We had a discussion with friend and we were talking about if the airplane is on approach and the one is landing with tailwind and the other one is landing with headwind, so both have the same value of IAS but the different value of ground speed. Sodoes it means that the airplanes produce in both cases same value of lift or will be the value of lift different? He said that value will be still the same and I was arquining with if you have tailwind you will have to increase angle of attack which means that parazite drag will increase so the value of lift will have to increase.

EDIT: Question is: If two same aircrafts are on same glidepath (descending on ils glide slope) with same IAS but different ground speed, because of headwind acting on one aircraft and the tailwind acting on the other aircraft, will the amount of lift produce by one of the airplanes different or will be both two valeus the same?

  • $\begingroup$ Why would you have to increase the angle of attack? The air speed is the same, the air flowing over the wings is the same, the only difference is how long it takes you to stop once you touch the ground. $\endgroup$ – Ron Beyer Oct 21 '18 at 19:34
  • $\begingroup$ As @RonBeyer states above, it's all dependent upon IAS (airflow over the wings). Lift will be the same. By the way, increasing the angle of attack, all other things being equal, results in an increase in "induced" drag, not parasite drag. $\endgroup$ – 757toga Oct 21 '18 at 19:48
  • $\begingroup$ @RonBeyer, actually, there is a very subtle tricky bit! (assuming both aircraft follow the same glide-slope!) $\endgroup$ – Jan Hudec Oct 21 '18 at 19:49
  • $\begingroup$ You need to specify exactly what you are holding constant. Angle of attack? Indicated airspeed? It sounds like need to be asking a much more basic question, like "if a given airplane flies at a given angle-of-attack, will the indicated airspeed stay the same as it circles with a constant bank angle in a steady wind?", or "are any adjustments to angle-of-attack required to keep the indicated airspeed from changing as an aircraft circles at a constant bank angle in a steady wind?" A one-sentence question like that is more concise than describing an argument you are having with a friend. $\endgroup$ – quiet flyer Oct 21 '18 at 20:19
  • $\begingroup$ So, question should be edited. Maybe too late now with all the answers already posted. $\endgroup$ – quiet flyer Oct 21 '18 at 20:20

EDIT: This question is really not about lift vectors. It is asking about the total amount or total value of lift needed with varying headwinds.

Everyone seems to be over complicating this very basic question which asks about two identical aircraft flying down an identical ILS glideslope with the exact same indicated airspeed. The only difference is the amount of headwind.

It is so simple. Rate of descent is a direct result of the amount of lift being produced.

It is a FACT that the aircraft with less headwind will need to produce LESS lift than the aircraft with more headwind, in order to maintain the same indicated airspeed, and follow the glide path.

The rate of descent with a headwind could be -500ft/min while the aircraft with a tailwind could be -1000ft/min.

The aircraft maintaining a -1000ft/min descent because of a tailwind will need to produce LESS lift than an aircraft maintaining a -500ft/min with a headwind.

  • $\begingroup$ Your answer would seem to imply that if the sink rate in case B is double the sink rate in case A, then the lift vector in Case B is very significantly larger than the lift vector in case B. See my answer which compares a sink rate of zero to a sink rate of non-zero (giving a ratio between the two cases of infinite or undefined, depending on who is on top) and finds only a few percentage points of difference between the size of the required lift vector in each case. I think this answer could be improved. $\endgroup$ – quiet flyer Oct 22 '18 at 2:18
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    $\begingroup$ Both aircraft will have exactly the same amount of total upward aerodynamic force, right? The only difference is that for the aircraft that's descending more quickly, relatively more of the upward force is drag and less of it is lift. $\endgroup$ – Terran Swett Oct 22 '18 at 2:19
  • $\begingroup$ That is true about the lift-drag-weight vector triangle. It doesn't speak in detail as to the relationship between angle-of-attack and airspeed. For a given angle-of-attack, meaning a given lift coefficient, if the lift vector is smaller, then the airspeed must be lower. For a given airspeed, if the lift vector is smaller, then the angle-of-attack must be lower. This doesn't conflict with the answer we are commenting on, except for the point that if the glide angle is halfway decent then difference between the power-on and power-off case is very small. $\endgroup$ – quiet flyer Oct 22 '18 at 2:25
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    $\begingroup$ WHY has my answer been voted down? My answer is Correct. If you think it is not, please explain. $\endgroup$ – Mike Sowsun Oct 22 '18 at 12:30
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    $\begingroup$ @MikeSowsun your answer describes the effects correctly. I can only assume that whoever downvoted it found it lacking an explanation for different sink rates requiring different amount of lift (and it may not be obvious for everyone - after all, descend is an unaccelerated flight state, so sum of forces must be zero, so why would you not need same lift for same weight?) :) $\endgroup$ – Cpt Reynolds Oct 22 '18 at 20:28

There is more to this question than meets the eye. The question is related to the question 'Gravitational' power vs. engine power and this answer will be related to the answer https://aviation.stackexchange.com/a/56040/34686 . That question used the phrase "exactly identical" and so a very precise answer was given. At the end of the current answer we'll look at whether or not these considerations make any PRACTICAL difference for flight in conventional general aviation aircraft.

