Touch-and-go turn, wing drop and the 62 KIAS stall warning
An instructor and a CPL candidate are conducting touch-and-go training in a single-engine piston aeroplane (MTOW 1,600 kg, flaps-up 1g stall speed 48 KCAS) at an airfield with density altitude 7,500 ft. After the second touch-and-go the aircraft climbs at 75 KCAS. At 400 ft AGL the instructor calls for a left clearing turn. The candidate banks 45° and, seeing the nose drop, applies firm back pressure while adding right rudder. Airspeed decays through 62 KCAS. The stall warning horn sounds, the left wing drops sharply and the nose pitches down. The candidate's instinct is to pull back harder and apply more right rudder. The aircraft is 60 kg below MTOW; the engine is at full power; no flap is selected. The instructor has not yet taken control. The candidate asks afterwards why the horn sounded when the airspeed was still well above the book stall speed.
1. Explain why the stall warning activated at 62 KCAS although the published 1g stall speed is 48 KCAS. Show the load-factor and stall-speed calculation for the 45° bank, and comment on the effect of density altitude and the candidate's back pressure.
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Stall warning is triggered by angle of attack, not by a fixed airspeed. In the 45° bank the load factor is n = 1/cos 45° = 1.414. The stall speed in the turn is therefore Vs(turn) = Vs(1g) × √n = 48 × √1.414 ≈ 57 KCAS. At 62 KCAS the margin is only about 5 kt, and the candidate's firm back pressure raises AoA further and can push the load factor above 1.414, producing an accelerated stall above the steady-turn stall speed.
Density altitude does not change the indicated stall speed for a given weight and load factor, because IAS reflects dynamic pressure. It does raise TAS for the same IAS, and the normally aspirated engine produces less power, so any height lost in recovery is more costly. The 60 kg below MTOW reduces stall speed only slightly: √(1540/1600) ≈ 0.981, or about 47 KCAS at 1g, which does not explain the event. The dominant cause is bank angle plus increasing AoA.
2. Evaluate the candidate's instinctive response (pull back harder and apply more right rudder). What does each input do to angle of attack, load factor and yaw, and why is this especially dangerous at 400 ft AGL?
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Pulling back harder increases AoA toward and beyond the critical angle, deepens the stalled condition and raises load factor, which further increases stall speed. At 400 ft AGL there is very little height in which to unload the wing and recover, so a secondary stall or a developed spin is likely.
Right rudder yaws the nose to the right. With the left wing already dropped and stalled, that yaw can increase the effective AoA on the descending left wing and promote autorotation. A skidding turn also raises load factor and stall speed, compounding the problem.
The correct first priority is to reduce AoA, not to fight the wing drop with more back pressure or to chase yaw with large rudder inputs while the wing remains stalled.
3. State the recovery sequence you would teach for this situation and explain the aerodynamic reason for each step, including why aileron is not the first control used to pick up the dropped wing.
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The recovery sequence is: simultaneously relax the back pressure, or apply positive forward elevator, to reduce AoA below critical; neutralize ailerons; apply rudder opposite to the yaw and wing drop; apply power as required; once the stall is broken and the wing is flying, roll level with coordinated aileron and rudder; then establish a climb only after safe flying speed and no buffet.
Reducing AoA is essential because the wing is stalled at a high angle of attack, not simply because airspeed is low. Opposite rudder stops the yaw/rotation without increasing AoA. Aileron is not used first because the descending wing already has a higher AoA; applying down aileron on that wing increases its AoA further and can deepen the stall or trigger autorotation. Aileron input can also produce adverse yaw. Roll is therefore corrected after the stall is broken.