Science students need a problem too solve.

[BULLDOG]

You went off track when you refused to accept the possibility that you are wrong. Possibly before then, but for sure by that time.

Honestly, I am willing to accept that I am wrong. In fact, I believe there is a 95% chance that I am wrong. But that 5% keeps me up a night and I really want a god nights sleet. All that needs to be done is for someone to point out what aspect or what design flaws there are that proves it will not work. As of yet; no one has done that.

I get a lot of, “it is a perpetual motion machine therefore it cannot work.” That response falls short of pointing what technical aspect that is wrong.

The design principles are simple enough. Just point out the flaw & if it is a sound described flaw I will end this merry-go-round and move on to something else.

OK?

I don't care if you want to spend the rest of your life thinking you solved the world's energy problem, but nobody cares. You have had it explained to you multiple times. It's up to you whether you accept reality or not. I suggest you calculate the continuous volume and pressure you would require to fill those buckets, and then go to an industrial rental company to see if they have a compressor to supply that volume at that pressure. Take note of the horsepower required to run that compressor, and compare it to the calculated output of your device.

 

[watchingfromafar]

I don't care if you want to spend the rest of your life thinking you solved the world's energy problem

It would be a dream come true, but no I don't.

, but nobody cares.

At least now I know where to go to find the one who knows what everybody cares about.

You have had it explained to you multiple times.

No I have not. I have received comments like, "it will not work" or " it is a perpetual motion machine" but as of yet no one has explained why it will not work. But if you have found someone that details why it does not work I will be very grateful to you for cutting and pasting the flaws in the design right here.

It's up to you whether you accept reality or not.

It is up to you to provide the evidence showing why it will not work

I suggest you calculate the continuous volume and pressure you would require to fill those buckets, and then go to an industrial rental company to see if they have a compressor to supply that volume at that pressure. Take note of the horsepower required to run that compressor, and compare it to the calculated output of your device.

That is a good suggestion. I am already working on getting the answers to your questions. I was hoping a poster or posters here could help in getting the numbers.
Do you BULLDOG know where can go to get the technical question you asked?

 

[BULLDOG]

I don't care if you want to spend the rest of your life thinking you solved the world's energy problem

It would be a dream come true, but no I don't.

, but nobody cares.

At least now I know where to go to find the one who knows what everybody cares about.

You have had it explained to you multiple times.

No I have not. I have received comments like, "it will not work" or " it is a perpetual motion machine" but as of yet no one has explained why it will not work. But if you have found someone that details why it does not work I will be very grateful to you for cutting and pasting the flaws in the design right here.

It's up to you whether you accept reality or not.

It is up to you to provide the evidence showing why it will not work

I suggest you calculate the continuous volume and pressure you would require to fill those buckets, and then go to an industrial rental company to see if they have a compressor to supply that volume at that pressure. Take note of the horsepower required to run that compressor, and compare it to the calculated output of your device.

That is a good suggestion. I am already working on getting the answers to your questions. I was hoping a poster or posters here could help in getting the numbers.
Do you BULLDOG know where can go to get the technical question you asked?

I already answered it weeks ago. and explained how to calculate pressure at depth. The volume in CFM depends on how fast you want to fill the buckets.

 

[JoeMoma]

That is a good suggestion. I am already working on getting the answers to your questions. I was hoping a poster or posters here could help in getting the numbers.

Why pump air underwater! Why not pump water up to the top of a tower lets say 100 meters tall? From the bottom to the top of the tower is a series of buckets similar to your under water device. Buckets are spaced 10 meters apart. So water is pumped 100 meters to the bucket at the top of the tower to provide 10 buckets of water pulling down. And to make the numbers easy to deal with, lets make the amount of water to fill each bucket 10 Kg.

So each time one bucket is filled at the top of the tower, there is the force of ten buckets of water pulling down. Wouldn't this have the same multiplier effect as your design. 10 Kg of water is pumped to the top of the tower, 10 buckets of 10 Kg pulling down.

The reason I am suggesting this model rather than your underwater model is that the energy/work calculations are straight forward. Because the volume of a given mass of air varies with pressure and temperature, the calculations are no where near as straight forward for your model. That being said, the model I'm suggesting would have the same effect of the force of 10 buckets pulling each time one bucket is filled.

 

[Grumblenuts]

You also might consider parking it under a waterfall, perhaps even using a Pelton wheel. Btw,
stop being such an idiot..

It is up to you to provide the evidence showing why it will not work

Your positive assertions are always your burden to work out and demonstrate.. never anyone else's job to "disprove" or negate.

 

[watchingfromafar]

You are so uneducated about basic physics that you don't recognize the proof when it's pointed out to you.

Point it out, goooo foooor it !!!!!

