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I have searched if any explanation that bind Bernoulli's principle in lifting airplane. There is one here which receive many appreciations, but still did not answer the question. So far, it was taught, explained, or mentioned in any aerodynamic or airplane principle, but never bound the Bernoulli's formula to the lift formula. As we know, there are pressure conservation as below: $$\mathrm{P}_{1}+\frac{1}{2} \rho \mathrm{V}_{1}^{2}+\rho \mathrm{gh}_{1}=\mathrm{P}_{2}+\frac{1}{2} \rho \mathrm{V}_{2}^{2}+\rho \mathrm{gh}_{2}$$

Due to height is considered are same below and above the wing (the different are very small), then the third part of the equation will cancel one each other. Then left only part one and part two of each side of the equation. As the velocity above (considered as V1) is said will be different below the wing which wind is faster in above, then pressure above the wing will be lower, which will lift the wing. Thats is commonly taught in every explanation. Then, the

$$V_{2}=\sqrt{\frac{2\left(P_{1}-P_{2}\right)-\rho V_{1}^{2}}{\rho}}$$ Meanwhile, lifting formula is expressed as below: $$L=\frac{1}{2} \rho v^{2} S C_{L}$$ V here is the airplane velocity, which is the wind hit the airfoil. Thence, V=V2 of the Bernoulli's equation above. V1 that hit upper side of the airfoils/wing, which is said faster than below, is unknown.

So, where is the Bernoulli's principle contributes in this case? How actually we calculate the lift force of an airplane?

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If I understand correctly you want to know how to get from the first formula to the second one. If that is the case, the answer is simple: you don't :)

The lift formula is a simplified one, which considers the wing as a whole. It doesn't use velocities on different points of the wing, it uses the speed of the airplane. All the factors such as profile, wing shape, etc are included in the lift coefficient. This formula is useful when you already know the behavior of the wing in different situations (the behavior is determined in wind-tunnel tests or with numerical methods).

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  • $\begingroup$ Yes, you got my point. So the question is, where is the bernoulli's principle in lifting the airplane? Is bernoulli's contributing in that thing? $\endgroup$ – AirCraft Lover Jan 14 at 0:50
  • $\begingroup$ @AirCraftLover No, not really. Bernoulli's principle is not a correct explanation for the creation of lift (grc.nasa.gov/www/k-12/airplane/wrong3.html). $\endgroup$ – Emil Jan 15 at 13:32
  • $\begingroup$ Bernoulli's principle itself is perfectly correct. Out of all of the explanations of lift that mention Bernoulli's principle, some of them are correct and some of them are incorrect. $\endgroup$ – Tanner Swett May 9 at 16:27
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1. Physical laws don't act, they only explain

The universe does not (seem to) calculate equation. It just works its way and we use equations to try to find some order in it that would allow us predict how it will react to what we plan to do.

2. Physical laws all hold at the same time

Physical laws don't describe parts of how the universe works that would be added one to another. Rather they each describe an aspect of how it works all the time. Therefore it is not so that one law (Bernoulli's principle) would contribute a bit of something (lift) and then another would contribute another bit. Rather it is their combination that tells us the phenomenon will occur.

After all, the laws are expressed by equations. Each law is one. But they usually have many free variables. One equation with many free variables constraints the solution, but you need as many equations as you have free variables to get a unique solution. Bernoulli's equation is not enough.

3. Applicability of Bernoulli's principle

Bernoulli's principle is just an expression of conservation of energy. We have all the reasons to believe that conservation of energy (of mass-energy) holds everywhere in universe, which includes around the wing generating lift.

But as said above, Bernoulli's equation is just one equation with way too many variables to produce solution on its own

4. Other laws

Due to the unbounded nature of the situation, the only way to get enough constraints to actually find a solution requires turning to the ultimate hammer of fluid dynamics, the Navier–Stokes equations. This is highly advanced equation that involves inertia and viscosity of air, both essential properties for generating lift. Inviscid flow does not produce any lift as can be tested in liquid helium. Massless flow wouldn't produce any either, but unfortunately there are no massless fluids to test it with.

