9

This is meant to be a question about terminology, not a request for an explanation of the underlying physics at play.

When an airplane is in constant-speed straight-and-level upright flight with the longitudinal axis horizontal, or when an airplane is at rest on level ground with the longitudinal axis horizontal, is the "normal acceleration" (nz), which is defined as the component of the linear acceleration of an aircraft along the body Z axis, considered to be 1 G, or 0, or does it vary depending on the context?

If the latter, how so? For example, what answer would a flight test engineer give?

Bonus question-- in cases where the "normal acceleration" (nz) is considered to be 0 G rather than 1 G in constant-speed straight-and-level upright flight with the aircraft's longitudinal axis horizontal -- if any such cases exist--- then what is the "normal acceleration" (nz) considered to be in a constant-speed 60-degree-banked turn with the aircraft's longitudinal axis horizontal, with the G-meter reading 2 G's? Is nz considered to be 1 G, or is nz considered to be 1.5 G? In other words, have we simply shifted all the nz values downward by 1 G, or have we switched to an entirely different method of calculating nz, based on the net acceleration acting on the aircraft, rather than the felt or non-gravitational acceleration acting on the aircraft? (For more context, see this related ASE answer.)

quiet flyer
  • 22,598
  • 5
  • 45
  • 129
  • 1
    Not a duplicate of https://aviation.stackexchange.com/questions/95313/aircraft-load-factor-and-body-normal-acceleration , because that question is so broad (contains many questions in one) that no answer is likely to decisively answer this question. – quiet flyer Oct 18 '22 at 17:16
  • I think the real answer is given here -- https://aviation.stackexchange.com/a/95531/34686 – quiet flyer Oct 21 '22 at 08:28
  • Highly relevant, esp comments from 10-27-22 onwards-- https://chat.stackexchange.com/rooms/140031/discussion-on-answer-by-sophit-terminology-is-the-normal-acceleration-nz-co -- (answer was apparently deleted) -- adopting the 0G convention for straight and level flight leads to real problems in other attitudes. – quiet flyer Oct 28 '22 at 15:23

3 Answers3

20

At zero G, you're floating (Vomit Comet ride). At 1 G, you have normal weight. At 2 G's, you seem to weigh twice as much as usual (60 degree banked turn).

The normal aviation convention is that 1 G is straight & level unaccelerated flight, same as sitting on the ground.

A typical aircraft G-meter:

enter image description here

Note that G's are not acceleration; they are force per unit mass. They may produce acceleration, but if the answer is expressed in G's, then the force produced by gravity counts. Even when it's being opposed by the push of the chair you're sitting in.

If the question is "what is your acceleration" (delta velocity) then the answer comes in units of meters per second-squared, and sitting in the chair on the floor, the answer is zero. If the question is, how many G's do you experience in that condition, the answer is in units of G's (or force per mass), and the answer is 1.0. If the question is "what do you weigh," then the answer comes in units of force (i.e. Newtons). If the question is "how much mass do you have," then the answer comes in units of mass, i.e. kilograms.

If the expected answer will have units of G's, then (on earth) when sitting still, the scaler part of the answer is 1. On the moon, it would be 1/6th of a G (even tho delta V would still be zero).