 

[watchingfromafar]

You went off track when you refused to accept the possibility that you are wrong. Possibly before then, but for sure by that time.

And still, even today I hear the same old reply, "it will not work"
But no one can point out why?
the reason is simple enough, no one knows why
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[BULLDOG]

You went off track when you refused to accept the possibility that you are wrong. Possibly before then, but for sure by that time.

And still, even today I hear the same old reply, "it will not work"
But no one can point out why?
the reason is simple enough, no one knows why
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Give it a rest. Lots of people know why. Nobody can make you, personally capable of understanding why is a more accurate statement.

 

[watchingfromafar]

This reminds me of the plan to raise the Titanic by filling it with ping pong balls. The physics of pumping air to a particular depth requires as much energy (assuming 100% efficiency) as can be recovered by the lift as it approaches the surface.

No one here seems to be able to grasp the simplest of ideas.

If you fill a one cubic foot container with air and submerge it into water, it will try to rise to the surface with an upward force of 67 pounds of force. Call this bucket #1

Repeat the above with another bucket and call the bucket #2
Repeat the above with another bucket and call the bucket #3
Repeat the above with another bucket and call the bucket #4
Repeat the above with another bucket and call the bucket #5
Repeat the above with another bucket and call the bucket #6
Repeat the above with another bucket and call the bucket #7
Repeat the above with another bucket and call the bucket #8
Repeat the above with another bucket and call the bucket #9
Repeat the above with another bucket and call the bucket #10
Now we have ten (10) buckets each having a lifting force of 67 pounds.

Connect the ten (10) buckets together and now you have (10)X67=670 pound of lifting force

Can we at least agree on the above?

 

[watchingfromafar]

I already answered it weeks ago. and explained how to calculate pressure at depth. The volume in CFM depends on how fast you want to fill the buckets.

Finally, we may be getting somewhere. In the drawing the buckets are 33 feet apart. The time between refills is therefor the speed it takes for each bucket to travel 33 feet.

It is that speed I am trying to resolve. I found an article that states an air bubble rises at a speed of three (3) feet per second. I have also found other articles that are a bit more complicated. Since the air in the buckets are expanding this creates a force of acceleration that I have not solved.

 

[BULLDOG]

This reminds me of the plan to raise the Titanic by filling it with ping pong balls. The physics of pumping air to a particular depth requires as much energy (assuming 100% efficiency) as can be recovered by the lift as it approaches the surface.

No one here seems to be able to grasp the simplest of ideas.

If you fill a one cubic foot container with air and submerge it into water, it will try to rise to the surface with an upward force of 67 pounds of force. Call this bucket #1

Repeat the above with another bucket and call the bucket #2
Repeat the above with another bucket and call the bucket #3
Repeat the above with another bucket and call the bucket #4
Repeat the above with another bucket and call the bucket #5
Repeat the above with another bucket and call the bucket #6
Repeat the above with another bucket and call the bucket #7
Repeat the above with another bucket and call the bucket #8
Repeat the above with another bucket and call the bucket #9
Repeat the above with another bucket and call the bucket #10
Now we have ten (10) buckets each having a lifting force of 67 pounds.

Connect the ten (10) buckets together and now you have (10)X67=670 pound of lifting force

Can we at least agree on the above?

67 pounds at what depth? You haven't even worked out the buoyancy at different depths. HINT - 1 one cubic foot of air at 1 foot depth is not the same as 1 cubic ft. of air at 100 feet of depth.
Do your research. Start with Archimedes principal.

 

[watchingfromafar]

67 pounds at what depth? You haven't even worked out the buoyancy at different depths.

Thanks for your reply. I have been using the same principle, a cubic foot of seawater weighs 67 pounds, therefore, according to Archimedes' principle the lifting force of a cubic foot of seawater is 67 pounds. That is the same principle I have been using.
Archimedes' principle states that the upward buoyant force that is exerted on a body immersed in a fluid, whether fully or partially, is equal to the weight of the fluid that the body displaces.
Archimedes' principle - Wikipedia

It seems we both agree on the lifting force, now we need to solve the rising speed and then the conversion to electric power or watts
or so it seems
BTW: how does depth apply here?
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[watchingfromafar]

I already answered it weeks ago.

If you have, I missed it.
BTW: I was wrong about the weight of a cubic foot of seawater.
Seawater weighs about 64 pounds per cubic foot—on average. "Average" fresh water weighs about 62.2 pounds per cubic foot. Through all the conversions, a gallon of seawater weighs approximately 8.556 or rounded to 8.6 pounds—on average. There's only one-tenth of a pound difference between the two.
How Much Does a Pint of Sweat Weigh?
I will correct my drawing once I return to my office
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[Grumblenuts]

Thanks for your reply. I have been using the same principle, a cubic foot of seawater weighs 67 pounds, therefore, according to Archimedes' principle the lifting force of a cubic foot of seawater is 67 pounds. That is the same principle I have been using.

