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I am bringing this drawing back into the discussion just to ask one last question.
In the drawing there are twelve (12) buckets on the right side. Just for discussion each bucket has a lifting force of 100-foot pounds. 12 buckets times 100 = 1200-foot pounds of lifting force.
1200-foot pounds of lifting force can produce more energy at any one moment in time than 100-foot pounds;
Once all the buckets are full and this machine is running, the process continues to produce 1200-foot pounds of force if you continue to fill one (1) bucket at the bottom in sequence with the rest.
YES or NO?

[ElectriciansForums.net] A new energy source-maybe--maybe-not
 
Thank you all for your contributions. Now that I understand the value of a spreadsheet, I still have a few details I will need before I can get the results I need.

I know the relationship between the size of an air bubble under different pressures.

I also know the lifting force of displaced water as in the lifting force that keeps a boat afloat.

What I don’t know is the speed of the rising bubbles which I believe is an increasing speed.

The output power will be a function of the drag put upon the rising force. If drag (X) is the amount of resistance that stops the rising force, then I need drag (X) minus some value less than drag (X) that allows the system to continue to rise. This drag will slow down the speed. The difference between its maximum speed potential and the drag applied to it will equal the useful energy from the system.

I have no idea how the create a spread sheet to determine its output power with so many unknown variables involved.

In short, I am dwelling about a system that I am unqualified to be contemplating.

This concept is better left to others to consider.

bye
 

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  • SEAPOWER12.pdf
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I need help in calculating the numbers
Help !!!!!!!!!!!!!

You previously said you were able to do the calculations (in Basic was it? - can't remember), but you no longer had that software. You asked for advice on a program which could do the calculations for you. The spreadsheet we advised about will able to do mathematical calculations.

However, if you don't know what calculations you are wanting to calculate then I can't see how you're going to progress.
 
I've not looked too deep (pun-intended),
but suspect your bubbles may not be as big , if the source "gasses are above sea temperature , they will contract some on cooling ..
(even though plenty of expanding will occur , also some chemical reactions with the sea water ,may dissolve some chemicals present, and pose some corrosion problems with matterials ) These factors are going to eat into you gains a little. (Nothing planet earth does is stable when dealing with lava !)
 
No one seems to want to directly address the design but instead dance around the subject.

The principle is simple enough so address the design/ principle behind the design.

Thanks in advance
 

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  • [ElectriciansForums.net] A new energy source-maybe--maybe-not
    seaengine -4.jpg
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I think you summed things up nicely in post #76.
Thank you for your reply.

Having said that, I am still not getting any feedback on the principle behind the design itself.

Principles to run the machine

[1] an enclosed container (X) of air submerged in water has a lifting force (Y) equal to the volume of the water displaced minus the weight of the container;

[2] connecting multiple containers one on top of the other creates a combined lifting force of (Y)+ (Y)+ (Y)+ (Y)+ (Y)+ (Y)+ (Y)+ (Y)+ (Y)+ (Y)

Which is a greater lifting force than (Y);

[3] the energy needed to fill one container is equal to the energy needed to sustain the combined lifting force of the 10 (ten) containers referenced above;

Formula used (ATM/V1) X V1 = bubble size

Output of this machine is 118,428 pounds of lifting force moving at 33 feet per second at any one moment in time
 
No one seems to want to directly address the design but instead dance around the subject.

The principle is simple enough so address the design/ principle behind the design.

Thanks in advance
If you know it all why are you on this forum. This is an electrical forum NOT rising bubbles forum. You’ve been on here before asking stupid questions. If you want advice on electrical then come back. These folks have better things to do than play your game
 
No one seems to want to directly address the design but instead dance around the subject.
I completely explained the operation of the machine in post #23.

Where’s @Lucien at when you need him
You missed my answer as well, so I'll say it again:

The 'SeaPower' machine is a pneumatic motor, a.k.a air-motor, in which air expands from one volume to another while doing mechanical work on a linkage. The SeaPower differs in constructional details from conventional air-motors by having a balloon (or inverted bucket) and liquid medium to confine the air, rather than a piston sliding within a cylinder or a rotor with sliding vanes mounted eccentrically in a cylindrical housing. The liquid serves as a mechanical linkage to transfer to the output shaft the force exerted by the boundary surface of the air, just as the piston and connecting rod, or vanes and rotor do, in the normal type of air-motor.

Intuitively, we can see that driving the shaft (using an external prime mover) so that the balloons descend, will cause the machine to function as a compressor. Absent any losses, it will compress the same volume of air to the same pressure as would be required to make an identical machine operate as a motor and deliver the same shaft torque and speed as output. Intuitively again, we can see that coupling the two machine shafts and air pipes together would result in equilibrium and all would remain stationary.

If, however, you want to ignore the obvious symmetry and prove the result numerically, you will first need to learn the basics of integral calculus (and I do mean just the basics.)
1. Learn calculus.
2. Find an expression in terms of depth for the force exerted on the belt by a balloon containing a unit volume of air.
3. Integrate this expression with respect to depth to obtain the work done on the belt by each rising balloon.
4. Compute the work done by a unit volume of air injected at arbitrary depth, and divide by the work done compressing that air against the head of water at that depth.
5. If the result is over unity, you probably made a mistake!

Anyone who is still reading at this point might have sufficient interest in water-pistons as to be familiar with the Humphrey pump. And if you aren't, you should be. I don't know where in Texas @justcurioustwo is located but one of the few Humphrey pump installations in the world was in Del Rio TX. The Chingford installation is just ten miles from here.
Humphrey Pumps Wikipedia article
 
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