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Thread: Converting G to degrees per second.

  1. #1
    Join Date
    Apr 2009
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    Converting G to degrees per second.

    Hi, I seem to be having a problem. I want to convert G at a given speed to degrees per second, but I don't know how. I've tried using v^2 / r = acceleration, but I'm getting extremely wonky figures of like 26 degrees per second comprising a 9G turn at Mach .6 at 15000 feet. Am I screwing up the physics somehow?

  2. #2
    Join Date
    Dec 2011
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    your forgetting bank angle- the g's generated depends on aircraft's bank angle.

    Rate of turn= 1,091 x tan(bank angle)/ V(knots)

  3. #3
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    Nah, if you're not interested in the actual lift force because you're just taking it for granted that the aircraft can generate the required lift coefficient and has sufficient thrust to overcome the resulting drag, there's no need to worry about bank angle. You're doing it right - what's so wonky about 26deg/s in that scenario?

  4. #4
    Join Date
    Apr 2009
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    Here are the equations I'm using, where a is acceleration, v is velocity, and p is pi.

    180a/(v*pi) = deg sec

    a = deg sec * (v*pi) / 180

    a*180 / (deg sec * pi) = v

    This roughly indicates that if the F-22 is G-limited at 9G, it can perform 28 degree / sec turns at up to 650 km/h, or mach .56 at 15000 ft.

    Other interesting results are that a 9G turn at Mach 1.5 at 15000 ft only gives you 10 degree / sec maneuver, or it'd take 18 seconds for you to make a full 180. This answers questions I raised in other threads: while subsonic maneuverability is about lowering your corner speed and attempting to get good sustained performance at low speeds, supersonic maneuverability is about your ability to decelerate and accelerate; i.e, you'd have to deliberately bleed energy in an attempt to increase degrees per second maneuverability, then use your acceleration ability (high T/W, low drag) to regain the energy you lost.

  5. #5
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    Quote Originally Posted by Trident View Post
    Nah, if you're not interested in the actual lift force because you're just taking it for granted that the aircraft can generate the required lift coefficient and has sufficient thrust to overcome the resulting drag, there's no need to worry about bank angle. You're doing it right - what's so wonky about 26deg/s in that scenario?
    Yeah, sure. I assumed by the question he was looking at comparing corner speeds of different aircraft.

    @ Inst [/QUOTE]This roughly indicates that if the F-22 is G-limited at 9G, it can perform 28 degree / sec turns at up to 650 km/h, or mach .56 at 15000 ft [/QUOTE]

    It probably can but you need the E-M diagram to be sure of the aircraft envelope. BTW- you really don’t need to know the equations anymore, just go to one of the online pilot references that have turn rate calculators.
    Last edited by FBW; 1st December 2017 at 02:49.

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