I'm a little confused about the definition of the load factor, n, as the definitions I found seem to contradict its application in force and moment balance analysis of an aircraft structure.

The definition given in my university script states that the load factor equals the ratio of all external forces acting on an aircraft minus its weight, to the magnitude of the aircraft's weight; this implies that the load factor is the ratio of all aerodynamic forces acting on an aircraft to the magnitude of its weight.

The Wikipedia definition of the load factor states that the load factor is the ratio of the sum total of all aerodynamic forces acting on an aircraft to its weight.

My script definition therefore tallies with the Wikipedia definition. All good so far.

Now, however, there is also a sentence in my script that states that the load factor is a means of including all inertial loads and gravity loads in observing the loading of an aircraft; this implies that weight is included in observing the loading of an aircraft: this makes better sense to me, since, when expressing the loading of an aircraft, it would be easier to say that the aircraft is now being loaded "this many times its weight", compared to this many newtons; this however, contradicts the definitions above.

And now to the force and moment balance analysis issue, I noticed that all components of an aircraft are expressed in terms of its load factor, which is fine, but then in performing the balance, the weight of the components are not considered. How can you ignore the weight force which is always acting on the aircraft when the load factor (per definition) excludes it? An added confusion is in that lift is considered in the moment balance, with all component's loading expressed in terms of load factor; how can you include lift, when it is already incorporated into the load factor of the components; this implies that the part is further loaded above its current loading.

I would appreciate if someone can straighten up what the load factor means, and how should it be applied in force and moment balance calculations.

  • $\begingroup$ I thought "load factor" was a business economics term for the percentage of revenue seats filled on a commercial aircraft (for a given flight). The load factors of multiple iterations of the same numbered flight are averaged, as in "Southwest Airlines' flight WN4604 has an average load factor of 89%". Perhaps I'm confusing this metric with another name... $\endgroup$
    – pr1268
    Commented Dec 3, 2017 at 14:37
  • $\begingroup$ Guha, weight is not a "real" force, it is an artificial force, like centripetal force, used to make the math work because we are doing our calculation on the earth, in an accelerated frame of reference. Weight does not act on the aircraft. Only aerodynamic forces do. But since we are in an accelerated frame of reference, the "No force" trajectory is opposite the acceleration vector of entire frame of reference. i.e, a ballistic arc constantly curving downwards - away from the acceleration vector (which is upwards.) $\endgroup$ Commented Dec 23, 2017 at 2:21
  • $\begingroup$ I looked at the definition of "load factor" on Wikipedia. They show 4 or 5 meanings for the term, two of which are associated with aviation, so you might consider changing the title of the question to reflect which one you want. $\endgroup$
    – Terry
    Commented Dec 23, 2017 at 20:29

6 Answers 6


The load factor is the total acceleration you feel, pointing downwards. In straight and level flight, the load factor is 1: you only feel the acceleration of gravity. So a load factor 1 equates to 9.81 m/s$^2$ (If gravity was higher, let's say 15 m/s$^2$. load factor 1 would equate to 15 m/s$^2$. But that's another story.)

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Bank the aircraft 60° and fly a co-ordinated turn, and you'll experience a downward acceleration of 2g. This is a case that is easy to understand since it is a static situation with constant velocities. If we have a dynamic sine wave for instance, the load factor would be a function of where we are in the cycle, if the aircraft is accelerating up or down. The actual acceleration is added to the gravity vector.

Load factor does not ignore the weight of the aircraft:

  • In steady horizontal flight, load factor is 1 because it is the same weight that would show up on a scale on the surface of the earth.
  • All dynamic accelerations are added to 1. If the aircraft accelerates downward with -1g, the load factor is zero and everyone on board is weightless.
  • Forces are vector entities. For the load vector, we're only counting the resulting accelerations in the z-axis of the aircraft. After summing the forces in 3 degrees of freedom, the z-axis force F drives the airplane up/downwards with an acceleration of a = F/m.
  • $\begingroup$ Yes, but can you provide an explanation from a structural point of view. Thank you. $\endgroup$
    – Guha.Gubin
    Commented Dec 21, 2017 at 18:46

Load factor is simply Aerodynamic Lift divided by aircraft weight, or more correctly, by aircraft inertial mass as normalized by 1 "G" - (32.2 ft/sec2 or 9.98 m/sec2. - CORRECTION (thx to @Michael Kjorling : 9.80665 m/sec2)

The units are dimensionless (if both Lift and weight are measured in the same units), but we commonly refer to the load factors in "G"s. It really doesn't matter what the angle of bank is. Even if you are upside down, if the wings are producing twice as much lift as the aircraft weighs, you have 2G's Load factor.

