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 11 Guests are viewing this topic.

mondrasek

Quote from: MarkE on March 06, 2014, 11:06:37 AM
This suffers from the tyranny of the:  N*X/N2 = X/N problem.
The other problem that you have is that it requires head to support the payload weight.  You have to pick a payload that is less than the initial uplift force, because the uplift force declines steadily towards zero with lift.  If you pick a payload weight that is the same as the uplift force then the thing never moves.  If you pick one slightly less as you propose then it moves only slightly.

MarkE, the initial payload is slightly below the maximum buoyancy Force in State 2.  And the payload also declines linearly as the system lifts up.  The payload "slug" of water does not sit on the piston at the end of State 3.  It has been allowed to run off unimpeded across the top surface shown that includes the bore.  The correct Energy equation reduces to .5*Fmax*S.

MarkE

Quote from: mondrasek on March 06, 2014, 11:28:21 AM
MarkE, the initial payload is slightly below the maximum buoyancy Force in State 2.  And the payload also declines linearly as the system lifts up.  The payload "slug" of water does not sit on the piston at the end of State 3.  It has been allowed to run off unimpeded across the top surface shown that includes the bore.  The correct Energy equation reduces to .5*Fmax*S.
You are still subject to the N*(X/N)2 tyranny.  Let's say that you set-up your payload such that it runs off at the rate of 64.649N/m to match the rate at which the riser lift force runs down, then the  Wpayload = 0.30078N - 64.649N/m.  Now perform the integral math on the work you impart doing the lifting:

E = integral( F*ds )
F = 0.30078N - 64.649N/m
integral from 0 to 4.653mm is:
0.30078N * 0.004653m - 0.5 * 64.649N/m * (0.004653m)2 = 0.7mJ

The internal change in energy from State 2 to State 3 was 1.024mJ.  So you are still stuck with a 30% loss.


TinselKoala

Quote from: mondrasek on March 06, 2014, 10:39:27 AM
TK, regarding your "peanut oil" comment earlier (sorry, but I'm too lazy to go back and find it).  I was wondering if you, or anyone else, had experience with Pentane?  I did a little bit of research on a low and high density fluid and had settled on Pentane and an Aqueous Sodium Polytungstate solution for candidates to use in a production level ZED system.  Only downside to Pentane though (that I can tell) is it is flammable.  Any thoughts?

http://en.wikipedia.org/wiki/Pentane

http://en.wikipedia.org/wiki/Sodium_metatungstate
Pentane is flammable, volatile, toxic and carcinogenic probably. Sodium polytungstate? Man, you are weird. How about using galinstan and kerosene?





Or just a big spring and a rock.

mondrasek

Quote from: MarkE on March 06, 2014, 11:36:05 AM
No, the payload is a fixed mass.  It does not decrease as it moves up.  A payload just less than the State 2 lift force barely moves.

Incorrect.  It is the case I explained and that you address below.

Quote from: MarkE on March 06, 2014, 11:36:05 AM
If you create some arrangement where you lift the payload slightly and then it falls off of a cliff, then you need to account for the loss in PE that occurs when you do that.  You cannot pretend that you lift what you drop, or that you lift what runs off.  No matter how you slice you have to perform the integral math.  And no matter how you slice it, when you convert potential energy stored in some device into potential stored in multiple devices, you get screwed by the N*X/N2 problem.

There is no need to be concerned with where the exiting Energy goes off to once it is properly calculated as leaving the system.  The water could "fall off a cliff" or sit on top of an infinitely large surface.  It does not matter.  The only thing that matters is that this method allows for a correct calculation of the Energy that leaves the system during the transition from State 2 to State 3.

TinselKoala

Quote from: minnie on March 06, 2014, 06:30:19 AM

   Question,
                 how much does it cost to pump a little bit of "virtual water"?
                                                      John.

Thanks for reminding me. I wuz gonna post this last night but my uplink went down and I fell asleep. Here's a Little demonstration of "virtual water" that I made back in 2012. Note that there is _real data_ being displayed here.

The term "virtual water" has been used for a long time, but in a different context. It refers to the water involved in producing and processing and delivering food to the table. Like how much water do the cows drink, how much water did it take to grow the grain to feed the cows, etc , per hamburger patty served up to your kids. It is a concept very familiar to those who actually work for a living, growing our food.   ;)

However, the term "virtual water" seemed like a very obvious way to describe what happens with these nested buoyancy things. It's a result of displacement, it is buoyancy, it is gravity, it is looking at the inside of the pod chamber instead of the outside. It is the key to the "inverted Travis effect". It's still a Red Herring, though.

My virtual water demonstration. Take note of the scale readings at each step, and the water levels. Uploaded August 13, 2012.

http://www.youtube.com/watch?v=qFjqBaH_NWU