4
$\begingroup$

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. Will the airplanes produce the same value of lift in both cases or will be the value of lift be different? He said that value will be still the same and I was argueing that if you have tailwind you will have to increase angle of attack which means that parasite drag will increase so the value of lift will have to increase.

EDIT: Question is: If two same aircraft 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 produced by one of the airplanes be different or will be both two values be the same?

$\endgroup$
  • $\begingroup$ 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
  • 4
    $\begingroup$ @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
  • $\begingroup$ It's kind of nice when a question has no answers with a positive vote score-- "community" keeps bumping the question up to the top of the "active" stack every few months and we get to see the question over, and over, and over... $\endgroup$ – quiet flyer Jun 2 at 15:21
1
$\begingroup$

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.

| improve this answer | |
$\endgroup$
  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 2
    $\begingroup$ @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
  • $\begingroup$ Mike has said "produce more lift" as in throttle. See how we are somewhere between glider and level with our "produced" lift? And how it affects our rate of descent at constant speed? $\endgroup$ – Robert DiGiovanni Oct 23 '18 at 9:47
  • 1
    $\begingroup$ In this answer we read "Rate of descent is a direct result of the amount of lift being produced." That sounds like an Aristotelian point of view. I.e. more upward force = faster upward velocity; less upward force = lower upward velocity or higher downward velocity. If that was not what was intended the answer could be clarified, for example by pointing out that the reduction in lift associated with a higher sink rate is real but extremely small, for reasonable ratios of L to (Drag minus Thrust). Also by pointing out that climbing also involves a reduction in Lift, compared to horizntl flight. $\endgroup$ – quiet flyer Oct 24 '18 at 2:14
-1
$\begingroup$

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 following the same glide path 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 have a larger value of (drag minus thrust) than the aircraft with a headwind. Since the angle-of-attack of each aircraft is almost identical (more on this below), this means that the aircraft with the tailwind must be flying at a lower power setting than the aircraft with the headwind.

See the related answer https://aviation.stackexchange.com/a/56040/34686. The vector diagrams in that answer can be adapted to the current question simply by replacing the label "drag" with label "drag minus thrust", and by recognizing that in a powered glide, the glide ratio is the ratio of lift to (drag minus thrust) rather than the ratio of lift to drag. The lift, (drag minus thrust), and weight vectors still must form a closed vector triangle, exactly as illustrated in https://aviation.stackexchange.com/a/56040/34686. Meanwhile the ratio of lift to drag is still governed by the angle-of-attack of the wing: every specific angle-of-attack corresponds to a specific lift coefficient, drag coefficient, and lift-to-drag ratio.

Due to the relationships explained and illustrated in the related answer https://aviation.stackexchange.com/a/56040/34686, including the relationship lift = weight times cosine glide angle, 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 the airspeed is the same in each case, 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 directly 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. For a given airspeed, only a tiny change in angle-of-attack would be needed to cause this change in the magnitude of the lift vector.

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 magnitude of the lift vector between 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, and in lift coefficient and lift force, 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. The aircraft with the tailwind, i.e. the aircraft with the lower power setting, is indeed flying at a lower angle-of-attack and generating less lift than the aircraft with the headwind, but the difference is extremely small.

The only practical way a pilot could detect the tiny difference between the magnitude of the lift vectors associated with the headwind and tailwind cases is by being aware of the fact that a change in the glide ratio (with respect to the airmass) always involves a small change in the magnitude of the the lift vector, as well as a much larger change in the magnitude of the (drag minus thrust) vector. When we are descending at either a given angle-of-attack or a given airspeed, if we reduce thrust and increase the (drag minus thrust) vector and degrade the glide ratio and cause the aircraft to glide in a more nose-down pitch attitude, this always goes hand-in-hand with a very slight reduction in the magnitude of the lift vector. If angle-of-attack is constant, then we've decreased the airspeed, and if airspeed is constant, then we've decreased the angle-of-attack. The change in the airspeed or angle-of-attack will typically be extremely small, but the change in the glide ratio through the airmass, and the resulting change in the aircraft's pitch attitude, is in fact a signal that we've slightly reduced the magnitude of the lift vector by slightly decreasing the airspeed, the angle-of-attack, or both.

A final note-- for simplicity, this answer has ignored the effect of propwash over the wings. In truth, the required difference in lift force between the aircraft with the headwind and the aircraft with the tailwind is so small that in many aircraft it could easily be fully accounted for by the higher power setting and increased propwash over the wings on the aircraft with the headwind. Therefore on any aircraft where there is a propwash over the wings, we can't say for certain that the angle-of-attack is slightly lower when descending with a tailwind than with a headwind at some fixed airspeed and some fixed glide ratio over the ground. But we can still say for certain that the lift vector is slightly smaller when descending with a tailwind than with a headwind at some fixed airspeed and some fixed glide ratio over the ground.