Ralph J
  • 51,356
  • 17
  • 157
  • 249
  • 9
    To be entirely technically correct, it should be noted that the units of g and of acceleration are the same. The SI units of force per unit mass are N/kg = (kg * m/s²)/kg = m/s². That is not a coincidence: g is an acceleration: the acceleration observed by an object in free fall near the earth's surface. In special relativity there isn't even a real difference between gravity and acceleration. Conventionally of course it can make sense to make a distinction. – Roel Schroeven Oct 19 '22 at 08:53
  • Re "but if the answer is expressed in G's, then the force produced by gravity counts." -- but does it really? Sitting in chair-- chair is pushing up on you at 1 G * mass-- hence the G-meter you are holding in your hand reads 1 G. Even though gravity is also pulling down at 1 G * mass, producing a net force of zero. Then the floor breaks and you are falling and gravity is still exerting a force on you, yet the G-meter reads zero. Related: https://aviation.stackexchange.com/questions/95313/aircraft-load-factor-and-body-normal-acceleration – quiet flyer Oct 20 '22 at 13:10
  • @quietflyer "G's" are force experienced, not force exerted. Free-falling & experiencing "0 G" in the Vomit Comet means you *feel* 0 G's, not that Earth's gravity no longer acts on you. If that weren't the case, we could never discuss an "zero G" situation. – Ralph J Oct 20 '22 at 16:49
  • Yes, but force "experienced" or "felt" by object is identical to the non-gravitational component of the net force exerted on object. (Or your could say, the mirror image of) That's why I've been talking about subtracting out the gravitational force from the net force. And that's why the G-meter tells you the aerodynamic force (per unit mass) acting in the body Z axis, which is essentially the same as the lift force (per unit mass) generated by the wings. (Except for the discrepancy due to lift vector not always being perfectly aligned with body Z axis.) – quiet flyer Oct 20 '22 at 22:54
  • Related: https://aviation.stackexchange.com/a/95531/34686 – quiet flyer Oct 20 '22 at 23:02
  • 1
    Gs *are* acceleration. It's just that G's measure the acceleration you would measure if you were measuring acceleration in an *Inertial* (non-accelerated) frame of reference. The surface of the Earth is an accelerated frame, exactly the same as if you measure G's in a rocket ship out in space that was accelerating at 1 "G". The entire frame of reference is accelerating at 1 G so the G-Meter reads the acceleration of the frame of reference. A body in free-fall is experiencing zero G and zero acceleration in an inertial frame of reference. – Charles Bretana Oct 21 '22 at 03:49
  • Technically, we now understand that Gravitation is NOT a force. It is just the tendency of all bodies, when no force is acting on them, to travel in a straight line in space-time. A body in free-fall is indeed traveling in a straight line in four-dimensional space-time. Even if you graph the motion of objects traveling at widely different speeds, in an accelerated frame, you discover that in a space-time graph, their curved paths all have exactly identical curvatures in space-time - the curvature indicates nothing more than the acceleration of the frame of reference. – Charles Bretana Oct 21 '22 at 03:55
  • @CharlesBretana "G" stands for "gravity" - 1G = 1.0 x the earth's gravity, 2 G's = 2x earth's gravity, and so on. Gravity is a *force*. That force can produce an acceleration, but gravity is a force, and a "G" is that force, normalized by dividing out the mass of whatever is being discussed (pilot, aircraft, rocket payload, etc). At 1G, there is no "delta V" happening. That one G is not "an acceleration". – Ralph J Oct 21 '22 at 03:59
  • All the confusion results from the WRONG assumption that there is a natural, normal, absolute frame of reference that we need to measure everything in, and we assume that the stable surface of the earth is that absolute frame of reference. Acceleration is no different from Velocity. in that it is *RELATIVE* The acceleration you will measure depends completely on what frame of reference you measure it relative to. if you measure it in an accelerated frame, you need to create a fictional psuedo force (gravity) to make the equations balance. – Charles Bretana Oct 21 '22 at 04:03
  • 1
    @Ralph, No it is not a force. It is a curvature in Space-Time. No such force exists. Einstein proved that some time ago. You are still thinking Newtonian. Science has moved past Newton. – Charles Bretana Oct 21 '22 at 04:06
  • This is, philosophically, identical to the issue of Centrifugal and centripetal forces. The 1 G you feel when on a rotating space station, like in 2001, or the Gs you feel in a centrifuge, are explained away by creating fictional psuedo force called centripetal. This is only necessary because we are measuring the acceleration in an accelerated frame of reference. We add the psuedo force to make the forces balance. – Charles Bretana Oct 21 '22 at 04:11
  • In normal Pilot Math & Engineering, gravity is a force, and it can cause an acceleration. If I drop an object, it accelerates down; that change in velocity is described by F=MA, and gravity acts like every other "force" we encounter. Relativistic physics & unobservable frames of reference and abstracting away "the force of gravity" is all well & good, but when you're discussing aviation with actual aviators, gravity most certainly IS a force that is recognized and understood, so that it can be safely overcome with lift and thrust and lots & lots of money. – Ralph J Oct 21 '22 at 04:31
  • 1
    Aviation works and is taught & understood in Newtonian physics. Besides, the question is about terminology & convention, *not* about physics. Injecting "curvature of space-time" into this discussion is unhelpful & unproductive to understanding why the G-meter points to "1" rather than "0" when at rest. – Ralph J Oct 21 '22 at 04:32
  • @RalphJ, sticking to Newtonian physics is making things more complicated here, not simpler, because you keep distinguishing gravity from acceleration. General relativity lumps them together (and you don't need to mention curvature of space-time, just take the basic postulate that “gravity is locally indistinguishable from effect of accelerating reference frame”) and now the G-meter simply shows the experienced inertial acceleration. Which is opposed by aerodynamic forces on the airplane as those are all the available real forces in flight. – Jan Hudec Oct 21 '22 at 06:06
  • @RalphJ, also note that the term ‘gravity’ does not actually mean the force due to gravitation, but includes the centrifugal force due to rotation of Earth. – Jan Hudec Oct 21 '22 at 06:08
  • I guess we're outta luck at the north and south poles, and one may muse that flying through Los Angeles smog may increase bouyancy. Abstractions aside, it is very important to differentiate gravity from the forces the aircraft must produce because gravity will always affect its flight path (until Starship leaves Earth's gravity well), and a specific amount of force must be produced to counteract it, along with the force required to counteract drag. – Robert DiGiovanni Oct 21 '22 at 11:51
  • Now throw in the amount of force needed to overcome inertia and turn the aircraft. That's your wing putting out 1.15 G in a 30 degree turn. No doubt the engine works a little harder too, to overcome increased drag. Stay coordinated, my friends, and go ahead, draw the force vector diagram, then include gravitational force. Viola, the aircraft is in a level turn. – Robert DiGiovanni Oct 21 '22 at 12:04
7

