081 Principles of Flight topic guide
Stall Speed and Load Factor
A stall is an angle of attack event, not a speed event. It happens the instant a wing's angle of attack passes its critical value and the airflow separates from the upper surface, and that critical angle is essentially fixed for a given wing and configuration. What changes from one stall to the next is the indicated airspeed at which that critical angle happens to be reached, and that airspeed is what the exam calls the stall speed.
Anything that forces the wing to fly at a higher angle of attack for the same speed pushes the critical angle closer and raises the indicated stall speed. Load factor does exactly that: a steepening turn, a pull into a climb, or a gust all raise load factor, and the ATPL exam expects you to turn a bank angle straight into a load factor and a new stall speed without hesitation.
Angle of attack first, airspeed second
Every aerofoil has a critical angle of attack at which the boundary layer can no longer stay attached to the upper surface. Below that angle, lift keeps rising smoothly as angle of attack increases; at that angle, lift collapses and drag rises sharply. The critical angle itself barely moves with speed, weight, or bank angle, which is why a wing can stall at any speed and in any attitude, nose high or nose low, provided it reaches that angle.
The indicated stall speed is simply the airspeed at which level, one g flight happens to need the critical angle of attack to support the weight. Raise the load on the wing, through weight or load factor, and the wing needs more lift at the same speed, which means a higher angle of attack at that speed, which means the critical angle, and the stall, is reached sooner, at a higher indicated airspeed.
Turning a bank angle into a load factor
In a level turn, the lift from the wings does two jobs at once: it supports the aeroplane's weight and provides the horizontal component that curves the flight path. Resolving the vertical component of lift against weight gives the standard result that the load factor n in a level, coordinated turn equals one divided by the cosine of the bank angle. At 0 degrees of bank, n is 1. At 60 degrees, the cosine is exactly 0.5, so n is exactly 2.
Stall speed scales with the square root of load factor, because lift depends on speed squared for a given angle of attack: doubling the load the wing must carry only needs the stall speed to rise by the square root of 2, not by a factor of 2. The same square root relationship applies to weight changes at a constant load factor, which is why a heavier aeroplane always stalls faster in an otherwise identical configuration.
- 0 degrees of bank: n = 1, stall speed unchanged.
- 30 degrees of bank: n is about 1.15, stall speed up about 7 per cent.
- 45 degrees of bank: n is about 1.41, stall speed up about 19 per cent.
- 60 degrees of bank: n = 2 exactly, stall speed up by the square root of 2, about 41 per cent.
Worked example
Worked example: stall speed in a 60 degree banked turn
Not quite. The correct answer is B.
Correct answer: B
- A. This multiplies the stall speed directly by the load factor of 2, skipping the square root that actually links load factor to stall speed.
- B. Correct: n = 1 / cos(60) = 1 / 0.5 = 2 exactly, and stall speed scales with the square root of n, so 100 kt x square root of 2 (1.4142) gives 141.4 kt, which rounds to 141 kt.
- C. This uses tan(60), which is 1.732, in place of the load factor formula 1 / cos(bank). Tan(bank) belongs to the turn radius and rate of turn relationships, not to the vertical load factor.
- D. This applies a flat 50 per cent increase regardless of the bank angle, effectively guessing rather than working cos(60) = 0.5 correctly into the load factor formula.
Step by step
- Find the level turn load factor: n = 1 / cos(bank). At 60 degrees, cos(60) = 0.5 exactly, so n = 1 / 0.5 = 2.
- Stall speed scales with the square root of load factor: new stall speed = old stall speed x square root of n.
- Square root of 2 is 1.4142, to four decimal places.
- New stall speed = 100 kt x 1.4142 = 141.4 kt, which rounds to 141 kt.
- Sanity check: doubling the load factor never doubles the stall speed, only the square root of 2 does, about a 41 per cent increase, the same scale used throughout this relationship.
Common mistakes
Multiplying stall speed directly by the load factor
Load factor scales the load the wing carries, but lift depends on speed squared, so stall speed only scales with the square root of load factor. Skipping the square root turns a 41 per cent increase into a wrongly doubled speed and drops the mark.
Using tan(bank) instead of 1 / cos(bank) for load factor
Tan(bank) belongs to the turn radius and rate of turn relationships, not the vertical load factor. Confusing the two formulas produces a plausible looking but wrong stall speed in exactly the kind of question this topic tests.
Forgetting that weight changes shift stall speed the same way
A stall speed computed at one weight does not transfer to another without applying the same square root scaling to the weight ratio, and stems that change both weight and bank angle in one question catch students who only adjust for one of the two.
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Last reviewed July 2026