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A load factor is defined as "the ratio of the lift of an aircraft to its weight", i.e., N = Lift/Weight.

Normal acceleration (nz) is defined as the component of the linear acceleration of an aircraft along the body Z axis.

So my questions is: does the load factor N is the same as the normal acceleration nz? I thought normal acceleration computation should be more complicated to include all the forces'(weight, lift, drag, thrust)components acting on the body z axis.

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    $\begingroup$ What was the actual question here? Are you asking if your understanding in the final paragraph is correct? $\endgroup$
    – Jamiec
    Oct 10 at 11:40
  • $\begingroup$ It would seem "the ball" gives an accurate picture opposite of the direction of net aerodynamic forces + the contribution of gravity This is why additional yaw must be added to center it on the z axis in a turn. So now you have gravity + lateral acceleration + yaw to consider. $\endgroup$ Oct 13 at 15:30
  • $\begingroup$ @RobertDiGiovanni -- no, the ball just shows the (mirror image of the) net aerodynamic forces (including thrust if applicable), or at least the component thereof that acts in the "plane" of the curved tube. Ball deflection = aerodynamic actual sideforce. But, best discussed in comments under one of several existing questions on the slip-skid ball, or in a new chat room. $\endgroup$ Oct 13 at 15:51
  • $\begingroup$ No, the ball shows "centripetal" and gravity forces as per this reference. Always happy to help refresh knowledge of even experienced pilots. $\endgroup$ Oct 13 at 17:55
  • $\begingroup$ @RobertDiGiovanni -- both statements are true. If you only specify the "centripetal" force, then you must also include gravity into the calculation to see where the ball will end up. If you only specify the net aerodynamic force vector, specifically it's direction relative to the aircraft (don't even need to know the magnitude), then that's all the information you need to know where the ball will be. Don't even need to know bank angle, or whether or not aircraft is doing a wingover or a partial loop. Don't need to know orientation of gravity vector (vertical) relative to aircraft. $\endgroup$ Oct 13 at 18:06

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EDIT: I modified my answer in order for it to match the modifications in the question.


Unfortunately load factor is almost always explained making the airplane doing some cool manoeuvre à la Top Gun: from there to picturing a sweaty Tom Cruise trying to avoid greyout, is just a short step. And to complicate the matter even more, gravity, Gs, apparent weight and other fancy terms are also drawn into play...

But load factor is just what you have written: lift to weight ratio. As simple as that.

The picture in your question (picture which is now gone and that I repost here under) is a very good example to answer your question:

steady turn

In this example of a steady turn, the load factor $N$, with $\phi=30°$, is:

$N=\frac{L}{W}=\frac{W/\cos\phi}{W}=1.155$

and that simply means that the wing is producing some 15.5% more lift than the airplane's weight (note: weight, not mass $\Leftrightarrow$ Newton, not kg).

does the load factor N is the same as the normal acceleration nz?

This should be better formulated: load factor is a pure number while acceleration is length/time², so they cannot be really compared. "Normal" is also ambiguous. As a consequence I suppose here that "normal" aka perpendicular is the direction going from belly to canopy and that the question is if load factor and "normal lift divided by weight" are the same.

Well, the answer is easy if you picture the airplane from the side at high AoA (image from Wikipedia modified by me):

F-16 at high angle of incidence

By definition, lift is perpendicular to the airflow and in this case its direction is not coincident with the airplane's perpendicular direction z. So the general answer is simply no.

I thought normal acceleration computation should be more complicated to include all the forces' (weight, lift, drag, thrust) components acting on the body z axis.

Theoretically yes but practically it depends on how mutually big their z components are. In the first picture for the steady turn, drag and thrust are basically aligned with the x-axis and do not contribute on the z-y directions and therefore there's no need to consider them in calculation on z or y.


Bonus material

  • Where does the 15.5% higher load factor of the first calculation go? It goes in the force $F_r$ which makes the airplane stay in a curved path.

  • Why is only lift considered in the calculation of N? Lift is by far the biggest aerodynamic force produced by the wing and therefore the highest source of stress by a structural point of view.

