We observe that lift force increases as speed increases on aircraft wings. Theoretically, there must be increase in pressure difference between top and bottom surface of wings.

What I don't understand is why pressure difference increases as speed increases? Air velocity increases on top of wings, but not only top of wings, it increases on bottom of wings too. So pressure on top decreases, but pressure on bottom decreases too. Why pressure on bottom decreases less than it does on top?

In this link, he proves lift equation but in 4.23, he removes delta V(V2-V1) as writes it as only V. Why he converted V2-V1 to only V ?


3 Answers 3


EDIT: Just realised that there's another big mistake in that video i.e. the "equal transit time" hypothesis. The higher speed on the upper surface in respect to the lower one is justified saying that air travels a longer distance on the upper surface and a shorter distance on the lower surface and afterwards it rejoins behind the airfoil... that's just plain wrong as explained for example here.

he removes delta V(V2-V1) as writes it as only V. Why he converted V2-V1 to only V?

Because he's making a very huge and basic mathematical mistake! At minute 3:50 he substitutes $(V_2²-V_1²)$ with $∆V²$ but by a mathematical point of view the latter correctly expands to:

$∆V²=(V_2-V_1)²=V_2²+V_1²-2V_2 V_1$

which is definitely not equal to $(V_2²-V_1²)$!

Air velocity... increases on bottom of wings too

That's another mistake! Normally the opposite is actually observed, i.e.:

  • an increase of speed on the upper surface and;
  • a decrease of speed on the lower surface.

Due to the momentum conservation (aka Newton's second law aka Bernoulli equation), an increase in speed is "compensated for" a decrease in pressure and viceversa. That implies that we get:

  • a lower than ambient pressure acting on the upper surface and;
  • a higher than ambient pressure on the lower side.

The upper surface is therefore "sucked upward" and the bottom surface is "pushed upward". The sum of these two effects is an aerodynamic force mainly directed upward (obviously).

Just as an example, the following picture taken from this paper shows the typical distribution of variation of pressure along the chord of typical airfoil (NACA 0012 @ 8° AoA):

pressure distribution NACA 0012

The upper curve is the plot for the upper surface and the lower curve is the plot for the bottom surface. Negative numbers mean that the local pressure is lower (but obviously not negative) than ambient pressure and viceversa.

  • $\begingroup$ According to Bernoulli equation, it's important V2^2 - V1^2. Why V2 (velocity on top) increases much more than V1 (on bottom) does? $\endgroup$
    – Jawel7
    Mar 20, 2023 at 9:50
  • $\begingroup$ There is no suction. There is only pressure. "Suction" is just a word we use to talk about positive pressures that are less than normal ambient atmospheric pressure, (14.7 psi), $\endgroup$ Mar 20, 2023 at 13:38
  • $\begingroup$ Even in this answer, the quote states that "Negative numbers means lower than ambient pressure and vice versa." The actual numbers are NOT negative. They are just less than 14.7 psi. $\endgroup$ Mar 20, 2023 at 13:50
  • 1
    $\begingroup$ The plot has both positive and negative numbers on the y-axis, no idea what you're talking about... $\endgroup$
    – sophit
    Mar 20, 2023 at 13:52
  • $\begingroup$ @Jawel7 For a given angle of attack, shape of airfoil and sufficiently low Mach numbers, the pressure distribution described by the chart that this answer shows is always the same. There is not really a "why" to it, that's just how air in the physical world behaves. We know that pressure is proportional to the square of velocity (again, that's just how it is), so you can see from the chart that the square of the velocity at the bottom of the airfoil is roughly 0.8 times the square of the freestream velocity (cp = 1 - u^2/u_inf^2 = 0.2, so u^2/u_inf^2 = 0.8) ... $\endgroup$ Mar 20, 2023 at 21:25

Because lift is not suction. "Suction" is just a way we describe a pressure that is less than some defined "standard" atmospheric Pressure. The air flow on the top of the wing is still pushing down on the top with positive pressure. It's just less than the pressure pushing up on the bottom.

So as velocity increases, the pressure increases on top and bottom equivalently. As we increase velocity, the pressure goes up by the same percentage on both top and bottom. But if it's 100psi on the top, and 105psi on the bottom, and both increase by 10%, then it becomes 110psi on the top, and 115.5psi on the bottom. The difference has also increased by 10%.

  • $\begingroup$ As seen in sophits answer, particularly for various portions of the upper and lower wing, there is indeed positive and negative pressure relative to freestream. What should be reviewed is upper wing pressure increases with velocity. $\endgroup$ Mar 20, 2023 at 16:44
  • $\begingroup$ Indeed, the upper wing vacuum pump becomes more effective, allowing lower pressures, as V increases, especially when Reynolds number exceeds around 500,000. $\endgroup$ Mar 20, 2023 at 16:49

It is amazing how we continue to struggle to understand the fundamentals of lift, but no need to be tearing apart somebody else's work.

In the video (rightly or wrongly) the Bernoulli equation for flow through a restricted pipe is used:

P1 + 1/2 $\rho$v1$^2$ = P2 + 1/2 $\rho$v2$^2$

F/Area = (P1 - P2) or $\Delta$P

Where he get a little shakey is when a division by $\rho$ step is not explained, leading to

P1/$\rho$ + 1/2 v1$^2$ = P2/$\rho$ + 1/2 v2$^2$

Now we have $\Delta$P/$\rho$ = 1/2 (v2$^2$ - v1$^2$) = 1/2 $\Delta$ (of) v$^2$

Plugging back in to $\Delta$P/$\rho$ = F/(A$\rho$) = 1/2 $\Delta$v$^2$

--> F or Lift = 1/2 $\rho$ × A × $\Delta$v$^2$, or, substituting the symbol S for area :

Lift = 1/2 $\rho$ × S × $\Delta$v$^2$

Although the math is correct: the issue is that $\Delta$v$^2$ in pipes is further dissected to v$^2$ of the freestream × coefficient of lift in aviation.

Coefficient of lift contains the factors of Angle of Attack and Airfoil shape and camber.

Any object moving through a fluid will create drag, but we can study the phenomena relative to the aircraft, so now we have air flowing over the stationary object, as in a wind tunnel.

Perhaps easier to understand by taking a fully symmetrical object, and pointing it straight into the wind: we have high pressure and lower velocity in front, and lower pressure and higher velocity on top and bottom.

In order to create lift, we must have an angle of attack to the freestream. Lift can be positive or negative, but to counter-act gravitational force, we need to produce a postive angle of attack$^1$.

Now, the higher pressure, lower velocity region grows in size and shifts to underneath the wing, and the lower pressure higher velocity "wake" of the wing is on top of the wing.

Naturally, we see an increase in drag as AoA increases.

But as speed increases, the ability of the wing to hold higher pressure below, and lower pressure above, increases because $\Delta$ pressure can be maintained faster than the natural tendency of air pressure to equalize pressure.

Here the energy of the thrust is literally running a vacuum pump and a pressure pump on the wing at the same time, creating the lifting force.

Another very important factor is Reynolds number. Above a certain limit, the efficiency of the pressure/vacuum pump increases greatly, allowing much higher Lift to Drag ratios.

Changes in airfoil shapes design can improve this further.

$^1$ asymmetrical wings can generate lift at 0, or even slightly negative, Angle of Attack

  • $\begingroup$ My way: $\Delta$P = F/A = 1/2 $\rho$(v2$^2$ - v1$^2$). $\endgroup$ Mar 21, 2023 at 15:56

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