It really depends on the convention. I've seen both answers used before, but most Avionics software I've seen assumes the 0g convention so that normal acceleration is a rough approximation for the amount of vertical acceleration.

ARINC for example defines multiple labels for IRUs. The standard one is Label 333 for Body Normal Acceleration (0 g on ground), but there's also for on some hardware label 370 unbiased normal acceleration (1 g on ground).

It seems intuitive to have normal acceleration be 0 when in constant-speed straight and level flight as this translates easily to other parameters like vertical acceleration and works better for math based on ratios and magnitudes. For example with this convention 0.75 g is just as bad as -0.75 g. On the other hand having normal acceleration typically be 1g makes it easier to translate into lift or load factors like Az/W. Whichever way, this bias to 0g usually doesn't account for roll or high pitch- the bias to normal acceleration should really be cos(pitch)*cos(roll)*(1g). So no matter the convention you'll likely need to subtract the gravity vector and there's no free lunch.

Cody P
  • 6,773
  • 2
  • 26
  • 55
  • Thanks for your answer. Interested in your input on the (newly added) last part of the question as well -- leave a brief comment if don't want to modify answer-- – quiet flyer Oct 20 '22 at 12:11
0

Wow, can terminology produce some lively discussion. No wonder the world has so many languages!

Here is a reference on where the term "normal acceleration" comes from and how can apply to the z axis of aircraft. Normal acceleration is actually a component of centripetal force perpendicular to the path of an object moving in a circle.

Tangential acceleration is a component of force acting in the path of the object affecting its velocity (operating against drag).

With aircraft, in straight and level flight, the "normal acceleration" of 1 G along the z axis (approximately)$^1$ balances the gravitational force of 1 G.

Sitting stationary on the ground, normal acceleration of the aircraft is 0. The aircraft is not producing any force.

Why we have G meters? They are for the human limits as well as the structural limits. It is important to understand the difference due to gravity between the "felt" loads and what the plane is actually doing. Back to our 30 degree turn. G forces are 1.15. The plane is actually generating 0.57 G by accelerating to the side, with 0 vertical acceleration (gravity balances the vertical force), and is also yawing to keep the nose aligned with the relative wind as much as possible (lowest drag).

The "ball" is also accelerating sideways at 0.57 G, but because it has no vertical acceleration, it also has 1 G of gravitational "pull" downwards. As a result, the ball will align itself with the z axis as long as the nose is kept coordinated.

If the nose is not properly aligned, then the glass tube/ball are no longer in the plane of the "normal force". This will show as the ball being too far outside in a skidding turn, and too far inside in a slipping turn.

$^1$ the vertical component of it, taking angle of attack into account

Robert DiGiovanni
  • 20,216
  • 2
  • 24
  • 73