To address the unnecessarily complicating issue of gravity, you must understand that considering gravity is only necessary to determine the motion or flight path because an aircraft in flight is moving in an accelerated frame of reference. It would be exactly the same as if your aircraft was in an enormous closed box, filled with air, in outer space, under acceleration from a giant rocket strapped on one side, that accelerated the entire box at 32.2 ft/sec2.

To hold a constant "altitude" in the accelerated frame of reference of this giant box, you would have to trim the aircraft so that the wings were producing lift equal to the "weight" (mass x 32.2ft/sec2) of the aircraft. The Lift on the wings (if it equals the weight, just produces an "upwards" acceleration of 32 ft/sec2, which matches the 32.2ft/sec2 that the box is accelerating, and keeps the aircraft the same distance above the "floor" of the box.

On the earth, we are in an identical 32.2 ft/sec2 accelerated frame of reference, except that is being caused by the gravitational field, and not by an external rocket.

  • $\begingroup$ Do you have a citation for the number 9.98 m/s², or can you at least clarify where it came from? That's quite far from the 9.81-9.82 m/s² commonly used as an approximation for Earth's gravitational accelleration. $\endgroup$
    – user
    Commented Dec 23, 2017 at 13:44
  • 1
    $\begingroup$ Ahhh, I was just remembering the value from when I was in undergraduate school. As it's been quite a few years (1967-1971) since then I am actually surprised I was as close as I was. But you are right. Wikipedia says (en.wikipedia.org/wiki/Standard_gravity) 9.80665 m/sec2. $\endgroup$ Commented Dec 23, 2017 at 15:33
  • $\begingroup$ The best answer above is from Charles Bretana But, additionally can be calculated from : 1/cos(bank angle) $\endgroup$ Commented Mar 22 at 20:01
  • $\begingroup$ User31584.., your answer is only valid when in level flight, i.e., when maintaining constant altitude. It is an a simplification. Very useful as it makes calcukation of load factor simpler, (under thst special condition), but True only when in that special condition. $\endgroup$ Commented May 7 at 13:12

If you are flying straight and level, the weight of your plane is in equilibrium with the lift, and the load factor is lift/weight = 1.

Imagine that you are flying s/l and –for some strange reason– the intensity of the earth's gravitational field, (usually expressed as the acceleration of gravity, g) suddenly becomes 3g. In order to keep flying straight and level, you will have to increase your airspeed, or to (prudently) increase the AoA of the wings. Under those conditions, your wings will be stressed to 3g. The 'load factor' will rise to 3g.

As we all know, the intensity of the gravitational field is quite stable, and you'll never find yourself flying in the conditions mentioned above, but if your plane follows a curved trajectory, the apparent weight of your plane will be increased by the said inertial forces. If the acceleration (normal to the relative wind) associated with those inertial forces is, for instance, 2g, your wings will be stressed by the load due to gravity plus 2g...

