Load Factor in a Coordinated Level Turn

In straight, level flight the wing lifts exactly the weight of the aircraft. In a banked turn the lift is tilted toward the centre of the turn, so only the part $L\cos\varphi$ is left to hold the aircraft up. To keep its height the wing must make $L=W/\cos\varphi$. The load factor $n=L/W=1/\cos\varphi$ is therefore the number of times its own weight the wing is carrying, usually quoted in G.

The variables are the load factor $n$ and the angle of bank $\varphi$, from 0 to 90 degrees. The formula has no speed or weight in it, so any aircraft in a coordinated, level turn at a given bank carries the same load factor. At zero bank $n$ is 1, and it climbs slowly at first and then steeply as $\cos\varphi$ shrinks: 1.1547 at 30 degrees, 2 at 60 degrees and 5.76 at 80 degrees. The Federal Aviation Administration's Pilot's Handbook of Aeronautical Knowledge states the last two, and the page also runs backwards: 2 G is 60 degrees of bank and 5.76 G is 80 degrees.

The load factor matters because of the stall. The handbook states that an aircraft's stalling speed increases in proportion to the square root of the load factor. The Airplane Flying Handbook gives the case: in a coordinated, level 60 degree turn the load factor is 2 G, and the airplane stalls at a speed 41 percent higher than its 1 G stall speed. A turn at the standard rate of 3 degrees per second at an airliner's cruise speed of 450 knots (231.5 m/s) needs $\tan\varphi=V\times\mathrm{rate}/g=1.24$, a bank of 51 degrees and a load factor of 1.6 (1.5890 by hand), which is why airliners turn at a lower rate at cruise speed.

The formula is for a level, coordinated turn at constant speed. The turn also needs enough thrust to hold the speed, because the drag rises with the lift.