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Force (and Velocity) Required to Hover

  1. Oct 23, 2011 #1
    Hello Everyone,

    I am by no means a physics expert, I have a basic/intermediate knowledge at best. I am a graduate-level computer science researcher working on building computer simulations of quad-rotor helicopters.

    For simplicity, assume that the helicopters move only with their rotors (i.e. no thrusters or jets), and all external forces (like air, etc) can be neglected. They are also flown indoors and at low altitudes. The helicopters weight 0.4kg and require about 3.9N of upward force to hover.

    The question:

    I know that as a helicopter hovers it has no net force acting on it (i.e. the force with which the rotors are pushing upward is equal to the gravity pulling down)... that also means that it has no acceleration. But, does it have velocity? Pitching the helicopter as it hovers will cause it to start moving, meaning that the downward (i.e. z axis) component of the velocity is non-zero... right?

    The problem I am having is relating the force required to hover and the upward velocity that must be provided to generate enough force to allow for hovering (I know that one cannot convert between force and velocity).

    Is there an equation or concept that will allow me to express the upward force required to hover as some sort of velocity measurement?
  2. jcsd
  3. Oct 24, 2011 #2
    I think they can also alter the pitch (angle) of the rotor blades, maybe even the shape.

    If so, the helicopter rises; otherwise it hovers.

    The force of gravity F = GMm/r2 must be offset, or exceeded, by the force generated from the helicopter blades....I guess what they call "lift". That is surely related to
    the velocity of the rotation of the blades: v =wr and of course their shape.

    Seems like vertical lift from a helicoper blade is closely related to the horizontal force from a traditional airplane prop??

    Maybe check here:

  4. Oct 24, 2011 #3
    If you only want to hover, only force sufficient to counteract gravity is required.

    The only velocity required is that which allows your mass to overcome the force of gravity (as in F=mV).
  5. Oct 24, 2011 #4
    OK, I see some people are in need of education concerning helicopters.

    So, I'll start a new thread in "General Engineering" entitled "The Principle of the Helicopter"

    See you there!
  6. Oct 24, 2011 #5
    It must exert the same force on the air as it's weight, mg.
    The mass of the air depends on the area of the helicopter blades multiplied by the distance it pushes it through. mair=∏l2*d, if d=distance, and l=length of helicopter blades.
    In the suvats, s=0.5(u+v)t, so if the air starts at 0ms-1, d=0.5vt
    Acceleration is velocity per unit time, so a=v/t.
    If you plug-in values of mhelicopter, g, d, l, and t, and solve the resulting equations, you can work out v, velocity of air dispelled by the helicopter.
  7. Oct 24, 2011 #6

    With your phrase "...velocity of air dispelled by the helicopter", you're describing, not the useful interaction between rotor blades and air, but cavitation, which is the phenomenon by means of which a rotating body (be it a ship's propeller, a helicopter's rotor, or an airplane's propeller) pushes the substance against which it must exert its force away from itself, thus depriving itself of traction. I'll never forget the time when I was at the stern of the Staten Island ferry in NYC when we were trying to pull away from the Manhattan terminal, and the engineer applied excessive torque to the shaft and the whole ship shuddered while the propeller just sent up bubbles and splashes without moving the ship. That's cavitation; that's what happens when your propeller "dispels" the substance it's designed to work against in order to apply it's power to that substance.

    A helicopter's blades do not "dispel" air, they grab into it to exploit its physical properties to provide lift and traction.

    Please read my "Principles of the Helicopter" thread under the "General Engineering" category.
  8. Oct 24, 2011 #7

    D H

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    That depends entirely on what one means by "gravity". This is wrong if you mean the gravitational force between the Earth and the helicopter. It is correct if by "gravity" you mean the vector sum of the gravitational force plus the fictitious centrifugal force due to the Earth's rotation. More below.

    This makes zero sense. mV does not have units of force.

    Let's look at the latter item first, the velocity required to hover. "Hovering" means that the velocity with respect to some fixed point on the surface of the rotating Earth is zero. A helicopter whose velocity vector has a non-zero vertical component is not hovering; it is climbing or descending. If the velocity vector has a non-zero horizontal component it is not hovering either.

    It is important to remember that the Earth is rotating. The forces acting on the helicopter in a frame of reference fixed with respect to the rotating Earth are
    • Lift, the vertical component of the force generated by the helicopter's blades;
    • Thrust, the horizontal component of the force generated by the helicopter's blades and by its tail rotor;
    • Drag, which is a function of the velocity of the helicopter with respect to the surrounding air;
    • Buoyancy, which is the weight of the air displaced by the helicopter;
    • Gravitation, which is given by Newton's law of gravity;
    • The fictitious centrifugal force, which is a function of the Earth's angular velocity and the distance between the helicopter and the Earth's rotation axis; and
    • The fictitious coriolis force, which is a function of the Earth's angular velocity and helicopter's velocity with respect to the rotating Earth.

    Both drag and coriolis force are zero when a helicopter is hovering in a stationary air mass. You can pretty much ignore buoyancy; it is a small effect. Gravitation and centrifugal force are often times collected as one single force, "gravity". (For example, the 9.80665 m/s2 figure often used as the acceleration due to gravity includes gravitational and centrifugal accelerations.)

    So, to hover you need lift to exactly counterbalance gravitational+centrifugal forces, and you need thrust to be zero. The latter is a bit tricky because the tail rotor will provide a lateral thrust. You need to lean the helicopter a bit sideways to counteract the lateral thrust from the tail rotor.
  9. Oct 24, 2011 #8


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    D_H, there is no need for all of that: aeronautical calcs like this rarely need suchthe detail and the OP explicitly stated a desire for simplicity!

    A hovering helicopter has two relevant forces on it: its weight and the force imparted on it by the air interacting with the rotors. And they are equal.

    Now the air is accelerated down from a standstill, so you can calculate the force from the change in momentum of the air.

    Or, you can use Bernoulli's equation and relate force to velocity pressure.
    Last edited: Oct 24, 2011
  10. Oct 24, 2011 #9


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    Then the rotors need to exert 3.9N of downward force onto the air, normally by accelerating some amount of mass of air per unit time (unless in ground effect). The amount of required acceleration is decreased if the rotors are larger, since the mass flow rate would be larger. Note that while in a hover, an induced wash is produced, accelerating the air from well above the rotor, so the "exit velocity" (the velocity of the affected air when it's pressure returns to ambient) will be higher while in a hover than when moving forward and level. The rotor speed and/or pitch will need to be higher for a hover than forward "flight".
  11. Oct 24, 2011 #10
    The rotors will in fact spin faster when pitching (i.e. during forward flight) because the helicopter needs to maintain the same upward force required when hovering (so it does not fall), but because it is at a pitch its downward force vector is no longer perpendicular to the ground. For the z-component of force it to remain the same, the rotors must speed up.
  12. Oct 24, 2011 #11


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    From wiki article:

    As a helicopter moves from hover to forward flight it enters a state called Translational lift which provides extra lift without increasing power.

  13. Oct 25, 2011 #12
    I'm not great at terminology, (but I am willing to learn) but that is what I meant. Is my general idea right though? I'll PM you my solution to the equations I wrote down, as I don't know if I am allowed to post it directly on the thread.
  14. Oct 26, 2011 #13

    D H

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    By saying "weight" you just did all that. What you meant by "weight" is what a (big) spring scale would read were the helicopter placed on the scale. This scale weight includes the force due to gravitation plus the centrifugal force plus buoyancy. It is not "weight" as defined in elementary physics texts.
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