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Take for example a jet fighter, when at low speeds vs high speeds, airspeed will differ a lot. As the turning moment is due to the lift force from the tail and lift increases quadratically with airspeed, why doesn't the jet fall out of control at such high speeds. To give an example, take a jet flying at 300 kts vs 1000kts. That means that the faster jet will have a turning moment 11 times stronger than the slower plane?

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  • $\begingroup$ Observe how much less you deflect your steering wheel at freeway speeds compared to half or less that speed on a slow road... $\endgroup$ Mar 26 at 17:24
  • $\begingroup$ You use the term "turning moment", I guess in this context that's probably the same as "pitching moment"? Just to be clear, that's not what actually determines turn rate; net pitching moment must be zero in a steady-state turn. $\endgroup$ Mar 26 at 17:51
  • $\begingroup$ Why do you assume the same control surface deflection is used at all speeds? $\endgroup$
    – DKNguyen
    Mar 26 at 22:57

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lift force from the tail and lift increases quadratically with airspeed

Correct

That means that the faster jet will have a turning moment 11 times stronger than the slower plane?

No, because the aerodynamic force depends also on how many degrees the control surface is deployed, more or less in a linear manner. That means that instead of deploying the control surface of say 11° as you do at low speed, you deploy it only 1° i.e. 11 times less in order to compensate for the 11 times higher speed's contribution.

Changes in altitude ($\rho$), comprimibility effects (Mach > 0.3) and structural deformations modify this simple linear relationship but the main idea still holds i.e. at higher speed you need lower deflections.

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