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I have a basic question and sorry if it is a well known fact.

I understand that the induced drag is due to the tip vortices changing the effective angle of attack (downwash).

I searched some website and they mentioned that for high span wings the disturbance caused by tip vortices is less and hence, less induced drag.

Now the confusion arises.

As per the lifting line theory, the tip vortices strength will be as same as the bound vortex.

Does this mean that the induced drag is reduced only for a high span wing when having same bound vortex strength as a low span wing?

[For simplicity we can assume a elliptical lift distribution]

DeltaLima
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Selva
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  • A high span wing of the same chord has more area than a low span wing, so it needs a smaller vortex for the same lift. – Peter Kämpf Jul 03 '16 at 16:08

2 Answers2

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You start from wrong assumptions, which explains your doubts. The line

the induced drag is due to the tip vortices

is as true as saying that wet streets cause rain. Also, the opinion that the

tip vortices strength will be as same as the bound vortex

is wrong. Unfortunately, many authors don't understand the topic themselves and copy what others have written before without thinking the issue through. Ideally, you would forget about all what you have heard about vortices and lifting lines, but since you ask I will try to explain potential flow theory a little.

In potential flow theory, lift is caused by vortices which are caused by the movement of a wing through air. These vortices run along a closed line: Within the wing they form the bound vortex, then they leave the wing backwards as trailing vortices and are connected at the point where the movement started by the starting vortex.

Now comes the important part which most authors conveniently leave out: There is no single vortex; instead, potential flow assumes an infinite number of infinitesimally small vortices which form out of nowhere when lift is increased or speed is reduced. Consequently, no single vortex leaves the wing at the tips, instead, a sheet of vortices leaves the wing at the trailing edge. The change in strength of the bound vortices over span is equivalent to the strength of the vortices leaving the wing, so the vortices fade out towards the tips.

My advice is: If you do not want to operate or to write a potential flow code, do yourself a favor and forget all that. It is much better to interpret lift as the consequence of a pressure field around a wing which accelerates the air flowing around this wing downwards. Induced drag is simply the component of the resulting pressure forces parallel to the direction of movement, while the perpendicular component is lift. Please make sure to follow at least the last link; it gives a very good explanation what induced drag really is.

Tip vortices are the consequence of air filling the void above the downward moving air behind the wing. They are not originating from the wingtips, but the consequence of the vortex sheet rolling up (if you want to stay in that picture). Note that the distance between the cores of the vortices is much smaller than the wingspan. For an elliptic wing of span $b$, it is actually only $\frac{\pi}{4}\cdot b$

A higher wingspan allows to capture more air for lift creation, so less downward acceleration is needed. Lower downwash speed also causes a less powerful trailing vortex. Note that the mass of air affected by the wing grows with the square of the wingspan!

