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In upward motion, drag and gravity is in the same direction (both in downward direction). So in this situation G is negative or less than 1 or something else?

In downward motion with thrust greater than drag (so we have acceleration), whats will happen for G?

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  • $\begingroup$ I believe you'll get a better answer on physics.stackexchange. $\endgroup$
    – sophit
    Commented Apr 15, 2023 at 13:37
  • $\begingroup$ I think the question is insufficiently constrained to clearly ask a single answerable question. Last two paragraphs of my answer (before footnotes) should explain why. $\endgroup$ Commented Apr 15, 2023 at 13:46

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A zero G flight is called like that because the people inside the aircraft do not feel any force relative to the aircraft during the zero-gravity maneuver, meaning there is no acceleration (or G force) relative to the aircraft and people therefore experience weightlessness.

However, that does not mean that the aircraft experiences zero G with respect to the Earth. The aircraft is still accelerated by the forces acting on it (like gravity).

The zero-gravity maneuver is achieved by flying on a specific parabola. Imagine throwing some object upwards and sideways. The object will move on a parabola because of gravity with a constant acceleration of 1 G in the downwards direction (ignoring drag). An aircraft can be intentionally flown on such a parabola, which then causes the zero G effect inside the aircraft. Aerodynamic forces (lift and drag) and thrust are not negligible here, so the aircraft must be carefully flown on a parabolic path with exactly 1 G acceleration in the downwards direction. This acceleration is maintained during the entire weightlessness period (both upwards and downwards part):

Zero-gravity maneuver

What is a parabolic maneuver?

The parabolic maneuver or ellipse arc during a Zero G flight is divided into three stages: the parabola pull-up, the parabola, and the parabola pull-out.

Pull-up

The pilot lifts the nose of the A310 Zero G airbus upward from horizontal flight to an angle of 50 degrees.
Passengers experience a pull of 1.8 times that of gravity on Earth. This is called hypergravity.
The parabola pull-up or “nose-up” lasts around 20 seconds.

Parabola

The parabola starts with injection: as the aircraft travels upward the pilots reduce engine speed and the Zero G aircraft follows a ballistic trajectory.
Weightlessness begins when the aircraft enters a parabola during which it is in free fall for 22 seconds.

Pull-out

The nose of the plane is tilted back downward 42 degrees, before the pilots gradually level off the aircraft as they increase engine speed.
Passengers again experience a pull of 1.8 times that of gravity on Earth.
After 20 seconds, the A310 Zero G returns to a horizontal trajectory.

(AirZeroG - How zero-gravity parabolic flights work)

Flying on such a parabola is not easy. The A310 by AirZeroG is flown by 3 pilots at the same time to carefully stay on the parabola:

Parabolic maneuvres piloting technique

Unlike commercial flights, the Airbus A310 Zero G is simultaneously piloted by three members of the flying crew during the parabolic manuevres.

  • One pilot controls the pitching (nose-up and nose-down angle).
  • A second pilot controls the roll movement (to keep the wings horizontal).
  • A third pilot, sitting behind them, controls the engine speed (he or she also monitors the flight parameters: warnings, temperatures and pressure).

Together, all three pilots maintain a near-zero acceleration level in the three axes to guarantee zero-gravity precision ± 0.02 g

Unlike in a standard aircraft, the pitching and roll control commands are dissociated. The two control column functions are dissociated by the use of an additional control column on one side and two small cables hung from the conventional control column.

AirZeroG cockpit

(AirZeroG - How zero-gravity parabolic flights work)

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  • $\begingroup$ 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. $\endgroup$
    – DeltaLima
    Commented Apr 16, 2023 at 17:54
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Key point: the force of gravity "acts from within"-- it's an internal force, not an external force. Gravity exerts an equal force per unit mass on every molecule of the aircraft structure and the aircraft contents (including the bodies of the crew and passengers). Thus if the only force acting on the aircraft and content is gravity, every molecule of the whole assembly accelerates together in a way that creates no stresses or strains of any kind on the aircraft structure and contents. Unrestrained bodies will float freely. This is "weightlessness".

The same is true if the net force acting on the aircraft is exactly equal to the force of gravity. In other words, if all the aerodynamic and thrust forces cancel each other out. To a first approximation-- specifically, assuming that the thrust line from the engines acts exactly parallel to the flight path-- this is the case whenever lift is zero, and thrust is regulated to be exactly equal to drag. Maintaining this condition with great precision is what the crew is doing with the interesting "split" control system described in another answer to this question.

The three-dimensional "G-loading"1 experienced by the aircraft is the vector sum of all the aerodynamic and thrust forces, divided by mass. When this is equal to zero, the only force acting on the aircraft and contents is the force of gravity, and the aircraft and contents experience "weightlessness", or 0-G flight.

In upward motion, drag and gravity is in the same direction (both in downward direction). So in this situation G is negative or less than 1 or something else?

