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I saw the answer to airspeed and rate of turn but this I think isn't what I want. I've seen it many times that aircraft will pitch up much slower when at higher speeds particularly fighter aircraft. What is the mathematical relation of these two characteristics if any?

I am trying to make a game about fighter aircraft and want to find a realistic solution to this problem.

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A constant speed pull-up is basically a circular motion, albeit in vertical plane:

$$q=\frac{(n_z-1)g}{V}$$

where $q$ is pitch rate, $V$ is true airspeed assuming no wind, $n_z$ is the load factor (lift to weight) and $g$ is gravitational acceleration.

The amount of load factor is limited by the airplane and its flight condition. As you can see, the faster the speed, the smaller is the obtainable pitch rate for a constant load factor.

Note: a constant speed pull-up is a theoretical idea that only approximates a pull-up at the very beginning, before the airspeed begins to decay. But the conclusion is valid.

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  • $\begingroup$ is g 9.81? or is it like g's where it can increase or decrease? Or better said, downward acceleration in circular motion. And the load factor I'm assuming is the weight in kilos of the aircraft? $\endgroup$ – Kiyo Feb 25 at 19:44
  • $\begingroup$ @Kiyo There is no weight or mass in this. g can be in whatever unit you want, as long as it jives with the rest of the equation (i.e. must be consistent with the units of q and V). $\endgroup$ – JZYL Feb 25 at 19:48
  • $\begingroup$ I see but when what exactly is the load factor? Where could I get this number from? $\endgroup$ – Kiyo Feb 25 at 19:49
  • $\begingroup$ @Kiyo See aviation.stackexchange.com/questions/46287/…. It's the instantaneous lift over weight. $\endgroup$ – JZYL Feb 25 at 19:51
  • $\begingroup$ I see, thank you. One last thing. what units is q in? Ideally, I'd like to be able to put this into torque. $\endgroup$ – Kiyo Feb 25 at 19:54
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If the aeroplane flies straight and level, and the pilot pulls on the stick in a step input, the elevator deflects => increased Angle of Attack => increased lift. The aeroplane climbs while continuing to rotate backwards, trading kinetic energy for potential energy.

From prof. Gerlachs' lecture book on Stability & Control

In the situation depicted above, the centripetal acceleration is $V \cdot q$, and hence: $$N - W = m \cdot V \cdot q$$. So for a given elevator deflection and therefore increase in N, at higher V the achieved q is lower.

As a guideline for maximum n = N/W, from this wiki page:

  • For transport category airplanes, from -1 to +2.5 (or up to +3.8 depending on design takeoff weight)
  • For normal category and commuter category airplanes, from -1.52 to +3.8
  • For utility category airplanes, from -1.76 to +4.4
  • For acrobatic category airplanes, from -3.0 to +6.0
  • For helicopters, from -1 to +3.5[7]
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