Nothing in this answer should be construed as a suggestion that the wings somehow directly "feel" the effects of a headwind or tailwind. The key is that the pilot is forced to choose a different power setting to stay on the glideslope at some given airspeed when flying with a headwind versus a tailwind. We would see exactly the same effects if we explored the effects of changing the power setting while maintaining some given airspeed while flying in a descending glide in still air.

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

| improve this answer | |
$\endgroup$
  • $\begingroup$ 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
  • $\begingroup$ Yes, the question is answered. $\endgroup$ – quiet flyer Jun 2 at 14:53
  • $\begingroup$ I am tempted to downvote this answer merely on the grounds of ineffectiveness of getting any points across, regardless of accuracy. Long form links, no diagrams, relying on links to other answers to make points. However, what actually earns my downvote is that you ignore that the thrust vector will not generally not be aligned with the relative wind. The necessary variation in thrust to vary descent rate will therefore add to (rarely subtract from?) "K". Therefore you cannot make the assumption to simplify to (drag-thrust) as you do. Hence my downvote. $\endgroup$ – J Walters Aug 6 at 11:52
  • $\begingroup$ @JWalters --- Can you explain exactly what you mean by "long form links"? $\endgroup$ – quiet flyer Aug 6 at 14:56
  • $\begingroup$ @JWalters -- What is "K"? $\endgroup$ – quiet flyer Aug 6 at 14:58
-3
$\begingroup$

At "steady state" conditions of constant IAS and trim an aircraft in a headwind, no wind, or tailwind theoretically "moves with the airmass" which produces the same lift, if the rate of descent is the same.

However, what needs to be done to hold a glide slope at constant IAS with variable headwinds is to vary rate of descent. Using throttle (for a staticly stable aircraft) is the easiest way to do it.

There is a slight difference in the required lift vector to maintain the same vertical lift vector because the plane in the tailwind will pitch down further, at lower power, than the plane with a headwind to get the rate of descent and ground speed to match the glideslope.

These differences are very slight, for example, changing pitch from 0 degrees to +/- 10 degrees (relative to the horizon) only reduces vertical lift by 1.5 %!

Vertical Lift = $cosine$ Lift vector.

The tailwind plane needs a bit more lift and this is where flaps come in.

It is the vertical lift vector, not the lift vector, that determines whether of not the plane deviates from its line of flight (accelerates upwards or downwards). This is why flaps are so effective for landing. At a given airspeed, they allow a higher rate of descent and maintain the same vertical lift by increasing the lift vector with an increased Lift Coefficient.

In the downwind case, flaps can help even more by allowing a lower IAS for the same amount of vertical lift.

But to answer the question, without reconfiguration, the plane with the higher rate of descent (tailwind) would require a fractionally higher amount of lift than the headwind plane to maintain the same vertical lift. This means a higher IAS for the downwinder, sans flaps. Go with flaps.

| improve this answer | |
$\endgroup$
  • $\begingroup$ Horrors, this may mean lift is greater than weight in a descent too! It might, think of where the nose is pointing, and what is needed to most efficiently create vertical lift. Reference: Vmin sink vs Vbg and Vy. $\endgroup$ – Robert DiGiovanni Jul 3 at 2:21
  • $\begingroup$ Re "Horrors, this may mean lift is greater than weight in a descent too"--well, see my answer for an opposing viewpoint. $\endgroup$ – quiet flyer Jul 3 at 15:45
  • $\begingroup$ @quiet flyer Figured I'd hear from you. As it turns out, running the cosine for various angles to the horizon (+,-), lost vertical lift due to pitch is relatively minor, and can be compensated by adding a few more knots. (Cosine 10 degrees is 0.985). Important to note this is only in the range where the wing is providing all the vertical lift, and only applies in steady state cases where there is no vertical acceleration. Above and below this pitch range, yours and others points are entirely correct. Enjoy your writings, and thanks for the response. $\endgroup$ – Robert DiGiovanni Jul 4 at 0:43
-5
$\begingroup$

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.

| improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ 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
  • $\begingroup$ Jan Hudec no, both airplanes, being equal weights, are in constant rate descents, therefore lift, by definition, MUST be equal in both cases. However the aircraft descending with a tailwind would need to use less power to maintain stabilized flight and descent. $\endgroup$ – Carlo Felicione Oct 22 '18 at 9:37
  • 1
    $\begingroup$ It sounds like you may be assuming that airplanes are held up entirely by lift. They're not; if an airplane is descending (relative to the air), then some of the force holding it up is lift, and some of the force holding it up is drag. $\endgroup$ – Terran Swett Oct 22 '18 at 13:47
  • 3
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.