I'm attempting to make a game similar to Ace Combat, and I'm still confused about banked turns, namely the speed of bank turning depending on the aircraft's roll.

Can I assume, for simplicity's sake, that it's something like $Rgy = K * Rz$, where $Rgy$ is the rate of turning on the global Y axis (the banked turn's angular speed), $Rz$ is the roll angle of the aircraft, and $K$ is a constant? And if I can do so, how do I calculate $K$?


1 Answer 1


It's computed with centripetal force:

$$F_C = m \cdot V \cdot \omega $$

  • m = aircraft mass [kg]
  • V = aircraft speed [m/s]
  • $\omega$ = rate of turn [rad/s]

Centripetal force is a function of bank angle $\Phi$ and lift, or mass and flight path. For a coordinated, level turn:

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Vertical lift $L \cdot cos(\Phi)= m \cdot g$ and turning lift $L \cdot sin(\Phi) = F_C$, so in a steady turn

$$ F_C = m \cdot g \cdot tan(\Phi)$$

$$\omega = \frac{g}{V} \cdot tan (\Phi) \tag{Steady Turn}$$

A more general equation, also valid for 90 degree bank angles and not concerned with the aircraft staying up in the air, can be found by considering horizontal forces only:

$$ \omega = \frac {L \cdot sin(\Phi)}{m \cdot V} \tag{always} $$

And then if you assume that lift L is still equal to weight $m \cdot g$ which means the aircraft is losing altitude in this turn, you're almost back at the Steady Turn equation but now with a sin instead of a tan.

The above is for SI units. For aircraft units with feet and knots, please check this answer, and here is one that introduces lift coefficient into the equation.

  • $\begingroup$ Thank you. Though if Φ is 90 degrees, won't ω be infinite, since tand(90) is +/-infinity? $\endgroup$ Aug 19, 2017 at 7:25
  • 1
    $\begingroup$ @Gensoukyou1337 It would, but you can't fly stationary turns at high bank angles. The aircraft would break apart $\endgroup$
    – Gypaets
    Aug 19, 2017 at 7:43
  • $\begingroup$ @Gensoukyou1337 Yes the equation is only valid for steady turns where the vertical component of the wing lift still compensates for weight. At 90 degrees there is no such component. A fighter jet might use thrust in a 90 deg turn, point the nose angled upwards while flying on its side, using fuselage lift as well. $F_C$ would be provided by all of the wing lift in that case. A more generic equation would use $L \cdot sin(\Phi)$ for the centripetal force. $\endgroup$
    – Koyovis
    Aug 19, 2017 at 8:16

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