  • $\begingroup$ Your definition of load factor here is not quite correct. If the strength of the gravitational field increases, an aircraft will still have a load factor of 1 in unaccelerated, straight and level flight. An airplane flying straight and level in Jupiter’s atmosphere with experience a gravitational acceleration 2.4 times that of an airplane flying straight ald level in Earth’s atmosphere, but both aircraft are subject to a load factor of 1. $\endgroup$ Commented Dec 3, 2017 at 14:16
  • $\begingroup$ @Carlo Felicione To a first approximation, 'inertial forces', i.e. those derived from an acceleration, are indistinguishable from gravitational forces. Hence, a pilot flying s/l can't tell the difference between an increment in the gravitational field and an 'inertial force'. $\endgroup$
    – xxavier
    Commented Dec 3, 2017 at 15:21
  • $\begingroup$ @Carlo, if you define load factor by Lift divided by weight as defined by the gravity field you happen to be in (using your Jupiter analogy) then yes, but this is really stupid, and would yield silly answers. Using this definition, if you were flying a 1000 kilogram aircraft in a giant box in outer space, filled with air, and this box is un-accelerated, an you pulled back on the stick hard enough to generate 1000 newtons of wing Lift, pushing you into the seat with a force of 5 G's, by your definition, (Since aircraft weight is zero in this box), the load factor would be zero? $\endgroup$ Commented Dec 23, 2017 at 2:12
  • $\begingroup$ No, Load factor should be defined as Lift divided by aircraft inertial mass (normalized) and inertial mass does not change no matter what planet you happen to be on. $\endgroup$ Commented Dec 23, 2017 at 2:13
  • $\begingroup$ ...and Romeo, NO, the G-meter on the aircraft on Jupiter would read 2.4 Gs. To keep the aircraft in level flight in a 2.4 G accelerated frame of reference, the wing would have to be generating enough lift equivilent to 2.4 x the aircraft's inertial mass. $\endgroup$ Commented May 7 at 13:17

As another answer stated, Load Factor is simply Aerodynamic Lift divided by aircraft weight. Note that if Aerodynamic Lift is known and weight is known, then one does NOT need to know whether or not the aircraft is accelerating to calculate the Load Factor. If Aerodynamic Lift is known and weight is known, then inertial loads do not need to be considered. If aircraft X weighs 5000 pounds and the aircraft is generating 5000 pounds of lift, then the load factor is 1. We can also call the load factor the G-loading-- or at least the component of the G-loading that we measure on a G-meter-- the component that acts in the up-and-down direction in the aircraft's reference frame. Anyway, if aircraft X weighs 5000 pounds and the aircraft is generating 5000 pounds of lift, then the load factor is 1, regardless of whether the aircraft is in straight-and-level flight, or is banked 45 degrees, or is inverted at the top of a loop. The aircraft's acceleration through space will be different in all these cases, and in all but the first case the flight path will be curving (accelerating) earthward, but the load factor will be the same.

We can also say that the load factor is the "felt" acceleration, or at least the component of the "felt" acceleration that acts in the up-and-down direction of the aircraft's reference frame. The total acceleration is equal to the sum of the "felt" acceleration plus the 1-G downward pull of gravity. Examples-- straight and level flight-- total acceleration 0 G, aerodynamic lift force = 1G * weight, "felt" acceleration 1 G, load factor = 1. Aircraft inverted at top of loop with 1 G showing on the G-meter-- total acceleration 2 G downward, aerodynamic lift force = 1 G * weight, "felt" acceleration 1 G, load factor = 1. So if we DO know the "felt" acceleration, or if we DO know the the total acceleration and the aircraft's attitude in space, then we can calculate the Load Factor without knowing the actual lift force in pounds that the wing is generating, and without knowing the aircraft weight.

Knowing the load factor alone does not tell us how much stress is being exerted on, say, the connection between the wing and the fuselage. This will be affected by how the mass of the aircraft is distributed. It also will be affected by whether or not the horizontal tail is generating a downward lift force, which requires the wing to generate more lift to achieve a given Load Factor. For more on this see this answer to the related question "How does an aircraft's weight affect the V-n diagram?" How does an aircraft's weight affect the V-n diagram?

