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in vertical climb , if velocity is constant (no acceleration) the pilot feels 1g gravitational acceleration that is equals for crusing flight. but in vertical climb lift must be zero so load factor is zero that means 0g

which one is correct and why?

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    $\begingroup$ Lift isn't zero, it just isn't being provided by the wings. $\endgroup$
    – Ron Beyer
    Oct 22, 2022 at 15:15
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    $\begingroup$ Lift is zero, thrust equals drag+weight, load factor is zero, aircraft would be weightless in free fall $\endgroup$
    – sophit
    Oct 22, 2022 at 15:37
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    $\begingroup$ Having been in an F-16 going vertical from 12k feet to ~17k feet, at a constant speed, I can assure you that were were not "weightless or "0g". My back against the seat was a constant 1g. $\endgroup$
    – WPNSGuy
    Oct 22, 2022 at 17:40
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    $\begingroup$ @RonBeyer: lift is zero in this case, by definition. Thrust against drag and weight is the right term. $\endgroup$
    – sophit
    Oct 22, 2022 at 18:38
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    $\begingroup$ @alireza: You are confusing load factor and Gs which are two different things. Load factor is a pure number (lift/weight), while G is measures acceleration i.e. length/time². Being one of the two zero does not imply that the other one is zero too. Maybe this answer can help. $\endgroup$
    – sophit
    Oct 22, 2022 at 18:43

2 Answers 2

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No, the aircraft is not weightless.

In level flight the pilot feels 1 G in the vertical axis, through the seat of the pants. The wings also "feel" 1 G because they are supporting the weight of the aircraft.

In pure vertical unaccelerated flight what the pilot feels is rotated 90 degrees to the rear, and he/she will feel 1 G against the backrest of the seat. The wings also "feel" 1 G to the rear, but they are unloaded and not supporting the weight of the aircraft as they do in level flight so you could say the load factor is zero.

The wings are not generating any aerodynamic lift, they are acting as simple fins, like on a rocket in vertical flight. The load is borne by the engines.

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but in vertical climb lift must be zero so load factor is zero that means 0g

Here is where you are going astray.

The best definition of the G-loading is the net force acting on the aircraft, minus the force of gravity, divided by (mass * G).1

In other words, the G-loading is the net "felt" force acting on the aircraft, divided by (mass * G).

In other words, the G-loading is the "felt" acceleration component, attributable to the "felt" forces acting on the aircraft. Gravity is not a "felt" force, and does not contribute to the "felt" acceleration.2. That's why we can find the net "felt" force by subtracting away the force of gravity from the net force.

In flight, the only forces acting on the aircraft other than gravity are the aerodynamic forces generated by the aircraft, and thrust.3 So in flight, the G-loading is the vector sum of the aerodynamic and thrust forces, divided by (mass * G).

Note that (mass * G) is just another expression for weight.

The point we're getting around to here, is that the G-loading may act in any direction in relation to the aircraft. It is not limited to pointing "upward" or "downward" in the aircraft's reference frame-- though that's the only component that we measure with a traditional panel-mounted mechanical G-meter.

So in a steady-state vertical climb, the G-loading is 1 G skywards in the earth reference frame, and 1 G nose-wards in the aircraft body reference frame.4 Because that is the vector sum of Thrust and Drag, divided by Weight. Or speaking in terms of scalar values, we can say that Thrust minus Drag = Weight.

Once we recognize that the G-loading can act in any direction relative to the aircraft, we have to abandon any conception of a fixed relationship between G-loading and wing loading. When we talk about a fixed relationship between G-loading and wing loading, we are really only talking about the component of the G-loading that acts parallel to the lift vector. This is, basically, the component of the G-loading that is measured by a traditional mechanical G-meter mounted on the aircraft instrument panel, or by a 1-axis electronic G-meter mounted in the same orientation.5 Because the maximum lift force that can be generated by the wings dwarfs any other aerodynamic force acting on the aircraft, and also dwarfs the thrust force, this is the normally only component of the G-loading that is of any real practical concern to the pilot, or to the aircraft designer.6 But in the broader aerospace context, we need to treat the G-loading as a vector that may point in any direction relative to the aircraft.

Footnotes:

  1. This is vector arithmetic. We can't just work with the magnitudes of the force vectors, unless they are lined all lined up to point in parallel (same or opposite) directions. Example: steady state vertical climb at constant airspeed, net force is zero, so if we subtract the 1G downward pull of gravity, we get a net G-loading of 1G upward. See footnote 4 for more on the sign convention at play here.

  2. The reason that gravity cannot be "felt"-- or detected by the G-meter or the slip-skid ball-- is 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.

  3. Strictly speaking, we should also consider apparent "centrifugal forces" arising from the rotation of the aircraft about its own CG. For example, if the aircraft is in an extreme flat spin with a very high yaw rotation rate, the longitudinal component of the G-loading, as measured by a fore-and-aft oriented accelerometer, would be different (opposite in direction) at the front and the rear of the aircraft. In most cases, this effect is negligible. Don't confuse this with the component of the G-loading that is associated with the curvature in the flight path, the curvature in the path followed by the CG of the aircraft. The latter is already fully accounted for, when we compute the net force acting on the aircraft.

  4. Some may argue that the direction of the G-load is reversed in this paragraph. That would be taking the viewpoint that when we stand at rest on the earth's surface, the stresses and strains we "feel" in our bones and tendons and muscles is due to the downward pull of gravity, so the G-loading should be considered to be 1 G downward. Actually, when we stand at rest on the earth's surface, the stresses and strains we "feel" in our bones and tendons and muscles is due to the upward push of the ground against our feet. Nothing would change in what we "feel" if gravity suddenly vanished, but the ground kept pushing up against our feet as usual (which would require that the ground suddenly be rapidly accelerating upwards.) Anyway, if we want to choose to adopt the convention that the +1 G load we feel when at rest on the earth's surface acts in the downward direction, then we have to recognize that we are really talking about an apparent load that is the mirror image of the actual "felt" component of the net force acting on a body, divided by (mass * G).

  5. "Basically", because the lift vector does not always act exactly parallel to the body z axis of the airplane, i.e. parallel to the direction of "up" in the aircraft's own reference frame.

  6. Except, of course, during a catapult launch -- or a crash!

Related ASE answers:

Does lift equal weight in a climb?

What is the definition of load factor and how do you apply it?

Aircraft load factor and body normal acceleration

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