# Amount of energy required to hover.

I've noticed a motionless kingfisher over a lake looking for prey and wondered what amount of energy does a bird, weighing 0.15kg, require to hover for 15s?

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In the physics 101 sense it requires no work (i.e. energy expended) to hover. Think about it (and see Why does holding something up cost energy while no work is being done?). So what you've got here is a question about biomechanics. –  dmckee Jun 2 '12 at 20:19
Is this so? Certainly not in a more general sense. Say it wasn't a bird but a helicopter. Does a biomechanics analysis allow me to determine the amount of gasoline consumed. –  user9590 Jun 2 '12 at 22:56
Understand that the amount of physics 101 "work" done holding the bird or helicopter or whatever in place is the same if it is hovering on wingpower or hovering rotor power or sitting on a pillar. And that is zero. This is not a fault in your understanding of how the world works, but a difference in the day to day meaning of "work" and the one where $W = \int \vec{F} \cdot \mathrm{d}\vec{x}$. Ron's answer shows the way out of this dilemma: analyze the forces on the thing doing the supporting. –  dmckee Jun 2 '12 at 23:05

I understand what it takes to hover with wings but how would an animal or a person (like Superman) hover without wings? I know that the definition of flying is falling and missing the Earth, but how do you hover over the ground when you don't have wings and you have gravity working against you. As you can probably tell I just watched Man of Steel recently and it is an okay movie, but I want to know the physics of how Superman can fly.

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The question didn't talk anything about wings at all. What are you talk about? –  hwlau 2 days ago
Here is a youtube video of a frog hovering without any wings. And he is not expending any energy to hover. (Ok this is cheating a little.) –  NowIGetToLearnWhatAHeadIs 2 days ago

If the mass of the bird is M, and it is modelled as a fan which is pushing air to velocity v downward constantly and continuously, then in any unit of time $dt$ it must push an amount of air down on average to get $Mgdt$ up-momentum. This means that the mass dm of the air it pushes down to velocity v in time $dt$ is such that it's momentum is $dm v = Mg dt$, so the amount of air pushed down per unit time is

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

The energy this air gets, assuming the air starts at rest is

$$dm {v^2\over 2}$$

So the work done is

$${dE\over dt} = {Mg\over v} {v^2\over 2} = {Mgv\over 2}$$

This assumes that all the air accelerated by the bird dissipates its energy, so that the energy is lost forever. This is not accurate, and the above is a simple estimate. For a bird of mass .1 kg, g=10 m/s^2, v=1 m/s (assuming the wing is 10 cm from top to bottom of the stroke and flaps 20 times a second), the work required is 1 Watt.

The parameter v is determined from the wing-speed, and the total mass of air you push per wing-flap is the area of the wing times the density of air times the period of a wing-flap. The gives a relation between the size of the bird and the wing-flap frequency. This is order of magnitude only, and it is more valid the more turbulent the air-flow is.

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Thanks somuch, esp. for the wing-flap displacement explanation. Just to clarify my (remedial) understanding of the units, did you mean the energy required is 1 W or 1 Watt-second? Would this mean that a hovering time of 15s would require 15W-s of energy? I think of the Watt as a unit of power, not energy. Again, thanks. –  user9590 Jun 3 '12 at 0:26
@user9590: Yes, stupid mistake, I'm sorry--- the power is 1W, and the energy in each second is 1 W-s. –  Ron Maimon Jun 3 '12 at 5:35