  • Why the ratio lift to weight is considered and not the actual lift? Because aerospace engineers like numbers, i.e. quantities without dimension: Reynolds number, Mach number, lift coefficient, drag coefficient, ... Dimensionless quantities cannot be mistaken (lb vs. lbf vs. ft$\cdot$lbf for example) and they make comparisons easier.

  • In the aerospace world there are at least 3 different reference systems which can be used to do calculations: the one attached with the earth (first picture), the one attached with the body (second picture) and the one attached to the airflow. The one attached to the body can also be further expanded according to where it starts and how it is oriented: it can for example originate in the CG and have the z-axis pointing downward; or it can originate somewhere in front and under the nose of the airplane and have the z-axis pointing upward.

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Normal acceleration (nz) is defined as the component of the linear acceleration of an aircraft along the body Z axis.

To a physicist1, the term "linear acceleration" in the above definition would mean the net linear acceleration.

The net linear acceleration is the acceleration due to vector sum of all the forces acting on the aircraft, including gravity.

In the context of coordinated flight at a constant airspeed, with a horizontal flight path, Thrust and Drag are equal and opposite, and the net linear acceleration is the acceleration due to the vector sum of Lift and Weight.

In this case, if we draw the closed vector triangle of Weight, Lift, and resulting net force, we obtain the following:

Lift = Weight / cosine (bank angle)

Net force = Lift * sine (bank angle) = Weight * tan (bank angle)

If we make the simplifying assumption that the wing incidence is such that the fuselage is exactly horizontal, so that the Lift vector is entirely aligned with the aircraft's body Z axis, then to obtain the component of this net force that acts parallel to the aircraft's body Z axis, we "project" the net force onto the body Z axis by multiplying by sine (bank angle). (Similarly, to compute the net "lateral" force-- the net force component acting parallel to the aircraft's body Y axis, we would multiply the net force by cosine (bank angle).)

To compute accelerations rather than forces, we simply substitute "1G" for "Weight".

This yields the following:

Lift = Weight / cosine (bank angle)

Net force - Lift * sine (bank angle) = Weight * tan (bank angle)

Net acceleration = 1G * tangent (bank angle). (Note that the net acceleration is purely horizontal.)

az = component of net acceleration acting in body Z axis = 1G * tan (bank angle) * sine (bank angle)

ay = component of net acceleration acting in body Y axis = 1G * tan (bank angle) * cosine (bank angle) = 1G * sine (bank angle)

This yields the following results:

0 degrees bank:

Lift = Weight, net acceleration = 0 G, az = 0 G, ay = 0 G

30 degrees bank:

Lift = 1.15 Weight, net acceleration = 0.577 G horizontal, az = 0.289 G, ay = 0.500 G

45 degrees bank:

Lift = 1.414 Weight, net acceleration = 1 G horizontal, az = 0.707 G, ay = 0.707 G

60 degrees bank:

Lift = 2.000 Weight, net acceleration = 1.732 G horizontal, az = 1.500 G, ay = 0.866 G

It is difficult to see why az and ay figures computed in this manner would be of much interest to anyone.

If we wanted to actually predict the aircraft's trajectory, we would work with the net force, not the body-Z and body-Y components of the net force.

And note that when we compute the az figure as above-- by simply taking the body-Z component of the net force acting on the aircraft-- then the resulting figure does not correlate to what we read on a G-meter (which is simply a 1-axis accelerometer, with the sensing axis aligned with the aircraft body Z axis.) (And likewise, the ay or "lateral acceleration" figure as computed above bears no relation to the deflection of a slip-skid ball-- assuming the fuselage is pointing directly into the relative wind, the ball would be centered in all of these turns, even though the ay figure is not zero.)

Note also that the acceleration component due to gravity creates no stresses or strains on the aircraft structure, or on the contents of the aircraft, because gravity "acts from within", creating an equal acceleration on every molecule of the aircraft and occupants.

The logical conclusion is that in the aviation / aerospace context, az as described above is usually not what we mean by the "normal acceleration" nz. In the aviation / aerospace context, the "normal acceleration" nz usually refers to the body-Z axis component of the "felt" acceleration, not the body-Z axis of the net acceleration.

