Coordinated turns and load factor

Bank trades vertical lift for turning force

A coordinated level turn needs enough total lift to retain its vertical component.

Resolve banked lift

L cosφ = W and L sinφ = mV²/R. Dividing gives tanφ = V²/(gR). Thus n = L/W = 1/cosφ in the stated steady coordinated level turn.

At 60° bank, n = 2. At 45°, n = 1.414. Bank angle alone does not fix n during every manoeuvre: a descending, unloading or uncoordinated motion has different balances.

Radius and rate

R = V²/(g tanφ), and angular turn rate ω = g tanφ/V in rad/s. At the same bank, doubling TAS quadruples radius and halves angular rate.

These are air-relative relations. Wind distorts the ground track and changes the ground-referenced radius pattern without changing the required air-relative centripetal acceleration in a uniform wind.

Solve a complete turn at two speeds

At TAS 100 m/s and bank 45°, load factor is 1.414, radius is 100²/9.81 ≈ 1,019 m and angular rate is 9.81/100 = 0.0981 rad/s, or about 5.62°/s. A 180° heading change in the steady air-relative model therefore takes approximately 32 seconds.

At TAS 200 m/s and the same bank, load factor remains 1.414, radius becomes about 4,077 m and turn rate about 2.81°/s. The heading change takes about 64 seconds. Keeping the same bank does not preserve either radius or rate as speed changes.

At zero bank, the laboratory reports straight flight rather than a finite turn radius. This is the limiting behaviour of R = V²/(g tanφ), whose denominator tends to zero. Use the nonzero-bank examples to compare numerical ratios, and treat the zero-bank case as a distinct interpretation rather than a meaningless extremely large rounded radius. Wind would require a separate ground-track construction.

Turning performance meets the stall boundary

More lift at the same speed requires a higher coefficient and angle of attack.

The load cost

Required CL in a level turn is n times the straight-level value at the same speed. Induced drag grows approximately as n² in the simple polar. Thrust and energy requirements therefore rise sharply with bank.

Stall speed grows as √n. A 60° bank level turn with a 90 kt 1 g stall reaches its ideal stall boundary near 127 kt, before adding any operational margin.

Limit and recovery decisions

A turn can be limited by stall, thrust available, structure or an aircraft-specific handling limitation. A transient turn sustained with height loss is not the same as a sustained level turn.

Reducing bank reduces the level-turn lift requirement; reducing α is essential if stalled. Apply the aircraft’s approved upset or stall recovery procedure rather than holding altitude at all costs.

Intersect the desired turn with available lift

With a 1 g stall EAS of 80 kt, the ideal stall boundary at 60° bank in a level coordinated turn is 80√2 = 113.14 kt. At 100 kt, the maximum load before the same CLmax is (100/80)² = 1.5625 g. The corresponding maximum level-turn bank is arccos(1/1.5625), about 50.2°, before any operational margin.

A 60° bank at 100 kt could still occur during a descending or unloading transient, but it cannot satisfy all the stated steady level, coordinated and unchanged-CLmax conditions. The incompatible assumptions must be identified instead of forcing the 2 g equation onto the observed bank alone.

Use the laboratory to calculate n and the stall boundary for a selected bank. Then compare the demanded condition with a separate selected speed or supplied flight condition. State which constraint fails first: maximum lift, structure, thrust or another type-specific limit. The turn geometry alone establishes none of the remaining aircraft margins.

Can you explain and apply the mechanism?

Four original questions. Work from the stated facts, then compare your reasoning with the explanation.