Hi Rizzi. Clausius has the closest correct answer here. This is the same affect as hydroplaning as mentioned before. The same principal is applied to the main crankshaft bearings in a car where oil is injected at a higher pressure and the crankshaft rotates on a cushion of oil.
The weight of the ball can be much larger than the water displaced if you have dynamic pressure under the ball. The ball doesn't need to be hollow, and it is not 'floating' - the pressure of the water isn't simply rho*g*h. The pressure is greater than that. Water is being forced in under pressure, creating a cushion of water. The water then escapes up between the ball and bowl and in so doing, there is an unrecoverable, frictional loss of pressure, so the pressure along the rim ends up being close to atmospheric pressure while the pressure at the bottom where the water is injected is much greater than Patm + rho*g*h.
To calculate the pressure and flow for a pump using this model would be bit more complex than I'd like to get into here. Maybe someone else has the derivation but I don't, and it wouldn't be particularly easy to derive for this case. Besides which you would need make a lot of assumptions such as the bowl and ball surface being parallel, which they aren't likely to be.
If you wanted to make your own fountain, I'd think most pumps would have sufficient pressure capacity, I doubt you need more than about 5 psi or so. The issue is how much flow.
- I think there's a trick to making this work well. Imagine the ball floating above a water filled cylinder, but the ID of the cylinder is slightly smaller than the OD of the ball so the ball doesn't quite fit into the cylinder but is very close. The weight of the ball then can be calculated, subtract the weight of water it displaces, and the remaining pressure needed to hold the ball above the cylinder is the pressure you need. The HEIGHT the ball floats at (above the cylinder) is determined by flow, the more flow the higher it goes up, but pressure would go up only slightly inside the cylinder.
To determine the 'height' above the cylinder rim the ball would float then, is simply a function of flow rate. The gap between the cylinder and ball times the circumference gives you a flow area. Then you need a single pressure drop calculation to determine flow velocity and you'll get the volumetric flow rate.
The ball in cylinder model would work a lot better than the ball in bowl model since it eliminates the need to do a pressure distribution calculation. You can make the bowl look like a bowl, but simply pinch down around the circumference. Such a design would eliminate the need to do some really nasty calculations. If you're serious about doing this I could provide you the equations.
