Gravity and water (oil) wheel 2010, lets consider...

[ChrzaniK]

Hello everyone 
I would like to present an old idea found on Youtube.com.

http://zapodaj.net/images/d270e598f928.jpg

Lets do some calculations:

Force-gravity   = Fg
Force-buoyancey = Fb
Force-pressure  = Fp
p=3,14159265   ~= 3,14
Pressure        =P


If Fg+Fb=Fp than stoped
If Fg+Fb>Fp than overunity
If Fg+Fb<Fp than emptying the tank

Density:

Density = q=m/V

Density of water at 20 C' = 998.2071 ~= 1000 kg/m3
Density of air at   20 C' = 1.2041 ~= 1 kg/m3

Balls:
Density of our balls should be greater than air and less than water.

qa < qb < qw
1 < qb < 1000

Lets set qb=500
Cedar has 485 Kg/m3
http://www.wzinternational.com/species.htm

Dimensions of the balls:

qb=500kg/m3
Rb=5mm=0,005m
Vb=4/3*p*Rb3
Vb=4/3*3,14*0,005*0,005*0,005= 0,0000005233(3) m3

Weight of the ball:
q=m/V
mb=qb*Vb
mb=qb*Vb
mb=500*0,0000005233(3)=~ 0,000262 kg = 0,262 g

Now forces:

Fg1=mb*g= 0,000262 * 9,81 = 0,002567 N of one ball
Fg=10*Fg1=0,02567 N of ten balls

Fb1=qb*g*Vb = 500 * 9,81 * 0,00000523 = 0,002565315 N of one ball
Fb=10*Fb1= 0,025653 N of ten balls

Fg~=0,025 N
Fb~=0,025 N

Force pressure:
P=Fp/S
Fp=P*S
P=qw*g*H
H=(from 10 to 11 of 2Rb)=~10*2*Rb=20*0,005= 0,1 m
P=1000*9,81*0,1=981 kg/(m*s2)
S~~=p*0,005*0,005=3,14*0,000025=0,0000785
Fp=981*0,0000785= 0,0770085 N

Fp~= 0,077 N

So Fg+Fb<Fp and it will emptying the tank

Waiting to show errors.
Thank You.

Greatings
Michael

[ChrzaniK]

In this situation Fp is 0,077 and it is constant,
but we can change the dimension of the balls.

Lets change the shape: 4/3*p*R3 to a roller p*R2*H.
H~=2R so Volume of the roller is Vr=p*R2*2*R= 2*p*R3

Roller and the ball is made of the same material.
Vb=4/3*p*R3
Vr=2*p*R3
4/3 vs 2       div / 2
2/3 to 1
The roller is greater then ball about 1/3, so the force is about 1/3 greater too.

Fr ~= Fp
And it STOP.

Thank You. 

[Cloxxki]

Take a watertank, with a tab significantly under the surface.
Or, just take a long hollow rod, till near the bottom of a bucket full of water. Does air flow naturally into the rod, bubbling up the water? No, it doesn't. How much effort does it cost you to get a bubble to come out, vs. what it could deliver in useful work?
I've seen and come up with several such systems (unless I misunderstand yours). You can't just enter the water blow the surface, there is pressure to deal with. You'll have to come up with something really smart to place any buoyant mass at the bottom of any tank.

[pese]

Possibly the balls will move the other way

Because the water pressure will press the in the tube
back an "out" the watertank WITH LOST of flowing Waer ! (!?)

Pese

[AB Hammer]

Possibly the balls will move the other way

Because the water pressure will press the in the tube
back an "out" the watertank WITH LOST of flowing Waer ! (!?)

Pese

Gustav

I do believe you are correct. It seems no one ever considers the fluid pressure when trying similar approaches.

Alan