Revision 3. Clarifications, as required by moderators, added. The main reason for asking the question provided.

Regarding the reverse command regime of a plane

Assuming a plane hovers (ground speed $V_{g,1}=0)$ against a headwind with a velocity $V_{w,1}$, is it possible for this plane to start moving ($V_{g,2}$ > 0) just because the wind intensifies increasing its speed from $V_{w,1}$ to $V_{w,2}$ > $V_{w,1}$?

It is known that once the plane hovers against the $V_{w,1}$ wind all its control surfaces and the power delivered to the propeller remain locked.

As a further clarification, the plane is not restricted to always stay at the same altitude. It can rise. However the power must remain unchanged.

In general, I am interested to find out if a plane can fly faster and faster as the headwind speed increases in the particular case the plane is in the reverse command regime (see definition: http://aviationglossary.com/region-of-reverse-command/ ). For the direct command regime I understand that this behavior is not possible.

Why do I ask this apparently strange question? Because it appears that a plane, in certain conditions which are not clear for me, can gain some ground speed only from the intensity of the headwind, as long as two aviation pioneers noticed this effect.

In 1904, the Wright Brothers started to test a new plane, Flyer II, somewhere near Dayton, Ohio where they managed to get permission to use a flat pasture for their experiments. The winds were light there and, in the beginning, they had no catapult to quickly accelerate their machine and throw it into the air. They simply started the engine of the airplane which began to move along a track (a runway) while a headwind of moderate intensity was blowing and finally they got into the air and flew slower if the headwind speed was lower and faster if the headwind was stronger (see letters 1 and 2).

Letter 1: Fragment from a letter addressed by Wilbur Wright to Octave Chanute, on August 8, 1904: "One of the Saturday flights reached 600 ft. ... We have found great difficulty in getting sufficient initial velocity to get real starts. While the new machine lifts at a speed of about 23 miles, it is only after the speed reaches 27 or 28 miles that the resistance falls below the thrust. We have found it practically impossible to reach a higher speed than about 24 miles on a track of available length, and as the winds are mostly very light, and full of lulls in which the speed falls to almost nothing, we often find the relative velocity below the limit and are unable to proceed. ... It is evident that we will have to build a starting device that will render us independent of wind." Source: Page 52 of Octave Chanute Papers: Special Correspondence--Wright Brothers, 1904 | Library of Congress

Letter 2: Fragment from the letter written by Wilbur Wright to Octave Chanute on August 28, 1904: "Dayton, Ohio, August 28, 1904. Dear Mr Chanute ... ... Since the first of August we have made twenty five starts with the #2 Flyer. The longest flights were 1432 ft., 1304 ft, 1296, ft. and 1260 ft. These are about as long as we can readily make on over present grounds without circling. We find that the greatest speed over the ground is attained in the flights against the stronger breezes. We find that our speed at startup is about 29 or 30 ft per second, the last 60 ft of track being covered in from 2 to 2 1/4 seconds. The acceleration toward the end being very little. When the wind averages much below 10 ft per second it is very difficult to maintain flight, because the variations of the wind are such as to reduce the relative speed so low at times that the resistance becomes greater than the thrust of the screws. Under such circumstances the best of management will not insure a long flight, and at the best the speed accelerates very slowly. In one flight of 39 1/4 seconds the average speed over the ground was only 33 ft per second, a velocity only about 3 ft per second greater than that at startup. The wind averaged 12 ft per second. In a flight against a wind averaging 17 ft per second, the average speed over the ground was 42 ft per second, an average relative velocity of 59 ft per second and an indicated maximum velocity of 70 ft per second. We think the machine when in full flight will maintain an average relative speed of at least 45 miles an hour. This is rather more than we care for at present. Our starting apparatus is approaching completion and then we will be ready to start in calms and practice circling. Yours truly Wilbur Wright." Source: Page 55 of Octave Chanute Papers: Special Correspondence--Wright Brothers, 1904 | Library of Congress

  • $\begingroup$ Assuming an airplane has a ground speed of 0, an increase in headwind would cause the airplane to move backwards over the ground. That is, if Vg1 = 0 and Vw2 > Vw1, then Vg2 < 0. $\endgroup$ Commented Aug 29, 2015 at 20:15
  • $\begingroup$ Note that the reverse command regime is unstable; you will not be able to maintain constant air speed without constantly manipulating power / control surfaces. $\endgroup$
    – DeltaLima
    Commented Aug 29, 2015 at 20:57
  • $\begingroup$ how can you be faster with headwind? $\endgroup$
    – Federico
    Commented Aug 29, 2015 at 22:07
  • 3
    $\begingroup$ Thank you for the edit - now I understand what you ask. The Wrights were flying at the back side of the power curve where induced drag is dominant. Induced drag falls with increasing speed, and the power of the Flyer II was so limited that it had only a small speed range over which flight could be sustained. By launching into a headwind, the initial acceleration plus wind speed got the plane above the minimum sustainable speed. Without wind, the acceleration along the rail was not sufficient, and once the plane had to support its weight with its wings, drag rose above thrust. $\endgroup$ Commented Aug 30, 2015 at 7:41
  • 1
    $\begingroup$ I upvoted, I voted for reopening - now we have to wait until more people vote for reopening. Meanwhile you might want to check a few other answers here to learn the needed equations. Or check out this. $\endgroup$ Commented Aug 30, 2015 at 18:41

2 Answers 2


How was the Flyer II launched into the air?

