How much force does a helicopter's tail rotor counteract?

Question

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?

Answer 1

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.

Answer 2

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.