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If I understand correctly, wind force increases quadratically as airspeed increases. Therefore, all other things being equal, lift should also increase quadratically. Does that mean, a plane going at 600 mph has 9x more lift than when it was going at 200mph? How can the flight be stable at both speeds? Surely it will either have too much lift or too little lift at one speed?

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    $\begingroup$ By trimming. That's what the trim wheel is for $\endgroup$
    – slebetman
    Commented Jun 5 at 10:09

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The lift generated by wings is proportional to the square of the calibrated airspeed and proportional to the angle of attack up to a point. As an airplane accelerates, it must reduce its angle of attack to remain in level flight and avoid climbing. So an airplane traveling at 300 knots of calibrated airspeed will fly at a much lower angle of attack, but still produce the same lift force, as an airplane flying at 100 knots of calibrated airspeed.

If an airplane slows down, the pilot must increase the angle of attack in order to generate enough lift to fly. If they continue to slow down, at some point the angle of attack required to maintain enough lift for level flight will cause a disruption of airflow over the top of the wings. This is the critical angle of attack beyond which the wing will suffer an aerodynamic stall. If the wing suffers an aerodynamic stall, the amount of lift drops off dramatically for an increase in angle of attack, and the pressure drag dramatically increases. At this point, the airplane can no longer continue to create enough lift in order to fly.

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    $\begingroup$ It's actually pressure drag which dramatically increases during stall. Lift can, but must not always, drop dramatically in post-stall condition. $\endgroup$ Commented Jun 5 at 5:49
  • $\begingroup$ Thanks for this. Is there some tool that simulates/calculates the lift of an airfoil as a function of the angle of attack? $\endgroup$ Commented Jun 5 at 8:46
  • $\begingroup$ @CaptainCodeman To a good first order approximation, lift is proportional to angle of attack at any given speed, linearly, lift = ka, where k is a constant, and a is the angle of attack, when a is between the positive lift and negative lift critical angles. Small deviations in that approximation are handled automatically by the pilot simply adjusting the angle of attack to give the required aircraft acceleration. There is so much variation in the speed squared term they have to handle, that a minor deviation from angle linearity is irrelevant. $\endgroup$
    – Neil_UK
    Commented Jun 5 at 9:25
  • $\begingroup$ @CaptainCodeman The relationship between lift and angle of attack is something that is commonly provided, although it's usually given in terms of lift coefficient. For example, take a look at airfoiltools.com - on every airfoil's entry you'll see a graph of Cl vs Alpha, where Alpha is a common synonym for the AoA. With the lift coefficient, you can calculate the lift of any given wing using that airfoil; Lift = wing area * air density * (airspeed squared / 2) * lift coefficient. That graph is usually determined by experiment or CFD simulation. $\endgroup$ Commented Jun 5 at 10:01
  • $\begingroup$ But doesn't a cambered airfoil provide lift even at a 0 angle of attack? How does that work with the lift=ka formula? Is the camber added to the angle? $\endgroup$ Commented Jun 6 at 6:31
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If I understand correctly, wind force increases quadratically as airspeed increases

Correct. Anyway we speak more correctly of "aerodynamic" force.

Therefore, all other things being equal, lift should also increase quadratically.

If everything remains exactly the same then yes.

Anyway this is not how an airplane deals with higher speed. The aerodynamic force depends not only on the speed but also on:

  1. the fluid density
  2. the dimension of the aerodynamic body
  3. the shape of the body
  4. the angle with which the airspeed hits the body, more or less in a linear manner. This angle is termed "angle of attack" and called $\alpha$. Comprimibility effects (Mach > 0.3) and structural deformations modify this simple linear relationship but the main idea still holds.

In general the equation expressing an aerodynamic force looks something like this:

$F=½ \rho V^2 S C_F$

Where:

  • $\rho$ is the fluid density - see previous point 1.
  • $V$ the speed
  • $S$ a surface which is used as reference area (for an airplane it is normally the wing area, for a car is the frontal surface) - see previous point 2.
  • $C_F$ is the part depending on the shape of the body and $\alpha$ - see previous point 3. and 4.

If for example we consider the lift, then that equation becomes:

$L=½ \rho V^2 S C_L$

Where $C_L$ (lift coefficient) has the following shape (underlined in blue):

Lift, drag and moment coefficient of NACA 0012 airfoil

As you can see, $C_L$ increases linearly with $\alpha$ (till a certain point called stall) and can be used to change lift as needed.

Let's do a simple example: if the speed doubles, its effect on $L$ becomes four time as big. If we reduce $\alpha$ from 10° to 2° then the $C_L$ decreases from 0.8 till 0.2 i.e. four times, compensating for the increase in $V$.

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Some aircraft deal with excessive lift by having a "negative flap" setting, where the flaps are positioned up a little bit. Some gliders do this to reduce drag caused in high speed flight by their high lift wings.

Reference: Do negative flaps increase glide ratio?

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Another aspect to consider is control sensitivity. Excessive lift at higher dynamic pressures usually increases load factor sensitivity to pitch inputs. According to MIL-STD-1797A Flying Qualities of Piloted Aircraft, there're requirements that must be satisfied when comes to handling qualities at all dynamic pressures. One of the parameters to look for is the $n$/α, or the steady-state load factor change per unit change in angle of attack for an incremental pitch control deflection at a certain airspeed. n-a requirements

When dynamic pressure increases, $n$/α usually increase as well because load factor $n$ = Lift / Weight, and $n$/α = Lift / (Weight * $\alpha$) = (CL$_α$ * α * dynamic pressure * S$_{ref}$) / (Weight * α) = CL$_α$ * dynamic pressure * S$_{ref}$ / Weight.

CL$_α$ is the slope of CL curve vs AOA that will also vary slightly w.r.t mach number.

In order to keep a satisfactory $n$/α at both low and high dynamic pressure, the gear ratio of stick input to elevator command is usually reduced with increasing dynamic pressure in CAS (control augmentation system) equipped aircraft, that reduces control sensitivity. An increase in stick force per g will also positively affect control sensitivity at higher dynamic pressures.

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