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Motorcycle cornering force on each tire?

  1. Jul 26, 2013 #1
    Thanks to anyone who cares to read or respond. I've never taken a physics class so really all I have is "it seems to me". But I would like what seems to me to make sense.

    Scenario. I have a motorcycle, in a steady-state corner(a circle), say 40 feet in diameter. And motorcycle is traversing the corner at a sustained 25mph. I'm trying to determine, what forces are acting on the bike. I really don't even know where to start with it. It seems to me that there is "camber thrust" from both tires, due to the fact the side of the tire is smaller than the centre. So that is doing something. And then the front tire is pointed to the inside of the turn, forcing the front of the bike to laterally accelerate in that direction as it goes around. And the back tire is providing forward motion.

    So it seems to me that the majority of the cornering load is going to be on the front tire, and the back tire is basically just following the front of the bike around. Is that a correct supposition? So the practical effect of this is that if you were to let go of the handlebars, and crank the throttle wide open, you would simply exit the turn and go straight(the bike would stand up).

    This also makes it seem like, you could lean the bike way over, open the throttle as much as you want, and as long as you don't change your lean angle, the back tire shouldn't slide out. Somehow I think there is less weight on the back tire when the bike is leaned over to allow for max throttle, but I can't figure out how.

    Sorry this is such a jumble!
     
  2. jcsd
  3. Jul 26, 2013 #2

    jbriggs444

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    The bike is cornering at a steady rate, right? So its rotation rate is constant, right? That means it cannot be under any net torque about a vertical axis.

    If we adopt a rotating frame of reference that is moving in a circle along with the motorcycle then there are three horizontal forces of interest.

    There is the inward horizontal force on the front tire. There is the inward horizontal force of the rear tire. And there is the outward centrifugal force that may be treated as being applied to the motorcycle's center of mass.

    If the center of mass is midway between the contact patches then the inward force on the two tires must be equal, otherwise there would be an unbalanced torque about the center of mass.
     
    Last edited: Jul 26, 2013
  4. Jul 26, 2013 #3

    rcgldr

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    The fact that the side of the tire is smaller than the center of the tire could result in a "cone" effect, but a rig with two cones, one in front of the other, and with parallel axis, tends to move in a nearly straight line.

    "camber thrust" is the lateral force related to linear lateral deformation at the contact patch, while "slip angle" is related to an outwards twisting deformation at the contact patch. I'm not aware of a term for the lasteral force related to "slip angle". Wiki articles:

    http://en.wikipedia.org/wiki/Camber_thrust

    http://en.wikipedia.org/wiki/Slip_angle
     
  5. Nov 13, 2014 #4
    A motorcycle in a steady turn has mostly camber force to keep the bike on the curved path. Camber force is determined by lean angle and results in no turning moment. You suppose that the rear tire is simply following along, and that's not true. There is a slip angle force pushing the rear wheel to the outside of the turn. Because this force opposes the camber force, the net effect is to reduce the side force in a turn. The front wheel can be steered to balance the rear force although some net turning moment may be needed because of the friction generated in turning. Because of the slip angles, it is necessary to lean farther to achieve the same lateral force.
     
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