I've been reading an article on the BBC Website about the 21/May/2024 incident on Singapore Airlines. According to this article:

"The rapid changes in G over the 4.6 seconds duration resulted in an altitude drop of 178ft (54m), from 37,362ft to 37,184ft," the TSIB report said.


The Transport Safety Investigation Bureau's (TSIB) preliminary findings found rapid changes in gravitational force (G) and the altitude drop ... likely injured those who were not wearing seatbelts.

My knowledge of physics tells me that the gravitational force is dependent on the masses of the two bodies (the aircraft and the Earth), the distance between them and 'G' a universal constant. Given the the Gravitational constant can not change and mass of the Earth is so much bigger than the other variables then in practical terms gravitational force should not be changing.

So what are they really saying is the problem?

  • 6
    $\begingroup$ G is not gravity in this instance but g-force: a force between a human body and another solid object: eg. your seat or seatbelt. For example if you drive your car and suddenly brake you will feel a strong force pulling you forward against your seatbelt. That's what G refers to. $\endgroup$
    – slebetman
    Commented May 30 at 9:04
  • $\begingroup$ This sounds like a poor translation $\endgroup$
    – mmathis
    Commented May 31 at 18:12

4 Answers 4


The TSIB preliminary findings and the BBC use "G" as a short-hand for g-force, which is not the same thing as "gravitational force". They are using the wrong terminology here.

The g-force is actually a measure of acceleration and not a true force at all. This acceleration doesn't necessarily come from gravity alone, it can include other accelerations.

The g-force or gravitational force equivalent is mass-specific force (force per unit mass), expressed in units of standard gravity (symbol $g$ or $g_0$, not to be confused with "g", the symbol for grams). It is used for sustained accelerations, that cause a perception of weight. For example, an object at rest on Earth's surface is subject to 1 g, equaling the conventional value of gravitational acceleration on Earth, about 9.8 m/s².

(Wikipedia: g-force)

The actual gravitational force did not change at all during this incident. The acceleration the aircraft experienced was due to "an area of developing convective activity" causing severe turbulence, i.e. a result of the interaction with the surrounding air, not gravity.


In short

BBC describes the force variation on persons and objects during the turbulent/windshear event. They say the magnitude was subject to rapid changes.

This force is sometimes called the g-force, it results from an acceleration of x g units (e.g. 2g). The confusion between unit of acceleration g and gravitational constant G is usual in aviation, and in other domains. But sometimes the correct unit is used, like in this profile of a parabolic flight:

enter image description here

(Source.) Note the force is always positive, it's a magnitude, not a vector.

The g-force corresponding to the regular weight is 1g. It's a unitary quantity (the force exerted on a unit of mass), and unless the direction is specified, e.g. $g_z$, it should be understood as a magnitude.

The force created by all accelerations on a body cannot be split between gravity and other causes. The resultant is perceived as a weight. If one says the force is 3g, it means something like "the magnitude of the force on any object of any mass is 3 times what it would be if the gravity were 3 times stronger". Perhaps this force is upwards.

This is rather informal and only aimed to provide an idea of the scale: A person would feel as if their weight was multiplied by 3, whatever their weight is. A person aboard the ISS is subject to 0g, and feels weightless (in the rotating reference frame neither the gravity nor the weight have actually changed, but another force, the centrifugal force, is cancelling the effects of gravity).

Details follow.

Some definitions

G constant: A constant which value is 6.674e-11 N.m²/kg². As we can see time is not part of the dimension, hence it's neither an acceleration (dimension m/s²) nor a force (N). This constant is used to calculate the attraction force between two bodies.

g: A unit of acceleration, a multiple of the regular SI unit (m/s²): 9.81 m/s². An italic is used to differentiate this unit from the SI unit gram.

Why this unit? When one of the two bodies is much larger than the other, like Earth compared to objects on Earth, it's convenient to see the attraction of gravity as resulting only from Earth. Seen like this, the force "exerted by Earth" has a value equal to 9.81 times the mass of the small object, and is called weight. Earth mass and distance have been removed from the equation. We say Earth is pulling the object at a rate of 1g. Technically the unit $g_0$ should be used to talk about Earth gravity acceleration, g should be used only when other accelerations are added.

