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NASA's X-57 electric plane has 12 wing-mounted propellors which drive air past the wing.

This video describes how the resulting increased air velocity causes an increase in the amount of lift the wing generates.

Suppose the propellors could generate enough airspeed over the wing to lift the entire plane -- could the plane then jump into the air without needing a runway?

This would require that sufficient lift is generated before the propellor thrust (minus wing drag) exceeds the craft's friction with the ground -- otherwise the propellor thrust would push the craft forward some before a liftoff could occur. And that leads to my secondary question: Is blowing air over a wing a more efficient means of achieving lift than would be directing those propellors downward rather than laterally over the wing?

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    $\begingroup$ You're talking about something like the Custer Channel Wing? $\endgroup$ – Rykara Mar 3 at 21:09
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    $\begingroup$ Isn't this a kite? $\endgroup$ – user3528438 Mar 3 at 22:32
  • $\begingroup$ The Hummingbird is (was, the page hasn't been updated since 2011) a concept for an aerobatic plane with a big ol' ducted fan in it. The plans say that it could be capable of VTOL. Some RC Models were made but I don't think they were VTOL capable. $\endgroup$ – mattrea6 Mar 4 at 9:12
  • $\begingroup$ "the resulting increased air velocity [from the engines] causes an increase in the amount of lift the wing generates." is how planes fly, no? Except for gliders. $\endgroup$ – user253751 Mar 4 at 11:01
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As a partial answer to your question, air moving over a stationary (relative to the ground) wing will generate lift. There have been several weather related incidences where aircraft were picked up off the ground by significantly strong winds. Videos of these incidents showed that it was lift and not force from form drag that lifted the aircraft off the ground. In the case that the planes were tied down, they would come back down to earth in place. Unfortunately, the force of the decent, in some cases, was still enough to damage the planes.

I am uncertain about what you mean by “This would require that the lift generated would exceed ground friction reacting the propellor thrust...“. Lift would be acting roughly perpendicular to the wings chord line, and the plane’s longitudinal, and lateral axes. On the other hand, any friction created by the relative wind would be acting roughly parallel to the relative wind and the plane’s longitudinal axes. The only ground friction would be that caused by the wheels touching the ground.

Thrust is a function of Newton’s Laws of Motion. The propellers turn chemical energy (in this case electrical energy) into thermal and mechanical energy. The mechanical energy is used to accelerate a mass of air. The mass of the air is accelerated by the propellers creating force. This force acts on the mass of the airplane to accelerate the airplane. The relative wind created by the propellers would act upon the airplane in the form of thrust and upon the wings to create lift regardless of how close to the ground the airplane was.

P.S.
Edited to include a new link pertinent to the discussion. Here is a new VTOL aircraft soon to be on the market.

Blackfly

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    $\begingroup$ Thank you for the response. I've rewritten part of the question as I realize my original wording did not make sense. The friction I meant to speak of was between the aircraft and the ground, which would be reacting to the propellor thrust (not lift), and the question I meant to ask was whether sufficient lift could be achieved without propellor thrust (minus wing drag) exceeding that ground friction. $\endgroup$ – mherzl Mar 3 at 19:56
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    $\begingroup$ @mherzl - Ah. Maybe. This is definitely a mathematical question. In a normal airplane, the propeller slipstream acts on such a comparatively small part of the wing’s surface. The thrust of several propellers along the wing would contribute much more slipstream to the entire system even if the combined thrust were equivalent to a normal plane. You can see the aerodynamic differences in current single engine versus multi-engine planes. $\endgroup$ – Dean F. Mar 3 at 19:59
  • $\begingroup$ It's "force of the descent", not "decent". If the descent is indecent, there may be an incidence of damage. $\endgroup$ – Monty Harder Mar 4 at 20:48
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Not sure if this is on point, but there are a few videos on youtube of parked unmanned aircraft lifting off in high winds - they weren't tied down, and off they went. They lift and kind of get blown backwards - probably flip over and get wrecked. Its the relative speed of the air going over/under the wings that generates the lift - how that relative velocity is achieved (propeller thrust or wind) is irrelevant. So to answer the question, yes, if you blow hard enough.

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    $\begingroup$ And this phenomenon isn't limited to airplanes; the same thing happens to tents, canopies, trampolines, etc. $\endgroup$ – Michael Hall Mar 3 at 23:50
  • $\begingroup$ And of course, kites and specialised gliders make use of it. The important thing is a tether to the ground! $\endgroup$ – nigel222 Mar 4 at 13:47
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(I am not a pilot, but...)

This would require that sufficient lift is generated before the propellor thrust (minus wing drag) exceeds the craft's friction with the ground -- otherwise the propellor thrust would push the craft forward some before a liftoff could occur.

The friction between two things is proportional to the force pushing them together - that is, weight minus lift. When lift equals weight and the plane can start moving upwards, the friction force is zero. Some time before that, the friction is less than the thrust.

So, no, you cannot take off vertically with horizontal thrust because at some point before the lift exceeds the weight, the thrust will exceed the friction.

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  • $\begingroup$ Although your answer provide useful information, I fail to see how it answers the question. I think it should be a comment so that the OP can improve its question. $\endgroup$ – Manu H Mar 4 at 12:00
  • $\begingroup$ @ManuH The answer to the question is "no". $\endgroup$ – user253751 Mar 4 at 12:01
  • $\begingroup$ your answer only address ground friction. This does not cover possible short or vertical takeoff $\endgroup$ – Manu H Mar 4 at 12:04
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    $\begingroup$ @ManuH You cannot take off vertically with horizontal thrust because at some point before the lift exceeds the weight, the thrust will exceed the friction. As I just said. $\endgroup$ – user253751 Mar 4 at 12:15
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    $\begingroup$ @user253751 - The point is still made. Truly vertical take off relative to the wind will not be achieved. There are scenarios where vertical take off relative to the ground can be. For instance, in calm wind, the aircraft will start moving in the opposite direction of thrust as soon as enough weight is lifted from the wheels. If the aircraft were tethered, this would not be an issue. If the aircraft were facing into the wind, at some point thrust and drag will be equal, causing no motion or acceleration. This would also aid in lift, requiring less thrust to be required from the propellers. $\endgroup$ – Dean F. Mar 4 at 14:30
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The variables in the lift equation (below, from wikipedia https://en.wikipedia.org/wiki/Lift_(force)) are the density and velocity of the air flow. How the air reaches that velocity is not relevant to the amount of lift generated.

$L={\tfrac 12}\rho v^{2}SC_{L}$

where

${\displaystyle L}$ is the lift force

${\displaystyle \rho }$ is the air density

$v$ is the velocity or true airspeed

$S$ is the planform (projected) wing area

$C_{L}$ is the lift coefficient at the desired angle of attack, Mach number, and Reynolds number

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