# How to Make a Granite Ball Float on Water?

1. Apr 17, 2006

### rizzi

Last week I went to Ripley's Believe it or not and saw this gigantic granite ball floating on water. I was wondering how much the water pressure is and what kind of calculation we need to find out the water pressure. I wanted to create a small fountain that would make the ball float.

Thanks

2. Apr 17, 2006

### Staff: Mentor

They had one of those at Disneyland in Anaheim. Very cool looking. The big rock ball thing (probably hollow) was about 5 feet in diameter or so, as I remember. Even if hollow, it was too heavy to influence much with your hands. You could get it to roll a bit, but that took some effort.

The water was barely flowing out through the small seam between the ball and the big opening below it. I'm sure it had to be a pretty good precision fit. As for pressure difference, the water pressure up on the bottom of the sphere has to be equal to the weight of the sphere plus the air pressure down on the sphere. Well, more accurately the air pressure down on the top of the sphere, symmetrically oriented to the water area below. But I suspect that the weight of the sphere is the dominant factor. Plus if you specify the water pressure in PSIG, then the atmospheric pressure component goes away, I believe.

So what would the water pressure have to be to hold up a 400 pound hollow sphere with the bottom water area being about PI * 0.5m^2? Like the way I mix units?

3. Apr 22, 2006

### Clausius2

Sounds to me like a Hydrodynamic Lubrication effect. You can lift a very heavy ball with a well calculated thin film of water, with an outwards flow and a variation of cross section in the two dimensional channel formed between the partial hemisphere of the ball and the support. It's the same effect than hydro-planning.

4. Apr 23, 2006

### bobbytkc

it is very easy, the total upward force on the ball is exactly its weight, since it is floating and in equilibrium.

The pressure acting at any point on the ball by the water is not related to the ball at all. it is given by the equation pressure= (depth of point on ball)x (density of the water)x (gravitational acceleration)

5. Apr 24, 2006

### Q_Goest

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.

6. Apr 24, 2006

### Tzemach

You can perform a similar stunt by floating a ball on the exhaust vent of a vacuum cleaner (as long as you have a circular outlet). A tennis ball will float some inches above a good flow, neat trick looks cool.

7. Apr 24, 2006

### DaveC426913

So, the engineering feat to floating a large granite ball doesn't require complex calculations, it requires
1] a precision-fit ball-and-bowl, and
2] a water pump of adequate pressure.

As long as the pump is able to overcome the weight of the ball, you're OK.

8. Oct 28, 2008

### kasilogan

So just say I am writing and I am trying to figure out a believable way to physicaly knock this ball off of its perch and send it crashing through downtown Gatlinburg...which is a slope...I need believable ideas. Please help. :-D

9. Oct 28, 2008

### mgb_phys

You don't need that much pressure. It's a classic problem - the pressure only has to equal the pressure of the ball over the area of the inlet ( or something - it's been a while).
The same hydrostatic bearings are used for telescopes and ships prop shafts.
You can push a 200ton telescope around with a 1/4Hp electric motor if the oil bearing is running.
They aren't that difficult to make, if you rub a roughly spherical ball in a roughly spherical base they will both polish to the same radius. It'sthe same principle as making spherical mirrors - except easier since you don't care about the final exact radius.

ps. How easy it is to push the ball off it's mount depends on how far up the sides of the ball the mount comes. YOu ar ebasically trying to tip it over a wall that high.

10. Oct 28, 2008

### DaveC426913

I love reading posts wherein I've posted to them soooo long ago that I don't even remember them.

11. Nov 8, 2008

### kasilogan

Is that it? :(

12. Nov 8, 2008

### montoyas7940

Break the mount.

13. Nov 8, 2008

### montoyas7940

A big hammer ought to do it.

14. Nov 10, 2008

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