# Descending on a given glide slope (e.g. ILS) at a given airspeed— is the size of the lift vector different in headwind versus tailwind?

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?

• 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. – Ron Beyer Oct 21 '18 at 19:34
• 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. – 757toga Oct 21 '18 at 19:48
• @RonBeyer, actually, there is a very subtle tricky bit! (assuming both aircraft follow the same glide-slope!) – Jan Hudec Oct 21 '18 at 19:49

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.

• 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. – Terran Swett Oct 22 '18 at 2:19
• WHY has my answer been voted down? My answer is Correct. If you think it is not, please explain. – Mike Sowsun Oct 22 '18 at 12:30
• @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?) :) – Cpt Reynolds Oct 22 '18 at 20:28
• 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? – Robert DiGiovanni Oct 23 '18 at 9:47
• 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. – quiet flyer Oct 24 '18 at 2:14

In this answer we'll start by making the simplifying assumption that the Thrust vector is parallel to the Drag vector and the direction of the flight path with respect to the surrounding airmass.

See the vector triangles of Weight, Lift, and Drag for gliding flight illustrated in these related answers (#1, #2). (The diagram in the second link is now reproduced at the top of the present answer as well.)

We can adapt these questions to a powered descent by simply substituting a vector labelled "Drag minus Thrust" in place of the vector labelled "Drag". We'll still have a closed vector triangle.

Once we make this simple substitution, the angle labelled "K" in the diagram at the top of this answer will be the aircraft's angle of descent at some particular angle-of-attack and L/D ratio and Thrust output, in still air. This is also the aircraft's angle of descent with respect to the surrounding airmass.

Note that Lift = Weight * cosine (K).

This means that Lift is less than Weight, and the steeper the still-air glide angle "K", the larger the difference between Lift and Weight.

To achieve the same descent path with respect to the ground with a headwind, we must change to a flatter descent path with respect to the surrounding airmass. Whether we increase Thrust, decrease Drag, or both, we must make the vector we've re-labelled "Drag minus Thrust" become smaller. This will decrease the size of angle K. This will increase the size of the Lift vector.

To achieve the same descent path with respect to the ground with a tailwind, we must change to a steeper descent path with respect to the surrounding airmass. Whether we decrease Thrust, increase Drag, or both, we must make the vector we've re-labelled "Drag minus Thrust" become larger. This will increase the size of angle K. This will decrease the size of the Lift vector.

Since Lift = Weight * cosine (K), the Lift vector must be larger when we are descending with a headwind than when we are descending with a tailwind, if we are achieving the same glide path with respect to the ground.

Note that this is true regardless of whether we're constraining the angle-of-attack to be the same in both cases, or the airspeed to be the same in both cases, or neither. This is also true regardless of whether we're descending on the "front side" of the power curve or the "back side" of the power curve. None of these things modify the fact that to stay on a fixed glide slope relative to the ground, angle "K" must be larger when we are descending with a tailwind than when we are descending with a headwind. Therefore the Lift vector must be larger when we are descending with a headwind than when we are descending with a tailwind.

Of course, as we vary angle K, the variation in the size of the Lift vector will be so small as to be negligible in the real world. For example, when gliding with a still-air glide ratio of 7:1, which corresponds to a still-air glide angle of about 8 degrees, the Lift vector is about .99 * Weight. If we were to change our glide path with respect to the airmass from a 7:1 glide ratio to a glide ratio of infinity (i.e. horizontal flight), the size of the Lift vector would only increase by a factor of about 1.01. Still, technically speaking, the flatter our glide angle with respect to the surrounding airmass (i.e. the larger our still-air glide ratio), the larger the Lift vector must be.

So the basic answer to the question is that the Lift vector is ever-so-slightly smaller when the aircraft is descending with a tailwind than when the aircraft is descending with a headwind, because to stay on the same glide slope with respect to the ground, the still-air glide angle must be larger (i.e. the still-air glide ratio must be poorer) in the former case.

Let's consider one specific idea in the original question--

I was arguing 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.

