Ultimate energy source

[Toddsterpatriot]

(5) times less than perfect efficiency still loses useful energy.

I do not deny that energy is consumed while the process runs. I do not deny that water resistance to the rising balloons consumes energy, etc.

What I have not conceded to is the following:

for example-

It takes (X) amount of energy to put air in one balloon

(X)+(X) +(X)+(X)+(X) = more energy output at any moment in time than (X)

Together with on other fact. As the balloons rise they will expand creating more lifting force
-

It takes (X) amount of energy to put air in one balloon

Yup.

(X)+(X) +(X)+(X)+(X) = more energy output at any moment in time than (X)

No. Your output is Z% of (X). With Z < 100.

Together with on other fact. As the balloons rise they will expand creating more lifting force​

You're right, the air takes more force to push down to that bucket, the deeper it gets.

​

 

[JoeMoma]

(5) times less than perfect efficiency still loses useful energy.

I do not deny that energy is consumed while the process runs. I do not deny that water resistance to the rising balloons consumes energy, etc.

What I have not conceded to is the following:

for example-

It takes (X) amount of energy to put air in one balloon

(X)+(X) +(X)+(X)+(X) = more energy output at any moment in time than (X)

Together with on other fact. As the balloons rise they will expand creating more lifting force
-

You don't get the energy it takes to fill the first 4 balloons for free. Also, the more balloons you have in the system, the quicker you have to fill each balloon at the bottom.

Each time one balloon releases its air at the top, you have to fill one balloon at the bottom. You are not getting 5 for one, You are getting one for one. By the time each balloon has made it's journey to the top, 5 balloons behind it have to be filled.

Also, you are still confusing force with energy.... they are not the same.

Perhaps you should think about how much energy you must put into the system and how much energy is output from the system in one complete cycle.

 

[Rigby5]

You will have to keep filling all the buckets, so all the compression of air to fill them will add up to more than you get out of it.

You have to fill all of the buckets (balloons) once, when they are all full all you need to do is fill one (1) at any moment in time to keep it running.

[yes] or [no]

-

Yes, you only have to refill one at a time.
But as buckets keep reaching the top and flipping, you will have to keep filling a new once that flips over at the bottom.
So you keep having to put energy into it, slightly greater than the energy you get out.

 

[JoeMoma]

You will have to keep filling all the buckets, so all the compression of air to fill them will add up to more than you get out of it.

You have to fill all of the buckets (balloons) once, when they are all full all you need to do is fill one (1) at any moment in time to keep it running.

[yes] or [no]

-

Yes, you only have to refill one at a time.
But as buckets keep reaching the top and flipping, you will have to keep filling a new once that flips over at the bottom.
So you keep having to put energy into it, slightly greater than the energy you get out.

slightly greater

Considering the complexity of this device and the viscosity of water, my vote is for "much greater".

 

[Rigby5]

Below is an old idea I had----
wrong drawing, I deleted it

Goethermal like that is very big in Iceland because they have so much volcanic actions so close to the surface, such as steaming gysers and hot pools of water right on the surface.
I do not know exactly what their mechanisms look like, but it would probably be worth studying what they already did, do find things to improve upon.

 

[Rigby5]

But don't ignore the ocean heat different engine to run a generator.
All you need is enough difference between warm and cold, like off the coast of Brazil, and you can run what essentially is a Sterling heat engine.

 

[watchingfromafar]

I understand that there are a lot of new innovative machines out there designed to create energy. Hopefully there are a few that create more energy than is needed to run the system.

I have seen the same argument here as the one my son presented to me and still, I refuse to concede.

Energy input verses output.

If I am using (X) amount of energy to keep the system running while at the same time I am getting out (X)+(X) +(X) +(X) +(X) amount of energy at any one moment in time; I am getting out “more” energy than I am putting in.

-

 

[Toddsterpatriot]

I understand that there are a lot of new innovative machines out there designed to create energy. Hopefully there are a few that create more energy than is needed to run the system.

I have seen the same argument here as the one my son presented to me and still, I refuse to concede.

Energy input verses output.

If I am using (X) amount of energy to keep the system running while at the same time I am getting out (X)+(X) +(X) +(X) +(X) amount of energy at any one moment in time; I am getting out “more” energy than I am putting in.

-

If I am using (X) amount of energy to keep the system running while at the same time I am getting out (X)+(X) +(X) +(X) +(X) amount of energy at any one moment in time;

You're not.

You put in 5(X) and get out less (much less) than 5(X).

 

[watchingfromafar]

If I am using (X) amount of energy to keep the system running while at the same time I am getting out (X)+(X) +(X) +(X) +(X) amount of energy at any one moment in time;

You're not.

You put in 5(X) and get out less (much less) than 5(X).

