Hydro Differential pressure exchange over unity system.

[LarryC]

I know this is repeating, I presume you have not read the whole thread - no problem.

For those that don't know,you can click on the mrwayne name in the left most area. This will bring you to the mrwayne screen, where you can click on the show post option and get a list of all of his post.
Larry

[Magluvin]

Every time I use spell check it deletes my post -
Hello Mags,
Not for 'just the patent search' - Those were my fee's so far - we still have much more to go.
I also did my own search  - when it came time to really sink my life into this, I had to make sure I was not treading on others work.
Thanks Wayne

I was just guessing. ;]  Im sure there are much higher prices than 40k for some patents, and some less, search n all.  I could only imagine. 
Do you only have 1 pat. ?

Mags

[KanShi]

I have physics/mathematical proof that harnessing "the Travis Effect" (AKA Archimedes' principle) can't generate energy that I posted over @ PESN:
Higher water pressure means lesser volume of air which in turn means higher net density which in turn means lower uplift. System where water exerts pressure on air (open chambers / flexible seal) is less efficient than a system where the chambers are completely enclosed so water cannot compress air at depth - so let's use closed chambers(cups) to maximize efficiency:

g = 9.80665 m/s^2;
ρ(empty cup) = 7850 kg/m^3 (carbon steel - using advanced alloys (1800 kg/m^3), you can get slightly better efficiency);
ρ(water) = 1000 kg/m^3;
ρ(air) = 1.2041 kg/m^3 (at 20 °C and 101 kPa);
-------------------------------------------------
V(air/water) = 1.0552 m^3, r(tank) = 0.63156 m (sinking tank);

m(air) = V(air)·ρ(air) = 1.0552 * 1.2041 = 1.27056632 kg;
m(cup) = V(empty cup) * ρ(empty cup) + m(air) = 0.18 * 7850 + 1.27056632 = 1414.27 kg;
m(displaced water) = 4000 kg;

d(cup) = 0.96 m;
d(tank) = 6.2231 m;
F(cup) = Fb - Fg = (4000 - 1414.27) * 9.80665 = 25357.35 N;
W(cup) = F·d = 25357.35 * 0.96 = 24343.06 J;

1) Using heavy tank, using energy to lift:
m(empty tank) = V(air)·ρ(water) - m(air) = 1054 kg;
W(tank) = m(empty tank)·g·d = 64323.26 J;
2) Using light tank, using energy to sink:
Fb(tank) = (V(air)·ρ(water) - m(air))·g = 10336.1037 N;
W(tank) = Fb(tank)·d =  64322.6 J;
3) On-demand air compression:
W = pB·vB·ln(pA / pB) = 151584 * 1.0552 * ln(101325 / 151584) = 64429.52 J;
where pB is water pressure @ 5m (151584 Pa), pA is ambient air pressure (101325 Pa) and vB is the compressed volume;

P.S.: without the piston/insert, the device would require about 4 times as much energy (over 244 kJ) each cycle which means less than 10% efficiency.
P.P.S.: too bad you can't just magically turn the buoyancy on and off without energy, it would work very well then
(actually, you can turn it off easily, but then you consume energy to turn it on again, and vice versa)

[mrwayne]

I have physics/mathematical proof that harnessing "the Travis Effect" (AKA Archimedes' principle) can't generate energy that I posted over @ PESN:
Higher water pressure means lesser volume of air which in turn means higher net density which in turn means lower uplift. System where water exerts pressure on air (open chambers / flexible seal) is less efficient than a system where the chambers are completely enclosed so water cannot compress air at depth - so let's use closed chambers(cups) to maximize efficiency:
P.S.: without the piston/insert, the device would require about 4 times as much energy (over 244 kJ) each cycle which means less than 10% efficiency.
P.P.S.: too bad you can't just magically turn the buoyancy on and off without energy, it would work very well then
(actually, you can turn it off easily, but then you consume energy to turn it on again, and vice versa)

Hello Kanshi,
This looks just like the post from another "user" early yesterday and then again under your name - on two of Sterling's links.

As I demonstarted in long detail on that site - your application do not correspond with our system.

To be very clear - We have built several fully functioning models that have been tested, third party.

It would be better to spend your time understanding the system - and impacting the world.

Thanks Wayne

[KanShi]

As I demonstarted in long detail on that site - your application do not correspond with our system.

Can you or one of your engineers point out exactly where the equations do not correspond? Because the equations support exactly what you show on those "Travis effect" videos, they just don't support your claim of overunity for the whole device.