20
$\begingroup$

Theoretically, could an aircraft, using any manner of self generated air flow over the (fixed) wing, ever gain enough lift in order to take off (or just simply lift off the ground) at zero forward velocity?

I am guessing that if an aircraft and its engines were light enough, you might be able to provide a high enough airflow across the wings using the propellers (akin to forward motion) - although obviously I am fully aware that most of the important control surfaces would not be functional. This is just theoretical.

To what end.. well... a more efficient understanding of self-generated lift (possibly?) might result from research into extreme blown flap/wing super-STOL utilization. High-lift, small transport aircraft operating out of small congested city air/heliports with very small runways.

I am aware of the limitations of such designs, such as inefficiency at higher speeds and, in the absence of variable form and geometry (which would add too much weight) the fuel usage would become prohibitive. It's purely an idea for speculation.

Improve this question
$\endgroup$
13
  • 1
    $\begingroup$ Maybe you should set some limits on your question. Does it have to be manned? Would bug or bird sized aircraft qualify? Would rotating wings (helicopter) qualify as "self generated air flow over the wing"? $\endgroup$
    – James
    Jul 30, 2015 at 11:26
  • 3
    $\begingroup$ I don't have the data to provide an answer to the question if a plane can fly by self-generating airflow over the wings. What I can say is, there are cases where light planes have become airborne because of very high wings blowing directly to the plane. Here is a video of a parked plane "taking off": youtube.com/watch?v=TlEKiSwttsc As you can see the plane doesn't generate the airflow over the wings by it self. It's the high-speed-wind that provides the airflow. - As you can see, the flight didn't last very long ;) $\endgroup$
    – jklingler
    Jul 30, 2015 at 11:56
  • 13
    $\begingroup$ Yes. It's called a helicopter. $\endgroup$
    – Ajedi32
    Jul 30, 2015 at 16:32
  • 5
    $\begingroup$ You seem to be suggesting some sort of engine blowing air across the wings to generate lift. That doesn't make sense. The wing is much, much less than 100% efficient at converting airflow into lift. You'd get massively more lift just by pointing those engines downwards. $\endgroup$
    – David Richerby
    Jul 30, 2015 at 19:12
  • 2
    $\begingroup$ @reirab Killjoy --> $\endgroup$
    – David Richerby
    Jul 30, 2015 at 19:29

7 Answers 7

Reset to default
20
$\begingroup$

Technically speaking, yes, it's possible, but wildly impractical.

Of course, as usual, Jan Hudec's explanation of the physics is completely correct. Whatever is blowing enough air over the wings to fly would generate an insane amount of forward thrust, meaning that you wouldn't stay at zero forward velocity for long.

However, it's completely possible to apply another force in the opposite direction equal in magnitude to what would otherwise be the net forward force produced by the engine(s) blowing the wings. This would yield a total net forward force of zero, thus no forward acceleration would happen, allowing you to take off at zero forward velocity. Of course, you'd want to make sure that whatever is applying this force isn't also blowing the wings in the opposite direction, otherwise you lose your lift.

Some possibilities for providing this force:

  • Normal jet engines mounted backwards
  • Normal propeller engines with negative pitch
  • Rocket engines mounted backwards
  • A (really strong) tether
  • Wheel brakes (this only works as long as you're on the ground, of course... you'd start accelerating forwards very quickly as the wheels break free from the ground, but you could technically leave the ground with zero forward velocity.)
  • A really big machine gun, such as the GAU-8 Avenger or Gryazev-Shipunov GSh-6-30
  • Lots of AK-47s ('lots' being a number roughly equal to the net forward thrust of the engines in pounds divided by 13)

This is roughly what the backwards-mounted rockets would look like:
C-130 Rocket Assisted Landing
C-130 Rocket Assisted Landing
Source: YouTube

Another possibility to technically meet your requirement of taking off at zero forward velocity would be to simply start off rolling backwards fast enough that you would accelerate forwards to zero forward velocity at the moment the tires leave the pavement. :)

Of course, as you've hopefully figured out by now, that all of the above are wildly inefficient ways to take off. They would, however, technically work. And, of course, in the case of the GAU-8, whatever is in front of you during the takeoff is going to have a very bad day.

