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Which corners faster - car or motorcycle?

  1. Feb 9, 2011 #1
    I've been turning this over in my head for a week & the answer is way beyond me, maybe y'all can help.

    There's a painted line that arcs 90 degrees over say 200'. A car follows that line with it's outside tires. A moto does the same. Which vehicle can travel fastest before the tires lose adhesion? How would someone even calculate this?

    The example can be tweaked into whatever is a more fair comparison, so if the car needs to straddle the line, etc. Also assume all other things are equal, like tire compounds etc. The question is really about how two wheels leaning into a turn differ from a 4-wheel flat car platform.

    Here's what I'm thinking. First, I'm no engineer, just thinking out loud. Adhesion is a result of mass (centrifugal force here) vs the tire's friction coefficient. I would think the size of the contact patch would be relevant too but it would increase with mass so I'm thinking it's a non-issue.

    So it seems the vector of the centrifugal force is the same with either a leaning bike or a flat car, right? And if so, they would both slide out at the same speed. Is this correct?
     
  2. jcsd
  3. Feb 9, 2011 #2

    berkeman

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    A car typically has aerodynamic downforce, which increases its traction capabilities versus a motorcycle.
     
  4. Feb 9, 2011 #3
    Assume downforce isn't an issue here.
     
  5. Feb 9, 2011 #4
    Hmmm... maybe I don't know what I'm talking about(certainly possible!!) but I thought that a motorcycle can turn much more acutely than a car.
    With a motorcycle, one has the potential benifit of "leaning into the curve" which is not possible with a car.
    A car without lean will flip over durring a sharp curve.
    A motorcycle, leaned properly, will not flip on that same curve.

    But again, I could be wrong.

    Edit: Note that a motorcycle leaning places the center of gravity(COG) closer to the gound, making it more difficult to "flip" This dynamic does not afford itself in a standard 4-wheeled vehicle.
     
    Last edited: Feb 9, 2011
  6. Feb 9, 2011 #5

    Drakkith

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    I'm not sure, but I think a motorcycle will lose traction first as it has much less surface contact with its tires. But don't quote me on that.
     
  7. Feb 9, 2011 #6

    Drakkith

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    Leaning into a curve is something you MUST do to avoid falling over as you turn. A car will flip before a motorcycle will because it cannot lean into the turn. Motorcycles typically lose traction before they can flip.
     
  8. Feb 9, 2011 #7

    rcgldr

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    Assuming no downforce, then it's an issue of coefficient of friction and dynamic issues related to the size of the contact patch and how "clean" the road is (like dust).

    Typically motorcycle tires used on sport bikes (the most common type of motorcycle) have more effective grip than the tires commonly used for sedan type cars, but not much more, if any, over the high performance tires used on sports cars. So sport motorcycles will out corner econoboxes and heavier sedans, but not the better sports oriented cars.
     
  9. Feb 9, 2011 #8
    Ah, indeed. Perhaps that's why some of the supersport motorcycles races on curved tracks have special tires with significant tread on the sides?
     
  10. Feb 9, 2011 #9

    turbo

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    Too many variables. Light, overpowered car with rear-wheel drive? You can drift a turn, letting the rear slide around and then hammer the throttle. Think MGA on this one, especially on loose road surfaces, like gravel. Change road surface, tires, power-balance, or geometry (or many other variables) and everything is up in the air.
     
  11. Feb 9, 2011 #10

    Drakkith

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    Yep. That's because those bikes lean SOOOO far over, and are going SOOO fast around a curve that they need as much tread on the side of the tire (which is where the tire is touching the ground in the turn) to avoid losing traction.
     
  12. Feb 9, 2011 #11
    Seriously tho, you have it right. Given similar tire compounds, no aero downforce and reasonable tire width to mass ratio, the cornering speeds and forces will be very similar. That happens to be around the 1.3 G mark for both.

    The dynamics are different but the physics isn't.

    P.S. "centrifugal force" doesn't enter into this unless the corners are banked. Then it will again be the same for both vehicles.
     
