Let’s consider the case of a plane flying s/l in a headwind with a positive gradient. A wing produces lift by inducing a bound vortex superimposed on the relative wind., and in the particular case of flying in a positive gradient, one can imagine an extra pseudo-vortex associated with the gradient itself, that re-inforces the bound vortex producing additional lift.

The question is now how to quantify that additional lift. From the variables that enter in the circulation theory of lift, one can guess that the variables involved are the density of the air rho, the airspeed v, the wing area A, the chord c and of course the wind gradient ß.

From dimensional analysis, it results that the extra lift L is:

L = k · rho · v · A · c · ß

where k is a constant to be determined. It can be probably done with a ‘material experiment’, but my question here is whether it can be derived from already known data or from a ‘mental experiment’.

  • $\begingroup$ See grc.nasa.gov/www/k-12/airplane/lifteq.html and grc.nasa.gov/www/k-12/airplane/liftco.html, describing some of the issues. $\endgroup$
    – Řídící
    Apr 16, 2017 at 14:08
  • $\begingroup$ @xxavier, do you mean the angle of attack when you actually say gradient? $\endgroup$
    – GHB
    Apr 17, 2017 at 9:20
  • $\begingroup$ @GHB No, of course I don't mean that... The gradient of the wind is the variation of wind speed with altitude. For example, if you have 20 m/s variation in an altitude difference of 100m, the gradient is 20/100 m/s/m = 0,2 s^-1 $\endgroup$
    – xxavier
    Apr 17, 2017 at 11:34
  • $\begingroup$ Why do you think that your formula is correct? There are a lot of unexplained assumptions in it. With the lift and drag equations of the first comment's link you should be able to get a formula using a physical model instead of guessing. $\endgroup$
    – Gypaets
    Apr 18, 2017 at 9:08
  • $\begingroup$ @Gypaets The variables are the relevant ones in the problem, in particular rho, v, A, and ß; with c added because of dimensional considerations. I have already found two possible solutions, one of them presumably exact (π/2) and the other is an approximate solution that differs by only 5%. But I'm not sure... Hence my question... $\endgroup$
    – xxavier
    Apr 18, 2017 at 10:28

1 Answer 1


The formula you have given looks strange to me, I have never seen it before. But possibly I can help you with the following brief explanation about kinematic relationships when flying under wind conditions.

Flight path velocity (VK) is the vector sum of aerodynamic velocity (VA) plus wind velocity (VW), see sketch. - When flying in a headwind condition aerodynamic velocity will increase and a pilot will need to decrease Alpha to keep the lift constant. It is the opposite way around in a tail wind condition. Now a pilot will have to increase Alpha in order to keep the lift constant and additionally will have to pay attention not to come too close to Alpha max. Or increase thrust.

If one uses the sketched kinematic relationships properly one does not need an additional formula for analysis. enter image description here

  • $\begingroup$ If I understand all this correctly, this answer appears to solve for an instantaneous flightpath velocity solution dependent on an instantaneous wind velocity. However, I understand that the question is asking about a dynamic lift solution dependent on a dynamic wind gradient. I do not see how this answers the question and believe it would fit better as a comment. Once you have sufficient reputation you will be able to comment on any post. $\endgroup$
    – J W
    May 23, 2017 at 15:46
  • 2
    $\begingroup$ @Cristoph An aircraft flies always with respect to the mass of air., and doesn't 'feel' the wind for lift. Lift is the same either in a 'headwind condition' or in a 'tailwind condition' because wind exists only in reference to a fixed frame, usually the ground. $\endgroup$
    – xxavier
    May 23, 2017 at 15:52
  • 1
    $\begingroup$ @Jonathan Walters Exactly... $\endgroup$
    – xxavier
    May 23, 2017 at 15:53
  • $\begingroup$ Lift is not determined by the sum of wind and aircraft velocities (measured relatively to the ground). Only relative velocity of air (air velocity relatively to the aircraft). So there is no alpha change according to wind velocity relatively to the ground (what you name $V_w$), and neither change according to aircraft velocity relatively to the ground ($V_k$). Or maybe I don't understand your (untold) definitions of these velocities. $\endgroup$
    – mins
    May 23, 2017 at 18:58

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