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Automotive Automotive physics

  1. Jul 1, 2007 #1
    Hi everyone ^ ^

    I was recently looking into the calculation to discover the turning radius of a vehicle and something peaked my interest about the way rear tires turn. I know it is common sense to just take for granted that in most FR cars these days, power is delivered to a differential which is allows the inside wheel to spin more slowly than the outside wheel.

    I'm not an expert in physics, but it just doesn't make sense to me that has a very specific direction in which force is applied can turn without having a force applied in the perpendicular direction.

    I do believe that the concept of centripetal force is explained in the reverse manner, which I am using as the basis for my ignorance, in which a given object which is spun in a circle actually only feels the effect of one distinct force which is changing direction, centripetal acceleration, ne? That only makes sense because, say you spin a ball attached to a string around you, you continuously produce this changing force.

    Again, the force for a rear tire in an FR car goes only one direction. Therefore I assume that it fights a drag force of some kind, but I am not sure exactly how that works. Anyone care to enlighten? Feel free to criticize too. ='P
     
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  3. Jul 3, 2007 #2

    berkeman

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    Staff: Mentor

    I think the key is that the tire rolls only in one direction, and resists forces with skidding friction force in the orthogonal direction. Does that help?
     
  4. Jul 3, 2007 #3

    russ_watters

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    Right, this doesn't have anything to do with centripetal acceleration/force. If you've ever driven a car with a solid rear axle (like a Jeep in 4wd mode), you can feel it shake and 'hop' in a tight turn. If the car doesn't have a differential gear, the differential force applied to the tires is considerable. With the gears, the differential in force is only that which is required to overcome rolling and internal friction forces. The torque/force applied by the engine is still nearly equal for the two wheels.
     
  5. Jul 4, 2007 #4
    I think I understand the answer to my question, and it is more easily seen when we look at a bicycle, but I would appreciate it if someone could confirm my belief.

    Okay, so if we look at a bicycle and turn the front wheel perpendicular to the frame and apply force to the front tire in the direction that it is going, then the rear wheel will pivot on the spot of where it rests which made me think that the rear tire will have to pivot when the bike turns normally.

    If you actually tired to ride the bike while the front tire is turned that way then it would stall because there is no force being applied a direction perpendicular to the frame (which gave rise to my initial question). However, I see that if you turn the front tire to form an angle less than 90°, say from N to NE at 45° then you can break the force applied to the bike in the N direction into its components and then component that is going in the direction of the tire (in this case NE), is the amount of force applied in that direction (I assume).

    Now, you can break the NE force down into it's components and discover the amount of force going perpendicular to your original N force. That E force times the distance of the frame will be the torque placed on the rear tire to get it to continuously pivot as you turn.

    Is that right?
     
  6. Jul 4, 2007 #5

    turbo

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    It's more complicated than that, unfortunately. If you have ever driven a RWD car with a high power/weight ratio (think MGA), you know that if you oversteer and punch the accelerator, you can drive the car into a drift by causing the resistance to forward movement (forward defined as the direction of the front wheels) of the front wheels to cause the rear tires to break traction. Once the car is swung around and is pointed in about the direction you want to go, you back off the accelerator as you straighten the wheels and punch it again when the car's tires grab. It's a quirky process, but once you get a feel for it, it allows you to enter sharp corners at much higher speeds than you would be able to without the maneuver, and still maintain control.
     
  7. Jul 4, 2007 #6
    Lmao, masaka. I appreciate your input about traction, but I gather that your initial response is "what you said is true, but in addition..." ?

    but let me observe your statement for a minute. You say that if the force in the direction of the rear tires exceeds that of the traction of the front tires, then regardless of which direction you turn, you will go in the direction of the F applied to the tires. I assume this also works with momentum since this is true when you brake and your tires slide.

    But in your scenario, you describe that the car has already begun a turn before loosing traction such that the momentum is still going in the direction you were initially going, but he car has already turned a large degree. I am still trying to imagine the forces at work here, since I know that you can pitch a car sideways.
     
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