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If we want to float an object of 1KG in the air near the surface of our earth (the object can move slowly but shouldn't move very fast like a flying plane), at least how much power do we consume? I don't know what method to float the object and I don't restrict this. We can use any proper methods, for example, install an engine in the object; but don't use a pole to support the object. It is the interaction between the object and the air that floats the object.

Note that the air buoyancy is neglected and not considered.

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  • $\begingroup$ Are superconductors allowed? Or should the object interact with the air to keep floating? $\endgroup$
    – NoMorePen
    Commented Mar 17, 2015 at 11:13
  • $\begingroup$ Balloons do not consume any power (except to produce the H2 or He to fill them with, and not counting hot air balloons). $\endgroup$
    – Jon Custer
    Commented Mar 17, 2015 at 14:12
  • $\begingroup$ How near the surface? If the surface is really flat, like a polished granite table, and if the weight rests on a plenum of wide area, with a 1-micron gap around the edge, then it will take very little energy. The only energy it would need is to keep up some pressure to replace the bit of air that escapes. $\endgroup$
    – Mike Dunlavey
    Commented Mar 18, 2015 at 0:57
  • $\begingroup$ @NoMorePen: Superconductors are allowed, because we have to lay superconductors on the ground and the whole system interact with the surface of the earth, not only with the air. That is, we only allow the object to interact with the air to keep floating. The helicopter is one allowed example. $\endgroup$
    – adamant07
    Commented Mar 18, 2015 at 3:03
  • $\begingroup$ @Jon Custer: Balloons should not be considered. $\endgroup$
    – adamant07
    Commented Mar 18, 2015 at 3:05

3 Answers 3

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A helicopter can soar using very little power, if its rotor is big and pushes down a large mass of air with very low speed. The limitation is related to strength of a large and low-mass rotor. For example, a helicopter (a quadcopter, actually) using human muscles only was demonstrated recently (https://www.youtube.com/watch?v=syJq10EQkog )

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  • $\begingroup$ This is a wonderful video. Based on akhmeteli's explanation, I think the final answer of my question is approximately 0. Anybody disagree? $\endgroup$
    – adamant07
    Commented Mar 18, 2015 at 3:16
  • $\begingroup$ The above is not a perfect answer, because the device is not feasible (too big). I hope the size of the whole system (the object + engine) is reasonably small, e.g., 0.01~0.1 m^2. $\endgroup$
    – adamant07
    Commented Mar 18, 2015 at 3:23
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I'm assuming that aerodynamic forces are the only allowed sources of lift.

Aerodynamically, lift is obtained by redirecting airflow, which is essentially to accelerate air molecules. When you accelerate a mass $dm$ to speed $v$, you gain a momentum $dp$ given by

$$dp = v dm$$

Divide both sides by $dt$ to obtain the force:

$$F = v \frac{dm}{dt}$$

In order to hover the force needs to equal the weight of the flying machine,

$$Mg = v \frac{dm}{dt}$$

Now consider the work done by the engine to accelerate the air molecules:

$$W = \frac{1}{2}dm v^2$$

$$\rightarrow P = \frac{1}{2} \frac{dm}{dt} v^2$$

I assume that $v$ and $\frac{dm}{dt}$ remain constant. If that bothers you, replace these quantities with suitably defined averages.

Now we can combine everything to find:

$$P = \frac{1}{2} Mg v$$

So as long as we design our flying machine with $v$ as small as possible (and, therefore, a mass flux as large as possible), we can make the steady state power as small as we like. The only constraints come from engineering considerations that limit how large a craft can be made at a given weight, material strength, parasitic drag which I've neglected, etc.

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If using an engine is allowed, then similar to the concept of an actual rocket, we can have Thrust just greater than or equal to weight. Thrust will be equal to rate of mass flow * exhaust velocity f gas from the engine.

T= dm/dt * v(exhaust) = mg

EDIT :- P = TV = (dm/dt)vv = (m(initial) -(dm/dt)*time)*g

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  • $\begingroup$ What I need is how much power (in Watt), not force or Thrust. I care the energy to be consumed. $\endgroup$
    – adamant07
    Commented Mar 18, 2015 at 3:19

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