use kg and meters

1000 * 0.5 * 9.81 = 4905

The buouyant potential energy created is (pV - container mass)gd.  Again in kg and meter it is

90 * 7.5 * 9.81 = 6621.75

All I'm doing is punching the figures into my spreadsheet.  So I'd like to know why this does not add up.  I haven't even done any of these kind of force equations in the engine.  First of all I'm just trying to calculate and prove that the engine can do a full cycle without using more energy than it is given.

Next I would try and prove that it can generate energy.

Next I'd like to prove why.

But the above calculation is, I assume, part of the effect I am trying to prove.

Normally I would assume that I got the calculations wrong.  But they are all above.  The size of the chamber and the pressure created by the piston are easy to calculate.  The pressure created is much greater than I need.  I don't see that 100l of water can resist 1000kg of weight unless I make the piston so large that the relative areas reduce the applied force by a factor of 5 or more.  The pressure created by gravity in the water is only 1.7bar.  I create just under 5 bar in the cylinder.

Honestly I did not expect those figures, was not relying on them and had not calculated them into the way my engine works.

So can someone cross check them for me?

100l container.
10kg weight of container
1000kg ballast weight pushing on it.
0.5m long container
100cm sq piston.
Start point 7m depth.
drop of chamber, 0.5m for the piston to traverse from the bottom to the top.



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Re: Open to destruction of an idea
« Reply #10 on: February 16, 2015, 03:53:08 PM »
Excuse me that i do not have time to comment on the whole machine you describe , but when reading your last posts about gravity i think you make a fundamental mistake.
When writing: "    There is a place on this planet where the effect of gravity is  mirrored."
and giving following example:  " If I take an empty flotation bag down 20m in the sea and inflate it with air, what direction is the Effect of the force of gravity?"
it is definitely bot gravity you are relating on but buoyancy.
Gravity is a force between two bodys which have a mass and a distance from each other and is dependend of the mass and the distance. if you consider this you will see, that you have to take the earth at mass 1 and your floating body as mass 2, respective the water surrounding as mass 2.
The force which is a result of gravity is directed towards the two bodys, NEVER into the other direction.
Buoyancy in your example creates a different force in the opposite direction which counters and gravity and may be stronger (in your example a balloon or a submarine), but it is NOT gravity.

Please note, that the units in your calculations are missing, please add these, otherwise it is difficult to understand.

i hope i will have time tomorrow to comment your machine, but not sure. I will do my very best.

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Re: Open to destruction of an idea
« Reply #11 on: February 16, 2015, 10:14:57 PM »
I do not see where you are including the buoyancy of the air tube attached to your chamber. Are you using compressed air to feed your system? You do not include a rate of work, will this be accomplished over 7 seconds, 7 min, 7 hours, or 7 life times? A lot missing.....
John

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Re: Open to destruction of an idea
« Reply #12 on: February 17, 2015, 12:20:37 AM »
"
Because I can tell you, for me that does not add up.

Did I get the calculation wrong.  Is it not possible for the 100kg weight to expel that 100l of water at 7m depth?  I calculate the pressure created as 5kg/sqcm

1000 (weight)/200 (sq cm piston)"

I believe that you will need a 2000 (sq cm piston) if you are moving it .5 meters. This will give you the 100000ml that you are trying to move. Then your pressure is reduced by a factor of 10!
John

NeilT

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Re: Open to destruction of an idea
« Reply #13 on: February 17, 2015, 01:14:31 AM »
Ok thanks this is the help I need.  Classic decimal point error which I now need to calculate back.  Hence 5m drop.  I still, now, want to work out forces but I'll have to show the diagram for that which I ran out of time for.

I'll be back tomorrow.

As for the effect of gravity.  It is gravity acting on the column of water which causes the buoyancy.  So the action of gravity is always on the water.  The effect of gravity acting on that column is buoyancy.  See my point?

Now I need to recalculate, I still have the balance action I need, but now the whole force equation becomes much closer and more important which is what I was trying to work out in the first place.

Time?  In what I'm trying to prove, time is not a factor.  Gravity does not donate it's force in limited time, it is constant and infinite.  It's not like I'm having to create the force to create weight (consider in a space station with rotation), gravity donates this gratis.