I'm assuming you are asking specifically about the upward portion of a 0-G trajectory. I hope this answer has helped make clear that during the actual 0-G portion of the 0-G trajectory, where the G-loading is zero, lift is zero and thrust exactly equals drag. The aircraft is decelerating (losing airspeed) but not due to drag-- the deceleration is entirely due to the component of gravity that is exacting parallel to the instantaneous direction of the aircraft. However, another answer has made clear that is preceded by an earlier phase where the flight path is curving upward and the G-loading is thus greater than 1. To a good approximation, the total G-loading in this earlier phase (given as 1.8 Gs in the other answer) is entirely due to the extra lift force generated by the wings to curve the flight path upwards.

In downward motion with thrust greater than drag (so we have acceleration), what will happen for G

In the portion of a 0-G trajectory where the aircraft is descending, thrust is never greater than drag, so it's hard to fit this part of your question into the context of the rest of your question. In the actual 0-G portion of descending portion of the 0-G trajectory, thrust is exactly to drag, and lift is zero. Airspeed is increasing not because thrust is greater than drag, but rather because the net force acting along the instantaneous direction of the flight path at any given moment, including the component of gravity acting in that direction, is greater than zero. In the subsequent recovery phase of the 0-G trajectory, lift is actually greater than the weight of the aircraft and contents, so the G-loading is greater than 1, as the flight path curves upward back toward horizontal.

But you seem to have shifted gears away from the 0-G situation to ask about a situation where the aircraft is descending and thrust is greater than drag. In any case where thrust is greater than drag, regardless of whether the aircraft is ascending or descending, the aircraft and occupants will experience a forward G-loading component. I.e. the crew and passengers will "feel" a force component pushing forward against their bodies from the back of their seats. This is true regardless of whether the net 3-dimensional G-loading is 1-G earthwards (think of a sustained steady climb at constant airspeed with the aircraft in a nose-high pitch attitude), or otherwise (consider a vertical dive with increasing airspeed), but it could never happen when the net 3-dimensional G-loading was exactly zero. So this condition would have nothing to do with 0-G flight.

As to the specific part of question that asks "what will happen for G", the answer can only be "it depends-- you haven't given us enough information to answer." You've told us that thrust is greater than drag, so we know there's a positive G-load in the fore-and-aft sense-- the back of the seats are pressing forward against the bodies of the crew and passengers. But just telling us that the flight path is "descending" isn't enough to know whether the lift vector from the wings is greater than 1 G positive, or less than 1 G positive, or zero, or negative. (Some of these cases would only be consistent with some rather aerobatic maneuvering!) So the magnitude and direction of the "up and down" component (in the aircraft's own reference frame) of the G-loading vector remains unknown, given the information contained in the question.

If we assume the aircraft is descending along a constant linear trajectory while accelerating2, then the "up and down" component (in the aircraft's own reference frame) of the G-loading vector must equal the cosine of the descent angle. In such a case, at steep descent angles, the total three-dimensional G-loading could be less than 1 G, even though we've specified that the fore-and-aft component of the G-loading is not zero. The question doesn't contain enough information to say more than that. Of course, the magnitude of the fore-and-aft component of the G-loading vector is a factor too-- if thrust is enough greater than the drag, this component of the G-loading alone could be greater than 1 G, in which case the total three-dimensional G-loading certainly would be as well!

Footnotes:

  1. Re "three-dimensional" -- sometimes G-loading is used in a one-dimensional sense, to simply mean the acceleration component acting in the upward direction in the aircraft's own reference frame. This is what a traditional mechanical panel-mounted G-meter measures. In many situations, this is only one component of the total three-dimensional G-loading.

  2. Note that it would not be necessary for thrust to be greater than drag for an aircraft to accelerate while descending with a linear trajectory. A roller coaster, with zero thrust, can accelerate along a linearly descending track, up to a certain speed where drag equals to the "forward" component of the gravity vector.

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  • $\begingroup$ Also related: aviation.stackexchange.com/questions/2890/… $\endgroup$ Commented Apr 16, 2023 at 14:28
  • $\begingroup$ It seems you are having a discussion with yourself. If you want to expand your own answer based on these comments, please do. I'll move this to chat. $\endgroup$
    – DeltaLima
    Commented Apr 16, 2023 at 18:03
  • $\begingroup$ 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. $\endgroup$
    – DeltaLima
    Commented Apr 16, 2023 at 18:04
  • $\begingroup$ @quiet Flyer, "acceleration component acting in the upward direction in the aircraft's own reference frame"? Acceleration, by definition, is the rate of change of velocity, which is the rate of change of position. Clearly, a moments thought makes it clear that in the frame of reference of the aircraft itself, the G-meter is not moving. A G-meter measures the inertial acceleration of the G-meter, i.e., the acceleration that would be measured in an inertial, (zeroz-G, free falling frame. $\endgroup$ Commented Apr 18, 2023 at 12:12

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