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Peter Kämpf
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    Thanks for the comments and I understand it. But it raises one more question. Since the vortex sheet will roll out into two tip vortices, how come a high span wing has lower vortex strength? Is it due to lower strength of smaller vortices (in sheet) caused by smaller change in span wise lift distribution? In other words, is it due to lower loading of the wing? That is a wing with span L will be loaded less than the L - dL span wing for same weight. – Selva Jul 03 '16 at 19:36
  • @Selva: A wing of higher span but the same area will have a smaller chord and consequently a reduced vortex strength at the same angle of attack. Wing loading (lift/area) will be the same, but vortex strength is proportional to chord length and has more span to act upon, so the same lift is produced by a weaker vortex. – Peter Kämpf Jul 03 '16 at 20:01
  • Ah.. Now I get it. Thanks a lot for your answers. – Selva Jul 03 '16 at 20:44
  • Your answers, as always, are impeccable. – slebetman Jul 29 '16 at 06:31
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    @Peter Kämpf, Peter, I seriously think you should write some of the ATPL books for pilots AND examiners (at least Principles of flight) since even the most popular ones (if not all) explain things in very illogical manner and contradicting common sence. Tip vortices is just really good example. Thanks. – Electric Pilot Dec 11 '17 at 13:09
  • So, even this NASA article describes wingtip drag as induced drag https://www.nasa.gov/centers/dryden/about/Organizations/Technology/Facts/TF-2004-15-DFRC.html Quote: "Winglets increase an aircraft's operating efficiency by reducing what is called induced drag at the tips of the wings." Is there a disagreement about what is and is not induced drag? See also https://www.grc.nasa.gov/WWW/K-12/airplane/induced.html – atlex2 Jul 25 '19 at 12:20
  • @atlex2: You shouldn't believe something that was probably hacked together by an intern. Just because it has a NASA URL doesn't mean it is the holy truth. Better judge for yourself whether that explanation makes sense and does not violate the laws of physics. The picture in the second link is plain wrong. – Peter Kämpf Aug 05 '19 at 19:51
  • @PeterKämpf What makes sense is that the energy that goes into making a wake vortex needs to come from somewhere, and accounting for it as part of drag makes the most sense. So far, people seem to call that induced drag. I'm still really scratching my head here, but you're right to say that an intern probably created that website. – atlex2 Aug 07 '19 at 22:22
  • If I may ask two things: 1) bringing your explanation about the rearward component of the total aerodynamic force to an extreme: imagine a door flying through air at 90 deg AoA; from your explanation, the total aerodynamic force formed by pressure difference on both sides will be facing backwards and thus in your terms fully equal to “induced drag”. But there’s no lift produced, so this “induced drag” explanation contradicts the common wisdom that induced drag only exists where lift is produced. – agronskiy Aug 14 '19 at 21:33
  • (Breaking in two) 2) How would you comment the Denker’s approach to explaining that (note he is careful with formulations onwhat is cause and what is consequence, saying “It doesn’t matter whether you consider the vorticity to be the cause or the effect of the descending air — you can’t have one without the other.”) https://www.av8n.com/how/htm/airfoils.html#sec-circulation His formulation is attractive because it operates in energy terms, saying that if certain energy was lost into tip vortices, there must exist a force that did the work. – agronskiy Aug 14 '19 at 21:34
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    @agronskiy: Induced drag is pressure drag, but not all pressure drag is induced drag. Denker is right when he says you cannot have descending air without vorticity, but to say that it doesn't matter which causes what is not rigorous enough. Clearly, air is accelerated downwards (that is where the energy is spent!) and that downward traveling air causes vorticity when the surrounding air fills up the space above it. Vorticity and friction is how that extra energy is dissipated eventually. – Peter Kämpf Aug 15 '19 at 08:14
  • Thx, I see! After reading tons of materials on that I start feeling that it is in the end slightly underdefined terminology, coming from a shady, vague border between pressure drag and induced drag: i.e. in my example about the door — where this pressure drag stops being “induced”? With loss of lift? But this division is somewhat arbitrary... I appreciate you pointing out that the energy is lost from downward acceleration (i.e. even by infinite wing), but this can be attributed to pressure drag, isn’t it? – agronskiy Aug 15 '19 at 09:11
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    @agronskiy: Yes, there are tons of material out there, but quality varies. Sometimes a lot. Now about the door: If no net down- (or up-)ward deflection of the air happens past the door, there is no induced drag. But still a lot of pressure drag. Please keep in mind that induced drag got its name from electrical induction: The Biot-Savart law allowed to calculate lift-dependent drag for the first time; quite some revelation in its days. But in the end the definition of induced drag is arbitrary; I prefer to see only friction (parallel) and pressure (normal) drag components. All further (cont.) – Peter Kämpf Aug 15 '19 at 16:36
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    @agronskiy: subdivision only makes things overly complicated. With more drag components you need to avoid double- or undercounting, so dividing drag into pressure and induced drag leads to more confusion, as you found out. Again: Ideally, you would forget about all what you have heard about vortices and lifting lines. – Peter Kämpf Aug 15 '19 at 16:38
  • Thanks a lot for your clarifications which offer some fresh view on the whole thing! – agronskiy Aug 15 '19 at 22:02
  • I have one question about wingspan effect on reducing induced drag - will increasing wing chord without change in wingspan also reduce induced drag? Because as well as increasing span, increasing chord will give greater wing area, so less angle of attack is needed to maintain required lift and so lower air down acceleration needed. Correct me if I’’ wrong please. – Konrad Jan 15 '21 at 16:07
  • @Konrad: Increasing chord means to increase reference area, too. The induced drag coefficient becomes smaller, but the induced drag force will be the same (plus some more friction drag). See here for a longer explanation. – Peter Kämpf Jan 15 '21 at 17:10
  • @PeterKämpf But increasing span doesn't increase reference area too? – Konrad Jan 16 '21 at 16:34
  • How I understand it, correct me if I'm wrong: induced drag will be lower if we accelerate more air apart from accelerating same amount of air more. Increasing wingspan allows to accelerate more air, and increasing chord will not change the amount of air accelerated, it will only increase the acceleration i.e. due to camber. And will induced drag be lower if we increase wingspan but we also reduce wing chord (if compared to increasing wingspan and staying with the same chord)? – Konrad Jan 16 '21 at 18:27
  • @PeterKämpf "induced drag is simply the component of the resulting pressure forces parallel to the direction of movement" But 2D airfoil also has pressure vector point back when do integration of static pressure? https://aviation.stackexchange.com/questions/98827/does-pressure-vector-point-back-at-2d-airfoil-in-viscid-flow – user707264 Apr 29 '23 at 07:39
  • @user207141 This line is only true for inviscid flow. I was sloppy. Friction adds mostly shear drag but also changes the pressure distribution such that some pressure drag is added, too. – Peter Kämpf Apr 29 '23 at 16:20
  • @PeterKämpf So induced drag is only drag that exist in inviscid flow. When I integrate static pressure around real wing(from static pressure taps), how we call this drag? – user707264 Apr 29 '23 at 20:31
  • @user207141 It is a mixture of induced and pressure drag, depending on your reference. Is x the airfoil chord? Or the direction of movement? Reality is messy. – Peter Kämpf Apr 30 '23 at 17:54
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A simple way to understand this which I've never seen anywhere is to consider the air pushed down by the wing as the aircraft passes. If aircraft 1 has a wider wingspan than aircraft 2 (but the same wing area), then aircraft 1 accelerates a wider swath of air downwards than aircraft 2. Lift is proportional to the momentum transferred to the air being pushed downwards, while the energy required to do the downwards pushing is proportional to the square of the downward velocity. Pushing down on the wider swath of air means the air doesn't have to move as fast as the smaller swath to give the same lift. Since momentum is the same for both the large and small swath, the energy imparted to the larger (more massive) swath of air is smaller because of the velocity-squared term.