What I'm trying to work out is that it is possible to create buoyancy at a depth, which can be recovered.  Unlike above the water, where it is not possible to just create weight at a height, it is possible to flood and sink a chamber at minimal cost, to a depth at which we want to work on it.

Air pressure is ambient at all times.  Yes there will be a weight cost to the air in the tube, but it should balance with the weight cost of the chambers so long as the tube is small enough.  Again, speed.  Slow release of water, slow input of air.

Diagram tomorrow. It's not polished but it is descriptive and shows where the air comes from.
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Re: Open to destruction of an idea
« Reply #14 on: February 17, 2015, 02:33:00 AM »
Would it be totally rude to ask your age?
"A force de chercher de bonnes raisons, on en trouve; on les dit; et après on y tient, non pas tant parce qu'elles sont bonnes que pour ne pas se démentir." Choderlos de Laclos "You struggle to come up with some valid reasons, then cling to them, not because they're good, but just to not back down."

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Re: Open to destruction of an idea
« Reply #15 on: February 17, 2015, 04:45:46 AM »
Time is absolutely a factor in gravity.  High gravitational fields bend time, have you ever heard of a black hole?  Standard acceleration due to gravity on Earth at sea level is defined as 9.80665 meters per second per second.  No offense but this idea is completely preposterous.  There's no energy in gravity.  Gravity is like a spring.  If you push it, it pushes back, and you can get out of it close to what you put in, minus friction.  But you can't use it to magically generate energy.

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Re: Open to destruction of an idea
« Reply #16 on: February 17, 2015, 06:37:06 AM »
I think jb is saying what most of us have been thinking.
"A force de chercher de bonnes raisons, on en trouve; on les dit; et après on y tient, non pas tant parce qu'elles sont bonnes que pour ne pas se démentir." Choderlos de Laclos "You struggle to come up with some valid reasons, then cling to them, not because they're good, but just to not back down."

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Re: Open to destruction of an idea
« Reply #17 on: February 17, 2015, 07:39:00 PM »
Time is absolutely a factor in gravity.  High gravitational fields bend time, have you ever heard of a black hole?  Standard acceleration due to gravity on Earth at sea level is defined as 9.80665 meters per second per second.  No offense but this idea is completely preposterous.  There's no energy in gravity.  Gravity is like a spring.  If you push it, it pushes back, and you can get out of it close to what you put in, minus friction.  But you can't use it to magically generate energy.

Even gravity, if memory serves, is not 100% efficient (though damn close). Therefore the laws of thermodynamics still ultimately bite you, even if you designed out all other inefficiencies.

[EDIT] Let me qualify what I supect is going to be a controversial statement above:

http://www.quora.com/How-does-the-earth-orbit-the-sun-without-losing-energy

Hence, my contention that gravity is not 100% efficient (not if viewed as a spring at least).

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Re: Open to destruction of an idea
« Reply #18 on: February 17, 2015, 09:41:38 PM »
Conservation of energy is an implicit outcome of how the universe works. For your system to gain energy, you would need to explain where it came from. A clever arrangement of mechanical things which in a thought experiment seems to gain energy from nowhere would indicate that some effort somewhere is being overlooked.

I would expect that the effort of compressing the air to produce the pressure needed to expand the object at depth would exceed the energy gained as that object floats back to the surface because when you release the air (even in a closed high pressure system) to sink again you will do it against a lower pressure background.

If the air was only lifting itself I would expect the energy to balance (compression to get to depth, expansion on rising).

Friction and entropy are much more nasty than many people realise, eg: the maximum efficiency for gas completion depends on the rate at which you compress it. (The cold of rapidly expanding gas, or the heat of rapidly composing gas is all energy lost to thermal effects).

Water is also quite a viscose liquid. If you want to extract every from a rising object that implies rapid rate of movement, with energy lost as turbulence.



On the conservation of every:


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Re: Open to destruction of an idea
« Reply #19 on: February 18, 2015, 12:41:47 AM »
I don't doubt that at all ccg.  Nothing is 100% efficient.

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Re: Open to destruction of an idea
« Reply #20 on: February 18, 2015, 05:51:15 AM »
Would it be totally rude to ask your age?
Not at all.  53.
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Re: Open to destruction of an idea
« Reply #21 on: February 18, 2015, 05:57:13 AM »
Sorry sleep intruded and I only got a little time with my spreadsheet.

there are two key parts to this.  Nothing discussed so far changes that.

