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I'm writing a sci-fi story in which levitation is a thing. Newton's Third Law still holds, so if a 10kg object with a $0.5m^2$ footprint was levitated, the floor below it would feel approx $mgA^{-1}\approx 296Pa$ of pressure when stationary.

The same levitation force would not hold this object in place over water, since the water would be displaced, energy being carried away by sideways movement of water. The water would move downwards due to the levitation force, and the levitating object would experience this as the floor dropping away.

The system is roughly equivalent to a hovercraft without a skirt (height above water would be approx 1m).

Obviously the interactions here are very complicated and turbulent, and an accurate answer would need a detailed physical simulation, but I would like to know how we might model this system to obtain a rough order-of-magnitude estimate for the following:

How much force/power would be required to levitate a mass $m$ object with footprint area $A$ over a body of water?

What kind of model is approximately valid for, say, $50kg<m<1000kg$, $0.1m^2<A<1m^2$?

How quickly will water be displaced (i.e. how much spray would we observe)?

How do the model and figures change when the water is flowing up to, say, a few metres per second?

A very rough estimate would be fine.

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  • $\begingroup$ This description is heap unclear. An object is levitated, OK. Then "the floor below it" would experience a force unclearly related to the weight of the object. (296 Pa acting on 0.5 m² is 148 N, which is about 15.1 kg-force.) What is this "floor below it"? Is it an object chosen by the magician? Is it somehow determined by the non-explained rules? Furthermore, you say that if the object is levitated above water, then "water would be displaced". What specific water would be displaced? The entire column of water down to the bottom of the Mariana Trench? Why does water count as a "floor"? $\endgroup$ – AlexP Nov 30 '20 at 0:34
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    $\begingroup$ Spraff, please remember you should be asking only one question per post. Multiple questions is a reason to close (Needs More Focus). Could you edit your question to focus on a single question? Note that over calm water, the levitation force is irrelevant. You're displacing as much water as the weight of the levitated object would if it were in the water itself. More or less levitating force doesn't change the issue of displacement. The question really isn't how much will be displaced but how it will be displaced. That depends on the design of the emitters. $\endgroup$ – JBH Nov 30 '20 at 0:47
  • $\begingroup$ I think you rprenise is wrong. It is like saying a helicopter will displace the water below it equal to its weight. Even if it is a thousand feet in the air. It is Newtons Laws on the helicopter itself, not the force of the air pushing on the water, that keeps the helicopter in the air. Same with a hovercraft. Same, I presume, with anything that is being levitated. The levitating media, reactive material, whatever it is, works against the object, not the water. $\endgroup$ – Justin Thyme the Second Nov 30 '20 at 4:16
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    $\begingroup$ @spraff Yes. SE's model is one-specific-question/one-best-answer. We're lenient, but answering this question as it deserves is an 8-10 page dissertation. Closely-related isn't relevant. In fact, querents often discover that the follow-up questions change when they get the answer to the first question because they thought they had everything covered --- but didn't. $\endgroup$ – JBH Nov 30 '20 at 20:41
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    $\begingroup$ I suspect a typo in that force calculation. 10kg would require a force of 98 Newton to support. Over a 0.5m² surface, this is 196 Pa, not the stated 296 Pa. Looks like a simple keyboard error to me. $\endgroup$ – PcMan Nov 30 '20 at 22:00
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Your object will float atop its magic skirt.

Consider the case for 1 meter squared area. The skirt is also a meter deep. This is a cubic meter of air which weighs 1.2 kg. That volume will displace water weighing 1000 kg.

Your object atop its skirt will settle down into the water until the water displaced by its air skirt is equal in weight to the object + a trivial amount for the air in the skirt. It is a boat, except the air volume giving it buoyancy is not enclosed by a hull but by magicks. It might rest right on the water; the case for the biggest object. It might stop descending at a height above the water.

You can figure out how deep it will stop by taking the mass of your object, then taking the volume of the same mass of water with the cross-sectional area of your skirt. That is the water it will displace.

As it settles down there would be a rush of water outward, just as when a boat is lowered into the water.

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    $\begingroup$ The magic hull needs to exert force sideways against the water to keep the hole in the water from collapsing. Assuming levitation cannot do sideways forces, only downwards, water would get pushed down and refilled by water from the sides, thus creating a convection like this which might qualify as "complicated and turbulent interactions". And while a boat doesn't do work, a levitation spell that creates a convection vortex does, so asking for a power estimate becomes meaningful. $\endgroup$ – nwp Nov 30 '20 at 9:44
  • $\begingroup$ @nwp - wouldn't the need for sideways force also apply when levitating over dry ground? Air would get pushed down and refilled with air from the sides. $\endgroup$ – Willk Nov 30 '20 at 16:37
  • $\begingroup$ "As it settles down" the premise is that enough force is produced to prevent it from settling down. $\endgroup$ – spraff Nov 30 '20 at 20:29
  • $\begingroup$ @Willk It doesn't work well with air. Air weighs about 1 gram per liter which means the 1 ton object has to create a vacuum of a 10m*10m*10m cube in size to levitate. That's a really big cube of no air. I would assume the levitation spell ignores air and just pushes against the ground directly without displacing any air and for that it doesn't need sideways force. $\endgroup$ – nwp Dec 1 '20 at 9:15
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Your levitating object is:

A BOAT

You have just replaced a physical hull with a "levitation force" hull, but the effect on the water will be exactly the same.

