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In this question: Why is induced drag less on a high span wing?

In the answer it was stated that wing tip vortices do not cause induced drag. If this is the case then what causes the induced drag.

Thinks about this for 10 minutes

What if the Induced drag occurs because of the tilt backwards of the wing. Do the vortices just make the induced drag worse but, not cause it.

So basically I am asking if the backwards tilt of the wing causes induced drag and how do the wingtip vortices affect the induced drag

This may sound like a duplicate of:

Is induced drag not caused by tip vortices?

but, It is not because I am asking if the induced drag is causes by the tilt of the wing. I am not asking if induced drag is caused by wing-tip vortices or not.

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    $\begingroup$ You are on the right track to think induced drag is a component of lift pointing backwards. $\endgroup$
    – STWilson
    Jun 19, 2017 at 14:25
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    $\begingroup$ While you are not talking about wingtip vortices, doesn't this answer liked from the one in “Is induced drag not caused by tip vorties?”, together with the three more linked from there, explain it? $\endgroup$
    – Jan Hudec
    Jun 20, 2017 at 20:02
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    $\begingroup$ Or perhaps what about this answer. The question Is induced drag essentially nothing more than a specific type of form drag? is much closer to a duplicate. $\endgroup$
    – Jan Hudec
    Jun 20, 2017 at 20:34
  • $\begingroup$ I don't understand the last paragraph of the question. It might just be the grammar or it might be me. $\endgroup$
    – Devil07
    Jun 27, 2017 at 4:37

3 Answers 3

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Although your explanation isn't entirely wrong, It isn't necessarily the backward tilt of the wing, but the backward tilt of the aerodynamic force.

I look at it from two different perspectives. The airfoil is designed to accelerate air thus creating the pressure differentials that make the plane fly. The higher pressure areas will try to push the wing toward the lower pressure areas. The overall sum of these forces is called the resultant force.

enter image description here

That resultant force will have an amplitude and a direction or vector associated with it. The goal of designing an airfoil is to orient those forces upward to counteract gravity. So the designer will make the low pressure areas on top of the wing and the high pressure areas on the bottom to make the vector point upward. A perfect airfoil would create a vector pointing directly upward, 180° from the ground and 90° from the direction of travel. In reality, nothing is perfect, so that vector is always pointed backward to some degree. So we break that resultant force into two components. The part we're trying to accomplish, which is 90° from the direction of travel is called lift and the remaining part which is 180° from the direction of travel is called the induced drag since it is induced by the creation of lift. Even a wing producing lift at a 0° angle of attack will still produce a certain amount of induced drag.

Now we come back to your explanation. Although any time a wing is producing lift it will also produce some induced drag, as you increase the angle of attack the vector of the resultant force tilts back with it. Not necessarily at exactly the same rate, but usually not far off. Since we are still defining lift as 90° from the direction of travel and drag as 180° from it, the ratio between the two changes. For every bit of lift produced there is a lot more drag the further you tilt the wing back.

enter image description here

The second way of looking at it is from the inertia point of view. The intent of the wing is to accelerate air downward. Once again, a perfect airfoil would accelerate the air straight down, but in reality it will always accelerate it slightly forward also. And as you tilt the wing backwards you will produce more forward movement and less downward.

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  • $\begingroup$ "in reality it will always accelerate it slightly forward also" - that's the clearest/most understandable explanation of drag I've seen! $\endgroup$
    – FreeMan
    Jun 20, 2017 at 12:42
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    $\begingroup$ @Freeman Thanks. I don't feel like I really understand something unless I can explain it to someone else $\endgroup$
    – TomMcW
    Jun 20, 2017 at 17:27
  • $\begingroup$ @mins Ugh, thanks for catching that vector. I'll have to go fix it in my image editor. As far as the forward part: the suction peaks that pull air downward remains in the same place with respect to the wing, so it moves with the wing. Since the suction peak is moving forward the downward accelerating air also gets dragged along with the wing somewhat. So Newton tells us that by pulling forward some on the air it also pulls backward some on the wing. Do I need to add that into the answer. Trying not to get too confusing $\endgroup$
    – TomMcW
    Jun 25, 2017 at 20:29
  • $\begingroup$ Now I understand what you mean. Your explanation was helpful for me. $\endgroup$
    – mins
    Jun 25, 2017 at 21:17
  • $\begingroup$ @TomMcW The first illustration, red arrow should state "drag", not "imduced drag". In this situation there is only friction drag and a bit of pressure drag. $\endgroup$
    – Koyovis
    Jun 26, 2017 at 0:36
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You're on the right track. Induced drag is caused by a rearward component of aerodynamic force. And to be certain, whenever there is lift, there is drag.

enter image description here

The more a wing "plows" at slow speed, vs "planes" as higher speed, it will create more induced drag.

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  • $\begingroup$ So, the wing-tip vortices just make the induced drag worse but, they are not the cause. They make it worse because they bend airflow down so the wing must be flown at a higher backward tilt. Right? $\endgroup$
    – Crafterguy
    Jun 19, 2017 at 14:36
  • $\begingroup$ Just know that wingtip vortices contribute to the aft pointing lift. You are right to think that the greater the angle of attack, backward tilt as you put it, the more lift will oppose the direction of flight. $\endgroup$
    – STWilson
    Jun 19, 2017 at 14:44
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    $\begingroup$ A wing at AoA zero has induced drag as well. Only a symmetrical profile doesn't. As soon as lift is created there is induced drag. $\endgroup$
    – Koyovis
    Jun 19, 2017 at 22:55
  • $\begingroup$ @Crafterguy, wing-tip vortices, of course, do cause all of induced drag. That is because the backward tilt of the lift force is what causes those vortices! See also Is induced drag not caused by tip vortices? and How does an aircraft form wake turbulence? and the other linked material. $\endgroup$
    – Jan Hudec
    Jun 20, 2017 at 20:09
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    $\begingroup$ @STWilson, “lift” is defined as the force perpendicular to the direction of travel. So saying lift is tiltied makes no sense—the aerodynamic force is tilted. And no, wing-tip vortices don't contribute to to backward tilt of the aerodynamic force, they are caused by it. $\endgroup$
    – Jan Hudec
    Jun 20, 2017 at 20:30
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You can say, as above, that the drag is the horizontal component of the tilt to the overall lift vector. You can just as easily say the tilt of the overall lift vector is due to the drag. So what causes the drag?

Lift is ultimately created by deflecting the airstream downwards (Newton's action/reaction law). As such, the wing is adding energy to the ambient air by deflecting it. Adding energy to the air requires energy from the aircraft. Energy is force times distance. That force is the drag.

You can get a given amount of lift by deflecting a small amount of air at a high velocity or a large amount of air at a small velocity - the change of momentum is the same (m*v is the same). But the change of energy (.5*mvv) is not - the high velocity option costs more energy so has higher drag.

So? A short wing deflects less air so does so at a higher deflection compared to a long wing. In other words, high aspect ratio wings (sailplanes) have a higher lift to drag ratio than low aspect ratio (stubby) wings, all else being equal.

All of which is why the vector points back a bit - and there's a drag component.

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