  • $\begingroup$ (Accidental duplicate) $\endgroup$ Commented Nov 1, 2018 at 1:43
  • $\begingroup$ Second paragraph, "Examples-- straight and level flight-- total acceleration 0 G, aerodynamic lift force = 1G * weight" would better read "Examples-- straight and level flight-- total acceleration 0 G, aerodynamic lift force = 1 * weight". Etc. $\endgroup$ Commented Nov 1, 2018 at 1:43
  • $\begingroup$ This is correct for the load factor at the center of gravity only. Now consider a long fuselage with pilots sitting up in front. Pitch motion and fuselage bending will add their own contribution to the load factor at the pilot position, which may be significantly different from what is measured at the center of gravity. $\endgroup$ Commented Nov 1, 2018 at 2:52
  • $\begingroup$ Thanks @Peter Kampf! Again a good example of non convention of terms turning physics into witchcraft. I would try to explain that applying elevator would create stress loads no matter what the net sum Gs were. Secondly, considering the case of perceived Gs, and applying this aerodynamic stress may be erroneous. Tried to show horizontal loop is constant aerodynamic Gs with constant speed, Vertical loop of constant speed would involve same aerodynamic loads, but gravity vector at top and bottom are reversed WRT pilot, creating different perceived Gs. $\endgroup$ Commented Nov 1, 2018 at 9:34
  • $\begingroup$ Peter Kampf, roger that re CG only; I guess my answers assume the pilot and G-meter are located at the CG; yes clearly pitch rotation rate will affect "felt" acceleration and G-loading at a station well ahead of or aft of the CG. $\endgroup$ Commented Nov 1, 2018 at 9:37

The load factor of an aircraft describes the mass loading of an aircraft as a multiple of its weight. The same definition applies if we are speaking about an aircraft component: the load factor, here called the local load factor, will be the mass loading of that component as a multiple of that component's weight.

First we must understand what mass loading is. The mass load of an aircraft is the inertial loads, as a result of the acceleration of the aircraft and the weight of the aircraft, as a result of the acceleration of gravity. Since the inertial loads always act opposite to the aircraft's acceleration, it is vectorially signed negative. Gravity, as it always points downwards, is in sync with our coordinate system which defines the downward direction as positive; hence, it will be vectorially signed positive. Summing the two vectors will give us the total mass loading of the aircraft. Dividing this total loading by the aircraft's own weight will give us an idea as to how much more that its own weight is the aircraft is currently being loaded by the maneuver it is performing, and by that, how structurally strong it is or needs to be.

  • $\begingroup$ Really boils down to load forces created by acceleration and gravity, separately considered, then applied to the orientation in the gravity field. Your answer provides a starting point by using the universal and unchanging gravity vector as a standard reference signed positive. $\endgroup$ Commented Nov 1, 2018 at 19:17

A lot of good answers up there.
After going through all of them there was one question that still lingered in my mind. Why do we neglect the vertical component of lift in considering the load factor? Also why should the weight be considered since it is anyway getting offset by the vertical component of the lift force. The answer to that question simply lies in the way we define load factor. Examining the definition of load factor carefully we understand that it is the net force that the wings of an aircraft have to provide to accomplish a certain kind of maneuver. Since the wings provide this amount of force they experience an equal amount of loading too and this is what we refer to as the load factor. In simpler terms the amount of force that the wings are supposed to provide is simply the amount of force they end up enduring. Consider the same in a coordinated turn. Keep the lift forces aside and examine the other forces that are required for the maneuver along with the weight (which is something that the aircraft structure anyway has to support). This includes the centripetal force and the weight for a simple banked turn of an aircraft. Thus the net force required for the turn is the vector sum of the weight and the centripetal force. This (by drawing the free body diagram of the aircraft) simply comes out to be the net lift force acting on the aircraft. Divide it by the weight of the aircraft and you get the load factor. If the bank angle is $\theta$ then the load factor comes out to be sec$\theta$. And that explains the graph tending to infinity as the bank angle approaches 90 degrees.

  • $\begingroup$ We neglect the vertical component of lift because the orientation of the lift has nothing to do with the force of that lift. Load Factor is just a representation of the magnitude of the force the lift is exerting on the aircraft. What direction the force is acting in is not relevant. The aircraft structure does not have to support the weight. It only has to support the force of the aerodynamic force exerted on it, i.e., by the Lift. $\endgroup$ Commented Mar 23 at 2:14
  • $\begingroup$ If you unload the aircraft (push the stick forward until the G meter reads zero), there is no load on the airframe and the load factor is zero. The weight has certainly not disappeared. The aircraft still weighs what it weighs. But the Lift is gone, so the load is zero. $\endgroup$ Commented Mar 23 at 2:18

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