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Car acceleration problem

  1. Jul 25, 2013 #1
    Hello there,

    Just got an old car, and I was looking at the specifications, and trying to teach myself Newtonian physics. But I cannot get it to work out correctly. Please tell me where I go wrong.

    The car weighs 900 kg. The engine has 200nm of torque, it accelerates from still to 27.7 m/s in 8 seconds. The radius from the center of the drive axle to the outside of the tire is circa 0.3m.

    [itex]F = m\,a[/itex]

    Assuming constant acceleration:

    [itex]F = m\frac{v}{t}[/itex]

    [itex]F = 900\mathrm{ kg}\frac{27.7\mathrm{ m/s}}{8\mathrm{ s}}[/itex]

    [itex]F = 3116\,\mathrm{ N}[/itex]

    So, to go from zero to 100km/h in 8 seconds takes 3116 N. This is not taking aerodynamic drag and drivetrain friction into account.

    This is where I go wrong I think. The car has 200 Nm of torque:

    [itex]\tau = 200\,\mathrm{Nm}[/itex]

    I have an arm of 0.3m (the distance from the centre of the wheel to the outside of the tyre)

    [itex]F = \frac{\tau}{r} [/itex]

    [itex]F = 666 N [/itex]

    The problem is [itex]3116\,\mathrm{ N} >> 666\,\mathrm{ N}[/itex]

    And this is without even taken aerodynamic drag and drivetrain losses/friction and all of that jazz into account.

    How come my maths is off by so much?

    Thank you for your time.

    Kind regards,

    NOTE: Looks like my numbers may be off by a factor of 2∏ ? Not sure where it should go though.
    Last edited: Jul 25, 2013
  2. jcsd
  3. Jul 25, 2013 #2


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    2017 Award

    Staff: Mentor

    You have to take the gear into account, it can increase the torque (and reduce the rotation frequency).

    Quick check: a typical wheel has a diameter of ~50cm. At 30m/s, it rotates with 1200rpm. That is not the range where you operate the engine to accelerate.
  4. Jul 25, 2013 #3
    These are the gear ratios of the car:

    1st 3.80 : 1
    2nd 2.06 : 1
    3rd 1.26 : 1
    4th .89 : 1

    final drive ratio 4.125 : 1

    Do you think I am right in multiplying the ratio of the gear with the differential final drive ratio?

    F = 200Nm * gear ratio * final_drive_ratio/0.3m

    Thank you for your time.

    Kind regards,
  5. Jul 25, 2013 #4


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    Staff Emeritus
    Science Advisor
    Homework Helper

    That's what a gearbox is for. You have a higher overall gear ratio when starting in first gear to keep the engine from stalling. Once the car is moving, you change gears to reduce the overall ratio until you reach the point where the wheels are rotating at a speed which won't cause the engine to stall, and the engine is turning at sufficient RPM to keep the car moving at a constant travel speed.

    When you are accelerating to speed, unless you do it in a single gear, the analysis in the OP is not valid.
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