Accelerated stall in a survey turn
You are ferrying a Cessna 172S over the Kaimai Range at 4,500 ft AMSL, tracking north at 65 KIAS in a 45° banked level turn to keep a survey line in sight. The aircraft weighs 1,040 kg, below the 1,157 kg MTOW; POH V_S1 at MTOW is 48 KCAS and V_A is 105 KIAS. The day is unstable, with scattered cumulus and moderate thermal activity; you have just passed through a strong updraught that pushed you 200 ft above your altitude. As you correct back down and hold the bank, the stall warning horn sounds and you feel light buffet. You glance at the ASI: 65 KIAS. Your passenger asks why the horn is on 'when we are doing 65 knots and the stall is 48'. You still have 2,000 ft of terrain clearance.
1. Calculate the load factor and the stall speed in the 45° level turn at 1,040 kg, and state the margin over the stall at 65 KIAS.
Read the worked answer
Load factor. In a level turn, n = 1/cos φ. At 45°, cos 45° = 0.7071, so n = 1.414.
1g stall at the actual weight. V_S1 scales with √W: V_S1(1,040) = 48 × √(1,040/1,157) = 48 × 0.948 = 45.5 KCAS.
Stall speed in the turn. V_S(n) = V_S1 × √n = 45.5 × 1.414 = 64.3 KCAS.
Margin at 65 KIAS. At this speed and altitude IAS is close to CAS, so the margin is about 65 − 64.3 = 0.7 kt, roughly 1%. That is inside normal ASI and gust variation, so the aircraft is effectively at the accelerated-stall boundary. The correct reading is not '65 is above 48, so we are safe', but 'in this bank the stall speed has moved up to meet us'. A 1–2 kt gust or a small pull is enough to stall it.
2. Explain why the stall warning is sounding at 65 KIAS when the POH stall speed is 48 KCAS, and identify the contribution of the thermal and the bank.
Read the worked answer
The 48 KCAS is the 1g stall speed at MTOW. A wing stalls at a critical angle of attack, not at a fixed speed. In a level turn the wing must produce lift equal to weight times n, so the stall speed rises by √n; at 45° that is 1.414, a 41% increase.
At 1,040 kg the 1g stall is 45.5 KCAS, so in the turn it is 64.3 KCAS. The thermal gust raises the local angle of attack further, and the correction to hold altitude adds load factor; both push the aircraft through the buffet boundary.
The bank is the main cause; the thermal is the trigger. An aft CG within limits would reduce static margin and make any wing drop more abrupt, but it does not change the load-factor arithmetic.
3. State the recovery and explain why pulling back is the wrong response, including the relevance of V_A.
Read the worked answer
Recovery: reduce bank to unload the wing, lower the nose to reduce angle of attack, apply power, and keep the ball centred. Do not pull.
Pulling increases angle of attack and load factor at the same time - the exact inputs that produced the stall - and from 45° bank can progress to a spiral or a spin. The stall warning is an angle-of-attack warning, and the only way to silence it is to reduce angle of attack.
V_A is the speed above which a full control deflection can overstress the aircraft; at 1,040 kg it reduces to 105 × √(1,040/1,157) = 99.5 KIAS. In the turn the safe band is therefore from the accelerated stall speed (64.3) to the applicable V_A (99.5). At 65 KIAS you are at the bottom of that band, so the correct action is to unload and accelerate away from the stall, not to trade speed for altitude.