Marble has a density of 0.098 lbs/in3. So a 500lb sphere has a volume of 500lb/.098lb/in3= 5102 in3. That gives a radius of (5102 in3 * (3/(4pi)))(1/3)= 10.68 in.
Let’s say the bowl it sits in has 2/3 the radius. So 7.12in. That makes the enclosed surface area pi*( 7.12in) 2= 160in2. So 500lbs/160in2 = 3.14 psi. How fast do you think the water will spray out at 3.14 psi? Without any drag, out of an ideal nozzle, it would be 6.5mph. However, the pressure will actually be higher in the middle and tapers off to 0 at the edge thanks to the flow diverging and drag. So really the pressure drop at the opening is even smaller.
There’s a lot of lifting area which means you don’t actually need much pressure. You can lift entire buildings with large water bags.
Pressure is measured across surface area. The same way buoyancy is measured by water displacement. A larger object has more surface area. 1 PSI of pressure won’t be very much if the object only has 1 square inch of surface. But it would be a lot of pressure if the object has 500 square inches. 30 PSI is enough to keep your car off the ground. A few PSI would be enough to lift a 500lb rock if the surfaces were prepared properly.
For every weight of ball (within a certain density limit. Won’t work with a black hole, for instance) there’s some pressure and volume of water that will balance the weight and size of the ball perfectly to create a thin layer around the entire surface. Once the ball becomes too dense, there’s no volume/ pressure combo that would do the same (the weight would require too much pressure by volume of water, so it would either stop it or fly full force by while barely levitating the ball). But I don’t think whatever density that would be is found in pressures you could find on earth
Now do a marble that weighs 500 lbs.
It’s gonna spray
Why ask a question if you think you know better
Marble has a density of 0.098 lbs/in3. So a 500lb sphere has a volume of 500lb/.098lb/in3= 5102 in3. That gives a radius of (5102 in3 * (3/(4pi)))(1/3)= 10.68 in.
Let’s say the bowl it sits in has 2/3 the radius. So 7.12in. That makes the enclosed surface area pi*( 7.12in) 2= 160in2. So 500lbs/160in2 = 3.14 psi. How fast do you think the water will spray out at 3.14 psi? Without any drag, out of an ideal nozzle, it would be 6.5mph. However, the pressure will actually be higher in the middle and tapers off to 0 at the edge thanks to the flow diverging and drag. So really the pressure drop at the opening is even smaller.
There’s a lot of lifting area which means you don’t actually need much pressure. You can lift entire buildings with large water bags.
This guy.puts his physics where your opinions are
Bigger marble, bigger hose. It’s the same principle scaled up.
This one weighs 9 tons
https://en.wikipedia.org/wiki/File:Reaching_for_the_Stars.jpg
The variables in this equation are:
Along with some friction outliers, these can be adjusted to find an equilibrium.
Pressure is measured across surface area. The same way buoyancy is measured by water displacement. A larger object has more surface area. 1 PSI of pressure won’t be very much if the object only has 1 square inch of surface. But it would be a lot of pressure if the object has 500 square inches. 30 PSI is enough to keep your car off the ground. A few PSI would be enough to lift a 500lb rock if the surfaces were prepared properly.
Lol
For every weight of ball (within a certain density limit. Won’t work with a black hole, for instance) there’s some pressure and volume of water that will balance the weight and size of the ball perfectly to create a thin layer around the entire surface. Once the ball becomes too dense, there’s no volume/ pressure combo that would do the same (the weight would require too much pressure by volume of water, so it would either stop it or fly full force by while barely levitating the ball). But I don’t think whatever density that would be is found in pressures you could find on earth
It’s all about the pressure; more weight needs more pressure but they tune the pressure to match the weight of the ball.