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I'm aware that headwind or tailwind does not affect rate of climb or descent but only the angle. I'm not sure if it affects the g force or not. I'd assume it affects g force because using trignometry, using SIN theta the vertical component increases as the angle increases. However a confirmation would be aappreciated.

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  • $\begingroup$ The question would benefit from just asking about climb angle and leaving takeoff out of it-- as it stands it's not clear whether you are asking about the G-load imposed as we transition from horizontal to climbing, or the G-load in the climb, or both-- though the answer ends up being the same in all cases. $\endgroup$ – quiet flyer Nov 2 '18 at 13:39
  • $\begingroup$ Or for that matter the G-load (horizontal) pushing the pilot back in his seat during the takeoff run-- is this being considered? See above comment, I think the question can be simplified. $\endgroup$ – quiet flyer Nov 2 '18 at 14:48
  • $\begingroup$ Wait a minute here, please clarify exactly what you mean by "using sine the vertical force component increases as the angle increases." What is theta? What vertical component are you saying increases as theta increases? See diagrams attached to this answer to related question aviation.stackexchange.com/questions/40921/… . If by G-load you are just looking at the component of total aerodynamic force that acts in up-and-down direction in a/c reference frame, this will be proportional to L in the diagrams. $\endgroup$ – quiet flyer Nov 2 '18 at 16:09
  • $\begingroup$ (Ctd) It almost sounds like you are suggesting that the G-load is GREATER the steeper the climb angle. I think this question can use some clarification-- at least explain what you mean by theta and what vertical force component you are saying increases with climb angle. $\endgroup$ – quiet flyer Nov 2 '18 at 16:10
  • $\begingroup$ sorry for the late reply. Yes i had assumed the vertical component is the g force. The theta is using sin theta to calculate the vertical component during takeoff $\endgroup$ – Shuyaib Abdullah Nov 3 '18 at 3:10
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In the takeoff run, the acceleration is less with headwind, but once in flight, the airplane flies within the mass of air. The fact that that mass may move in relation with the ground doesn't affect the magnitude of the forces involved in flight, so accelerations are not affected by the wind.

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    $\begingroup$ Tell this to the albatross bird that can gain velocity by flying through a changing wind field. If acceleration wouldn't be affected by the wind speed this would not be possible. (dynamic soaring) $\endgroup$ – Jan Nov 2 '18 at 7:25
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    $\begingroup$ Dynamic soaring is an extremely fascinating phenomenon but has nothing to do with maneuvering within a UNIFORM airmass. There is a website available that claims otherwise but please let's not contaminate our minds with those thoughts. Anyone already contaminated by said website may wish to formulate a separate question for Aviation SE on the topic. $\endgroup$ – quiet flyer Nov 2 '18 at 13:33
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No--

For flight at any given angle-of-attack, the aircraft's pitch attitude in space is tied to the climb or descent angle in relation to the airmass, not in relation to the ground.

The climb angle achieved with respect to the airmass does NOT change depending on whether the aircraft is facing upwind or downwind, and therefore the aircraft's pitch attitude in space doesn't depend on whether the aircraft is facing upwind or downwind.

Just as a glider's pitch attitude in space doesn't change as it circles with a given angle-of-attack and airspeed, even in the presence of a very strong tailwind that drops its groundspeed to zero at times.

So even if we recognize that the component of the G-load that acts in the "up and down" direction in the aircraft's reference frame is reduced when the aircraft is in a nose-high (or nose-low) pitch attitude, we won't see a difference in this value when we climb upwind versus downwind.

Note-- it is a bit ambiguous as to exactly what "G-load" means. Is it what we read on the G-meter-- i.e. just the component of "felt" acceleration that acts in the up-and-down direction in the aircraft's reference frame? If so, this is just equal to the component of the net aerodynamic force that acts in the up-and-down direction in the aircraft's reference frame. Essentially, the magnitude of the lift vector, divided by the aircraft weight. The steeper the climb angle, the smaller the lift vector -- see related answer Does lift equal weight in a climb? .

Or by G-load, do we mean the total "felt" acceleration, including the component that acts in the fore-and-aft direction in the aircraft's reference frame? If so, this is just equal to the net aerodynamic force the aircraft is generating, divided by the aircraft weight. Since in a stabilized climb with constant airspeed and constant direction of flight path, the net aerodynamic force generated by the aircraft IS exactly equal to weight, the G-loading by this definition would always be "1" in a stabilized climb, regardless of climb angle.

At any rate, by neither definition of "G-loading" do we see a difference when climbing upwind versus when climbing downwind. Nor do we see a difference in the aircraft's pitch attitude.

(Nuances-- this answer assumes that EITHER the pilot and G-meter are located at the aircraft CG in the fore-and-aft sense, or the pitch rotation rate is zero. Otherwise the relationship between G-meter reading (and "felt" acceleration) and aerodynamic force is influenced by the pitch rotation rate, as has been recently been pointed out in comments to other related answers. But even considering such added complications, no difference is caused by climbing upwind versus downwind.)

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  • $\begingroup$ Note that as we initially pitch up into the climb, there's an INCREASE in G-load-- we must exert a "centripetal" force to make the flight path curve skywards. For typical light planes this extra force is small, but it cannot be zero. That didn't seem to be what you were asking about, and it would not be changed depending on whether we were taking off w/ headwind versus tailwind, for a given airspeed and a given rate of pitching up into the climb. $\endgroup$ – quiet flyer Nov 2 '18 at 16:04
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Key here is to realize that head wind, tail wind, climb, descent, Wright Flyer, sailplane, jet, balloon, the physics is the same regarding acceleration and non-accelerated flight. The four forces, simple as they seem, can be combined in an infinite number of orientations to achieve zero acceleration.

This is NOT necessarily just standing still, it can be an equilibrium of forces at a given velocity and direction. This, obviously but critically, applies to heavier than air flight. "If you ain't moving, you ain't flying". There for, in steady state flight, regardless of orientation or velocity, G force is only from gravity, and it is 1.

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An airplane flying into a headwind and pitching up will feel the same G As flying in stationary air, but the angle of climb will increase.

It will increase with the ratio of arccos (V_plane - Vwind)/V_p.

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  • $\begingroup$ I assume you are not trying to suggest that there will be any change in the aircraft's pitch attitude? $\endgroup$ – quiet flyer Nov 2 '18 at 21:48
  • $\begingroup$ @quietflyer. No change in pitch flying into a headwind even if the plane is accelerating. Same pitch as normal take off pitch. That's why we land or take off into the wind keeping the same pitch, unless we have cross wind. $\endgroup$ – kamran Nov 4 '18 at 19:21
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When you are only looking for the static acceleration that the pilot feels in his/her seat then the g-force is only a factor of the pitch attitude (theta). That is if we assume a stationary climb or descent, which by definition has a netto acceleration of zero. So 1g pulling the aircraft towards the ground, the more pitch attitude we have the more of the gravity has to be compensated by engine thrust to maintain an unaccelerated flight. This leaves less a smaller vertical g component. At 90deg pitch up the vertical component (from your pilot view) is going to be zero but the acceleration pushing you into the seat is now 1g (and engine thrust must equal aircraft weight). So in your fighter jet at 90 degrees nose up you will just feel as if you are lying on your back and you don't feel any vertical acceleration anymore, regardless of the wind speed.

  • Ergo the vertical acceleration felt by the pilot in stationary flight is cos( theta ) * g, longitudinal acceleration on the pilot is sin( theta ) * g.
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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – Federico Nov 5 '18 at 8:46

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