lift coefficient

I understand the lift coefficient shown above. But does lift increase with speed, and how?

I understand I can't apply this to the graph as it is a coefficient, but if I had an angle of attack of 10 degrees and a speed of 0 knots, we would obviously have no lift assuming there isn't any wind, but if we increased our thrust and speed, we would gain more lift, and if we increase AOA even more, we get even more lift and the same with thrust.

But what about 0 degrees with symmetrical airfoils, does lift also increase with speed, and does it work similar?

  • $\begingroup$ Note that if you want no vertical acceleration (constant V/S, zero is also a constant), your lift must remain the same (Newton explain that better than me). Thus if you change your speed, you must change AoA to keep the same lift. $\endgroup$
    – Manu H
    Sep 4, 2019 at 5:01

4 Answers 4


The lift equation is $$L = \frac{1}{2} C_L · v^2 · S · \rho$$

where $S$ is wing area, $C_L$ coefficient of lift, $\rho$ air density, and $v$ airspeed. Using units of measure $\text{m}^2$ for $S$, $\text{m}/\text{s}$ for $\text{v}$, and $\text{kg}/\text{m}^3$ for $\rho$, the result is lift in $\text{N}$ (newtons).

Concerning symmetrical airfoils, they work quite well too, and the equation given above is perfectly valid for them. If it’s true that at zero degrees their $C_L$ is zero, they give zero lift regardless of the speed (zero times anything is zero).

  • $\begingroup$ so if we were to increase angle of attack, that increases the lift, and increasing the angle of atack and increasing the power, increases the lift even more, and if we wer e at zero angle of attack, we still get lift from an asymmetrical airfoil that too increases with speed? $\endgroup$ Sep 3, 2019 at 14:24
  • 2
    $\begingroup$ @JamesDavis Changing the AoA changes the CL (Lift coefficient). For an AoA where CL is not zero, the total lift will be proportional to the square of the velocity. In your example chart CL goes to zero at an AoA of -5 degrees. If you maintained an AoA of -5 degrees, then, the wings would generate no lift, regardless of airspeed. Anything above that AoA will generate lift proportional to the square of the velocity. $\endgroup$
    – J...
    Sep 3, 2019 at 17:04
  • $\begingroup$ The lift equation is idealized. The coefficient of lift versus angle of attack curve changes as speed and/or wing chord change. This is why air foil calculators like XFOIL also need Reynolds number, in addition to the air foil information to generate polars (lift vs aoa, drag vs aoa). Beyond some critical speed, due to flow separation, CL versus AOA decreases. $\endgroup$
    – rcgldr
    Sep 3, 2019 at 20:50

If you have a fixed aerofoil in a wind tunnel, lift increases with the square of the air speed.

However a real aircraft usually only needs enough lift to balance its weight, so as it flies faster it will also decrease the angle of attack to keep the lift constant.

Conversely, if a pilot wants more lift in order to make a tight turn, he will usually increase the angle of attack rather than increase speed.

Lift also increases with air density and wing area (as per xxavier's answer), but those are usually outside the control of the pilot!


It's difficult (impossible, really) to predict what happens based solely on air speed.

To get a meaningful result, you need to go by Reynolds number. Normally, you'll have a completely separate graph for each tested Reynolds number. At a different Reynolds number, not only with the lift change, but the basic shape of the Cl/alpha line is likely to change (e.g., as the Reynolds number drops, it's pretty routine to get a much "sharper" stall--that is, where your graph shows a nice, smooth roll off in lift as the AoA increases, at a lower Reynolds number, it might easily have only minimal loss of lift, then drop much more quickly. For example, here's a Cl/AoA graph for a NACA 6409 airfoil:

enter image description here

The gold line is at Re=1,000,000, the teal at Re=50,000. These are both quite low, but at least give the general idea. Note in particular that the curves have substantially different shapes. At Re=1M, the lift increases almost linearly with AoA, right up until it approaches stall (at which point it rolls off quite smoothly). At Re=50K, the increase is much less linear, and Cl increases sharply shortly before stall, then drops like a rock, with almost no warning at all.

But also note that it's all about Reynolds number. Low air speed with a large chord acts much the same as a much higher air speed with a much smaller chord.

For a concrete example, the airfoils used in the impellers of a jet engine operate at high air speeds, but have extremely small chords. This gives a very low Reynolds number despite the high air speed. Even for a supersonic jet, the impellers are often operating at a lower Reynolds number than the main wing of something like a Piper Cub or a Cessna 172.

  • $\begingroup$ At Reynolds numbers > $10^7$, at fully turbulent flow, the lift curves would be pretty close though. In fact so close, that the idealised lift equation of @xxavier’s answer can be used for practical computation of lift. $\endgroup$
    – Koyovis
    Sep 3, 2019 at 22:42

Welcome to Aviation Stack Exchange James! A simplified answer would be:

Lift is produced by airflow around the wing (expression for this is given in xxavier's answer). If you increase the speed, you will increase the airflow around the wing. Given everything else stays constant, you will get more lift. (IRL the consequences of increased speed and lift are numerous, but beyond the scope of this question)

For a symmetric airfoil at zero AoA the Cl would be 0, and speed would have no effect on lift: regardless of speed, lift would be zero.


You must log in to answer this question.

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