# Energy required to keep a mass in the air

Take for example an electric helicopter - to keep it suspended motionless in the air takes a force from the propeller(s) equal to the force of mass times gravity.

I would like to know the energy required per unit of time to keep this helicopter in the air, assuming 100% efficiency.

I'm not sure if this is even possible - I know that a watt is 1 joule per second, and that a joule is the force required to move 1kg 1 meter. However, since my object is at rest the equation W=FD states that there is no work done - and that there is no power (watts) required to keep it in the air.

However something has to counteract the force of gravity, my question is - is there a way to calculate the energy required to exert that force?

Thanks guys, this is just a theoretical question I was thinking about earlier.. my first post here too :D

A.T.
However something has to counteract the force of gravity, my question is - is there a way to calculate the energy required to exert that force?
You can calculate the kinetic energy put into the air by the rotor.

Ahh, that makes sense - I forgot to consider the air.

However, how would you calculate the energy per time unit?

Say the helicopter weighs 10N, to counteract that 1kg of air would have to be accelerated 10m/s^2 (F=ma).

Since K=1/2mv^2, would I take this in one-second chunks and say that every second 1kg of fresh air is accelerated to 10m/s, therefore 50J of work per second (50 watts)?

russ_watters
Mentor
Well, since no work is being done on a stationary object, how much power is required to keep it aloft can vary widely. For a helicopter hovering (but not suspended by a cable!), power is a function of the kinetic energy change in the air. But force is due to change in momentum. And since momentum and kinetic energy are not proportional to each other, you can increase one while decreasing the other. So the power required can be highly variable.

To put a finer point on it: if you use a larger rotor, you can move more air at a slower velocity, requiring less power for the same change in momentum (force).

CWatters
Homework Helper
Gold Member
Ahh, that makes sense - I forgot to consider the air.

However, how would you calculate the energy per time unit?

Say the helicopter weighs 10N, to counteract that 1kg of air would have to be accelerated 10m/s^2 (F=ma).

Since K=1/2mv^2, would I take this in one-second chunks and say that every second 1kg of fresh air is accelerated to 10m/s, therefore 50J of work per second (50 watts)?
If that were correct then you could also use a larger but slower rotor and accelerate 10kg to 1m/s^2 as that also produces 10N..... So every second 10kg is accelerated to 1m/s.

K = 0.5 * 10 * 1^2
= 5Joules = 5W

so clearly this approach is not correct, or at least not perfect. You need to know what velocity the air is actually doing. Even then it assumes all the energy put into the air is lost.

sophiecentaur
Gold Member
2020 Award
If that were correct then you could also use a larger but slower rotor and accelerate 10kg to 1m/s^2 as that also produces 10N..... So every second 10kg is accelerated to 1m/s.

K = 0.5 * 10 * 1^2
= 5Joules = 5W

so clearly this approach is not correct, or at least not perfect. You need to know what velocity the air is actually doing. Even then it assumes all the energy put into the air is lost.
Yes - what it's missing is the power that you need to supply to a rotor to produce the 5W of 'useful' power.

I looked up the spec for a random model helicopter. It is 450g and has a 9Wh battery. If it can hover for 10min on a charge (pretty optimistic I reckon, as you can get 50% more flight time when not hovering) it would be using 54W. That compares with the couple of Watts that your simplified approach would yield. There are clearly a lot of factors involved!

sophiecentaur
Gold Member
2020 Award
Looks like the model helicopter is operating at about 100W/Kg.

Figure 2.6 and the text at the following link says that full size helicopters operate up to about 5kg/kW = 200W/Kg. That's only a factor of two difference which is less than I expected.

I found that article too and that's where I got my 50% extra consumption for hovering, I think.
There is another factor which we haven't mentioned and that is the scaling. If you go down another order of magnitude or two (i.e. bumblebees and the like) they find that the viscosity of the air allows them to use much less 'fuel' for flying and hovering. I wonder whether this could affect a model aircraft. Perhaps not.

CWatters
Homework Helper
Gold Member
For medium speed aircraft the "Reynolds number" is a constant that attempts to account for scaling effect.

http://www.mh-aerotools.de/airfoils/howdoi.htm

Does indeed effect model aircraft as they operate at low Reynolds numbers (due to small size) and man powered aircraft (due to slow speed) compared to full size aircraft.

Hi.
No energy is required to maintain the height of your helicopter in cases:

- the weight of helicopter is so tiny that its density is same as that of air.

- top lid of a cylinder, the role of the helicopter plays. No air leak away from the cylinder thus the pressure of air inside the cylinder sustain the height of the helicopter.

- the shape of helicopter is thin sphere and big enough to cover whole earth surface. no air leaks again so that air pressure sustain the height.

Regards.

Take for example an electric helicopter - to keep it suspended motionless in the air takes a force from the propeller(s) equal to the force of mass times gravity.

This force the helicopter exerts on the air.

The energy used by the helicopter is the force times the distance that the air moves.

10000 toy helicopters keep their 5000 kg mass suspended in the air more effectively than 1 real helicopter that has a mass of 5000 kg. Air moves faster - less efficient, air moves slower - more efficient.

sophiecentaur