Note in aviation field, g is often misspelled G.

g-force: It's the force pulling the object, created by the sum of all accelerations. Some of these accelerations are due to gravity, but other are due to body displacement. Related to the load factor: A load factor of 2 means a g-force of 2g.

Let's say we are in a lift going up with a velocity increasing by 9.81 m/s each second. Our body senses the weight due to gravity, and also a force from the lift floor which magnitude is equal to the weight. This force is due to the acceleration of the lift, not to gravity. Still we feel like our weight has doubled. We say the g-force is 2g.

Applied to turbulence

In this case air had an unusual vertical velocity in the vicinity of a jet-stream, creating a strong erratic wind (windshear). This can happen from 6,000ft below the tropopause to 3,000ft above. This is called clear-air turbulence (CAT), in contrast with the usual turbulence occurring in visible clouds.

The aircraft still fly within this moving air at its constant cruise airspeed, but w.r.t. the ground, the aircraft velocity is the vector sum of both velocities. This velocity is erratic, meaning the aircraft is accelerated, like our previous lift. The direction of this acceleration is also erratic, so the g-force exerted by the aircraft on passengers and objects is erratic, pulling/pushing passengers and objects in erratic directions.

So the mechanism is not the "The rapid changes in G over the 4.6 seconds duration resulted in an altitude drop".

  • G has nothing to do here, it should be the g-force or the total acceleration (which includes gravity, but also accelerations not related to gravity).

  • The g-force doesn't create the change in altitude. It's the contrary, the change in altitude creates the g-force.

  • The change in altitude is due to the vertical windshear associated with the turbulence.

Acceleration magnitude

The BBC article mentions 54m in 4.6 seconds. Assuming the velocity changed at a constant rate, this is an acceleration of -5.1 m/s² or -0.52g. The resulting g-force was equivalent to 0.48g.

People were injured by floating objects, because the acceleration changed a lot during the 4.6s, possibly in a range of several g. The danger is from persons and/or objects meeting with different velocities.


The intensity of CATs is predicted to increase as atmosphere temperature increases, and could be a real problem for aviation in the future.

  • $\begingroup$ If g-force is a force then it cannot be expressed in units of g. F = ma, and only a can be expressed in units of g because both a and g are in m/s^2 or ft/s^2, so that g-unit = a/g = ma/mg = F/W. $\endgroup$
    – LJQCN101
    Commented May 31 at 3:51
  • 1
    $\begingroup$ @LJQCN101: As indicated g is an acceleration but g-force is a force. When a doctor says the blood pressure is 70 mm of Hg, they are expressing a pressure (which unit is Pa), not a length. This is the same here. Some equivalence is used. It's not very scientific and it's confusing. $\endgroup$
    – mins
    Commented May 31 at 10:08
  • $\begingroup$ "The force created by all accelerations on a body cannot be split between gravity and other causes." Acceleration does not create force. It's the contrary, force creates acceleration. And forces on a body from its center of gravity can definitely be split between weight and other forces. By simply dividing the forces by mass, you get gravity and other accelerations caused by those forces. $\endgroup$
    – LJQCN101
    Commented Jun 4 at 12:46

g-force or load factor $n$ = (total acceleration minus the acceleration components due to gravity) / g. It is an acceleration expressed in g-units, or multiples of 9.81 m/s^2.

In the equation $n_z$ = Lift / Weight, lift does not include weight ( = mass * gravity acceleration) component, so when people are using this equation of load factor, they automatically exclude the gravity component.

Why the exclusion from total acceleration? Acceleration due to gravity and inertial forces do not create or create very little internal stress at any internal point of an object, as the forces pulling each internal point are equal. While load factor is about motions that create internal stress in an object, so that such accelerations must be excluded when calculating load factor due to motions.