You say both aircraft are on the same glide path-- meaning, no doubt, relative to ground. This means the aircraft with a tailwind is following a steeper glide path relative to the airmass within which it is flying. This means that the glide path through the airmass is aimed more steeply down, relative to the horizon, in the case of the aircraft with a tailwind. This means that the aircraft with a tailwind must be flying at a lower power setting than the aircraft with a headwind.

Due to the relationships explained (and now also illustrated) here https://aviation.stackexchange.com/a/56040/34686 , the lift vector is slightly SMALLER in the case where the lift vector is tilted forward more, i.e. where the flight path through the airmass is aimed more steeply downward. This describes the aircraft flying at the lower power setting-- the aircraft with the tailwind.

If angle-of-attack were the same in both cases, the aircraft with the tailwind would be flying at a slightly lower airspeed-- that is the only way it could be producing less lift, for the same angle-of-attack.

Since you specify that airspeed is constant, the aircraft with the tailwind must be flying at a slightly lower angle-of-attack than the aircraft with the headwind. (This introduces the complication that the L/D ratio will not be exactly the same in both cases-- but the basic ideas explored in the answer https://aviation.stackexchange.com/a/56040/34686 will still apply.)

It is very unlikely that a pilot would be able to detect this difference in actual practice in most cases. In actual practice, the angle-of-attack would probably seem to be the same in both cases. For example, consider an aircraft with a L/D ratio of 8:1 at idle power at some given angle-of-attack. In still air, the aircraft can glide 8 feet forward from every foot of altitude it loses. The glide angle at this angle-of-attack is equal to arctan (1/8) = 7.1 degrees, and the lift vector is equal to weight * cosine (glide angle) = weight * cosine (arctan (1/8)). This works out to weight * .9923. So there is a less than 1% change in the magnitude of the lift vector between the idle-power case, and the case where aircraft is generating enough power to fly horizontally at that same angle-of-attack. This means the airspeed need only change by a factor of (square root of (.9925)). or .9961 .

If we are talking about an aircraft with the glide angle of a brick-- like an aircraft with a L/D ratio of 3:1 or 2:1 or less -- then the difference in the power-on and power-off cases would be more significant, regardless of whether it is the airspeed or the angle-of-attack that we are constraining to stay constant.

So the PRACTICAL answer to your question is, that for most things that fit our definition of an "aircraft", there is essentially no difference in angle-of-attack required to fly down a given glide path (defined relative to the ground) with a tailwind or headwind, with the same airspeed in both cases.

By the way, anyone unclear on more basic concepts such as the idea that even in the presence of a strong wind, no changes in angle-of-attack or power setting are needed to hold a constant altitude and airspeed while "circling" at a constant bank angle (and here we mean that the flight path through the airmass, not the ground track, is a round circle) should probably focus on those basic concepts rather than the nuances explored in this answer.

See related answers to related questions:

"'Gravitational' power vs. engine power" -- https://aviation.stackexchange.com/a/56040/34686

"Does lift equal weight in a climb?" -- https://aviation.stackexchange.com/a/56476/34686

"What produces Thrust along the line of flight in a glider?" -- https://aviation.stackexchange.com/a/56371/34686

  • $\begingroup$ You didn't answer the question: "will the amount of lift produce by one of the airplanes different or will be both two valeus the same?" $\endgroup$ – Mike Sowsun Oct 22 '18 at 20:31
  • $\begingroup$ Are you talking to me Mike? I think I did answer the question. See here : "Due to the relationships explained here aviation.stackexchange.com/a/56040/34686 , the lift vector is slightly SMALLER in the case where the lift vector is tilted forward more, i.e. where the flight path through the airmass is aimed more steeply downward. This describes the aircraft flying at the lower power setting-- the aircraft with the tailwind." $\endgroup$ – quiet flyer Oct 23 '18 at 2:46
  • $\begingroup$ PS there are at least three typos in your comment, please be a little more careful $\endgroup$ – quiet flyer Oct 23 '18 at 2:54
  • $\begingroup$ Please visit updated answer aviation.stackexchange.com/questions/56023/… to related question for more. $\endgroup$ – quiet flyer Oct 27 '18 at 12:07

Another glider vs powered question, this one with a tail-wind twist. The basics, 2 identical air craft at the same IAS need the same amount of lift to fly a straight path. So what is different? Use of POWER (throttle). How to fly the same flight path angle (W.R.T. ground)? No wind - more throttle Head wind - even more throttle! Tail wind less throttle.

