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Test track's profile

  1. Dec 9, 2014 #1
    Why test tracks are banked parabolic why not elliptical or circular or hyperbolic? what are advantages of parabolic curve over another curves?
     
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  3. Dec 9, 2014 #2

    Ranger Mike

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    The key word here is TEST...automobile manufactures must test the vehicle over a number of different operating conditions. One symetrical oval would not tell much about the handling of the vehicle. Many different turns with differing radius and banking, short straights long straights all add to real world operating conditions. Just about all the test tracks have road course as well, and changing elevation to evaluate brakes on differing grade or slopes..
     
  4. Dec 9, 2014 #3
    Thanx Mike..i agree with ur point as varying angle with the curve will help to counteract the centrifugal forces which vary with changing speeds.. but why specifically parabola why not ellipse or hyperbola, what will be in case of ellipse or hyperbola.. i need some geometrical reason to support why we choose this curve.
     
  5. Dec 9, 2014 #4

    Ranger Mike

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    the curves are not symmetrical and have various diminishing radius to replicate the off ramp of many highway exits. Nor are they constant. And you are incorrect thinking centrifugal..the proper term is centripetal force
     
  6. Dec 9, 2014 #5
    thank you mike. yeah i might be wrong !! and always get confuse between two but can you refer some book or reading material which will give m a more broad idea about same and i am very specifically talking about banking.. it would be great if u provide any study material for designing track if u know one.
     
    Last edited: Dec 9, 2014
  7. Dec 9, 2014 #6

    Ranger Mike

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    i just race on um.. i don't build um..but read race car suspension class the above post..here is what i just wrote..


    When you are in a race car turning in at corner entry, you feel like you're being pushed toward the outside of the turn. Most of us refer to this as the centrifugal force. WRONG. There isn't any force pushing you outward. Centrifugal force is what physicists call a pseudo or a fictitious force, because it doesn't really exist. More specifically, in Newtonian mechanics, the term centrifugal force is used to refer to one of two distinct concepts: an inertial force (also called a "fictitious" force) observed in a non-inertialreference frame, and also the equal and opposite reaction to a centripetal force.
    So is the centrifugal force isn't real, why do you feel like there's something pushing you out the right-side window when you make a high speed left turn? The answers lie in Newton's laws of motion. An object going straight will keep going straight unless a force makes it change speed, direction or both. When a driver is bombing down the straightaway and starts to turn, the centripetal force makes the car turn and, because he's buckled tightly into the car, he turns also. The force he feels is because his body is trying to keep going straight. The seat and shoulder straps, lap belt and sub straps tied to the car are all exerting a force on him toward the inside of the turn while he's trying to go straight. The net result is that the driver perceives a force to be acting outward, but it is actually acting inward.
    Got it?
    The force that makes a car turn is called the centripetal force. Centripetal literally means "toward the center". Imagine you had a rubber ball with a string attached to it. Whirl the ball over your head in a horizontal circle. What makes the ball go in a circle instead of flying away from you is the force the string exerts on the ball, which pulls the ball in a circle.

    A race car doesn't have a string attached to make it go in a circle but it does have TIRES. The tires contact the pavement exerting force toward the center of the turn. Engineers talk about lateral force. The lateral force is perpendicular to the direction the car is going at any moment.

    The size of the centripetal force is given by multiplying the mass of the car by the speed of the car squared, and then dividing by the radius of the turn.

    F= MV2 / R where centripetal force equals the mass of the car, v is the speed of the car and r is the turn radius.

    Without going into a lot of math , the faster you go, the more force you need to be able to turn. Tighter turns require more force. Just like Aerodynamics, the force isn't linearly dependent on the speed. If you double your speed, the force needed to turn goes up by a factor of four. If you triple your speed, the force increases by a factor of nine.

    pls re-read post # 691 on page 34 on weight jacking.
    When a race car goes into a turn three things can happen and two are bad.
    1. Tires don’t have enough down force and will slip.
    2. Tires have too much down force and will overheat the right front tire and eventually will slip.
    3. The car completes phase one turn entry and enters mid turn phase two.

    The key to this event is to keep maximum tire contact during the dive and roll. This is why the right front tire goes negative camber and the left front tire goes positive camber in the turn. We want both front tires to carry the same amount of load when turning. This is why we bias the car with left side weight. We purposely offset the weight up to 60% static when we place the car on the weight scales. We do this knowing that this need to be done to counter weight transfer during cornering.
    WRONG! No “weight” is transferred. The tires react like weight was transferred but what we are really dealing with is FORCE as described above.
    Back to the race car racing down the back straight at 90 MPH. When we go in to turn entry phase one we change both speed and direction via the tires. The car wants to continue going straight. The suspension and tires are the only tools we have to deal with this force. During the turning event the body will roll to the right side in the typical left turn. It rolls through the front and rear Roll Centers (RC). Some of the momentum Force is scrubbed off by the coil springs ,ARB (sway bar) and dampers (shock absorbers) compressing and converting the force to heat. Once the body has taken a set the tires are left to deal with the rest of the force. If we look at the post # 691 on page 34, we see the force vectors of straight sideways lateral force shearing the tires and the right front tire contact patch countering the body roll force. If we have the front roll center located too far to the right side we start to lift the left front tire in a jacking effect. If the front roll center is located too far to the left there is not enough leverage angle to counter body roll and the force shears the tire contact patch. Ifin have the front roll center located properly, we have the maximum down force possible to stick the right front tire and provide maximum tire adhesion to counter the force and we beat he other race cars out there. Savvy?
     
  8. Dec 9, 2014 #7
    i want a mathematical reason why a parabola profile is used in super elevation of testtrack like bloss curve or clothoid geometry or whichever curve it takes by profile i mean u have a side view of path the angle changes with change in distance, whichever equation we choose what is the reason, lets say if eccentricity contributes to it then how eccentricity of circle 0 makes it perfect or is the reason for not chosing it or how eccentricity of parabola is 1 which is good for it. (i'm giving an assumption of curve property which might be wrong or btr say i don't have idea.) Untitled.png
     

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  9. Dec 9, 2014 #8

    Ranger Mike

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    good luck with that
     
  10. Dec 9, 2014 #9
    never mind , thanx 4 ur warm wishes
     
  11. Feb 6, 2015 #10
    Nuetral banking angle is the key phrase.
    If you want to travel a bend at speed with a minimum or no sideways slipping (drifting) forces, you calculate the banking angle to suit, if you want it to suit a range of speeds on one corner you end up with various banking angles added end to end or, if your really cool you could create an exponential or parobolic which gives you infinite variation over the cornering speed range.
    Have a look at the data sheet for nailing the nuetral angle:
    ive added the excel sheet, try a few different speeds.
     

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