But, but, but it only takes one (1)X to keep it running at any one moment in time
look at it another way, I am consuming 100 watts while my output is putting out 500 watts
-

 

[Toddsterpatriot]

If I am using (X) amount of energy to keep the system running while at the same time I am getting out (X)+(X) +(X) +(X) +(X) amount of energy at any one moment in time;

You're not.

You put in 5(X) and get out less (much less) than 5(X).

But, but, but it only takes one (1)X to keep it running at any one moment in time
look at it another way, I am consuming 100 watts while my output is putting out 500 watts
-

But, but, but it only takes one (1)X to keep it running at any one moment in time

Yup. And your output is always less (a lot less) than your input.

 

[JoeMoma]

I understand that there are a lot of new innovative machines out there designed to create energy. Hopefully there are a few that create more energy than is needed to run the system.

I have seen the same argument here as the one my son presented to me and still, I refuse to concede.

Energy input verses output.

If I am using (X) amount of energy to keep the system running while at the same time I am getting out (X)+(X) +(X) +(X) +(X) amount of energy at any one moment in time; I am getting out “more” energy than I am putting in.

-

No, there are not any innovative machines designed to CREATE energy that actually work. There are innovative machines that tap into energy sources already there. Machines do not create energy, they consume energy.

Refuse to concede all you want, you will never be able to build a machine that produces more energy than it consumes.

 

[JoeMoma]

I understand that there are a lot of new innovative machines out there designed to create energy. Hopefully there are a few that create more energy than is needed to run the system.

I have seen the same argument here as the one my son presented to me and still, I refuse to concede.

Energy input verses output.

If I am using (X) amount of energy to keep the system running while at the same time I am getting out (X)+(X) +(X) +(X) +(X) amount of energy at any one moment in time; I am getting out “more” energy than I am putting in.

-

You are very confused because you have 5 balloons going up at a time while you are filling one balloon. That does not matter. What matters is that the rate at which you move (pump) air down to the bottom of your device is at least equal to the rate the air travels back up to the surface, otherwise you device will run short of air.

Also, you will not get any energy out of your device until you harness it to some type of load such as a generator.

 

[watchingfromafar]

You are very confused because you have 5 balloons going up at a time while you are filling one balloon. That does not matter. What matters is that the rate at which you move (pump) air down to the bottom of your device is at least equal to the rate the air travels back up to the surface

There is one factor that I believe you are missing or ignoring.
As these balloons rise they wil expand and this expansion will displace more water giving it more lifting force.

Also, you will not get any energy out of your device until you harness it to some type of load such as a generator.

I agree but I can only deal with the theory, not the technocial inner works.
I have no idea of what these balloons will be made of; nor the type of cable..,.. etc.
If the mythology of the system is valid then others with far more experience will pick up the flag and bring it to the finished line.

It won’t be me
-

 

[JoeMoma]

You are very confused because you have 5 balloons going up at a time while you are filling one balloon. That does not matter. What matters is that the rate at which you move (pump) air down to the bottom of your device is at least equal to the rate the air travels back up to the surface

There is one factor that I believe you are missing or ignoring.
As these balloons rise they wil expand and this expansion will displace more water giving it more lifting force.

Also, you will not get any energy out of your device until you harness it to some type of load such as a generator.

I agree but I can only deal with the theory, not the technocial inner works.
I have no idea of what these balloons will be made of; nor the type of cable..,.. etc.
If the mythology of the system is valid then others with far more experience will pick up the flag and bring it to the finished line.

It won’t be me
-

Kinetic energy = (1/2)MV^2 and Force = MA. You keep going down the wrong rabbit trail with this "force at the moment" stuff.

 

[watchingfromafar]

JoeMona,
I’m at home (without a backyard) and I have decided to take 15-20 bills to put me to sleep.

See you on the other side, it’s been fun

-

 

[watchingfromafar]

Air Bubble

Air is a gas

gases compress with pressure

Assuming constant temperature, Boyle's - a pressure of 19 atmospheres (1 atmosphere at sea level + 18 atmospheres for being 600ft under water)

One square food of water at 1 ATM weighs 64 lbs.


One square foot of water will compress to 1/2 square foot at 2 ATM

EXAMPLE—

Surface volume (X) = 300 CF

Upper force (Y) = 64 lbs.

Volume at 2 atm (X)/2 = 150 CF = 150 X 64 = 9,600 pounds of upward force
Volume at 3 atm (X)/3 = 100 CF = 100 X 64 = 6,400 pounds of upward force
Volume at 6 atm (X)/6 = 50 CF = 50 X 64 = + Y X CF = 3,200 lbs.
Volume at 9 atm (X)/9 = 33.33 CF; upward force = Y X CF = 2,133 lbs.
Volume at 12 atm (X)/12 = 25 CF; upward force = Y X CF = 1,600 lbs.
Volume at 15 atm (X)/15 = 20 CF; upward force = Y X CF = 1,280 lbs.
Volume at 18 atm (X)/18 = 16.66 CF; upward force = Y X CF = 1,066 lbs.
Total upward force = 25,259 lbs.