This is a flawed thesis. To fly with the same airspeed, why would you need a higher angle-of-attack when you have a tailwind? We've seen that-- disregarding any effects from a tilted thrust line relative to the direction of the flight path-- the Lift vector actually must be slightly smaller when the still-air glide path is steeper, which describes the tailwind case. To keep the airspeed constant, the aircraft must be flown at a slightly lower angle-of-attack when descending with a steeper still-air glide path (i.e the tailwind case) than when descending with a shallower still-air glide path (i.e. the headwind case).

Now for another twist--

If we complicate the picture by recognizing that the Thrust line may be tilted to point somewhat upward or downward of the direction of the Drag vector and the flight path through the airmass, this may dwarf the effects described above. The original question constrains the airspeed to be the same in the headwind case as the tailwind case, so the aircraft must be flying at nearly the same angle-of-attack in each case. Imagine a case where we are flying our approach "on the back side of the power curve", at a high angle-of-attack, with the engine pointed significantly upward in relation to the direction of the flight path through the airmass. The component of the engine's Thrust vector that acts perpendicular to the flight path effectively counts as part of the "Lift" vector in our vector diagram, meaning that the wing's actual Lift vector must be reduced to smaller than Weight * cosine (K). In a case like this, reducing the power setting to make the glide path steeper may well force the wing's lift to increase. On the other hand, if we are flying our approach at a low angle-of-attack, well on the "front side" of the power curve as per normal practice in a light airplane, it is no longer so obvious which direction the Thrust vector is pointing in relation to the direction of the flight path through the airmass. Since many light aircraft have some built-in down thrust, the Thrust vector may well point slightly downward in relation to the direction of the flight path, so that the wing must produce more Lift when the engine's power setting is increased. This would magnify the effect we discussed earlier in this answer-- the tendency for the Lift vector to be larger when the still-air glide angle is flatter, due to the relationship Lift = Weight * cosine (K).

At the end of the day, all these effects are so small that it's unlikely a pilot could detect the change in the magnitude of the wing's Lift vector for a given airspeed at various power settings and still-air glide angles. But in theory there is some variation in the size of the Lift vector as the power setting and still-air glide angle are changed. And for the simple case where we are assuming that the thrust line acts parallel to the Drag vector, the nature of the variation is such that the Lift vector is slightly larger when the still-air glide path is flatter (i.e. at higher power settings, such as when trying to stay on glide slope with a headwind) than when the still-air glide path is steeper (i.e. at lower power settings, such as when trying to stay on glide slope with a tailwind).

Can we show through simple geometry rather than formulae or graphs that the best glide ratio occurs at the maximum ratio of Lift to Drag? -- a concise answer with clear diagrams

Can we show through simple geometry rather than formulae or graphs that the best glide ratio occurs at the maximum ratio of Lift to Drag? -- a longer answer that emphasizes many concepts essential to the present answer

Why is the L/D ratio numerically equal to the glide ratio? -- similar to link immediately above

'Gravitational' power vs. engine power -- the variation in the size of the Lift vector as the still-air glide ratio changes comes into play here, exactly as it does in the present answer

Does lift equal weight in a climb?

• 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?" – Mike Sowsun Oct 22 '18 at 20:31
• Yes, the question is answered. – quiet flyer Jun 2 at 14:53
• 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. – J Walters Aug 6 at 11:52
• By long form links I refer to your practice here of inserting the entire hyperlink as plain text, rather than as an embedded link. Using "the answer aviation.stackexchange.com/a/56040/34686" instead of "this answer". – J Walters Aug 6 at 16:59
• "K" was the variable used in one of your linked answers to name the vertical component of lift. I should have been more specific since it has no meaning outside of that context. – J Walters Aug 6 at 17:00

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.

• 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. – Robert DiGiovanni Jul 3 at 2:21
• @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. – Robert DiGiovanni Jul 4 at 0:43

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.

• 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. – Jan Hudec Oct 22 '18 at 5:10
• 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... – xxavier Oct 22 '18 at 5:58
• 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. – Carlo Felicione Oct 22 '18 at 9:37
• 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. – Terran Swett Oct 22 '18 at 13:47
• 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? – Terran Swett Oct 22 '18 at 20:28