Improve this answer
$\endgroup$
6
  • $\begingroup$ Zero forward velocity doesn't mean a negative (backward) velocity is ok. It should mean simply this: zero horizontal velocity. $\endgroup$
    – user7241
    Jul 30, 2015 at 19:05
  • 1
    $\begingroup$ @jjack That paragraph was a joke. What I described in that paragraph would still have zero horizontal velocity at liftoff. It would just be non-zero prior to liftoff, thus still technically meeting the OP's requirement (as he didn't say it couldn't have a non-zero horizontal velocity before takeoff.) $\endgroup$
    – reirab
    Jul 30, 2015 at 19:10
  • $\begingroup$ I assume simply canting the wings far enough back that much of the lift generated acts to counter the thrust forward has been discarded as a possibility for good and sufficient reasons? $\endgroup$
    – Nathan Tuggy
    Aug 2, 2015 at 21:37
  • $\begingroup$ @NathanTuggy Any AoA that doesn't result in the wing being stalled will still leave a rather large forward force. I was assuming from the way the OP asked the question that he wanted the wing to be generating lift in the normal way (i.e. not stalled.) Blowing the wings just below the critical AoA would be certainly be helpful, but it wouldn't completely eliminate (or even come close to completely eliminating) the forward force. $\endgroup$
    – reirab
    Aug 3, 2015 at 0:49
  • $\begingroup$ @reirab: Fair enough, although I would like to see any detailed calculations/diagrams you can dig up. $\endgroup$
    – Nathan Tuggy
    Aug 3, 2015 at 0:52
17
$\begingroup$

What you're asking for is basically a helicopter: rather than blowing air over the wing it's more efficient to attach airfoils to a vertical axle and generate relative movement (and hence lift) by rotating them.

Improve this answer
$\endgroup$
2
  • 1
    $\begingroup$ I would say, your answer should be: "Yes it's definitely possible - It's called a helicopter" :) Seems to be the best answer IMHO. +1 $\endgroup$
    – Alexus
    Jul 30, 2015 at 22:44
  • 1
    $\begingroup$ I know all of the above, it was merely a speculative poser. It's an absurd question, but I guess I was just wondering whether enough of a pressure differential could be created using self generated airflow over a fixed wing, perhaps tethered to prevent forward motion. $\endgroup$
    – Russ Boys
    Jul 30, 2015 at 23:27
7
$\begingroup$

No, it would not.

To take off vertically you need to apply vertical force equal to the weight of the aircraft and no or just a little horizontal force to possibly start moving.

Now the total force on an aircraft (any aircraft) has opposite direction to (average) change of velocity of the air flowing around the aircraft, and magnitude proportional to magnitude of the velocity change and amount of the air.

So to lift off, the aircraft has to accelerate air down a lot. In normal rolling take-off, the air is moving horizontally around the aircraft, so the wings need to only bend it down a little to generate sufficient downward acceleration and corresponding lift force. But in vertical take-off the air is not moving, so it simply needs to be accelerated downward.

But a wing is not capable of bending the flow by 90°, only maybe 30° or something like that, so to give the air enough vertical velocity you'd also give it even more horizontal velocity, generating a lot of thrust.

For all that thrust you would need to provide energy with the engines. It means thrust/weight ratio at least 2:1. No aircraft has that. Conventional aircraft can do with 0.2, VTOLs (including helicopters) need 1 and a bit. 2 is absurdly inefficient. And it would not be true vertical take-off as the aircraft would accelerate horizontally very fast.

With circular shape Coandă effect can bend the air by more, but that does not really count as wing any more. And it still does not have any advantage over rotating the thrust vector itself. The advantage of blowing engine exhaust over the wing is not that it would be more efficient, it is that it is simpler than rotating the engines.

Improve this answer
$\endgroup$
8
  • $\begingroup$ The last point is the important one - rotating the thrust vector itself is far easier. So I would agree if the question was if it was reasonable to do that. But concerning the possibility, I have three concerns: a) Why would an aerodynamic surface with a 70°-80° deflection angle not be called a wing? b) There are turbofans with a thrust/weight ratio of around 6, so a system thrust/weight of around 2 is not unthinkable (it would look very silly, admittedly). c) You could canel out the horizontal thrust with two counteracting engines (again, silly, but possible). $\endgroup$
    – JulianHzg
    Jul 30, 2015 at 14:26
  • 2
    $\begingroup$ @jjack: You have to remember that all physical laws hold at the same time. Newton's third law requires that if the air applies upward force on the aircraft, the aircraft must be applying downward force to the air and the Newton's second law requires that the air is accelerated downward proportional to the force acting on it since it is free to do so. The forces are created by the pressure differences, but that is totally irrelevant to where the air ends up moving. Newton's laws require it to be down, so down it is. $\endgroup$
    – Jan Hudec
    Jul 30, 2015 at 19:34
  • 2
    $\begingroup$ @RussBoys, it is still disputed because it is a total misconception that still does not want to die. There is no some lift generated by 3rd law reaction and other lift created by something else. There is just aerodynamic lift and a bunch of physical laws that apply to it and one of those laws is 3rd law: all force the air applies to the aircraft must have reaction force of the same magnitude and opposite direction that the aircraft applies to the air. And since the air does not have any solid support, it accelerates according to the second law. $\endgroup$
    – Jan Hudec
    Jul 31, 2015 at 5:14
  • 1
    $\begingroup$ @RussBoys, the second law does not require the mass of the reaction air to be comparable to the mass of the aircraft, it just requires that force equals mass time acceleration. The air is accelerated a lot. And it also is quite a lot of air: all air within the wing span and to comparable height. both below and above the wing. Yes, the air above the wing is also reaction mass and it is also accelerated downwards. It is also applying force (more than the air below the wing, in fact) and the laws of motion apply to it, so it is accelerated. $\endgroup$
    – Jan Hudec
    Jul 31, 2015 at 5:17
  • 1
    $\begingroup$ Jan is right. Whether you look at the lift as a pressure difference or as the equal and opposite reaction of the force applied to the air (both above and below the wing,) you get the same numbers. Both are just different ways to describe the same thing. There are tons of questions on this SE about how this works. $\endgroup$
    – reirab
    Jul 31, 2015 at 13:18
5
$\begingroup$