An air bubble rises at 2 feet a second

2 X 60 = 120 feet per minute
120 X 60 = 7,200 feet per hour
5,280 feet in a mile

7,200/5,280 = 1.36 miles an hour
TOTAL force 25,259 lbs. traveling at 1.36 miles an hour
globule

Now all I have to do is convert the above to electric energy
-

 

[fncceo]

Now all I have to do is convert the above to electric energy
-

I convert this to electricity every month...

 

[watchingfromafar]

Foot-Pound : The foot-pound (symbol: ft•lb) is a measurement unit of energy which is equivalent to 1.3558179483314 joules. It is defined as the amount of energy expended in applying a force of one pound-force through a displacement of one foot. In one second

A Joule = The foot-pound (also and originally known as foot-pound force) is a traditional English unit of work. It is equal to the work done by one pound of force acting through a distance of one foot. For example, when James Watt determined that a horse could lift 550 lbs. at a rate of one foot per second, he declared it one horsepower. The SI or international equivalent of the foot-pound is the Joule (J).

 

[Toddsterpatriot]

Air Bubble

Air is a gas

gases compress with pressure

Assuming constant temperature, Boyle's - a pressure of 19 atmospheres (1 atmosphere at sea level + 18 atmospheres for being 600ft under water)

One square food of water at 1 ATM weighs 64 lbs.


One square foot of water will compress to 1/2 square foot at 2 ATM

EXAMPLE—

Surface volume (X) = 300 CF

Upper force (Y) = 64 lbs.

Volume at 2 atm (X)/2 = 150 CF = 150 X 64 = 9,600 pounds of upward force
Volume at 3 atm (X)/3 = 100 CF = 100 X 64 = 6,400 pounds of upward force
Volume at 6 atm (X)/6 = 50 CF = 50 X 64 = + Y X CF = 3,200 lbs.
Volume at 9 atm (X)/9 = 33.33 CF; upward force = Y X CF = 2,133 lbs.
Volume at 12 atm (X)/12 = 25 CF; upward force = Y X CF = 1,600 lbs.
Volume at 15 atm (X)/15 = 20 CF; upward force = Y X CF = 1,280 lbs.
Volume at 18 atm (X)/18 = 16.66 CF; upward force = Y X CF = 1,066 lbs.
Total upward force = 25,259 lbs.

An air bubble rises at 2 feet a second

2 X 60 = 120 feet per minute
120 X 60 = 7,200 feet per hour
5,280 feet in a mile

7,200/5,280 = 1.36 miles an hour
TOTAL force 25,259 lbs. traveling at 1.36 miles an hour
globule

Now all I have to do is convert the above to electric energy
-

Take the amount of energy it took to move the air to the bottom bucket and divide by 2. Or 3.

 

[fncceo]

Air Bubble

Air is a gas

gases compress with pressure

Assuming constant temperature, Boyle's - a pressure of 19 atmospheres (1 atmosphere at sea level + 18 atmospheres for being 600ft under water)

One square food of water at 1 ATM weighs 64 lbs.


One square foot of water will compress to 1/2 square foot at 2 ATM

EXAMPLE—

Surface volume (X) = 300 CF

Upper force (Y) = 64 lbs.

Volume at 2 atm (X)/2 = 150 CF = 150 X 64 = 9,600 pounds of upward force
Volume at 3 atm (X)/3 = 100 CF = 100 X 64 = 6,400 pounds of upward force
Volume at 6 atm (X)/6 = 50 CF = 50 X 64 = + Y X CF = 3,200 lbs.
Volume at 9 atm (X)/9 = 33.33 CF; upward force = Y X CF = 2,133 lbs.
Volume at 12 atm (X)/12 = 25 CF; upward force = Y X CF = 1,600 lbs.
Volume at 15 atm (X)/15 = 20 CF; upward force = Y X CF = 1,280 lbs.
Volume at 18 atm (X)/18 = 16.66 CF; upward force = Y X CF = 1,066 lbs.
Total upward force = 25,259 lbs.

An air bubble rises at 2 feet a second

2 X 60 = 120 feet per minute
120 X 60 = 7,200 feet per hour
5,280 feet in a mile

7,200/5,280 = 1.36 miles an hour
TOTAL force 25,259 lbs. traveling at 1.36 miles an hour
globule

Now all I have to do is convert the above to electric energy
-

Take the amount of energy it took to move the air to the bottom bucket and divide by 2. Or 3.

You're wasting your time ...

It's been explained to him ... he's not listening.

I'm curious why he hasn't run down to the Patent Office with his plan.