What is happening here? 1. Same air speed, so same lift. How do we play it to stay on the same glide path W.R.T. ground? 2. By changing our rate of descent and allowing the wind to compensate, holding IAS constant. With a tail wind, you will cover the distance faster W.R.T ground so you must descend faster holding constant IAS. How we do this is well known - 1. Chop power and glide 2. Add flaps

And finally, I might see if another runway is available with a more favorable wind direction.

  • $\begingroup$ Please explain down vote. Relative wind is not different. Unless wind changes, at constant IAS, whether gliding or powered, AOA and lift is same. Please also review Carlo's answer. The flight path, W.R.T ground, is different, aerodynamicly, they are identical. Down vote the down vote! $\endgroup$ – Robert DiGiovanni Oct 22 '18 at 21:28
  • $\begingroup$ And the tailwind compensates increased rate of descent. Turn it around to a headwind. Was in Amarillo once and saw a pilot take 10 minutes to do a final approach into a 35 mph headwind. Adding throttle for sure! Need to take Jan flying! $\endgroup$ – Robert DiGiovanni Oct 22 '18 at 21:35
  • $\begingroup$ No, not same lift, see my answer. $\endgroup$ – quiet flyer Oct 23 '18 at 2:48
  • $\begingroup$ See Capt. Reynolds comment. We are after the truth here. That's all. Keep in mind some one actually flying these approaches needs to know the safest way, which is CONSTANT IAS SET WITH PITCH TRIM AND LEFT ALONE. VARIATIONS IN WIND ARE COMPENSATED WITH THROTTLE . ATTEMPTING TO CHANGE PITCH MAY CAUSE STALL. This is the safe way. $\endgroup$ – Robert DiGiovanni Oct 23 '18 at 8:33
  • $\begingroup$ This, after all the frill is taken away, is a rate of descent question. But the academics are enjoyable. $\endgroup$ – Robert DiGiovanni Oct 23 '18 at 8:42

If both aircraft maintain a constant rate of descent, both aircraft will be generating the same amount of lift, as in both cases the motion is steady state ie not accelerating and therefore lift will be equal to gravity. The aircraft descending with a tailwind will be doing so at a greater rate of descent, due to the higher ground speed and will require less thrust to maintain this state than the airplane landing in a headwind or still air, as more of the the potential energy can be converted into kinetic energy in the descent than in the case of the airplane descending with a headwind down an identical glideslope.

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    $\begingroup$ Lift is, by definition, the component of aerodynamic force perpendicular to relative wind. Since the descent rates, and thus the relative wind angles, will be different, the magnitudes of lift will be different for the vertical component to be the same. $\endgroup$ – Jan Hudec Oct 22 '18 at 5:10
  • $\begingroup$ The descent rate, in both cases, in relation to the air mass is the same. The descent rate in relation to the ground may be different, but relative wind angles are obviously determined by the motion in relation to the airmass, not to the ground... $\endgroup$ – xxavier Oct 22 '18 at 5:58
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    $\begingroup$ See the vector diagram here-- aviation.stackexchange.com/questions/56023/… -- K is the same magnitude as Weight, but Lift is smaller than K. Explored further in my answer to the present question. $\endgroup$ – quiet flyer Oct 22 '18 at 11:48
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    $\begingroup$ I'm looking at the vector diagram of an airplane descending (I assume you mean the one here: aviation.stackexchange.com/questions/56023/…). What I see on the diagram is that K (total aerodynamic force) is equal in length to W (weight), but L (lift) is shorter than K, so the diagram is indicating that lift is less than weight. Do you see something about the diagram that disagrees with what I just said? Also, I think that, since the drag force is not horizontal, drag contributes to holding the airplane up; do you disagree? $\endgroup$ – Terran Swett Oct 22 '18 at 20:28
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    $\begingroup$ @Robert DiGiovanni , there no basis whatsoever for saying that the vector diagram referenced by myself, Tanner Swett, etc is "a glider descending at a speed lower than Vbg." $\endgroup$ – quiet flyer Oct 23 '18 at 18:06

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