First is that this engine relies on the pressure created by gravity to vent water from a series of chambers.

Crucial to the idea is that the energy sequestered in one chamber can be added to the force of the original starting force, i.e. 1tonne weight above the water and a chamber of air, now vented of water, added together.  The extra force of the buoyant chamber acts on the system in direct opposition to the force of pressure at depth, balancing out the effect.

Second.  When equilibrium is reached, i.e. I have 1 tonne of weight and 1 tonne of buoyancy, the two forces are working in opposition to each other.  Therefore any chamber to be vented is now being acted on by two tonnes of force, not one.

So, it is critical to understand whether the first premise is correct.  I have to update my diagram based on the correction of my admittedly shoddy maths.  It doesn't change the idea or negate the idea, just needs me to be significantly more precise.

Sorry, I know, I know, diagram needed to explain.  I will get there even if it is only at the weekend.  Work is pressing.
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NeilT

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Re: Open to destruction of an idea
« Reply #22 on: February 18, 2015, 06:07:16 AM »
Time is absolutely a factor in gravity.  High gravitational fields bend time, have you ever heard of a black hole?  Standard acceleration due to gravity on Earth at sea level is defined as 9.80665 meters per second per second.  No offense but this idea is completely preposterous.  There's no energy in gravity.  Gravity is like a spring.  If you push it, it pushes back, and you can get out of it close to what you put in, minus friction.  But you can't use it to magically generate energy.

OK, so you are telling me that if I put a chamber full of water into a container of water with a piston on it and I put a 10kg weight onto the piston, given that the mechanism creates more pressure than the pressure of the water around the chamber, then the rate at which the water flows out of the chamber (time), is critically important?

Because, as far as I understand it, the weight falls the same distance venting the same amount of water.  If I use a large vent for the water, allowing a high flow (reduced time), or a small vent, allowing in a low flow (increased time), the same pressure will be created, the same water will be vented and the weight will have travelled the same distance.

The only difference will be in the w/h, not in the newtons consumed.

If, of course, I was having to use an engine, which consumed fuel, to provide the pressure in the chamber, then time would be critical. As it is when launching satellites into orbit.  If you don't reach escape velocity then you can burn all the fuel In the universe (ok exaggeration but not much) and gravity will still be holding you there.

As I don't use fuel other than gravity, then, as I see it, time is irrelevant.

Please tell me if I'm wrong, but I don't see it.  I'm using the force of gravity.  Time only comes into play once balance is overcome and movement with energy is required.  Then and only then is time required to measure work.

If it takes one hour to vent a chamber instead of 1 minute should not matter in that respect?

And, again, I know I need the diagram.
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NeilT

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Re: Open to destruction of an idea
« Reply #23 on: February 18, 2015, 06:23:48 AM »
I would expect that the effort of compressing the air to produce the pressure needed to expand the object at depth would exceed the energy gained as that object floats back to the surface because when you release the air (even in a closed high pressure system) to sink again you will do it against a lower pressure background.

I'm not compressing air.  I'm drawing in air at ambient pressure on the pressure side of the piston.  I know, diagram. But my revelations of getting my math wrong has made me work more on the design.  I need a flexible piston because I need to create a pressure bubble (force not air), in the chamber to let air in at 1 atm.  Once I have air behind the piston at 1atm, I only have to deal with the pressure at depth.

Again.  No pressurisation. The chambers are rigid, filled with water at the pressure of 5m depth (arbitrary depth and may be revised).  Although I'm going to have to assume a plastic composition which is neutral buoyancy, not the 10kg I assumed for a metal one.  Small change in the overall, it just makes the calculations easier.

When a chamber is vented of water, the piston draws ambient air behind it from the surface.  Think small, light, pipe connected to the bottom of the chamber.

I know diagram needed.  Think oil drum fixed to a weight above the surface by a frame, valve on top, piston in the bottom, air pipe connected to the bottom of the drum and reaching the surface.  There is a rod, sealed, in the bottom of the drum, connected to the piston. When the rod strikes a fixed, unmoving surface, the weight above the water drives the piston and empties the chamber.  Air at 1atm is drawn in through the pipe and adds 1atm of pressure below the piston.  When the water is vented, the chamber is now filled with air at 1atm and is buoyant to the weight of the volume of the water displaced by it.

