The 60-Degree Turn on Final
Cessna 172, landing configuration, 1,050 kg. AFM stall speed in the landing configuration is 45 kt CAS and Va is 105 kt. Normal approach speed is 1.3 Vs, about 59 kt. Wind is 240°/15 kt gusting 25 kt and the runway in use is 24, paved, 900 m, with trees 15 m high 100 m before the threshold. On final at 400 ft AGL you are about 150 ft high and half a wingspan left of the centreline. Instead of going around, you consider a continuous descending turn at 70 kt IAS and 60° of bank to lose height and re-align.
1. Calculate the load factor, stall speed and speed margin in the proposed 60° bank at 70 kt.
Read the worked answer
Load factor n = 1 / cos φ.
| Bank | n | Stall speed (45 × √n) | Margin at 70 kt |
|---|---|---|---|
| 0° | 1.00 | 45.0 kt | 1.56 Vs |
| 30° | 1.15 | 48.4 kt | 1.45 Vs |
| 45° | 1.41 | 53.5 kt | 1.31 Vs |
| 60° | 2.00 | 63.6 kt | 1.10 Vs |
In a 60° bank only cos 60° = 0.5 of the total lift acts vertically, so to hold level flight the wing must produce twice the weight - the wing is working at 2g.
At 70 kt in a 60° bank the aeroplane is flying at 1.10 Vs. That is not a margin; it is inside the band where a small increase in angle of attack stalls the wing. The normal approach speed of 1.3 Vs exists precisely to absorb turbulence, shear and pilot technique; in this turn that buffer has been spent before you have even started.
2. What do the gusty surface wind and the low altitude do to that margin?
Read the worked answer
The wind is 240°/15G25, runway 24 - a headwind of 15 gusting 25, so 10 kt of gust spread. The danger is not the gust arriving, it is the headwind being lost: a 10 kt shear or lull takes 70 kt IAS to 60 kt, which is below the 63.6 kt stall speed for that bank angle. The wing stalls in a banked attitude with a live rudder.
Low altitude is the multiplier:
- At 400 ft AGL, in a bank, with the aeroplane pitching down, a stall is followed by a wing drop and yaw - the classic stall/spin entry on final, from which the height loss is roughly 200–300 ft in the incipient phase alone.
- The trees 100 m before the threshold remove the option of a low, flat recovery; they also generate mechanical turbulence and a wind gradient over the last 200 ft.
- Bank angle also steepens the effective approach path, so the pilot instinctively pulls to arrest the sink rate. Pulling at 2g and 1.1 Vs is exactly the input that produces an accelerated stall.
Va is not the issue here - 70 kt is well below 105 kt, so structural overload is unlikely. The hazard is purely aerodynamic, and being below Va offers no protection against it.
3. Compare the 30° bank option and state your decision with reasoning.
Read the worked answer
A 30° bank gives n = 1.15 and a stall speed of 48.4 kt, so at 70 kt the margin is 1.45 Vs - comfortable, and a 10 kt loss of headwind leaves 60 kt, still 1.24 Vs in the turn. But a 30° bank turns slowly: to displace the flight path enough to correct half a wingspan of offset at 400 ft you will need a wider, longer turn, and it will not remove 150 ft of height without a pitch change that costs airspeed. In any case, a banked descending turn at 400 ft AGL in a 10 kt gust spread, over trees, is a manoeuvre with no error budget, whatever the exact bank angle.
Decision: go around. The quantitative case is decisive - in the proposed manoeuvre the aeroplane is at 1.10 Vs, in gusty conditions, at 400 ft AGL, with a sink rate to arrest and an obstacle ahead. The go-around costs about five minutes and a circuit. The alternative has a small but real probability of a fatal outcome, and no amount of skill converts a 1.10 Vs margin into a safe manoeuvre.
Then fix the cause: the approach was unstable - high and offset at 400 ft. Adopt a stabilised-approach gate (by 500 ft AGL: on centreline, on speed, on path, sink rate under control) and treat any breach of it as an automatic go-around, so the decision is made before the situation is flattering.