Modified Heron's Fountain

[johnny874]

tk,
u r just like alan. I try discussing something and he jumps the thread.

adding; since u say u don diablo dont wont mi work, y u bugin' on me fool ?

[TinselKoala]

tk,
u r just like alan. I try discussing something and he jumps the thread.

adding; since u say u don diablo dont wont mi work, y u bugin' on me fool ?

Every time you mention my name or initials or direct a flame comment at me, Jimpbo, until you apologise and retract your lying libellous accusations and your ridiculous mischaracterisations of me, I will "jump" in and point out that you are a liar and that you have indeed libelled me, outside this forum, with no provocation. You will face the consequences of that, one way or another, you can be sure of that.

adding; since u say u don diablo dont wont mi work, y u bugin' on me fool ?

Are you drunk, or high on drugs, or what? You really need to settle down, Jimpbo, and get a grip on yourself.

[johnny874]

  tk,
I thinm your response is quite lucid to the knowing mind.

    Jim

[johnny874]

Priceless. Surface area is now measured in cubic inches. Who knew?

Now we have the secret of overunity.

>>  You cannot support your claim with any evidence and in fact the EVIDENCE indicates that you probably saw my videos before you even thought of your "idea".   <<

pg. 6,   
Re: Heron's Fountain  « Reply #85 on: Today at 02:10:36 PM »   http://www.overunity.com/12640/herons-fountain/75/     


In reality everyone, if you have a surface area of 4 square inches with 14 psi of atmospheric pressure acting on it and the output is 2 square inches of surface area in the compression chamber, the potential of the pressure head becomes 28 psi.  And if the opposing force on the other side of the barrier is less that 28 psi, according to hydraulic theory, the pressure head with the larger surface area should win out.  What does need to be considered is the KE of the pressure head, ie., how much does it weigh ? might not matter much, after all, if 14 pounds were a tube 1 inch in diameter, it could only generate 28 psi. it's own mass plus atmospheric pressure.  Of course, some people don't care to do the math because they find ridiculing people to be much easier, that just requires a keyboard and internet access.  Of course, with 28 psi, the area of the compression chamber needs to be considered. if it has 3 square inches of surface area, then the compression chamber should have about 18 psi of air pressure. That should be able to pump water continuously. There would be about 4 psi (the difference between atmospheric air pressure and the air pressure in the compression chamber) to play with.  Of course, when Stefan bans me, then the trash will have been removed   

edited to add; the consideration for this idea came about by considering how hydraulic theory applies to Heron's Fountain. He readily demonstrates that by seperating pressure heads that the more voluomous pressure head performs work and the less voluomous pressure head responds by creating Heron's Fountain.
I merely substituded a lower elevation for a greater (larger) diameter for the more voluomous pressure head and in doing so was able to consider both pressure heads working in the same compression chamber.

[Pirate88179]

Jim:

It is painfully obvious from your post that you do not understand the gravity of the situation.

Bill