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This question already has an answer here:

At first I thought that the lift force of an airplane wing came from the buoyant force of the air. Thus we can calculate it's lift force using $$ F_b={\rho}gV$$

However, that does not seem to be true since this would yield a smaller number than the weight of the plane consequently the plane would be in free-fall which is not the case. Then I tried reasoning with Bernoulli's equation.

$$P_1+{\rho}gh+0.5{\rho}v_1^2=P_2+{\rho}gh+0.5{\rho}v_2^2$$ Where $v_1$ is the speed of the wind above the wing and $v_2$ is the speed below the wing. Now if I assumed that the thickness of the wing to be negligible I can simplify the equation as such $$P_1+0.5{\rho}v_1^2=P_2+0.5{\rho}v_2^2$$

Algebraically moving things around

$$P_2-P_1=0.5{\rho}(v_1^2-v_2^2)$$ Lastly, using $\Delta PA=F$ $$(0.5{\rho}(v_1^2-v_2^2))*A=F$$ Where $F$ represents the lift force on the airplane wing. Can I set this F equal to ${\rho}vg$? Resulting in this $$(0.5{\rho}(v_1^2-v_2^2))*A={\rho}Vg$$ $$(0.5(v_1^2-v_2^2))*A=Vg$$ Would it make sense setting the lift force equal to the buoyant force? Since I keep thinking that it must be the buoyant force that lifts the wings. Otherwise there does not seem to be any other force that could hold up the plane.

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marked as duplicate by sammy gerbil, stafusa, knzhou, Kyle Kanos, Qmechanic Apr 26 '18 at 11:55

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ None of the equations you are attempting to use here are intended to account for lift generation in wings. I suggest you move this post to the aviation stack exchange. there are several accomplished aerodynamicists there who can furnish you with the correct equations to use. $\endgroup$ – niels nielsen Apr 25 '18 at 0:57
  • $\begingroup$ "...Otherwise there does not seem to be any other force that could hold up the plane." Never stuck your hand out the window of a moving automobile? Travelling at freeway speeds, you can experience noticeable "lift" just by holding your flat palm at different angles to the wind. You can try it with something bigger---a stiff piece of cardboard, a DVD case, etc.---but hold on tight if you try that experiment, or at least wait until there's nobody following close behind you. $\endgroup$ – Solomon Slow Apr 25 '18 at 1:15
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    $\begingroup$ Look at aviation.stackexchange.com/q/16193 on the Aviation SE and the numerous links therein to see that Bernoulli is not enough. $\endgroup$ – Farcher Apr 25 '18 at 4:57
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    $\begingroup$ Possible duplicates: physics.stackexchange.com/q/290/2451 and links therein. $\endgroup$ – Qmechanic Apr 25 '18 at 4:59
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    $\begingroup$ Possible duplicate of What really allows airplanes to fly? $\endgroup$ – sammy gerbil Apr 25 '18 at 15:18
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The two answers provided here are wrong. The equal transit hypothesis is easily refuted by simulations. Having a longer path alone does not require air molecules to somehow move faster (why should they need to match up at the end?). Also, the theory breaks down in explaining why airplanes can fly when they're upside down, with the longer side of their wings facing the ground.

The reason for lift simply results (mostly) from a turning of the air stream as it goes over the surface, whether that surface is a wing or any other parts of the plane. So long as a horizontal air stream ends up having some downward velocity component after interaction with the surface, by Newton's third law there will be an upward force on the surface. This is also where angle of attack and stalling arises. The tendency for air streams to follow surface profiles (their detachment leading to a sudden and possibly catastrophic loss of lift) is key here. NASA has decent diagrams (but pretty bad explanations imo) on their website: https://www.grc.nasa.gov/www/k-12/airplane/presar.html.

Note: this answer is not meant to be a proper answer on lift, as Neils has suggested in the question comments, the aviation sx should have a proper discussion on this. I'm just here to correct the misconceptions expressed in the other two.

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  • $\begingroup$ Wait what? I did not say that speed pressure has no relation. I'm saying that the shape of the wing does not really matter that much. The angle of attack is the reason why the top flow is accelerated, and the bottom is decelerated. If the surface is angled upwards, the top flow is forced to go over and yet still follow the top surface profile because of atm pressure, while the bottom flow is slowed/compressed by the incoming bottom surface. The shape of the wing merely aid this following, and it also reduces the likelihood of stalling (allow for a larger range of angle of attack). $\endgroup$ – Melvin Apr 25 '18 at 5:25
  • $\begingroup$ Which is essentially described in the top answer and comment to qmechanic's link in question comment (just saw it). $\endgroup$ – Melvin Apr 25 '18 at 5:28
  • $\begingroup$ You are correct that the 'longer path' explanation is not accepted. I have not been able to find the verity of the explanation starting in your second paragraph though. Perhaps just the way you wrote it. $\endgroup$ – Dlamini Apr 25 '18 at 12:01
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    $\begingroup$ @Melvin: Even Einstein believed the 'longer path" explanation & had a wing built & tested - with disastrous results! wrightstories.com/einsteins-wing-flops $\endgroup$ – D. Halsey Apr 25 '18 at 22:36
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You would be well advised to leave the buoyancy force out of it. It is negligible for an airplane since the density of air is low.