The equation $n_z$ = lift / weight can also be interpreted as, aircraft or person feels X times the weight of itself opposite to the direction of lift. Consider an aircraft in a 2g coordinated turn with constant speed and altitude, then $n$ = $n_z$ = 2g as lift is pointed vertically in the aircraft body Z axis, while $n_y$ = 0g because no side acceleration in a coordinated turn. The person feels the acceleration downwards to the floor of the aircraft, opposite to the lift direction. In this case it is not to the direction of gravity nor centrifugal force.

With an object resting on the ground, the total acceleration is 0 m/s^2, and the acceleration excluding gravity component is 9.81 m/s^2 upwards due to the opposing force of weight when an object is contacting the ground according to Newton’s third law, so $n$ = 9.81 upwards / 9.81 constant = 1g.

With an object in a low orbit at ISS, the total acceleration is mostly centripetal acceleration, which is the gravity, as the object is not moving in a straight line, but in an orbit. And the acceleration due to gravity is excluded from the equation, so that $n$ = (g - g) / g = 0g. Inertial forces like centrifugal force is not relevant here.

It's also very important to clarify which direction the acceleration it is referring to. In the case of acceleration directly sensed by the accelerometer, they're body axis acceleration, or $n$ w.r.t the aircraft body: $n_x$, $n_y$ and $n_z$.

The vertical acceleration, or normal load factor ($n_z$) is widely displayed in the cockpit instruments or used as a feedback by longitudinal control systems. This answer provided some examples of the usage of $n_z$ in certain fighter aircraft.

There's also side acceleration ($n_y$) that is not displayed to the pilot, but can be used by directional control systems to provide turn coordination.

A decomposition of $n_y$ and $n_z$ is provided by NASA TM-107601

ny and nz

In which:

  • g is the constant acceleration due to gravity of 9.81 m/s^2.
  • q, p, r are body axis pitch rate, roll rate, yaw rate respectively.
  • u, v, w are body axis velocity in the body X, Y, Z axis respectively.
  • the dot above u, v, w means derivative of velocity, or acceleration in the body X, Y, Z axis.
  • lyz, lzz are vector transformation from the local (Earth) axes to the body axes.
  • glyz, glzz are the components of g in the body Y and Z axis respectively.

So if not considering angular rates, $n_z$ can be expressed as (w_dot - glzz) / g, or (body Z axis acceleration - body Z axis gravity acceleration component) / gravity acceleration.

It is a/g in short, so $n$ can also be calculated using Force(=ma)/Weight(=mg), which is, total forces excluding weight, divided by the weight of an aircraft.

That’s why $n_z$ can be expressed as Lift / Weight.

  • 1
    $\begingroup$ "g-force = acceleration excluding gravity component" Do you mean the g-force in the ISS is not 0g (micro-gravity) but -1g, the value of the centrifugal acceleration? $\endgroup$
    – mins
    Commented May 31 at 10:15
  • $\begingroup$ @mins It’s still 0g if excluding gravity component. Check the NASA equations. An aircraft flying straight and inverted experiences $n_z$ of -1g. $\endgroup$
    – LJQCN101
    Commented May 31 at 11:09
  • 1
    $\begingroup$ "It’s still 0g if excluding gravity component" To get 0 g I need to sum centrifugal and gravity accelerations which are opposite. For the inverted flight in straight line, adding centrifugal acceleration 0 g and gravity -1 g (in the aircraft frame), also works. Both examples contradict your premise of excluding gravity acceleration from the sum. $\endgroup$
    – mins
    Commented May 31 at 12:04
  • $\begingroup$ Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on Aviation Meta, or in Aviation Chat. Comments continuing discussion may be removed. $\endgroup$
    – Ralph J
    Commented Jun 1 at 16:39

Relativity states that the so-called force of gravity is not a force, but an acceleration due to the distortion of space time.

There is no physical difference in between gravity and acceleration, or free-fall and zero G.This is known as the equivalence principle.

So if you believe the physics, yes, gravity as experienced by the people inside the plane changed, but the article incorrectly describes this as a force rather than acceleration.


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