The first thing to consider is the take-off method used by the Wrights when flying on Huffman Prarie in 1904. The plane was set on a wheeled dolly which moved along a single wooden rail, the end of which was slightly elevated to give the aircraft more pitch angle and some vertical speed upon leaving the rail. This answer has some details on the rails, but centers on the catapult which was added later in 1904.

Drag components

The next is the general composition of drag in subsonic aircraft. Two main components can be distinguished, one growing with the square of airspeed and caused by friction (called zero-lift drag), and one dropping with the square of airspeed, caused by the downward acceleration of air when lift is created (called induced drag). The plot below shows them and the total drag as a green line. The drag minimum is attained at some intermediate speed. As long as the aircraft flies slower than at the speed of minimum drag, speeding up will lower the total drag.

Plot of the drag components over speed

Note that I took the liberty of adding some extra drag at the slow end of both drag curves to simulate the additional pressure drag caused by the beginning flow separation at low speed. When looking at wind tunnel data from tests on models of the 1903 Flyer (taken from Fred Culicks paper), the flow separation is apparent by the gradual decline of the lift curve slope shortly before maximum lift is reached.

Lift and Drag of the 1903 Wright Flyer

I added a red line on the drag curve on the left. It shows at which lift coefficient the ratio between lift and drag is minimized, and this again confirms that the lowest drag is possible at some intermediate lift coefficient. The lift coefficient of an airplane is proportional to the inverse of the airspeed squared, so this shows again that the lowest drag is attained at some intermediate speed.

Drag increase after launch

The attitude of the airplane on the track was approximately 0°. Note that the polar above gives an angle of attack between 2° and 5° for the flight condition. This means that the aircraft produced less lift while resting on the dolly and consequently less induced drag at the same airspeed. The friction drag of the dolly should be an order of magnitude smaller than the saving in induced drag due to the reduced angle of attack during acceleration. Only when the Flyer II pitched up at the end of the track it gained the lift needed to stay it in the air.

Another effect to consider is ground effect. It lowers the induced drag component, and this effect should have helped to accelerate the Wright Flyer more while it was running along the track. Once it was airborne and had climbed some feet into the air, the speed would drop due to the climb and drag would increase due to the decreasing ground effect.

Propeller thrust over speed

The second important force is the propeller thrust, of course. In case of the Flyer II, two pusher propellers were driven by an internal combustion engine of maybe 16 HP. The power output of an IC engine is constant regardless of the speed of the aircraft and increases linearly with RPM. When driving two large, fixed-pitch propellers, the engine speed will settle at the point where the torque output equals the torque to drive the propellers. This should give the Flyer II engine a tendency to run at higher RPMs the faster the aircraft flies. However, at some point the increased aerodynamic drag of the propellers will limit this RPM increase. Around this point, the propeller thrust will change with the inverse of airspeed, since propeller thrust is power divided by speed.

The plot below is pure guesswork, but should look similar to what the Flyer II was capable of. Thrust was marginal.

Drag and Thrust over speed

Generic plot of drag and thrust over speed with fixed propellers. The plane has two trim points, one unstable point close to stall and one stable point beyond the drag minimum.

Wind speed

The last effect to consider is wind speed. When launching the Flyer II into a headwind, the airplane will have a higher airspeed at every point along the track and leave it at a higher speed than when being launched in calm wind. After launch, the aerodynamic drag increases suddenly as soon as the Flyer flies, and will increase again a little when it climbs. At these speeds, the thrust produced by the propellers will be roughly constant over airspeed - the details depend on the gearing and the engine characteristics. This means that the Flyer would be launched into the air at a point of its drag polar where drag in absolute terms was lower the more headwind was available, and the propeller thrust was not sufficient in calm air to sustain level flight. Only with the added speed possible in a headwind the Flyer could produce enough thrust to overcome the now lower drag and stay airborne.

In case of the lower headwind of 12 ft/s the plane lifted off with an airspeed of around 42 ft/s, just a bit above the slow trim point, and when it landed at the end of the field half a minute later, it had only accelerated to 48 ft/s, and the average speed over ground was only 33 ft/s. On another day with 17 ft/s headwind, the plane accelerated on the rail to maybe 47 ft/s and was well above the lower trim point at launch, so it could accelerate to the fast trim point reasonably quickly, which was maybe at 65 ft/s. Now the average speed over ground was 42 ft/s, giving the illusion that headwind made the aircraft faster. Had the flights taken longer, the speed difference between both days would had been smaller because the transient effect of the launch speed would had been smaller.