An aerofoil (e.g. aircraft wing) is pushed upwards by the difference in pressure between the bottom and the top surfaces. The wing is shaped and then angled during flight in such a way that the air pressure on the top surface of the wing is lower than the pressure on the bottom surface. The higher pressure at the bottom results in a net force is upwards. Airspeeds are higher

There is an image of the observed reality below, from https://www.skybrary.aero/index.php/AP4ATCO_-_Aerofoil_Terminology The pressure-speed causality explanation gets a bit involved, as attempted in a Cambridge University article here. The figure does capture the experimentally observed reality (airspeeds and pressure at the top and bottom surfaces).

The shape of the trailing edge of the aerofoil is also a factor in generating lift. From the reference quoted in the figure:

"Leading edge slats, leading edge Krueger Flaps and trailing edge (Fowler) flaps, when extended from the basic wing structure, literally change the aerofoil shape into the classic concave form, thereby generating much greater lift during slow flight conditions.".

Increasing the angle of attack, which is, tilting the aerofoil so that the leading edge is higher than the trailing edge, does increase lift, for the mere fact that the there is a resultant upward force on the aerofoil as the wind pushes on the underside. The downside of this is that it generates significant drag as a side effect, so it is generally not desirable for cruising. Beyond a critical angle (often found to be around $15^o$), the lift effect collapses, known as stalling.

There is a popular 'longer path' explanation for the lower pressure over the top of the aerofoil, but this is considered incorrect by sources at NASA and the Cambridge University article mentioned above.

enter image description here

Further description after a question by @coderhk.

For a fluid, Bernoulli's equation states that:

$P_1 + \rho v^2 + \rho gh = constant$,

i.e. the sum of pressure energy, kinetic energy, and potential energy is constant. In this equation, $\rho gh$ is the potential energy(per unit volume) of the fluid at a particular point as it moves around the aerofoil, and not the average buoyancy pressure experienced by the aerofoil. The other two terms (pressure and kinetic energy) are also not averages, but point properties as the fluid moves around the aerofoil. To calculate the buoyancy force due to this complex fluid (air in motion), you would need to do something like "calculate the average potential energy pressure ($\rho gh$) around the wing", and multiply that by the volume of the wing.

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  • $\begingroup$ Since the force is change in pressure time area why can't we substitute the pressure for pgh? Thus F=A(P2-P1)=A(pgh2-pgh1)=Apg(h2-h1)=pgV?Therefore F=pgV=A*0.5*p(v2^2-v1^2) Thanks $\endgroup$ – coderhk Apr 25 '18 at 2:57
  • $\begingroup$ @coderhk I have added text to explain why the $\rho gh$ term in Bernoulli's equation cannot automatically be used to calculate the buoyancy force on a body (aerofoil in this case). $\endgroup$ – Dlamini Apr 25 '18 at 4:55
  • $\begingroup$ The longer-path-explanation is simply not true. It would imply that the flows above and below the wing somehow stay in sync, meeting up at the trailing edge again. That is nonsense and refuted by experiment and simulation. Typically the portion of the flow above the airfoil arrives at the trailing edge much earlier. But in order to understand why the flow is faster you'd need to look at the Navier-Stokes equation. And that is simply one contribution to lift. The other comes from momentum conservation as the air is deflected downwards at the trailing edge. $\endgroup$ – Nephente Apr 25 '18 at 7:38
  • $\begingroup$ The pressure difference idea described here is wrong. Lift is the reaction force caused by deflecting the airstream downwards. But, don't take my word for it, take NASA's grc.nasa.gov/www/k-12/airplane/wrong1.html $\endgroup$ – Oscar Bravo Apr 25 '18 at 9:30
  • $\begingroup$ @Nephente the long path explanation is clearly not appreciated by references I have quoted in the corrected answer. The pressure and speed realities remain unchanged. Thank you for the correction. $\endgroup$ – Dlamini Apr 25 '18 at 11:57
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Another part of what creates lift is that wings are often angled upwards so that more air hits the bottom of the wing than the top which pushes upwards on the wing creating lift as well.

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  • $\begingroup$ This pressure difference idea is wrong. NASA have a proper description: grc.nasa.gov/www/k-12/airplane/lift1.html $\endgroup$ – Oscar Bravo Apr 25 '18 at 9:39
  • $\begingroup$ I have changed the answer not to include 'longer path' because some credible sources (whom I have quoted in the corrected answer) are not convinced of its verity. $\endgroup$ – Dlamini Apr 25 '18 at 12:04

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