This offers an intuitive explanation for why induced drag drops as speed rises: at low speeds the mass of air pushed downward is proportional to the wingspan times the distance covered per second. At higher speeds the distance covered per second rises, giving the aircraft more mass to push against. Double the air mass by doubling the speed and the downward air velocity drops by half, which cuts the energy required by a factor of root two.

This extends to explain why an infinitely long wing produces zero induced drag: the accelerated air mass is infinite, so the air velocity imparted by downwards acceleration is zero, which costs no energy.

In a real aircraft the air doesn't go straight down, so you get vortices. But the momentum/energy relationship holds, as does the main principle: the larger the mass of air, the less you have to accelerate it downward to get the upward lift you want.

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  • so the air velocity imparted by downwards acceleration is zero afaik, not true. source? – Federico Nov 22 '16 at 10:10
  • @Federico: This is for an infinite wing only. And true. – Peter Kämpf Nov 22 '16 at 13:09
  • If you read @JanHudec's and my answers, you will find plenty of mention of your view of lift creation. +1 – Peter Kämpf Nov 22 '16 at 13:11
  • @PeterKämpf I noted the "infinite wing", I still don't see how it is true from the equations. I reiterate my request for a source. – Federico Nov 22 '16 at 13:12
  • @Federico: Consider an infinite wing of constant angle of attack, airfoil and chord. All bound vortices remain in the wing, there is no free vortex leaving it and, consequently, no downwash. This is really only a thought experiment and, if followed through, means that lift is also zero. – Peter Kämpf Nov 22 '16 at 13:15
  • @PeterKämpf ah, that's what was confusing me. by reading it I understood that the answer claimed that an infinite wing would produce lift with no downwash. ok, if you say that there is also no lift I get it. – Federico Nov 22 '16 at 13:37
  • @Federico: The infinite wing will produce infinite lift with the slightest circulation. The easy solution is to let it produce no lift by giving it an infinitesimally small chord. – Peter Kämpf Nov 22 '16 at 16:26