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How much force does a helicopters tail rotor counteract?

What are the formulas for determining this? How much rotational torque is produced by spinning the blades?

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    $\begingroup$ Multiply engine output with a efficiency factor of about 3/4 to get the power at the main rotor, divide by rpm you get the torque at the rotor, divided by the length of the tail you get the force at the tail. If you get the efficiency number right this should be pretty accurate. $\endgroup$ – user3528438 Jan 18 at 17:32
  • $\begingroup$ @user3528438 But don't forget that only a small part of the power absorbed by the rotor is used to overcome the drag of the rotor blades, since most of that power is consumed by the lift generated... It's only the blade drag that causes the torque reaction that has to be counteracted by a tail rotor... $\endgroup$ – xxavier Jan 18 at 18:41
  • $\begingroup$ @xxavier there is power required to overcome induced blade drag as well. $\endgroup$ – Koyovis Jan 19 at 2:29
  • $\begingroup$ @Koyovis You're right. The part of the power related to lift is accounted by the induced drag... $\endgroup$ – xxavier Jan 19 at 13:26
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The lift-to-drag ratio of the main rotor blades is around 10, so the tail roter has to provide a force equal to one tenth of the aircraft's weight.

Of course you have to take into consideration torques and lever arms, but if we assume that most of the lift and drag is created by the outer portion of the main rotor and that the tail rotor's lever arm is approximately the same, we can simply ignore them. On most helicopters the tail boom is a bit longer than the radius of the main rotor, so the force might be 20-50% lower. On the other hand, the tail rotor is often less efficient, especially on some newer models with small, ducted fans, so the tail rotor might consume a bit more than 10% of the power.

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    $\begingroup$ A very clear and simple explanation, directly from first principles... $\endgroup$ – xxavier Jan 19 at 17:42
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The tail rotor on helicopters indeed provides anti-torque thrust, plus thrust for manoeuvring in the yaw axis. A quote from Principles of Helicopter Aerodynamics by j. Gordon Leishman section 6.9:

The magnitude of this thrust, as well as its power consumption, depends on the reaction torque from the main rotor, $Q_{MR}$, and the location of the tail rotor from the centre of gravity (i.e. the moment arm $l_{MR}$). In addition, there are inertial effects that the tail rotor must overcome during yawing manoeuvres. In this case, the tail rotor thrust can be found from $$Q_{MR} + I_{ZZ} \ddot{\Psi} = T_{TR} \cdot l_{TR}$$ where $\ddot{\Psi}$ is the yaw acceleration and $I_{ZZ}$ is the mass moment of inertia of the helicopter about the yaw axis.

The main rotor torque $Q_{MR}$ can be computed from the installed net power and the main rotor RPM. For instance for the Bell 212, for which we have done a simulator aerodynamic model:

  • Installed derated TO power $P_I$= 962 kW
  • Rotor RPM = 320 => $\Omega$ = 33.5 rad/sec
  • Main rotor torque $T = P_I / \Omega = 962 * 10^3 / 33.5$ = 28,716 Nm

With a tail rotor moment arm of 8.82m, the tail rotor thrust would then be 28,716/8.82 ≃ 3,260 N

And indeed the above method must account for power losses:

  • The power to drive the tail rotor is between 10-20% of the installed power and is lost completely, unless the tail rotor is canted.
  • Transmission between engine and rotors causes friction losses.
  • Installed hydraulic & electrical systems require engine power.

A complete method for computing tail rotor performance is quite broad in scope, and can be found for instance in Helicopter Performance, Stability and Control by Ray Prouty, chapter 3.

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    $\begingroup$ Now if we think about it this way, the tail rotor and main rotor perform very similarly (assume so for a guesstimate) and tail takes 10%-20% the power, then the force if the tail rotor would be 10%-20% of the weight of the helicopter plus some room for maneuverability. $\endgroup$ – user3528438 Jan 19 at 2:44

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