# What types of maneuvers increase the load factor on the aircraft?

Out of turns, stalls and climbs, which of the basic maneuvers increases the load factor on an airplane as compared to straight and level flight?

• Aug 18, 2015 at 7:58

As to which maneuvers generate more loading, it's completely dependent on the design and behavior of the airplane, as well as how aggressively the pilot has provided control input. I think many people would think that a stall provides more loading than a climb, but I've stalled a piper cub so gently that you could barely feel it - but if you pull a high performance airplane into a steep climb you can easily exceed 5g or more.

The load factor of an aircraft is given as the ratio of the lift to the weight:

$$n = \frac{L}{W}$$

So, a maneuver that changes any of these two forces acting on the aircraft causes a change in the load factor. Usually, it is the lift which is considered the variable.

Consider an aircraft in a level turn. In addition to the lift and weight, the aircraft experiences centrifugal force, which is counteracted by the horizontal component of the lift.

Courtesy: www.rainierflightservice.com

For a bank angle of $$\theta$$, it can be shown that the load factor can be given as,

$$n = \frac{1}{\cos\theta}$$

So, as the bank becomes steeper and the turn 'tighter', greater lift is required to keep level flight and the load factor increases. A load factor greater than one will cause the stall speed to increase by the square root of the load factor.

As a result, as the load factor is increased, the aircraft's minimum speed should be increased to prevent stall.

For aircraft in steady climb, the principle is essentially the same.

Picture Source: classicairshows.com

Here, the lift is greater than the weight and as such, the load factor is (slightly) greater than one.

As the aircraft stalls in level flight or unaccelerated straight climb, the lift decreases rapidly and so does the load factor.

• The 2nd example is incorrect. In the banking example, a g-meter will show a g-load of 1.15g. In your unaccelerated climb example, the g-meter would read 1.0g once the climb is established. A climb requires an increase in power to sustain and that increase in power has a vertical component and horizontal component. There's a momentary increase in load at climb entry. When the vertical component of increasing thrust is equal to the lift deficit caused by the aftward cant of the lift vector, the plane is in a steady state unaccelerated climb at 1g. Aug 6, 2022 at 18:52
• "So, as the bank becomes steeper and the turn 'tighter', greater lift is required to keep level flight and the load factor increases. " & "As a result, as the load factor is increased, the aircraft's minimum speed should be increased to prevent stall." The second statement doesn't seem correct. Because the aircraft's minimum speed is increased to maintain level flight and prevent stall, the load factor is increased, and not the other way about. (If you just plain roll, load factor doesn't increase - you will start stalling, as the magnitude of total lift will be the same, that until you increa Sep 21, 2022 at 6:16
• @GroundSchooler your underlying premise is fundamentally incorrect. If I bank the plane 30 degrees, the amount of lift I need to produce to remain in level flight increases by 15% to 1.15Gs. If I am producing that lift by pulling back on the stick to increase AOA, I will bleed off speed continuously, eventually stalling. Or I can produce that lift by increasing my speed 7.46%, at which speed I produce enough additional lift that I am not relying on increased AOA, and thus not increasing drag, and thus not decelerating. The load factor was not introduced by the addition of the speed. Sep 21, 2022 at 18:07