What is the definition of load factor and how do you apply it?

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.

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.

• 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... – pr1268 Dec 3 '17 at 14:37
• 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.) – Charles Bretana Dec 23 '17 at 2:21
• 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. – Terry Dec 23 '17 at 20:29

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.)

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.
• Yes, but can you provide an explanation from a structural point of view. Thank you. – Guha.Gubin Dec 21 '17 at 18:46

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...

• 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. – Carlo Felicione Dec 3 '17 at 14:16
• @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'. – xxavier Dec 3 '17 at 15:21
• @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? – Charles Bretana Dec 23 '17 at 2:12
• 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. – Charles Bretana Dec 23 '17 at 2:13

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.

• 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. – a CVn Dec 23 '17 at 13:44
• 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. – Charles Bretana Dec 23 '17 at 15:33

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?

• (Accidental duplicate) – quiet flyer Nov 1 '18 at 1:43
• 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. – quiet flyer Nov 1 '18 at 1:43
• 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. – Peter Kämpf Nov 1 '18 at 2:52
• 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. – Robert DiGiovanni Nov 1 '18 at 9:34
• 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. – quiet flyer Nov 1 '18 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.

• 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. – Robert DiGiovanni Nov 1 '18 at 19:17

Okay.
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.

Another case where definitions and language make for very interesting discussion. So we will start with the universal and unchanging force vector, gravity, and how it loads the plane at rest (parked on the tarmac).

Strap in and let's go up! G "felt" by pilot=1. Aerodynamic load on wing=0. Gravity load on wing= 1 Total load= 1

Go to steady straight level cruise flight. G "felt"=1 Aerodynamic load=-1 (it is a flying wing as this is easier for now) Gravity load=1 Total load = 0. Lift and Gravity cancel out.

Top of 1G loop. G "felt"=1. Aerodynamic load=-2 (to create 1G inverted for pilot) Gravity load=1 Total load = -1G (+1G WRT inverted pilot and plane).

Now, at top of loop, go Immelmann and level out inverted. G "felt"= -1 (pilot and plane inverted) Aerodynamic load=-1.

Roll to level and land.

The structural stress of the aircraft is the sum of gravity and aerodynamic loads. This is the important concept as the orientation of the aircraft in the gravitational field can change.

The books can be confusing, approaching from reality can make understanding a bit easier.

Answer has been modified to use positive gravity vector as written by @Guha.Gubin and comments from @quiet flyer.

Invite comments from aerobatic flyers!

• Robert, I hate to keep disagrng w/ you, but in case where a/c is parked, I think you forgot to consider the force of the earth pushing up through the tires. The total acceleration in this case is zero G, while the "felt" acceleration is one G. Just like in level flight. The rest of your answer, more or less seems right. Except to say that the structural stress on the a/c includes gravitational loads. I'd say that the structural stress relates only to the "felt" acceleration. – quiet flyer Oct 31 '18 at 20:09
• Which is the same thing as saying the structural stress relates to the actual acceleration minus the gravitational load. Hmm recipe for an argument there, two ways to say the same thing. – quiet flyer Oct 31 '18 at 20:10
• Yes, I guess if wheels were under wings, true! Wings are hanging down however. Appreciate your attentiveness, I was expecting to be jumped for the top of the loop because I did not consider turning (centripetal) force. Could also try to figure stress loads on specific part of aircraft. I just feel for folks when they get confusing definitions, so I continue to wade forward. – Robert DiGiovanni Oct 31 '18 at 20:21
• So, hopefully, Jan, Peter, Federico, Tanner, and others, if you are out there, what about that loop? – Robert DiGiovanni Oct 31 '18 at 20:23
• A key point is that "wing loading" does NOT mean the net force pulling the wings up or down relative to fuselage-- if it were, in the tarmac case we'd have to know whether the wheels were on the fuselage or the wings. In the in-flight cases, we'd also have to subtract the weight of the wings (multiplied by the G-loading) away from the aerodynamic force generated by the wings to get the wing loading. That is not the standard usage of "Wing Loading" in aviation as I understand it. – quiet flyer Oct 31 '18 at 20:59