# How much lift must a helicopter rotor produce?

How much lift must a typical helicopter rotor produce? For example, consider the rotor of what seems to be a common helicopter: the Robinson R22. Is the lift produced equal to one of the following:

• The weight of the helicopter;
• Double the weight of the helicopter; or
• Simply some value greater than the weight of the helicopter?

For example: for a 700kg helicopter, what maximum lift will the helicopter rotor need to produce?

• In order for it to fly, it has to produce more lift than weight... – Ron Beyer May 16 '17 at 18:22
• This may be a little too broad. Depends on the phase of flight. In a hover (ignoring ground effect and some other factors), the lift produced must exactly equal the weight of the craft. For a climb, the lift produced must be greater, how much so depends on how fast you want to climb. Conversely, to descend, lift should be less than weight, again the extent depends on how fast you want to go down... – Ron Beyer May 16 '17 at 20:48
• This may be better suited for physics.se – Burhan Khalid May 17 '17 at 5:43
• The QE already knows that the (maximum) weight of the helicopter must be lifted. Now, the question is what is the minimum required vertical acceleration performance, and what margins are required. What other parameters play a role, max. service altitude, etc. ? The question is not that broad. – bogl May 17 '17 at 7:47
• My answer was wrong, because I forgot a simple concept which is that the initial force required would be more (from rest, to lift the craft) that it would be to further lift the craft. – Burhan Khalid May 17 '17 at 8:18

## 1 Answer

In the hover, it must produce a lift force equal to the weight of the helicopter, in order for the helicopter to remain at constant altitude. A helicopter with a mass of 700 kg must produce a lift of 700*g = 6,867 N to hover at constant altitude.

In forward flight at constant altitude, the rotor must produce more thrust with the thrust vector tilted forward at angle α. Vertical thrust component = lift must be equal to weight for the helicopter to fly at constant altitude. As the helicopter picks up speed, less power is required to provide the same amount of thrust due to reduction of induced drag - if power is not reduced, the helicopter starts to climb when transferring from hover to forward speed.

If in the hover the rotor produces more lift than the weight, let's say 710*g,it accelerates upwards, resulting in a vertical velocity component that results in an additional aerdynamic damping force, proportional to vertical speed, caused by change of rotor blade Angle of Attack and fuselage vertical drag. So for a 700 kg helicopter with a rotor that exerts 710*g = 6,965 N, the extra 98 N is used for overcoming aerodynamic drag associated with vertical speed.

Assuming that by "lift" only the vertical component of thrust is meant: more lift than weight results in a constant upwards velocity. Less lift than weight results in a constant downwards velocity, even in a free fall (terminal velocity).