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Factors affecting motorcycle grip when turning

  1. Jul 19, 2012 #1
    Hi everyone.

    I've been having this discussion with a pal regarding what affects the grip of a motorcycle's tires when turning and I was hoping someone in here could enlighten us :)

    So, we have a motorcycle that is making a turn at a constant speed and leaned in an angle.
    How do you calculate if the bike is able to make the turn or if the tires will skid and the bike fall? Clearly, making everything else constant, there will be a velocity cutoff that the bike will not be able to make it.
    From wikipedia I got this equation that gives the lean angle with respect to the velocity and curve radius:
    ∂ = arctan( v^2 / (gr) )

    Because the tires are not spheres, the bike can not lean indefinitely. Assuming the lean is still inside the tire limits, what affects the grip?

    To me, this problem is solved as follows:
    We have a centripetal force that is causing the bike to turn.
    We have the weight of the bike equal to the normal force of the ground.
    therefore, let
    Fn be Normal force,
    Ff be Friction Force,
    μ be the static friction of the tire/asphalt
    m the mass of the bike
    r the radius of the turn
    ac centripetal accelaration
    v velocity


    we have:
    ac = v^2/r
    (fricion force is the friction coeficient times the normal force) (=) Ff = μ Fn
    (on the vertical axis, the resulting force is 0) (=) Fn - mg = 0 (=) Fn = mg
    (on the horizontal axis the bike is turning) (=) Ff = m ac (=) μ Fn = m v^2/r

    if we solve this for v, we get v=√(μ g r)

    So, this means that
    a) the mass of the bike is irrelevant
    b) the lean angle is irrelevant. The Friction does not increase with the lean angle

    However, I've seen in some places people say that the friction actually increases with the lean. How is this possible?
    (e.g. http://genjac.com/BoomerBiker/Two Wheeled Physics.htm
    How is this possible?

    Thank you!
     
  2. jcsd
  3. Jul 19, 2012 #2

    jbriggs444

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    The quoted passage appears to use the idea that the coefficient of static friction is the ratio of frictional force to total force. As such, it is flatly wrong.

    If that sort of notion were supportable then a coefficient of static friction above 1.0 would ensure that a tire could never detach from the pavement under any cornering speed or lean angle.
     
  4. Jul 19, 2012 #3
    Wouldn't the motorcycle's centrifugal force push sideways: parallel to the ground? I don't see how it would add to the weight and thus the grip in the tire.

    Unless the road was banked.
     
  5. Jul 19, 2012 #4
    That is my idea.
    The leaning is irrelevant to the friction. The weight of the bike does not change because it is leaning, therefore the Friction force must remain the same.
    I might be missing something though..

    Thank you for your answers
     
  6. Jul 19, 2012 #5
    The way this problem is stated it appears we are dealing with centripetal force. I dont see anything to indicate we are using a rotating frame of reference.

    Imo these types of questions are not meant to play around with the fact that the tire might flatten and change shape or adhere with a diff. part of the tread to change the μ.

    Imo this is a classic physics problem that requires you to have a good grip on forces involved in uniform circular motion. If they had banked the pavement as well they would have required you to have a very good grip on forces and angles and how to reconcile them so the net force was always in towards the center of the circular path taken. But they were nice.

    Notice that the mass of the vehicle does not change—if it weighed 600 lbs. on a straightaway, it still weighs 600 lbs. in a curve.........?

    The sentence above gives me a hint that the source of this quote is not so good. Looks like they are confusing mass and weight. Weight is a force. Mass is not. Maybe they meant it differently than I read it. And then again...

    Cornering causes centrifugal force to press the tires downward into the asphalt????

    Whaaatttttt?
     
    Last edited: Jul 19, 2012
  7. Jul 19, 2012 #6
    As your equation states, to increase the force of friction you need to increase v, increase g, or decrease r or any combination of these. So go faster in a tighter circle on a more massive planet and you will change the force of friction. And if μ is too small then have a good time skidding off the road.
     
  8. Mar 30, 2014 #7
    Grip is proportion to weight and coefficient of friction. Best theory we have come across is in Vittore Cossalters 'Motorcycle Dynamics'. Grip clearly depends on tyre compound, temperature, road surface etc, and tests have shown that grip with cope with about 60 degrees of lean dry road good surface with modern soft sports tyres.

    It is hugely complex due to the dynamic nature of the loads and various theories.

    The trick to turn a motorbike as quickly as possible is to get as much weight as you can onto the front tyre - which is pointing around the corner and is offset to the outside at a larger radius (the rear tyre points straight ahead). Because all else being equal weight = grip. So riders try and trail the front brake leaving it on as they lean into the corner, reducing the amount of braking as the bike leans and more grip is used up in coping with lateral cornering forces. This is not sensible on the highway as if you lose grip from the front tyre you will very likely fall.

    You also need a lot of physical strength as the front wheel (as it is offset in a corner) will try and move in line with the rear tyre, when as the corner tightens you need more not less steering angle.

    The recommendation on the road is to use the front brake when upright, and to use the rear brake in a corner if you feel you are going too fast.

    The other issue is that tyres slip sideways at a few degrees (as opposed to skidding) when lateral loads are applied, which tends to make bike understeer under braking, and oversteering under power (both while banked)

    Hope this helps - but have a look at Vittorre's book which we use with permission for our theoretical basis on which our training is based. Mike - British Superbike School
     
  9. Mar 31, 2014 #8

    CWatters

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    Homework Helper

    The suspension and damping system also effects grip on uneven surfaces. Leaning a bike reduces the effectiveness of the suspension because the swing arms move vertically relative to the bike, they no longer move vertically relative to the road.

    So leaning the bike make the suspension less optimal and makes the bike more susceptible to the effects of an uneven road surface. This is one reason racing motorcyclists climb off the bike in corners, it helps keep the bike and suspension system nearer to the vertical than it would otherwise be.

    I did wonder if anyone has designed a racing motorbike that kept the wheels and suspension vertical at all times?
     
  10. Apr 1, 2014 #9
    Have a look at information available on the Pacejka model, it is a good empirical model of the behaviour of the pneumatic tyre, and accounts for vertical load, camber, tractive force, slip angle and longitudinal slip.

    Cheers,
    Terry
     
  11. Apr 1, 2014 #10
    It depends on the design of the tyre, but side force generated by a tyre can depend on the camber angle.

    Cheers,
    Terry
     
  12. Apr 2, 2014 #11

    rcgldr

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    It's camber deformation, not camber angle. The force at the contact patch can result in two types of deformation: camber deformation refers to an inwards deformation, while slip angle refers to a twisting deformation. These deformations coexist with a lateral force on the tire while leaning. The force related to camber deformation is sometimes called camber thrust.
     
    Last edited: Apr 2, 2014
  13. Apr 8, 2014 #12
    OK. I was referring to the wheel camber angle, one of the variable parameters that tyre manufacturers measure and report in their tyre performance data.

    Cheers,
    Terry
     
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