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I'm currently studying aerodynamics. I have realized that a sudden encounter with a headwind during descent will result in the aircraft generating more lift, thus angle of descent decreases. I haven't studied the descent in a constant headwind, but through research I've come across that a constant headwind during descent will increase the angle of descent. This does not make sense. Shouldn't the angle of descent decrease in both situations? Will appreciate it if someone could explain this.

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  • $\begingroup$ you will find this a lot of the times in aerodynamic books/formulas, that people tend to neglect a very important aspect: aerodynamic formulas/principle are usually for steady-flow-state.. that is in unaccelerated,unchanging conditions. Adding gusts, control surface changes or accelerations will sometimes produce cotradictory results (like in this case). Just wait until you tackle elevator/trim changes... used to drive me mad $\endgroup$ – Radu094 Sep 8 '18 at 14:08
  • $\begingroup$ For an aircraft descending from a high altitude, the angle may also change during a single descent with constant headwind. But I see that that's a different effect than the one you were asking about. $\endgroup$ – David K Oct 12 '18 at 21:37
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You have to consider the difference between rate of descent and angle of descent.

The rate of descent is expressed in distance/time, usually feet/second. This is determined by your airspeed and trim configuration. The ground speed does not affect this, only airspeed.

The angle of descent is the angle the plane is flying with respect to the ground.

enter image description here

When flying into a steady headwind the ground speed will be lower at the same airspeed. But the descent rate will be the same. So if you're descending at the same rate, but moving forward at a slower rate then the angle will be steeper.

enter image description here

On the other hand, a transient gust of wind will temporarily increase the airspeed. When the gust hits it will take a moment to overcome the inertia and slow the aircraft. So momentarily your airspeed will increase but your ground speed stay the same. An increase in airspeed will mean more lift, thus a lower descent rate at the same forward ground speed. A lower descent rate at the same ground speed will yield a shallower angle.

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  • $\begingroup$ Thank you. I thought no one would answer it and had asked a captian about this and his answer was similar to yours. He had said that the descend in a headwind will occur later compared to nill wind and the "destination point" will be the same, thus the headwind will mean a steeper descend angle. I always took the angle with reference to the straight line ontop of the airplane instead of the ground reference, but your answer gives more clarity. Much appreciated. $\endgroup$ – Shuyaib Abdullah Sep 11 '18 at 23:24
  • $\begingroup$ Good artwork. Illustrates the problem very nicely! $\endgroup$ – PJNoes Sep 12 '18 at 18:52
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Consider descending in zero wind, trimmed to an airspeed of 80m/s (288km/h) and a sink rate of 2m/s. You won't touch stick, trim tabs or throttles during the descent, but you will encounter a headwind at a later time. For now, ground speed is equal to airspeed and the angle of descent is 1:40.

You suddenly encounter a headwind of 20m/s. Ground speed remains at 80m/s due to inertia; Airspeed rises to 100m/s. The rise in airspeed increases lift and reduces sink rate to 1m/s. The angle of descent is now much shallower, at 1:80.

After a couple of seconds, airspeed bleeds off and the aircraft resumes its trimmed state at an airspeed of 80m/s and a sink rate of 2m/s. This time, the ground speed is only 60m/s due to the headwind and the angle of descent is now 1:30.

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  • $\begingroup$ But that's only if we refer that angle to a ground observer. For example, a ground-based observer with a wind of 40 m/s will always measure an angle of descent of 90º for an airplane gliding upwind with an airspeed of 40 m/s, irrespective of the glide ratio of that aircraft... However, if the frame of reference is the mass of air, as is usual in aerodynamics, the angle of descent is unaffected by the wind... $\endgroup$ – xxavier Sep 8 '18 at 11:20
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    $\begingroup$ @xxavier - You only talk in angles when flying an approach or takeoff (obstacle clearance) or in a glider. In all these cases, glide angle over terrain is what matters, so angles are always referenced to ground. I agree that if it weren't so, this question would be pointless. $\endgroup$ – Rainer P. Sep 8 '18 at 11:27
  • $\begingroup$ @ Rainer P. Yes, in the case of a glider flying low, the reference is, for obvious reasons, the ground, but that seems to me a very special case, not mentioned, even indirectly, by the OP... $\endgroup$ – xxavier Sep 8 '18 at 11:46
  • $\begingroup$ @xxavier - The only instances where I needed to calculate such angles were like: If I encounter headwind on ILS, will my aircraft deviate up or down? If I'm in a glider 40km out, will I make it home? Will I clear that hill at the end of the runway? Angles are always relative to ground. $\endgroup$ – Rainer P. Sep 8 '18 at 12:00
  • $\begingroup$ Not in the OP's query. He doesn't mention a ground reference, not even when he speaks about the wind, since it's obvious that a plane, when crossing a wind shear interface, would suffer a transient perturbation in airspeed and descent rate, but things will soon become normal again, with all flying conditions unaltered in the new mass of air. The OP seems to believe that that transient perturbation persists, something that is wrong... $\endgroup$ – xxavier Sep 8 '18 at 12:11
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Apart from transient effects, that are short-lived as such, for example in that 'sudden encounter' that you mention, the angle of descent of a gliding aircraft is not affected by the wind...

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    $\begingroup$ That is correct as observed in the aircraft (i.e. a flightpath vector in a HUD), but from a stationary outside observer's viewpoint, the descent angle absolutely does change. Or, in a very practical application, if you need to cross a given fix at a particular altitude and airspeed, the effects of headwind and tailwind on your descent angle in space play a significant role and have to be accounted for. $\endgroup$ – Ralph J Sep 8 '18 at 11:52
  • $\begingroup$ @Ralph J The question, as stated by the OP, is not related to practical flight, but just to aerodynamic considerations. Hence, any ground reference is out of place... $\endgroup$ – xxavier Sep 8 '18 at 12:07
  • $\begingroup$ except for an obscure certification climb requirement in still wind (which exists only for funny political reasons) can you find any usefullness of calculating a climb/descent angle using the moving mass of air as reference instead of the ground? $\endgroup$ – Radu094 Sep 8 '18 at 14:03
  • $\begingroup$ @ Radu094. It's not a matter of usefulness, but of answering the query of the OP in its terms. Besides, yes, calculating the descent angle in relation to the mass of air is quite useful. It's exactly the angle determined by the L/D of the aircraft... $\endgroup$ – xxavier Sep 8 '18 at 15:52
  • $\begingroup$ I've never heard of anyone ever referring to an angle of descent that did not concern a ground reference. In your theoretical concept where the "angle" is referenced to the mass of air there is no such thing as headwind. The word "headwind" presumes a ground reference. $\endgroup$ – TomMcW Sep 11 '18 at 0:41

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