67 pounds at what depth? You haven't even worked out the buoyancy at different depths. HINT - 1 one cubic foot of air at 1 foot depth is not the same as 1 cubic ft. of air at 100 feet of depth.
Do your research.

 

[BULLDOG]

67 pounds at what depth? You haven't even worked out the buoyancy at different depths.

Thanks for your reply. I have been using the same principle, a cubic foot of seawater weighs 67 pounds, therefore, according to Archimedes' principle the lifting force of a cubic foot of seawater is 67 pounds. That is the same principle I have been using.
Archimedes' principle states that the upward buoyant force that is exerted on a body immersed in a fluid, whether fully or partially, is equal to the weight of the fluid that the body displaces.
Archimedes' principle - Wikipedia

It seems we both agree on the lifting force, now we need to solve the rising speed and then the conversion to electric power or watts
or so it seems
BTW: how does depth apply here?
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No, we don't agree on the lifting force.
You truly have no idea what you are trying to do. You better add Boyles law to your study of Archimedes principal.

Intro to Boyle's Law
Suppose you had a balloon measuring one cubic foot at the surface of the water. This balloon is under 1 ATM (14.7 psi) of pressure. If we push the balloon underwater, and take it to a depth of 33 feet, it is now under 2 ATM of pressure (29.4 lbs) - 1 ATM of pressure from the air, 1 ATM of pressure from the water. Boyle's Law then tells us that since we have twice the absolute pressure, the volume of the balloon will be decreased to one half. It follows then, that taking the balloon to 66 feet, the pressure would compress the balloon to one third its original size, 99 feet would make it 1/4 etc.

 

[Grumblenuts]

The upward force on the cube is the pressure on the bottom surface integrated over its area. The surface is at constant depth, so the pressure is constant. Therefore, the integral of the pressure over the area of the horizontal bottom surface of the cube is the hydrostatic pressure at that depth multiplied by the area of the bottom surface.

Similarly, the downward force on the cube is the pressure on the top surface integrated over its area. The surface is at constant depth, so the pressure is constant. Therefore, the integral of the pressure over the area of the horizontal top surface of the cube is the hydrostatic pressure at that depth multiplied by the area of the top surface.

Hint: Does not apply here!

 

[Grumblenuts]

99 feet would make it 1/4 etc.

So dividing the static upward force exerted by Watching's presumed 67 lb. 1 foot cube by 4 then yields 16.7 lbs.. Then subtracting the roughly 16.7 lb. static downward force being exerted from 3 in. above yields approximately zilch.. So then ignoring silly stuff like container weight, friction, dynamics, etc.

Repeat the above with another bucket and call the bucket #10
Now we have ten (10) buckets each having a lifting force of zilch.

Connect the ten (10) buckets together and now you have (10)X0=0 pound of lifting force

Can we at least agree on the above?

(Hey, two can easily play dumb when one insists!)

 

[watchingfromafar]

The upward force on the cube is the pressure on the bottom surface integrated over its area. The surface is at constant depth, so the pressure is constant. Therefore, the integral of the pressure over the area of the horizontal bottom surface of the cube is the hydrostatic pressure at that depth multiplied by the area of the bottom surface.

Similarly, the downward force on the cube is the pressure on the top surface integrated over its area. The surface is at constant depth, so the pressure is constant. Therefore, the integral of the pressure over the area of the horizontal top surface of the cube is the hydrostatic pressure at that depth multiplied by the area of the top surface.

Hint: Does not apply here!

Finally I am learning again.
From the above I see
fluid pressure = (fluid density) X (acceleration due to gravity) X (fluid depth)

Can I equate 118,428 pounds = fluid pressure?
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[watchingfromafar]

So then ignoring silly stuff like container weight, friction, dynamics, etc.

Container weight cancels itself out because as one container is rising a container attached to the cable on the opposite side is going down.
one going down.
one going up.
Right opposite each other but going in opposite directions cancel their weight making this a mute problem.
friction does need to be considered. I look at it this way, it is a drag reducing the theoretical output which causing the system to slow down. tag this as drag as (D1).
I do not think friction will have much of an effect.

Grumblenuts, tell me a little more about dynamics, etc
thanks in advance
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[watchingfromafar]

So dividing the static upward force exerted by Watching's presumed 67 lb.

Sorry that I threw that 67 number around all this time. When I designed the SeaEngine I used 64 pounds, which is the correct number.
I just needed to correct my earlier mistake. I will correct the drawing & numbers later.
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