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Calulating brake moment/torque for an airplane...

  1. Jan 4, 2017 #1
    1. The problem statement, all variables and given/known data
    I'm working on this to further my understanding of parts used in an aircraft braking system.

    Using known weight, and constant acceleration determine braking moment for a tricycle gear airplane with two main landing gear brakes.
    Mass- 12,292 kg
    Acceleration: -1.4 m/s^2
    Tire radius: 0.4445m
    Brake Key Radius: 0.1016 (10 keys)

    Is it possible to determine brake moment at wheel- to- brake keys (a specified radius inside the wheel) using the above information? This would be similar to determining the braking moment on a car wheel lug during deceleration.



    2. Relevant equations
    F=ma

    3. The attempt at a solution
    The aircraft has a deceleration force of 17,264.6 N, how do I use this information to determine braking moment, at a specified radius, ignoring wind, rolling resistance ect?

    Googling I found the following equation: T = (BFw)(R)/ r
    Where
    T = Torque (Nm)
    BFw= Brake Force wheel
    R = Static laden wheel radius
    r= Speed Ratio between wheel and brake

    Is it correct to divide the linear force in half, for two wheel brakes, and use the result for BFw?
    If so I come up with
    T=16793.3 Nm converted= 12,385.36 Ft lbs
    BFw= 8632.3
    R= 0.4445m
    r= 0.22857 (35" inch tire diameter to 8 inch "brake key" diameter )
     
    Last edited: Jan 4, 2017
  2. jcsd
  3. Jan 4, 2017 #2

    mfb

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    You know the deceleration force and the tire radius. What can you say about torque at the wheels?
     
  4. Jan 5, 2017 #3
    Thanks for the response, I think what you're getting at is to simply take one half the force, for one of the two brakes, and do the following:
    Bt=Fd,
    F= 8632.3 N
    d= (17.5" tire radius - 4" brake key radius, gives 13.5" lever arm, converted into meters = 0.3429m)
    Bt=(0.3429m)(8632.3 N)
    Bt(at key radius)= 2960.15 Nm
    Bt(at key radius)=2183.2 lb ft
     
  5. Jan 5, 2017 #4

    mfb

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    I don't understand what you are calculating here.

    Forget the brakes for one step. There are the wheels, they exert a force on the ground. What is the corresponding torque at the wheels?
     
  6. Jan 5, 2017 #5
    I'm using the force of the airplane slowing at a constant rate; F=ma, using the calculated deceleration of the airplane and the mass, dividing this number by two for two brakes, then using the result to find torque, t=Fr using the tire radius and the brake diameter.
     
  7. Jan 5, 2017 #6

    mfb

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    The brake diameter (the size of the brakes) is not even given. Why do you want to include anything brake-related to calculate the torque the wheels get from the contact force with the ground?
     
  8. Jan 8, 2017 #7

    CWatters

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    Heloman30....

    The torque on the wheel is the same as the torque produced by the brake. The force on tyre and brake will be different because they have different radii but you asked about torques not forces.
     
  9. Jan 9, 2017 #8
    I don't know the torque of the brake, but I do know the deceleration and weight of the airplane; I'm attempting to work backwards using this information. I appreciate your help, thank you!
     
  10. Jan 9, 2017 #9

    mfb

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    See post 4. The next step is independent of the brakes.

    You have the wheels with a known diameter and a known force at the outside, you have to convert this to a torque.
     
  11. Jan 11, 2017 #10
    OK:
    Bt=Fd
    Bt= (17,264.6 N)(0.3429 m)
    Bt=5920.03 Nm (total-2 wheels)
     
  12. Jan 11, 2017 #11

    CWatters

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    That's not correct.

    At what radius does the deceleration force "F" act?
     
  13. Jan 13, 2017 #12
    The tire/ground contact point
     
  14. Jan 13, 2017 #13

    CWatters

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    So the distance from the center of rotation to the point where the force acts is equal to the radius of the tyre = 0.4445m (not 0.3429m).
     
  15. Jan 13, 2017 #14
    I'm not sure what to do here, I'm trying to do what you asked: You said forget the brake in step 4, and step 9 again you said focus on the wheel and the force, so I did. We're clearly not on the same page. I appreciate your help but I don't know where your trying to lead me. I wish could find a similar example, I've been searching the internet and all my old class books, but I can't find anything close enough.
     
  16. Jan 13, 2017 #15

    mfb

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    Sorry, I have no idea how to break down this into smaller substeps.

    You have a force of 17,264.6 N acting at the wheel diameter of 0.4445m. What is the torque?

    You can find the answer to this question without taking anything else into account, and it is not even half a line long.
     
  17. Jan 14, 2017 #16

    CWatters

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    The solution can be broken down into three steps.

    1) Calculate what the friction force must be to achieve the required deceleration (done = 17,264.6 N )
    2) Calculate the torque that the friction force must be applying to the wheel (force of 17,264.6 N acting at 0.4445m = ??).
    3) Calculate the torque that the brake must be applying to the wheel.

    If you want to know the force acting on the brake caliper then there is another 4th step.
     
  18. Jan 19, 2017 #17
    Thank you CWatters, and mfb! Your responses along with another post I found has hopefully got me back on track. I found the following post about the force generated at a motorcycle caliper bracket,it's quite similar to what I'm trying to do:

    https://www.physicsforums.com/threads/force-generated-at-caliper-bracket-while-braking.699932/

    Using F=ma I found the airplane has a deceleration force of 17,264 N, dividing this by two, for two braked landing gear wheels, I'm left with 8632.3N per wheel.
    Torque at the axle:
    T=F * r = 8632.3N * 0.4445m= 3837.057 Nm

    Force at the radius I'm interested in, brake key radius: 4 inches= 0.1016m (I believe this is similar to CWatters 4th step?)
    F=T/r = 3837.057 Nm/ 0.1016m = 37766.31 N
     
  19. Jan 19, 2017 #18

    mfb

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    Correct.
     
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