You still have to throw in conservation of energy, i.e. Bernoulli's equation, along with conservation of mass, and for larger pressure differences also the general gas equation and the equation for adiabatic process to get enough equations to restrict all the free variables.

The result is a set of partial differential equations that don't have any useful analytical solutions and need to be numerically integrated over sufficiently large volume of space surrounding the wing and sufficiently long time period.

Now you'll have huge dataset describing the flow at each point in space and time with some granularity. If you plot the total lift over sufficient set of boundary conditions, and try to fir a simple equation to it, you'll get the famous lift equation. Approximately—it is just fitting to a bunch of points!

5. “Contribution” of Bernoulli's principle

Bernoulli's principle does contribute to the explanation by holding in the situation. You wouldn't have enough constraints to get a unique solution otherwise. But there is no way to specify what the contribution meant in the resulting equation. All you can say is that it is needed to calculate the exact points that can be approximated with the lift equation.

Note: there are decent qualitative explanations of the phenomena described by the Navier-Stokes equations, but you've already seen them as it is the answer you linked in the question. No point in repeating them here.

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  • $\begingroup$ "inviscid flow does not produce any lift" perhaps an apter way of saying that is "inviscid flow cannot get any circulation?" $\endgroup$ – Meatball Princess Jan 25 at 4:27
  • $\begingroup$ @MeatballPrincess, yes, but that would mean introducing another term and what it has to do with lift. $\endgroup$ – Jan Hudec Jan 25 at 6:04
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The statement "Because of continuity, so the air that flows in from the left side must flow out at the right, the upper flow must get there at the same time. But because that line is curved, air has to go faster to get there at the same time." is not necessarily correct. See Chapter 3.2 from this e-book, excerpted here. https://www.av8n.com/how/ The author simulated a wind tunnel, and documented air flow patterns around a wing, with different colored chunks of air in different colors. He demonstrates that air above and below the wing do not arrive at the same time. The book also discusses quite well in my opinion how downwards airflow at the back of the wing really contributes to the lift created, more so than any low pressure area above the wing. Air being pushed down = weight of aircraft being pushed up. (The old "opposite and equal reaction".)enter image description here

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  • $\begingroup$ So, can I say that there is no Bernoulli's countribution in lifting process? I agree with you, the air is forced downward by the curved airfoil will act as agaist gravity (mg). $\endgroup$ – AirCraft Lover Dec 13 '18 at 16:18
  • $\begingroup$ I don't know about No contribution, but I would suggest perhaps not the main player anyway. See chapter 3.6 of the same e-book. av8n.com/how/htm/airfoils.html#sec-airplane-air $\endgroup$ – CrossRoads Dec 13 '18 at 16:35
  • $\begingroup$ Probably that the best way to conclude, that not the main part. $\endgroup$ – AirCraft Lover Dec 13 '18 at 18:45
  • $\begingroup$ downwards airflow at the back of the wing really contributes to the lift created Not that it doesn't contribute, but most of the lift is created at the front of the wing. The lion's share of the downward airflow at the back of the wing was turned downward by the front of the wing. $\endgroup$ – TomMcW Dec 13 '18 at 18:56
  • $\begingroup$ Then the question is, "why is the airflow turned downward?" It's turned downward because of the low pressure above the wing. Both are happening simultaneously $\endgroup$ – TomMcW Dec 13 '18 at 18:57
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See, here is the conundrum of introductory aerodynamics/fluid mechanics books. Lift is hard. There is simply no simple way of explaining lift. Why would there be? It is only fair that you need quite some math to figure out the pressure distribution, which is the pressure field around an arbitrary body in airflow, and as you can imagine, that is no easy task. Who gets to say that just because lift is essential to flight, it must be readily understandable?


First off $V_{\mathrm{free stream}}\ne V_1\ne V_2$, and $P_{\mathrm{atmosphere}}\ne P_1\ne P_2$, so here is your answer.

Secondly in a more accurate analysis $V_1$, $V_2$, $P_1$, $P_2$ can not be assumed constant at either under or over the wing.

