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How much of torque do you need to start wheels slipping?

  1. Mar 27, 2013 #1
    So first of all, hello! I have a question - how much of a torque a car needs to produce that a wheels would start slipping? Lets say i have a car which weights 3200 kg, and has a wheel radius of 0.4394m. So i've been told that to calculate that, i need to do the following procedures(thats very basic):

    1.) Calculate normal force:
    F = 3200 * 9.81 = 31392 N

    2) Multiply normal force by a friction coefficient(lets say its 1) and wheel radius:

    T = 31392 * 1 * 0.4394 = 13793,64 Nm

    3) Divide by number of wheels:

    T/4 = 3448,41 Nm.

    And we get the result - to slip a wheel you need to send to the wheel 3448,41 Nm of the torque. Is that true? How accurate is this method if its true? What about if a car has a rear wheel drive? A wheel will still slip if it reaches over calculated torque? Or you need 2x amount of that(for both wheels)? I'm just confused here..

    I need that because i'm calculating a set of gears, and i've calculated that a car at it's peak can send a total amount of 9000 Nm(after all gearing), and thats just crazy amount of torque - gears will be just too big if i would try to calculate them to withstand that. So i know that a wheel will start slipping much earlier and a wheel will never reach 9000 Nm, and thats what i need to calculate - how much of a theoretical torque a wheel can get in a worst case scenario with an LSD differential(i've been told that theoretically in a worst case scenario, one rear wheel of a rear wheeled car with an LSD differential could get a full torque from an engine).
     
    Last edited: Mar 27, 2013
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  3. Mar 27, 2013 #2

    mfb

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    Within the assumptions made here, it is.
    Assuming a flat surface and a symmetric car, as accurate as the given numbers.
    A real car will have a different load on the front and rear wheels, so they have different required torques.
    If we take the actual weight distribution of the car into account, right.
     
  4. Mar 27, 2013 #3
    Thanks for replying, it looked so simple so i thought maybe it is only a very very basic formula, but if i only need to take into account actual weight distribution to make it more accurate that is great. But maybe can you explain a little bit? I understand why we need normal force and a friction coefficient, but where and how does wheel radius fits here?
     
  5. Mar 27, 2013 #4

    SteamKing

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    What are you hauling around in your car? A BMW 3-series sedan has a curb weight of about 1550 kg.
     
  6. Mar 27, 2013 #5

    mfb

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    Wheel radius just relates the force (at a distance of the wheel radius, perpendicular to the radius) to a torque.
     
  7. Mar 28, 2013 #6
    I'm projecting a portal axles on a car.

    Yeah but i mean, how's that the wheels over-slipping occurs? Because the way i'm thinking now, i've just calculated how much of the force i need to simply start moving an object(you need to overcome normal force and a friction) but not that the wheels would start spinning(with spinning i mean that i need to calculate that a car would not just start simply moving, but that the wheels would start over-spinning when starting to move) at a start..Or thats not right?


    EDIT:
    So i'm thinking that to get vehicle moving, you need that each wheel would get 3448,41 Nm of torque. If vehicle is at still and not moving, and only one wheel gets 3448,41 Nm of torque,the wheel starts moving and rotating, but a whole vehicle is still not moving because that's just not enough of torque. Am i right? But that would mean that to get car moving after all gearing you need to have a 13793,64 Nm of torque - and after calculations(divided by 1st gear ratio * diff gear ratio) i get that an engine would need to produce a total 736 Nm of torque - and my selected engine doesn't produce even half of it(at its peak)..I'm confused again. What i am not getting here? Plus, if a car has a rear wheel drive(just 2 wheels can have torque), each wheel would need 13793,64/2 = 6896,82 of torque to get car moving, but that contradicts my calculations - both wheel would just start spinning. I don't get something here..
     
    Last edited: Mar 28, 2013
  8. Mar 28, 2013 #7

    mfb

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    The coefficient of friction you used is the static friction, and relevant for slipping.
    To get the car moving, you just need to counter the rolling resistance, which is significantly lower.
     
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