    Last edited: Feb 9, 2011
  13. Feb 9, 2011 #12
    Lean has nothing to do with it; centre of gravity does.

    Given the average C of G location on a typical passenger car, that cornering force would have to exceed 2 gs before being in danger of flipping from the cornering force. That or trip on a curb.
     
  14. Feb 9, 2011 #13

    Drakkith

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    Leaning changes your center of gravity on a bike.
     
  15. Feb 9, 2011 #14
    Read the quote to see exactly what I was addressing (car).

    The centre of gravity of a bike is always acting through the centre of the tire contact patch when cornering steady state.
     
  16. Feb 9, 2011 #15

    Drakkith

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    I saw the quote, I didn't realize you were only referring to the car. I was just throwing that out there.
     
  17. Feb 10, 2011 #16
    Thank God someone read the question :tongue:

    I was thinking that a square-shaped tire on a car that doesn't lean still has the same forces applied to it, that is a combination of 2 vectors - downward force from mass & lateral force from...something... centrifugal, centripetal...I don't know all the terms.

    A leaning moto seems like an excellent illustration of exactly how those two vectors affect each other. More lean means more lateral force, etc. But the car, while not leaning, sees the same vectors affecting the tire, it just ain't visible the same way.

    One last question, how does the ratio of mass to contact area affect tire adhesion?

    So for example, two identical bikes & riders with a weight total of 500# each but one has tire contact patches twice the size of the other due to wider tires. Assuming everything else being the same, would they both lose traction at the same speed?
     
  18. Feb 10, 2011 #17

    rcgldr

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    The larger contact patch results in less load per unit area of the tire, and less deformation of the contact patch. Tires have a load senstivity, where the coefficient of friction decreases with the load, so a larger contact patch increases grip, but there's a limit to that where there's almost no gain with further contact patch size increase.

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

    The other reason for a larger contact patch inherent with a tire with more surface area is heat dissipation and wear.
     
  19. Feb 10, 2011 #18
    rcgldr answered your main question but I'd like to add some clarification.

    If the tire pressure is the same for both bikes, the area of the contact patch will also be the same. Divide the weight by the tire pressure to get the contact patch area.

    500lbs/(25 lbs/in^2) = 20 square inches

    There will be no difference in area but the shape of the contact patch will change due to different construction of various tires.

    Long narrow patches don't work as well as short but wide ones for a few reasons, the main ones being that wider contact patches are better controlled and the percentage of the contact patch that is experiencing slippage is reduced. That results in a higher level of grip.

    To get the "wide tire" effect on motorcycle tires, the sidewalls have a larger radius than the centre - a parabolic shape as opposed to circular if you were to look at the tire from the back of the bike. As the bike is leaned over from vertical, the contact patch changes shape from an ellipse with the major axis in the fore-aft direction to one in the side to side direction. That mimics what the contact patch of a wide car tire looks like and gives the same benefits.

    Here's a traction circle for a bike:
    http://www.sportrider.com/riding_tips/146_0909_traction_circle_riding_skills/photo_02.html
    Given what I said above, can you guess why the bike has less traction in a straight line when exactly vertical, as mentioned in the footnote and demonstrated by the graph?
     
    Last edited: Feb 10, 2011
  20. Feb 10, 2011 #19
    I'm guessing b/c a parabolic shape with a narrower vertical profile yields less actual contact patch than when the bike's off-center & into the meat of the parabola - no?

    So (tall & narrow) < (short & wide) since the area of slippage vs area of controlled contact is proportionally smaller, as mentioned above.

    Thanks for the thoughtful answers Mender, much appreciated.
     
  21. Feb 10, 2011 #20
    I'm curious about something: When a bike is steeply leaned in a turn, the radius of the outermost portion of the tire that is in contact with the road is somewhat longer than the radius if the innermost portion of the tire that contacts the road. This implies that the inner and outer portions of the tire are skidding a bit in the turn, and only a small portion of the tire in the middle part of the tire track is not skidding. Given that static friction is greater than sliding friction, would this not be a disadvantage to a bike vs. a car?
     
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