A more accurate portrayal of lift is most easily achieved by simplifying the airflow first to 2D-potential flow, i.e. $\exists \varphi, v=\nabla\varphi$ then assuming $\rho=\text{const}$, then we would get $\nabla^2\varphi=0$ the Laplace equation. The Bernoulli's equation is used here to link $p$ singlesidedly to $v$ hence $\varphi$, if you apply Bernoulli's equation right from the beginning, how do you even know $V_1>V_2$ without making it an assertion?

(A ramification of making the flow a potential flow is we automatically set $T=\text{const}$ as well.)

Now we have only a single variable, namely $v$, and there are multiple interesting ways to solve the Laplace equation. But perhaps the way that provides the most insight into lift generation is through a conformal mapping, i.e. $f\colon\mathbb{C}\mapsto\mathbb{C}$ analytical and $f^\prime\ne0$. Conformal $f$ has the property that if $\nabla^2\varphi=0$ then $\nabla^2(\varphi\circ f)=0$ and $\nabla^2(\varphi\circ f^{-1})=0$.

As you can see where this is going, we study the potential flow $\varphi_0$ around a cylinder then find a $f$ that maps the cylinder to the wing profile and $\varphi=\varphi_0\circ f$ automatically yields the potential flow around the wing.

There are three kinds of basic solution to the flow around the cylinder: rectilinear, vortex and doublet. And per the linearity of the Laplace equation, any superposition of the three solutions is also a solution. Similarly, any flow around a wing can be seen as superpositioning of three respective $\varphi_0\circ f$.

Below is the visualization of the solution of the pressure field of flow around a cylinder (red, purple = high pressure, green, blue = low pressure, free stream flows from right to left):

enter image description here.

Note that just like I said, there are three basic solutions, the pressure shown above is the result of changing the coefficient of combining the solutions.

Here is the airfoil version, note the high degree of similarity

enter image description here.

A most useful result of this approach is a direct proof of Kutta–Joukowski theorem. This theorem states that the lift, defined as the component of net force acting on a body immersed in rectilinear airflow that is perpendicular to the free stream's velocity vector, of an arbitrarily shaped body in an inviscid potential flow is given by $$L=\rho V_{\infty}\Gamma,$$ where $\Gamma=\int_C v\,\mathrm{d}s$ for any route $C$ that surrounds the body. This tool, not the Bernoulli's principle, is the real workhorse of aerodynamicists.


Speaking of potential flow and Bernoulli's equation, here is an interesting fact:

From the differential form of the momentum equation $$\rho\dfrac{\mathrm{D}}{\mathrm{D}t}v+\nabla p=0$$ with the assumption that $\rho=\text{const}$, $\dfrac{\partial(\cdot)}{\partial t}=0$ (steady flow) we get $$\rho\dfrac{\mathrm{D}}{\mathrm{D}t}v+\nabla p=\rho(v\cdot\nabla)v+\nabla p=\nabla\left(\rho\dfrac{v\cdot v}{2}+p\right)=0,$$ which suggests the total head $$H=\rho\dfrac{V^2}{2}+P$$ is constant not only on streamline, but everywhere in the whole domain! This is a stronger version of Bernoulli's Law, implicit in the Newton's second law.


Please do notice I have only mentioned lift under the ideal circumstance of inviscid potential flow, and the solution given by this theory deviates from real life in a significant way. For example, you can tell from experience that there is no way a cylinder can stand in water flow and not feel any drag, yet the solution to flow around cylinder says so. This is called d'alembert paradox. The answer to this paradox is viscosity of water. The viscosity of water prevents a full pressure recovery on the rear half of the cylinder, and the flow would separate near the top and bottom of the cylinder. Viscosity is also important for airfoil, firstly because it is the primary reason why wings stall, secondly, it has an intricate relation to exactly which solution will be established around the wing, and this determines through K-L law the lift, i.e. frictional drag actually dictates pressure drag and lift!!!


EDIT: The lift equation is $L=\frac{1}{2}\rho C_LAv^2$. As you would have guessed it, the circulation $\Gamma$ is proportional to airspeed. But the reason why increasing airspeed increases circulation is even more arcane and too long for one answer.

EDIT: The lift equation is derived from the K-L law mentioned above. $C_L$ is defined as $\frac{L}{1/2\rho v^2 A}$, not obtained theoretically from the airfoil shape and $A$, as the equation would have you erroneously believe.

