Accelerated Stall Margin in Airframe Icing
A light twin is cruising at 12,000 ft, mass 2,300 kg, in visible moisture at -5°C with anti-ice systems on. The AFM clean 1g stall speed is 84 kt at this mass. The aircraft is in a 30° bank turn; the autopilot is holding altitude, and the crew is hand-flying at 108 kt IAS. Airframe icing has accumulated, and the manufacturer's icing supplement states that CLmax is reduced by 15% in this configuration. There is a 15 kt gust spread and light turbulence. The aircraft has no stick shaker, only a stall warning horn. The crew is 40 nm from a suitable airport with an ILS; the freezing level is 9,000 ft and the cloud tops are 16,000 ft. The crew must decide whether to continue at 108 kt, accelerate, or change altitude. A 45° bank may be needed for a deviation around a cell.
1. Calculate the clean 30° bank stall speed, the icing-adjusted stall speed, and the current margin.
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For a level turn, load factor n = 1/cos(bank). At 30°, n = 1/cos30° = 1.155. Stall speed increases with the square root of load factor: √1.155 = 1.075.
Clean 30° bank stall speed = 84 × 1.075 = 90.3 kt.
With a 15% reduction in CLmax, the stall-speed multiplier is √(1/0.85) = √1.176 = 1.085. Icing-adjusted 30° bank stall speed = 90.3 × 1.085 = 98.0 kt.
At 108 kt, the margin is only 108 - 98 = 10 kt. That is a small margin in turbulence, with a 15 kt gust spread and an autopilot holding altitude. The autopilot increases angle of attack to hold altitude in the turn, and the gust loads can momentarily increase load factor. The aircraft is not operating with a comfortable accelerated-stall margin.
2. What happens if a 45° bank is entered, or if a gust loads the wing to 1.5g?
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At 45° bank, n = 1/cos45° = 1.414, and √1.414 = 1.189. Clean stall speed = 84 × 1.189 = 99.9 kt. Applying the icing factor: 99.9 × 1.085 = 108.4 kt. At the current 108 kt, the aircraft would be at or below the icing-adjusted stall speed in a 45° bank. There would be essentially no margin, and the stall warning could occur with little or no buffet.
For a 1.5g gust load, √1.5 = 1.225. Clean stall = 84 × 1.225 = 102.9 kt; icing-adjusted stall = 102.9 × 1.085 = 111.6 kt. At 108 kt, the aircraft would already be below that stall speed. In practice, a gust plus aileron input or a pull to arrest a descent can produce the critical angle of attack very quickly.
The lack of a stick shaker matters: the only warning is the horn, which may be less effective in identifying an approaching stall. Tailplane icing can also reduce tailplane CLmax and produce a sudden nose-down pitch before the wing stalls. The crew should not treat the 10 kt margin as adequate.
3. Recommend recovery and strategic actions.
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The immediate priority is to reduce angle of attack. Lower the nose, add power smoothly, and roll the wings level; do not pull to maintain altitude. If the autopilot is holding altitude, disconnect or reduce its authority before it trims the aircraft into a higher angle of attack. Avoid abrupt elevator inputs, especially with tailplane icing, because the tailplane may stall suddenly.
Strategically, leave the icing environment. The cloud tops are at 16,000 ft and the freezing level is 9,000 ft, so a climb or descent may be possible, but the crew must consider aircraft performance, terrain, and oxygen. The suitable airport is only 40 nm away; diverting and landing is the most reliable way to remove the icing threat. If the aircraft must remain in icing, follow the AFM icing procedures, keep speed above the icing-adjusted stall margin, and avoid banking more than necessary. A 45° bank should not be attempted at 108 kt in this condition. Reducing bank angle and increasing speed are the key defences.