Overunity.com Archives is Temporarily on Read Mode Only!



Free Energy will change the World - Free Energy will stop Climate Change - Free Energy will give us hope
and we will not surrender until free energy will be enabled all over the world, to power planes, cars, ships and trains.
Free energy will help the poor to become independent of needing expensive fuels.
So all in all Free energy will bring far more peace to the world than any other invention has already brought to the world.
Those beautiful words were written by Stefan Hartmann/Owner/Admin at overunity.com
Unfortunately now, Stefan Hartmann is very ill and He needs our help
Stefan wanted that I have all these massive data to get it back online
even being as ill as Stefan is, he transferred all databases and folders
that without his help, this Forum Archives would have never been published here
so, please, as the Webmaster and Creator of these Archives, I am asking that you help him
by making a donation on the Paypal Button above.
You can visit us or register at my main site at:
Overunity Machines Forum



Mathematical Analysis of an Ideal ZED

Started by mondrasek, February 13, 2014, 09:17:30 AM

Previous topic - Next topic

0 Members and 67 Guests are viewing this topic.

mondrasek

Quote from: MarkE on March 04, 2014, 05:40:18 PM
Let's start with your stipulation that in State 1 there is no uplift.  Is that or is that not your stipulation?

Of course.  There is no head difference between the ID and OD surface on any riser.  The pod has no water in contact with it at all.  So all risers and the pod are being acted on by zero buoyant Forces.  Also, the sum of the buoyant Forces on all the risers and the pod is exactly zero.  There are zero Forces acting on the system and therefore zero motion would occur.

BTW, this is not a "stipulation."  This is a physical fact derived from the geometry and the assumption of incompressible fluids.

MT

Quote from: TinselKoala on March 04, 2014, 01:26:36 PM
[size=78%]Not only does the two units of fluid on the right support the entire 13 units on the left, when you remove the two units from the right.... the liquid level only goes down a fraction of that input starting head height. Therefore you have a "net" production that does not reduce the "input" by nearly the same amount.[/size]
Hi TK,
I like your OU U-tube but where is OU? You compare force (input head) with work (net production). Recovering initial head will cost same energy you gained.
have a nice day,
MT

mondrasek

Quote from: webby1 on March 04, 2014, 06:16:53 PM
MarkE is referencing the surface area of the bottom of the risers being a place where force can be exerted, as well as the volume that the down-tube of the risers occupy.

Webby1, there is no resultant lift Force due to the water pressure on the bottom surface of the risers.  If you tried to push a riser upwards from the bottom surface of the risers you would be defeated by the requirement that the volume of one of the fluids in the system would have to change.  Since that is impossible, there is no resultant lift Force.  The system as drawn is in complete equilibrium.

Magluvin

Quote from: mondrasek on March 04, 2014, 05:55:50 PM
The incompressible air has an ASSUMED Specific Gravity = 0.



What air is being used that is 'incompressible?? ???

Mags

MarkE

Quote from: Magluvin on March 04, 2014, 07:24:05 PM
What air is being used that is 'incompressible?? ???

Mags
The "air" that Mondrasek stipulated as an incompressible, massless fluid is being used in the model.