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The Bernoulli equation is defined between 2 points in a flow field. It has limitations, mainly that it's valid only for incompressible flow, so only valid for low airspeeds. As you noted accurately, the height term is usually ignored, but it's more to do with the low density of air than the size, as for water or mercury it would make a large difference even with small lengths.

So the equation you derived (apart from rho being at the wrong place, it should be p/rho) is accurate between any 2 points in the flow field. Lift is caused by the pressure difference between the two sides of the airfoil, so if you wanted to calculate the lift, you would need to evaluate pressures at all points on the upper and lower side of the airfoil. As the Bernoulli equation states, where the flow speeds up, the pressure drops, and this is the upper side of the airfoil. See picture from wikipedia, the bottom streamline is almost straight, so its shorter than the curved one on the top. Because of continuity, so the air that flows in from the left side must flow out at the right, the upper flow must get there at the same time. But because that line is curved, air has to go faster to get there at the same time. Faster flow = lower pressure. enter image description here

So you can use the Bernoulli principle to estimate the pressure at every point, provided you know the speed at the given point. Once you do that, you get pressure distributions over the airfoil, something like this: enter image description here

Note that these distributions are not calculated by hand, but rather with some CFD code, or measured in wind tunnels. Once you have the distribution, it can be summed to get the total forces. To make life easier, engineers measured all these forces as a function of altitude, speed (Re number) and angle of attack, and made it available for you as a CL-alpha curve. Then the lift equation is a similarity law essentially, where you turn the non-dimensional pressure distribution, eg the CL into actual force that you can use to calculate.

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  • $\begingroup$ @D Sziroczak, sorry for the wrong equation. Hope now it is correct. $\endgroup$ – AirCraft Lover Dec 13 '18 at 13:44
  • $\begingroup$ Dear Sziroczak, I quote this: "the air that flows in from the left side must flow out at the right, the upper flow must get there at the same time. But because that line is curved, air has to go faster to get there at the same time". I heard that one very often. But there is no evident confirming that claim. Also NASA in their website still wrote that the Bernoulli's contribution in lift is still in debate. $\endgroup$ – AirCraft Lover Dec 13 '18 at 13:58
  • $\begingroup$ From this statement: To make life easier, engineers measured all these forces as a function of altitude, speed (Re number) and angle of attack, and made it available for you as a CL-alpha curve. Then the lift equation is a similarity law essentially, where you turn the non-dimensional pressure distribution, eg the CL into actual force that you can use to calculate. Can I say that by using coefficient of lift in that lift formula as I wrote above, I have include the Bernoulli's contribution in that lifting force? Please, I need your feedback, Sir. $\endgroup$ – AirCraft Lover Dec 13 '18 at 14:01
  • $\begingroup$ Because of continuity, so the air that flows in from the left side must flow out at the right, the upper flow must get there at the same time.  What says it has to get there at the same time? It doesn't. In fact the upper flow generally gets there before the lower flow. $\endgroup$ – TomMcW Dec 13 '18 at 18:49
  • $\begingroup$ AirCraft Lover: The Bernoulli equation is a way to describe properties at 2 points in the flow space. It does not "contribute" to anything. The lift is generated because of the pressure differences. Individual pressure differences can be calculated using the Bernoulli equation (keeping in mind the limitations). Lift coefficient is a simplification based on similarity law, so you don't need to calculate pressure distribution all the time, but assume that under similar conditions, pressure distributions are similar. $\endgroup$ – D Sziroczak Dec 14 '18 at 17:17
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Easy to illustrate. Hold a piece of paper loosely by the corners so it is curved down hanging away from your face. Blow gently across the top of it and see it lift. There is no "push from below", only a suction from above.

As a 700+ hour pilot I enjoy watching my wings get 'sucked' into the wild blue yonder by the lower pressure above the wings and the relatively higher pressure below - also known as 'Lift'

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  • $\begingroup$ As I have explained in the question body with the two formulas, there is no equation which bind the bernoulli to lifting force. $\endgroup$ – AirCraft Lover Jan 13 at 16:37

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