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Wheel rotation force

  1. Nov 29, 2009 #1
    hello

    can you tell analyze please the force that is used to round the wheel of a car? I mean, how is this force called? is it torque?

    also, how does wheel characteristics affect the needed force to maintain a specific rotation speed? eg wheel weight, dimensions (both diameter and thickness of the wheel, which thickness indicates the distance from the horizontal distance of some points of the wheel from the rotation axis, etc)

    thanks
     
  2. jcsd
  3. Nov 29, 2009 #2
    The power P (watts) to propel a car is the torque T (Newton-meters) x angular velocity (radians per second). This may be written as

    P = T x 2 pi RPM/60

    The wheel torque in turn is equal to the wheel radius R times the horizontal azimuthal force F of the wheel pushing on the road. The vertical force of the wheel on the road does not enter the equation until the horizontal force approaches the static friction limit.

    Bob S
     
  4. Nov 30, 2009 #3
    thanks for your reply

    however how can we predict how wheel characteristics will affect the needed force to maintain a specific speed?

    I am talking about wheel diameter and wheel weight

    thanks
     
  5. Nov 30, 2009 #4
    The biggest retarding force on a car at highway speeds is air drag. The power to maintain constant speed to overcome air drag is given by

    Pdrag = torque (Newton-meters) x angular velocity (radians per second).

    This power is very roughly 30,000 watts, and varies as the cube of speed. This applies to the engine, or to the drive wheels. The only difference (neglecting power loss in the transmission) is that the drive wheels have a ~ 4 times lower RPM and 4 times higher torque.

    This may be rewritten

    Pdrag = torque x 2 pi RPM/60

    At highway speed, the drive wheels are spinning at perhaps ~900 RPM, and varies inversely with wheel radius. So for 30,000 watts, the wheels torque is (are, for two drive wheels)

    Torque = 30,000 x 60/(2 pi RPM) = 318 Newton-meters
    (sum for both drive wheels, two wheel drive)

    Suppose the wheel radius is 0.3 meters

    Then the horizontal force at the roadway propelling the car forward is

    F = 318 newton-meters/0.3 meters = 1061 Newtons (both wheels)

    This is a horizontal force at the roadway by the circumference of the tire(s).

    The weight of the wheel, and of the vehicle itself, is a vertical force pushing the tire down on the road. For a 1500 Kg car, this downward force is about 1500/4 x 9.81 =~ 3700 Newtons per wheel. As long as the downward force is more than twice the horizontal force (except on sand or ice), the wheel will not slip.

    The weight of the wheel itself is mainly a concern for the suspension system, in terms of th suspension spring constant (F = -kx) and the needed damping (shock absorber) to damp resonances and keep the wheel on the road.


    [added] here is a tire calculator:
    http://www.carforums.net/tirecalc.php [Broken]

    [added #2] Power can be represented by force x velocity. So a car with a forward horizontal wheel force of 1081 Newtons and a speed of 27.75 meters per second is using a wheel power of

    P = 1081 Newtons x 27.75 m/s = 30,000 watts

    Bob S
     
    Last edited by a moderator: May 4, 2017
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