Now we have a force below and a force above.  Disconnect first chamber from the weight above the water.  Second chamber has a rod right through the top, sealed at both ends.  Connect first chamber to the rod. When the rod on the second chamber strikes an unmoving surface, the force felt at the piston is the weight above the water plus the buoyant force of the first chamber (oil drum full of air).

It's not a difficult design so far.  I know picture speaks 10,000 words.  Will get there honest.  Just work intrudes.

What I'm trying to do is get to the point where equilibrium is achieved in force.  Then and only then can any real work be done.

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Re: Open to destruction of an idea
« Reply #24 on: February 18, 2015, 05:13:09 PM »
I only ask because I too was pretty darn sure that I could figure out a perpetual motion machine and had a number of designs. It probably helped me get passed that stage that I had a number of friends who were more skilled in math and physics than I and could point out problems with each.

That was when I was about 13, iirc. (Those friends have gone on to be professors in math and physics.)

So I understand the urge, but I would also suggest listening closely to your wise friends here and elsewhere.
"A force de chercher de bonnes raisons, on en trouve; on les dit; et après on y tient, non pas tant parce qu'elles sont bonnes que pour ne pas se démentir." Choderlos de Laclos "You struggle to come up with some valid reasons, then cling to them, not because they're good, but just to not back down."

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Re: Open to destruction of an idea
« Reply #25 on: February 18, 2015, 05:24:25 PM »
OK, so you are telling me that if I put a chamber full of water into a container of water with a piston on it and I put a 10kg weight onto the piston, given that the mechanism creates more pressure than the pressure of the water around the chamber, then the rate at which the water flows out of the chamber (time), is critically important?
Yes.
Quote
Because, as far as I understand it, the weight falls the same distance venting the same amount of water.  If I use a large vent for the water, allowing a high flow (reduced time), or a small vent, allowing in a low flow (increased time), the same pressure will be created, the same water will be vented and the weight will have travelled the same distance.

The only difference will be in the w/h, not in the newtons consumed.
If nothing else, higher turbulence through the smaller hole will be less efficient than the larger hole and rob energy from the system.  So no, the energies will not be equivalent.

Quote
If, of course, I was having to use an engine, which consumed fuel, to provide the pressure in the chamber, then time would be critical. As it is when launching satellites into orbit.  If you don't reach escape velocity then you can burn all the fuel In the universe (ok exaggeration but not much) and gravity will still be holding you there.

As I don't use fuel other than gravity, then, as I see it, time is irrelevant.

Please tell me if I'm wrong, but I don't see it.

We've been telling you this over and over again but you won't listen.  Are you truly open to the destruction of your idea?

Quote
I'm using the force of gravity.  Time only comes into play once balance is overcome and movement with energy is required.  Then and only then is time required to measure work.

If it takes one hour to vent a chamber instead of 1 minute should not matter in that respect?

And, again, I know I need the diagram.

The diagram won't help you.  Perpetual motion is impossible.  Gravity is a spring, not an energy source.  What you're describing here is akin to hooking up the output of a solar panel to a lightbulb and then counting on that lightbulb to power the solar panel and getting free light and electricity.

It's obvious from your response above that you lack the understanding of math and physics necessary to even comprehend our replies to you.  If you learned enough material to be able to understand our responses, you wouldn't even be asking this question in the first place, because you would understand why it is impossible.

You are the one making extraordinary claims, so the burden of proof is on you.  The nice thing about the truth is that it's true whether one believes and understands it or not.  Come back when you have a working prototype.

If you are truly open to the destruction of an idea, you can call it quits, it's been destroyed.

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Re: Open to destruction of an idea
« Reply #26 on: February 18, 2015, 08:46:17 PM »
So I understand the urge, but I would also suggest listening closely to your wise friends here and elsewhere.

Did occur to me one could use buoyant chambers to leverage the rise and fall of water during regular tides. No magic creation of energy involved of course. And not sure if it would do any better really than extracting wave energy which is far more frequent than tides.