  • $\begingroup$ 1) Thank you for your answer. 2) There should be other more recent and documented planes than Flyer II which, for a certain range of airspeeds, fly faster relative to the ground against stronger headwinds. If you or somebody else find such an aircraft tell me about it. $\endgroup$ Commented Aug 31, 2015 at 23:36
  • $\begingroup$ Most airplanes will fly faster over ground in a stronger headwind if you arrange for it. The behavior of the Flyer was caused by the launch method - without it the whole effect disappears. Every underpowered airplane with slowly progressing flow separation close to stall will show the same behavior in the same circumstances. You should not look for specific aircraft, but specific scenarios. A carrier launch without catapult would be such a scenario. $\endgroup$ Commented Sep 1, 2015 at 20:42
  • $\begingroup$ @ Peter Kämpf, I need the simplified equations that model Flyer II or another plane behaving in the same manner. I have tried to write these eqs. but I do not get anywhere. The plane looks to be always slowed down by the headwind. (1) T - 1/2 * Ro * S * Vx^2 * (Cd0+Cdi) = m * ax where Cdi = k * Cl^2, (2) 1/2 * Ro * S * Vy^2 * Cl - m * g = m * ay. Help me write these equations in the correct manner. I need a simple Mathcad model that can simulate the flight against headwinds of different intensities. $\endgroup$ Commented Sep 2, 2015 at 1:42
  • $\begingroup$ Your equations look fine, and the conclusion is correct. The aircraft will fly slower over ground in a headwind. See here for an answer about the equations of motion. What you need to add is the nonlinear behavior of the lift and drag coefficients at higher angles of attack. $\endgroup$ Commented Sep 2, 2015 at 6:31
  • $\begingroup$ I have already added that nonlinear dependency either as Cd = Cd0 + k * Cl^2 or extracting Cd=Cd(Cl) from the diagram: "Lift and Drag curves for a typical airfoil" (see: en.wikipedia.org/wiki/Lift-induced_drag ). The non-linearity of the functions Cd = Cd(alfa) and Cl = Cl(alfa) does not seem to speed up the plane as the headwind intensifies. $\endgroup$ Commented Sep 2, 2015 at 16:10

The question is not valid as asked because of a couple of factors. First, an aircraft in flight operates within a given airmass, and is unaffected by the 'wind'. Second, the 'reverse command regime' refers to an angle of attack beyond which (given constant thrust) increasing the angle of attack results in decreased net lift. In other words, the aircraft will begin to descend (or rate of descent will increase) when the stick is pulled back, i.e. the reverse of the 'commanded' climb.

If there were a wind-shear (relatively rapid increase in wind-speed), there would be a temporary increase in aircraft performance (lift and/or airspeed) due to the inherent increase in airspeed, but this will only be sustained for the duration of of the change in wind-speed. This does not appear to be what you are describing, but rapid changes in wind velocity are the only phenomena that would affect aircraft performance. In no case would airspeed be affected by a constant wind, regardless of the velocity.

Imagine an airplane flying at 100mph within an airmass that's moving at 200mph in the opposite direction; that aircraft is traveling backward (relative to the ground) at 100mph. Meanwhile, another aircraft is also flying at 100mph in the opposite direction (with the wind), resulting in a ground speed of 300mph. Both aircraft have the same airspeed; they both 'feel' the same 100mph relative wind. The movement of the airmass within which they operate is not relevant to their performance.

  • $\begingroup$ Because you talked about a "rapid increase in wind-speed" that can induce a "temporary increase in aircraft performance (lift and/or airspeed)", do you think it is possible that a plane flying horizontally with Vg1 against a Vw1 headwind to rise, even for a short time, its ground speed to Vg2 > Vg1 when the wind intensity suddenly gets higher changing from Vw1 to Vw2 > Vw1? $\endgroup$ Commented Sep 1, 2015 at 0:03
  • $\begingroup$ No. An increasing headwind gradient will result in a net decreasing groundspeed. $\endgroup$ Commented Sep 7, 2015 at 19:28
  • $\begingroup$ "Second, the 'reverse command regime' refers to an angle of attack beyond which (given constant thrust) increasing the angle of attack results in decreased net lift."-- this is inaccurate. As we decrease airspeed, we get into the region of "reversed command" well before we actually see a decrease in lift coefficient, which only happens at or near the stall. The region of reversed command is characterized by, as we slow down, a need for increased power to maintain level flight. If we don't increase power, we find that we increase the sink rate, as we move the yoke aft and slow down. $\endgroup$ Commented Feb 8, 2022 at 18:02

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