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Tq required to wheelie

  1. Feb 19, 2012 #1
    Its been way too long from my switch from Engg student to biz owner .. can you help me out with this?

    If I haven't given enough info below - the bike is a 2009 KTM 690 SMC.

    I'm trying to figure out the instantaneous torque required to wheelie my bike in order to develop gearing proper to roll on 2nd/3rd gear wheelies.

    In first - it will wheelie after 4k rpm with ease. Bike makes 63hp @ 7500, 47 lb*ft tq @ 5900. At 4k, its around 40 ft lb. With the gearing - 1st gear is 2.5:1, and final drive is 2.625:1. So first gear peak tq is 328 lb ft assuming no drive train losses.

    Bike weighs 320lbs, I'm guessing CoG is 24" high. I weigh 180 with gear, my CoG on bike would be about 48" high. So together - roughly 30" high you have 500lbs CoG.
    Wheel diameter is 24.56" - so from center of wheel to CoG(total) is 24" across, and 18" higher.

    From this can you calculate the req'd tq to wheelie and show me how you did the calculations?
    Thanks a bunch!
  2. jcsd
  3. Feb 20, 2012 #2


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    Looks to me like we need to know horizontal CoG positions for both you and the bike with respect to the rear axle in order to do anything.
  4. Feb 20, 2012 #3

    jack action

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    Check the theory at the bottom of this page. (Look for: Theory --> Longitudinal acceleration --> Accelerating --> Axle load distribution)

    Equation 14 from that page can be simplified to (assuming no aerodynamic lift):

    [itex]\frac{W_{f}}{mg} = \frac{l_r}{L} - \frac{h}{L} \left( \frac{F_t}{mg} - f_r \right)[/itex]

    For a wheelie, [itex]\frac{W_{f}}{mg} = 0[/itex], so the needed traction force at the tire contact patch ([itex]F_{t}[/itex]) is:

    [itex]\frac{F_{t}}{mg} = \frac{l_r/L}{h/L} + f_r[/itex]


    [itex]mg[/itex] is the total vehicle's weight;
    [itex]l_r/L[/itex] is the portion of the vehicle's weight on the front axle;
    [itex]h/L[/itex] is the CG-height-to-wheelbase ratio;
    [itex]f_r[/itex] is the rolling resistance coefficient (may be ignored as well);

    In your case (This is basically the equation found in post #13 of this thread):

    [itex]\frac{F_{t}}{mg} = \frac{l_r}{h} = \frac{24"}{30"} = 0.80 [/itex]

    Which means your traction force must be equal to 80% of your bike's weight or 400 lb. With a wheel diameter of 24.56", this means a wheel torque of 409 lb.ft is required or an engine torque of 62 lb.ft (based on your 1st gear ratio). Which is more than your 40 lb.ft you estimated. That may mean that your CG position is not where you think it is (If I didn't make any mistake).
  5. Feb 21, 2012 #4
    Thank you sir.

    I really was estimating based on visualizing the bike from the side, and estimating my CoG sitting on the seat with feet on the pegs.

    Based on the fact that it will wheelie from 4000 rpm, it gives us a 'check' of sorts for your formula to actually be able to estimate the CoG. Now, the one thing that adds to the problem is the front fork suspension - when you accelerate, the forks help "push" up the front wheel, and the rear suspension resists this until a certain point.

    At about 4k - the stock dyno figures are showing roughly 60Nm vs the peak 68Nm. So 44 lbft x 2.5(1stG) x 2.625(Final) = 289 lbft at the wheel, assuming no drivetrain loss.

    So if I look into changing sprockets, or adding power - I'll use 300 lbft at the wheel as a general wheelie reference for roll on.. of course clutching it up would make it easier - but I don't like the consistency of that.

    Big thanks!
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