This article discusses how loss of power (e.g. an engine failure) is handled in a helicopter. Like a fixed wing aircraft, loss of engine power is not a death sentence. It’s possible to “glide” to a safe landing zone and land a powerless helicopter via a process called autorotation. This process is described below.
Upon losing power there are three key maneuvers a pilot must perform: autorotation entry, steady autorotation and the flare/landing. Autorotation entry covers the transition from normal, powered flight to steady autorotation. Once the entry has finished and the pilot has stabilized the helicopter in autorotation, he can look for a landing zone and maneuver the helicopter appropriately towards that landing zone (LZ). When nearing the LZ, the pilot will need to arrest both the forward speed and descent rate to land safely, which is done by flaring, or pitching the aircraft nose up. More details on each of these phases is provided below.
Upon losing power two abrupt changes are (1) the loss of “reaction torque” which had supplied a yaw moment opposite to the direction of rotor rotation and (2) the slowing of the rotors which are no longer being kept at operational speed by the engine. The former requires the pilot to add significant right pedal and the latter requires the pilot to lower the collective control. If the helicopter had enough forward speed, aft cyclic may be applied to pitch the nose up, increasing airflow up through the rotor and further mitigating the loss of rotor speed. After these actions the helicopter will accelerate down towards the ground until it reaches a steady descent rate (around 2000 ft/min for some helicopters).
Once the pilot obtains a steady descent rate and airspeed, he must look for a safe LZ while ensuring the rotor speed remains in a safe range (typically 85 to 105% of normal operating NR). The pilot can lower collective to increase rotor speed and increase collective to decrease rotor speed. In this condition the main rotor is powered by the air flowing up through it, due to the helicopter’s descent. The tail rotor is geared to the main rotor so that the main rotor is effectively powering the tail rotor as well.
In this flight regime, a pilot must also monitor airspeed. Two key airspeeds are (1) the airspeed that minimizes autorotational descent rate and (2) the larger airspeed the provides maximum glide distance. The airspeed that minimizes descent rate is approximately the same as the airspeed that requires minimum power (max loiter time) in level flight, around 60kts for some helicopters. The airspeed that provides maximum glide distance is approximately the optimal cruise speed for the helicopter, where the ratio of forward speed to power is largest (often around 90kts).
Approaching the LZ the helicopter must reduce its forward speed and descent rate to near zero. Flaring, or pitching the nose up, does these two things simultaneously. When the aircraft pitches up more air flows up through the rotor (e.g. 60kts is much larger than 2000’/min), which increases rotor thrust, delivering both a vertical and aft force. This pitch must subsequently be removed in order to land flat on the ground. Forward cyclic is then applied to pitch the nose down as collective is increased to keep the descent rate small and soften the landing. Rotor speed decreases with the collective increase, and little upflow through the rotor. Timing is critical here – if the flare is done too soon the rotor speed will subsequently decay, the helicopter will accelerate down, and result in a hard landing. If the flare is done too late the helicopter will contact the ground before the descent rate and/or forward speed is sufficiently arrested, also resulting in a hard landing.
The landing is typically sliding one, with some forward speed when contacting the ground. Immediately lowering the collective full down helps the helicopter come to a stop quicker. In some cases, the terrain may not allow for a sliding landing. In such cases, pilots aim to reach 0 ground speed just before touching down, called a "0 0 autorotation" (0 speed at 0 altitude above ground) This is more difficult and typically results in a harder ground impact.
The acceleration of the rotor is proportional to the net torque acting on the rotor (this follows from Newton’s equation F=ma). Normally the engine supplies (positive) torque to counter (negative) torque from rotor aerodynamic forces. The result is zero net torque and constant rotor speed. Without the engine, aerodynamic forces alone must result in zero net torque. We’ll see that the air flowing up through the main rotor (due to the aircraft descent), along with the aerodynamic shape of the blades, facilitates this. The main rotor is geared to the tail rotor and effectively powers the tail rotor in autorotation.
If we zoom in on sections of a rotor blade, each by itself may provide a nonzero net torque. For example, aerodynamics on the innermost and outermost portions of a blade typically would slow the rotor in autorotation. However, a portion of the blade between these two regions has aerodynamic forces that contribute positive torque, facilitating zero net torque and constant rotor speed. This direction of torque contribution is shown with arrows in the picture above. We’ll investigate cross sections of the blade in more detail below.
A cross section of a blade is shaped like an airfoil shown in the diagram above. A key property of these airfoil shapes is high lift relative to drag. Lift is the aerodynamic force acting perpendicular to the relative air velocity \(V\), while drag is parallel. Since the aircraft is descending at a high rate and the rotor is spinning, the relative air velocity to the blade is at an angle shown in the diagram above (i.e. mostly right here but with an upward component associated with aircraft descent). Because of the upward component, the lift force is tilted slightly left, into the direction of rotor rotation. The drag mostly acts against rotor rotation. Since the lift is so much larger than the drag, the forward component of the lift (dotted horizontal blue line) has the potential to more than offset the aft portion of the drag. This results in positive torque that allows the rotors to maintain speed without engine power.
You may wonder why the inboard and outboard sections of the blade do not contribute positive torque. The speed of a blade section due to rotor rotation is proportional to the distance from the center of the rotor. This is very small for inboard sections and very large for outboard sections. This results in a large angle of incidence \(\alpha\) inboard and much smaller \(\alpha\) outboard. Above a threshold \(\alpha\), drag increases significantly and lift decreases. This is called stall, and in this context prevents an inboard section from producing positive torque as shown in the diagram above. Outboard the small \(\alpha\) shrinks the lift relative to the drag and orients the lift more upward (less forward), together resulting in a negative torque contribution as shown below.
How do things change in a turn / maneuvering? When the helicopter turns it rolls to a blank angle, say, \(\phi = 35deg\). At the same descent rate, the up flow through the rotor drops about 20% (from \(w\) to \(w\cos 35^o \)). Thrust and rotor speed cannot be maintained in this condition. The helicopter accelerates towards the ground, increasing flow up through the main rotor. This persists until vertical aerodynamic forces, primarily from the main rotor, can support the helicopter’s weight. At this point, main rotor thrust will be significantly larger than in level autorotation, because the thrust vector is tilted about \(35^o\). There is a large amount of thrust parallel to the ground, in the direction of the turn (that’s what’s facilitating the turn). To reach this condition the flow through the main rotor (normal to the main rotor) must be significantly larger than in level flight autorotation. This means the rotor speed will increase in a turn and collective is increased to compensate for this.