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Thread: Fokker 100 Brake Disc Actuator Main Landing Gear

  1. #1
    Associate Engineer
    Join Date
    Jan 2016

    Fokker 100 Brake Disc Actuator Main Landing Gear

    Dear users,

    I have a problem with a calculation. I have got an assignment. For this assignment I have to redesign the brake actuators for the brake discs of the main landing gear of th fokker 100. I have forgotten how to calculate a force that goes sideways. The sideway forces are created when the lineair actuator pushes the brake disc on the wheels. For the calculation I have a few values.

    - Total brake disc mass: 237,84 kg (90% worn brake, CS25.109i)
    - CS-25 braking speed is= -3,0 m/s^2
    - Assumed distance from neutral position actuator to the wheel= 5 cm
    - Maximum take-off weight Fokker 100: 44.450 kg
    - Kinetic energy brake discs: 162 MJ
    - Maximum actuator force for each actuator: 40.000N
    - Specific Heat of he brake disk material: 800 J/kg/K
    - RTO test speed: 85m/s, 166 knots
    - Effect of air resistance will be ignored, trust reverser and speedbrake on the wing will not be applied.

    I have a few questions:
    1 - Which formula do I need to determine the force that the actuators need to push the brake discs?
    2 - How much actuators do I need in total on each wheel?
    3 - What is the Work of each actuator?
    4 - What is the Power of each actuator?

    To be clear, I DON'T need to know the maximum kinetic energy capability of the brake disc, these questions are aimed only for the actuators. I need to know how much power the actuator need to produce and how much of them I need them on 1 wheel.
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    Last edited by Kelly Bramble; 01-15-2016 at 08:59 AM.

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