Just figure out how a physical hull with the shape and form of your levitation field will look. If the levitation field has a hard distance-boundary, then use a hard hull. If the levitation force is a more squishy force, spreading out and decreasing with distance, then envision your hull as a partly inflated balloon shape.

The Volume of water displaced will exactly match the total mass of your vehicle. At a very easy ratio, 1 metric tonne of vehicle displaces 1 cubic meter of water. (very slightly less if seawater)

Just because it's not made out of matter, does not mean you cannot treat it exactly the same way.

As for spray.. If your levitation mechanism only pushes against the WATER, there will be absolutely no spray at all. The water will not gush away, it will push out once to a distance where it buoyantly supports the weight of the vehicle, and then just sit there.
If your levitation mechanism pushes against everything below it, both air and water, there will be some spray. Possibly 5% as much as for a hovercraft.
Only if your levitation is achieved by pushing only against the air, will you get a glorious spray of water, identical to a hovercraft. (which, if unskirted, requires some serious engines and generate a ludicrous spray)

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I'll attempt to explain some of the issues with the model posed in the question and some potential workarounds.

I'm writing a sci-fi story in which levitation is a thing. Newton's Third Law still holds, so if a 10kg object with a 0.5$m^2$ footprint was levitated, the floor below it would feel approx mgA−1 ≈ 296 Pa of pressure when stationary

The assumption here is that the "floor" is supplying the force(s) which keep the object in the air. IRL the floor supplies the force on the object which counteracts the gravitational force through very short range contact forces.

Levitation on the other hand does not use contact forces. In fact it can be produced by several forces IRL. Thrust for example involves a force produced by using Newton's Third Law in which mass is accelerated in the opposite direction to the desired force reaction. Thrust is the reasons for which rockets work both in the atmosphere and in a vacuum. Note if thrust is used to keep the object levitated it is irrelevant what the exhaust is hitting since the exhaust itself is what keeps the object levitated, not the action of the exhaust hitting anything.

Aircraft on the other another hand rely on forces which result from the pressure differentials produced by moving airfoils through fluid-like media. Helicopters, for example hover over land and water just fine; they are held up through the action of the blades rotating through air which results in net force upwards on the blades which keeps the helicopter from falling. They are not kept in the air from the surface exerting a force upwards. Stand under a helicopter which is hovering overhead at 8000 ft and you will certainly not feel the weight of the helicopter.

Buoyant forces on the other hand are produced from the displacement of a volume of fluid caused by gravitational force on an object placed within the fluid. Buoyant forces are essentially contact forces in fluids. The magnitude of buoyant force is equal to the magnitude of gravitational force on the mass of the displaced fluid which depends on the density of the fluid.

So the question is, which force is causing the levitation?

The same levitation force would not hold this object in place over water, since the water would be displaced, energy being carried away by sideways movement of water. The water would move downwards due to the levitation force, and the levitating object would experience this as the floor dropping away.

This would only be relevant for buoyant-like forces. But buoyant forces rely on contact forces, so there must be some mechanism by which this "action at a distance" behaves like a contact force. If this is the case see @PcMan's answer, the object would behave like a boat, but for some reason be floating in the air above compressed water. There wouldn't be water spray do to the very short range contact forces involved.

If the levitating force behaves like thrust or a generated pressure differential in the air then what the water does below the object is not a concern.

How much force/power would be required to levitate a mass m object with footprint area A over a body of water?

Same minimum force required to levitate an object anywhere (mg). The power required depends on the mechanism which needs to be described to be calculated:

If contact forces are used then the power is zero (a box on top of a building does not expend any energy to stay there, neither do boats expend energy to float). If thrust is used then the energy required to produce the thrust which can vary greatly depending on the mechanism (see jet engines vs rockets). The lift produced by airfoils is dependent on the speed of the airfoil through the air and the angle of attack. The energy required to create these conditions depends on a lot of factors (see helicopters and airplanes). Note that hovercraft rely on built up pressures and thus are effectively are held up by contact forces.

Conclusion

Without knowing the mechanism by which the object levitates, the dynamics of the system can't be described. Levitation puts constraints of the force which the object must experience via Newton's second law, but does not explain how the force operates and how that force will affect the surroundings.

From what it sounds like you want (from the talk of water spray etc.) however, is a system in which a thrust like force results in levitation. To see the problems this poses, for a really simple model, the "exhaust" might be evenly spread over an area directly below the object. For example, if air one meter under the object is "magically" pulsed downward by the object to produce thrust with sufficient force to keep the object levitated, the acceleration of the air would calculated as follows:

$\rho V a = mg \quad \rightarrow \quad a_{air} = \frac{mg}{\rho V} = \frac{10 \cdot 10}{1 kg/m^{3} \cdot 0.5 m^{3}} \approx 200 \frac{m}{s^{2}}$

Clearly, using air would be a bad idea and would cause quite a splash (note this displacement of air would have to happen continuously for the object to float). You could increase the density of some downward ejected material but where would the material come from?

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