Maybe.

You would need a very special kind of wing, one which controls the flow direction at the trailing edge by blowing. Such a concept had been tested by AeroVironment in the 1980s and in a model, but to my knowledge no man-carrying airplane using this concept exists.

The cross section of the wing's airfoil is a circle, with an exhaust slot at the back. The exhaust stream of a gas turbine is channeled through the tubular spar of the wing and exits through a spanwise slot in a collar around the spar. By rotating the position of the slot and pumping enough air through it, you can control the amount of circulation this wing creates.

Cross section through tubular wing with rotating exhaust

The exhaust position of the inboard wing can be used to control lift, like flaps do on a "normal" wing, and the outer part takes over the function of ailerons. By pointing the exhaust straight down and with a sufficient air mass flow, this wing can indeed take off vertically. Efficiency will be nothing to write home about, however.

A blown flap in the conventional sense needs external flow, because its own mass flow is far too small to create sufficient lift. All it does is to invigorate an old, spent boundary layer, so it will stay attached over the contour downstream of the blowing point. Without speed, no external flow exists which could be influenced, except for a small quantity dragged along by the airstream of the blown flap.

A similar, albeit less radical, idea was used in the British research airplane Hunting 126. It used the exhaust gasses from its fuselage-mounted Orpheus engine to blow air over its flaps. Additionally, it had small nozzles at the end of the fuselage and the wing tips for low-speed control, just like the Harrier Jump Jet. However, it needed some forward speed to create enough lift to fly.

Hunting 126 in flight

Hunting 126 in flight (picture source).

Improve this answer
$\endgroup$
4
$\begingroup$

Using the Coandă effect, it is theoretically possible for an aircraft's engines to blow enough air over the wing to generate lift.

However, in a real aircraft the engines are placed so as not to place the complete wing within their outflow. Otherwise, a large amount of drag would be constantly produced during forward flight. Engine outflow also does not have sufficient cross-section to affect the complete wing, unless a very stubby (small aspect ratio, not very efficient) wing was used.

Applications have been seen in prototypes such as the Boeing YC-14 or production models such as the Antonov An-72, however these were STOL aircraft, not VTOL.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Interesting. Thank you. I didn't consider that. But useful to know for further research. $\endgroup$
    – Russ Boys
    Jul 30, 2015 at 23:28
1
$\begingroup$

JulianHzg asked if a surface with a large (70-80 degree) deflection angle could be called a wing. The Harrier has such a surface. If we are willing to call the exhaust nozzles "wings" then we have a working example.

enter image description here

Improve this answer
$\endgroup$
1
  • $\begingroup$ Exhaust nozzles aren't wings (or airfoils at all,) though. A helicopter is closer to what OP is asking for, since it actually does cause wind to flow over airfoils (the rotor blades) even when it's not moving horizontally, but OP was asking about a blown fixed wing. $\endgroup$
    – reirab
    Jul 31, 2015 at 14:39
1
$\begingroup$

I'd say vertical takeoff in a non-rotating wing, not deflected thrust machine was obtained in the Custer Channel Wing models, you have some videos in YouTube https://youtu.be/-Sn5JL9t_C4 http://www.flightglobal.com/pdfarchive/view/1952/1952%20-%200044.html http://www.angelfire.com/va3/bythefire/ NASA Technical Report Server -NTRS- has a L53A09 full size wind tunnel report of this arrangement http://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19930087470.pdf and links to the Custer follower's sites exist in Wikipedia

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .