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Downforce, banked turns, and friction

  1. May 25, 2015 #1
    In motorsports, everyone talks about getting the greatest amount of aerodynamic downforce on a car in order to get the most speed.

    However, every derivation I see for a formula regarding the maximum velocity a car can take around a banked turn, with friction, ignoring aero downforce, removes the normal force. This makes me think that aerodynamic downforce isn't important when in a banked turn, but that doesn't make sense to me at all, since it can be important on 0° and 90°+ turns.

    If I were to consider downforce, would this formula change at all? And if so, what would change?
     
  2. jcsd
  3. May 25, 2015 #2

    billy_joule

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    Science Advisor

    That is only true when friction is the limiting force. Downforce and it's associated increase in drag generally reduce top speed & acceleration in a straight line.

    Yes the formula would change. Try it out and see what you get.
     
  4. May 25, 2015 #3
    Is this all valid? Above the line is the non-downforce variant. I can take it from here if it's good.

    http://imgur.com/U1HBWYJ
     
  5. May 26, 2015 #4

    cjl

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    The reason that normal force is removed in most derivations of the speed a car can take around a corner is that the centripetal/centrifugal (take your pick) force to go around a corner at a given speed is proportional to mass, and the normal force is also proportional to speed, so they cancel out. Downforce increases the normal force without increasing the required centripetal force, so it will not cancel out in the derivation (and, in fact, it will also make it so the mass no longer cancels out as well).
     
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