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.)


3 Answers 3


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).

  • 9
    $\begingroup$ 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. $\endgroup$ Oct 19, 2022 at 8:53
  • $\begingroup$ 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: aviation.stackexchange.com/questions/95313/… $\endgroup$ Oct 20, 2022 at 13:10
  • 1
    $\begingroup$ 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. $\endgroup$ Oct 21, 2022 at 3:49
  • 1
    $\begingroup$ @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. $\endgroup$ Oct 21, 2022 at 4:06
  • 1
    $\begingroup$ 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. $\endgroup$
    – Ralph J
    Oct 21, 2022 at 4:32

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.

  • $\begingroup$ 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-- $\endgroup$ Oct 20, 2022 at 12:11

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

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Jamiec
    Oct 21, 2022 at 12:16

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .