Buoyancy/gravity wheel - another approach?

[Low-Q]

Hi,

I have made a drawing of an idea I had the other day. If it was possible to shield water in the right part of a container so it can be air in the left side, how would a buoyancy wheel act then?

I know there is harder to push ball E into the water than it is for ball A, due to the difference in pressure. But will the buoyancy of the balls F, G and H (Or more if more balls was available) be enough to force ball E (minus force of ball A) into the water?

Vidar

[onthecuttingedge2005]

B, C, D, E will 'resist' going into the water, this falls under the path of least resistance law which will always center itself out and settles into a zero kinetic potential.

[onthecuttingedge2005]

If you really wanted a wheel that runs on gravity then you must also understand that Gravity is not a constant around the world, different places have different gravitational influences.

if you built a rather large ordinary wheel that borders the blue gravity area vs. a heavy gravity environment in red then one side of that wheel will be heavier than the other side causing the wheel to perpetually rotate to the path of least resistance.

on the gravity map, in the blue areas objects fall slower than objects dropped in the red area.

32ft per sec squared only counts in the area it was calculated from, it does not hold true entirely around the world, there is a spot in the Indian ocean that has the least over all ratio of gravitational influence.

after you have fumbled with this concept in your head then tell me why it won't work as planned, 'remember the path of least resistance law'

Jerry

[Low-Q]

Only position E will be the counter energy with an average energy of the balls D_max x cos45^o x Depth_E-A

If the spheric balls was 1 litre. The maximum area is 250 cm^2. If Delta_E-A = 1m, the maximum counterforce is approx 250N when ball E is half way in. The average force is 250N x cos45 = 177N for the whole ball to enter the water. The distance D_max it must travel is 0,178m to be totally in the water. The total energy is then 177N x 0,178m = 31,5.

Then the three other balls F, G, H are travelling the same distance, but ball F and H are at average 45 degrees angle. What is the energy they provide?
If ball F and H are raising each approx: 0,178m x cos45 = 0,125m
In sum this is equivalent to 10N (About 1 litre) in 0,25m.
Ball G is raising approx 0,178m. The weight is 10N.
The total buoyancy energy is then (0,25m x 10N) + (0,178m x 10N) = 4,3

Ball E wins with 31,5 - 4,3 = 27,2 (something)

If we store that exsess energy in a fast spinning metal disc (Like the ones in some toys), should that energy be enough to keep the wheel to rotate in the opposite direction than I intended? ...so the balls at position F, then G, then H could provide energy each time they left position E and into the air?

I guess I'll not get a nice sleep tonight...too much thinking

Any thoughts @onthecuttingedge2005?

[onthecuttingedge2005]

lets say you weigh 200 pounds, take one soccer ball and one pool, the object of this mission, try to get the soccer ball down to the 12 foot level and touch the bottom of the pool.

don't hold your chin out to far or the ball might smack you a good one!

Sincerely
Jerry