We all know that on takeoff you need right rudder in order to balance out the "torque" of the propwash hitting the tail. Is there an exact crosswind that would cancel out this torque?


3 Answers 3


It is reasonable to believe, (as it is with most hypothetical questions of this sort...) that the answer is yes. However, there are a few things to consider:

  1. It would need to vary to the degree that the throttle is advanced.

  2. It would need to vary during the acceleration as aerodynamic effects change.

So, if you think you might be able to listen to ATIS and discern that you won't need rudder on the takeoff roll, it isn't that simple.

  • $\begingroup$ Then there was gyro precession effects on taildraggers. When I was part owner of a C-180, when it was on wheels, if you tried to raise the tail early, with a bit too much gusto, the propeller precession would swing it left pretty hard. If there was a strong left xwind, you'd find yourself running out or rudder. Much safer to hold the tail down with elev and let it come up gently after past about 30kt. $\endgroup$
    – John K
    Jan 10, 2021 at 22:31
  • $\begingroup$ Good point. I consider adding this to my answer, but the OP asked specifically about a 172. $\endgroup$ Jan 11, 2021 at 16:50
  • 1
    $\begingroup$ Yeah I know I was just story telling. $\endgroup$
    – John K
    Jan 11, 2021 at 16:52

NOTE: This is not meant to be a scientific analysis. In older 172's such as a 172M, when there is a 2-3 degree crosswind perpendicular to the path of the airplane it seems to cancel the torque out on departure. In the newer models like the 172S, it is more like 4-5 knots.

  • $\begingroup$ Fascinating experiential answer! $\endgroup$
    – Fattie
    Jan 11, 2021 at 16:20

Yes, but not really.

If the the 172 were traveling at a constant velocity and with a constant prop speed (rpm), then there would be a linear combination of crosswind direction and wind speed that would exactly cancel out the torque. For example, a 90 degree crosswind would need a lower velocity to counter the torque, whereas a narrow angle crosswind would need a much higher velocity.

But on takeoff, the aircraft velocity will be increasing over time. Whatever crosswind direction and velocity numbers you chose which fit the example above would not work for all other aircraft velocities. As the velocity increases, the effective force on the vertical stabilizer (and fuselage) from air will change.

To make this easier to imagine, consider a craft moving at 500kts with a 20kts 90 degree crosswind vs a craft moving at 5kts with the same crosswind. The faster the craft is moving forward, the less off-axis wind forces will matter.

So when the aircraft is motionless, and the engine is set for takeoff power, the crosswind direction and velocity combination that will perfectly counter the torque will be one set of numbers, but as the aircraft increases speed the perfect counter numbers will change. Ignoring the complexities of fluid dynamics, this could be represented in a 3D graph (crosswind direction, crosswind velocity, and aircraft velocity). It would look something like a tilted triangle.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.