The "felt" acceleration is the acceleration resulting from the net vector sum of the non-gravitational forces acting on the aircraft, not the net sum of all the forces acting on the aircraft.

At rest on the ground, the acceleration "felt" by an aircraft is the acceleration component due to the force of the earth pushing up against the landing gear, with a force of 1 G per unit mass.

In flight, the acceleration "felt" by an aircraft is the acceleration component due to the vector sum of all the aerodynamic and thrust forces, divided by mass.

The "felt" acceleration, not the net acceleration, is what of interest if we are looking at the loads transmitted through the aircraft structure and the resulting stresses and strains.

The body-Z component of the "felt" acceleration, not the net acceleration, is what is registered by the G-meter.

(And the ratio between the body-Y and body-Z components of the "felt" acceleration, not the net acceleration, determines the displacement of the slip-skid ball. The slip-skid ball is centered whenever the body-Y component of the "felt" acceleration is zero, which is also when aerodynamic sideforce is zero.)

In the context of banked, turning flight at a constant airspeed, with a horizontal flight path, with the wing incidence is such that the fuselage is exactly horizontal, with zero sideslip, all the "felt" acceleration is parallel to the aircraft's body Z axis. In this case, nz defined as the body-Z axis component of the "felt" acceleration is exactly equal to 1G * (Lift / Weight).

(Naturally, if the flight path is horizontal but the fuselage is not exactly horizontal, then the Lift vector will no longer be aligned with the aircraft body Z axis, and nz as defined above will be less than (1G * Lift / Weight). As illustrated in this answer.)

References from outside sources:

Here is an example of a technical engineering paper where "normal acceleration" in straight-and-level flight is treated as 1 G rather than 0 -- "A Method for Correcting The Error in Indicated Normal Acceleration Due to G-Sensor Location" by Mi Yi and Chen Mingtai of the Shanghai Aircraft Design and Research Institute-- (link to download PDF)

Note: the designations "az" and "ay" were introduced for convenience for this answer; there is no intent to represent that they are standard terminology.

Footnotes:

  1. A physicist would also object to the use of the word "normal" to mean "body Z axis component of". "Normal" means "perpendicular to the trajectory". (See for example this link.) What force component should properly be considered the "normal force" in a wings-level skidding turn, performed essentially with the rudder alone (plus opposite aileron as needed to keep the wings level)? Or in a control-line model airplane, racing around in a circle with the nose pointing directly into the airflow and relative wind, and with the wings completely level?
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  • $\begingroup$ Thanks, so in a coordinated turn, assuming the body axis align with the wind axis(or the alpha is small), lift could be considered as the dominate external force acting on the body Z axis, so we could approximate the sensed nz as Lift/mass*g. I think approximating sensed nz by lift/weight should be applied to other cases like aircraft is not in a levelling turn but in a climb or descent turning, not sure if it is still accurate if the turn is not coordinated when there is a side force acting on the body y axis. $\endgroup$
    – VvV
    Oct 19 at 15:30
  • $\begingroup$ @VvV -- I would say, given the constraints in first part of your comment, sensed nz is exactly 1G * Lift /Weight. In a climb or dive, however, the lift vector is less than in level flight. See for example aviation.stackexchange.com/questions/40921/… . $\endgroup$ Oct 20 at 12:54
  • $\begingroup$ @VvV -- when the turn is not coordinated, this definitely affects nz. For example in a "slip" there is a sideforce that partially opposes gravity and supports some of the aircraft weight, unloading the wing, compared to a coordinated turn at the same bank angle. Extreme case: sustained, linear knife-edge flight (90-degree bank angle). Lift from wings is zero, nz is zero. Conversely in a "skid", nz is decreased, and the wing is unloaded, compared to a banked, coordinated maneuver that would yield the same turn rate. So -- yeah, it's a little complicated! $\endgroup$ Oct 20 at 13:16
  • $\begingroup$ @VvV -- note that in slip or skid, Lift vector and nz are both affected in same way, so relationship between them stays same-- $\endgroup$ Oct 20 at 14:06

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