The simplest free energy system ever overlooked

[angryScientist]

Here is what I have figured so far.

One liter of water weighs about 2.2 lbs. That means for ever liter of water displaced we are going to have a force of 2.2 lbs. pushing the hydrogen to the surface. We now have a force.

Work is the product of the force and the distance through which a body is moved by that force. It is expressed in joules, ergs, foot-pounds, BTU, and kilowatt-hour. One foot-pound of work or energy equals .0003766 Watt-hours of energy.

If we take 1 liter of hydrogen gas at 1 atm. and put it under 32ft. of water we have doubled the pressure. Our hydrogen is now under twice the pressure and it's volume will be half of what it was on the surface. Our original liter of hydrogen is now under 2 atm. of pressure is half the volume and will displace 1.1 lbs. of water.

If it travels upward for one foot it produces 1.1 ft.-lbs. of energy. Now it's volume is greater because it is under less pressure. The pressure is going to decrease by 3.125% per foot. So after one foot we are at 31 feet under water and 1.9375 atm. The volume is .516129 liters. For our next foot we can expect 1.35 lbs. of force. Etcetera, etcetera.

Calculating it this way I've figured that you can get about 48.25 foot-pounds of energy, which is about .01817 watt-hours. This is a much more accurate figure than the 36 ft-lbs. I calculated earlier. (I don't know calculus so I have to learn as I go.)

I've read that modern electrolyzers can produce 1 liter of hydrogen with 1 watt-hour of energy.

I don't think that one could achieve over unity this way as I think the pressure would have to double 35 times. (2^35 = 34 billion atm.)

With that said, you can still get more energy than you would other wise with strait electrolysis. Also you could use a heavier liquid. With mercury you would only need
28.25in. instead of 32ft. You could also bubble the H2 through a tube, like a percolater to lift your liquid.

Anyway... I'm tired. Good night all.

[d3adp00l]

Hey angry, good to see you again. I love it when equations are somehow wrong, let me give you an example, a 12v 5a motor outputs 1 foot pound. Say you run it for 1 second, it would output 1ft*lbs/1sec of power= watts (according to online conversions) so 1ft*lbs/s = 60watts?!?!?! And yet ft*lbs = watthours (according to online conversions) so the same motor would yield 1 = .0166667watthours ((60w*1hour)/3600 seconds). WOW I made 60=1=.016667  I am a math genius. Something is  wrong with those equations. how can ft*lbs/second=watt and ft*lbs=watthours  . So I will have evaluate this further.

Here is what I have figured so far.

One liter of water weighs about 2.2 lbs. That means for ever liter of water displaced we are going to have a force of 2.2 lbs. pushing the hydrogen to the surface. We now have a force.

Work is the product of the force and the distance through which a body is moved by that force. It is expressed in joules, ergs, foot-pounds, BTU, and kilowatt-hour. One foot-pound of work or energy equals .0003766 Watt-hours of energy.

If we take 1 liter of hydrogen gas at 1 atm. and put it under 32ft. of water we have doubled the pressure. Our hydrogen is now under twice the pressure and it's volume will be half of what it was on the surface. Our original liter of hydrogen is now under 2 atm. of pressure is half the volume and will displace 1.1 lbs. of water.

If it travels upward for one foot it produces 1.1 ft.-lbs. of energy. Now it's volume is greater because it is under less pressure. The pressure is going to decrease by 3.125% per foot. So after one foot we are at 31 feet under water and 1.9375 atm. The volume is .516129 liters. For our next foot we can expect 1.35 lbs. of force. Etcetera, etcetera.

Calculating it this way I've figured that you can get about 48.25 foot-pounds of energy, which is about .01817 watt-hours. This is a much more accurate figure than the 36 ft-lbs. I calculated earlier. (I don't know calculus so I have to learn as I go.)

I've read that modern electrolyzers can produce 1 liter of hydrogen with 1 watt-hour of energy.

I don't think that one could achieve over unity this way as I think the pressure would have to double 35 times. (2^35 = 34 billion atm.)

With that said, you can still get more energy than you would other wise with strait electrolysis. Also you could use a heavier liquid. With mercury you would only need
28.25in. instead of 32ft. You could also bubble the H2 through a tube, like a percolater to lift your liquid.

Anyway... I'm tired. Good night all.

[angryScientist]

I thought I was wrong once but I was mistaken. 

I have been wrong many times before and it has taught me one thing; There is no doubt  that I can be wrong again.

My TI-85 says 1 ftlb/s = 1.35581794833 Watts and is POWER.

And 1 ft-lb = .000376616096759 W-hrs and is ENERGY

This may not make much sense to me, right now, but I believe ENERGY does not have a time component and POWER does. By the way, I don't understand how W-hrs does not have a time component. Although, Watts does have time built into it... (time + time = no time) (I don't understand)

Anyway I need to get some energy because tomorrow I have to go fight a fire. Good luck all.

[angryScientist]

Sorry, one more.

My calc says 12V*5A = 60W = 44.2537289566 ftlb/sec

Foot pounds
Foot pounds per second


Or this one...
32ft. per second squared

How do you square time!?!?!?!?!?

[d3adp00l]

squared time is actually easy. miles= distance    miles/time= speed    miles/time2=acceleration
but there is a massive problem with voltagesXAmps=ftXlbs/time
in order to have a functioning equation all units have to be balanced.
there is a difference between a primary equation  (distance/time=speed)
and a conversion equation like we are dealing with miles=kilometers distance=distance
miles/second=kilometers/second I don't understand this like I said I will have to run the complete conversion step by step.
But at 1.65 lbs ave lift for 33 feet = 54.45 ftlbs in 33 seconds = 1.65 ftXlbs/seconds X 20 buckets = 33ftXlbs/seconds=44.7watts X 33 seconds = 1476watt seconds or 161084.75 watthours Using their equations