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Maximal acceleration of a car in a curve

  1. Apr 4, 2010 #1
    I want to compute the maximal acceleration of a car in a curve.

    [tex]
    a_{total}=a_{c}^{2}+a_{tan}^{2}
    a_{c}=\frac{v^{2}}{r}
    [/tex]

    Now, I am not so sure. F=ma, then a= F/m. The force that propels the car is the friction force, which is equal to mu*N, where N = downforce + mg
    We have [tex]a=\frac{\mu(Downforce+mg)}{m}[/tex]
    I don't know what is this acceleration.. is it the tangential or the total one? All I need to do is figure out which acceleration it is and solving for the other, that would give me the maximal acceleration of the car.

    Is that right?
    Thank you
     
  2. jcsd
  3. Apr 4, 2010 #2
    If the curve is not banked, then the centripetal force (mv²/r) is provided by the friction between the tyres and the ground.
    This friction is equal to μN where N is the normal reaction.
    The normal reaction will be mg plus any extra downward force on the car provided by its aerodynamics; spoilers, for example.
    The limiting case that gives the maximum acceleration, is where this centripetal force is equal to the maximum frictional force. Above this and the car starts to slip.
    The aerodynamic downward force will depend, in some way, on the speed of the car.
     
  4. Apr 4, 2010 #3

    rcgldr

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    In real life,tires have a load sensitivity factor, and the coefficient of friction decreases with load or downforce:

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

    Also, the maximum grip force of a tire is affected by how much of the acceleration is lateral versus longitudunal.
     
  5. Apr 4, 2010 #4
    There is no one formula that perfectly describes this, since tire dynamics is a very complex and non-exact issue. The "Magic Formulae" by Mr. Pacejka might be the closest approximation that can be used for a lot of different tires.

    Some of the things that affect the outcome are:
    * Road banking
    * Camber angle
    * Tire lateral stiffness
    * Tire longitudinal stiffness
    * Longitudinal Slip
    * Lateral Slip

    It all depends on how exact you want to be. The computer game industry has put a lot of effort into creating as accurate calculations as possible, without spoiling game speed. Some variation of the Pacejka formulas seems to be most common.

    /Ricky
     
  6. Apr 4, 2010 #5

    rcgldr

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    It needs to be tweaked a lot and/or there are newer methods that more closely approximate reality. This was discussed at a game web site here:

    http://www.lfsforum.net/showthread.php?t=67133

    Which includes a link to a thesis:

    http://www.control.lth.se/documents/2007/jsvenPDH.pdf

    And this short paper:

    http://www-m6.ma.tum.de/~turova/html/TireSensor.pdf
     
    Last edited: Apr 4, 2010
  7. Apr 4, 2010 #6

    jack action

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    [tex]a = \sqrt{a^2_{longitudinal}+a^2_{lateral}}[/tex]

    The acceleration is proportional to the friction coefficient in any direction. Read about http://www.auto-ware.com/setup/fc1.htm". (Some purist will tell you that it is actually a friction "ellipse" because a tire have a slightly bigger µ in the longitudinal direction than in the lateral direction, but making them equal is good approximation)
     
    Last edited by a moderator: Apr 25, 2017
  8. Apr 4, 2010 #7

    rcgldr

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    It's an approximation that off enough to make racing games that use such simplifcations feel unrealistic when driven at the limits.

    Also most of these models don't take into account the hysteresis effect, when transitioning from static to dynamic friction versus transitioning from dynamic back to static friction.
     
  9. Apr 4, 2010 #8

    jack action

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    When conceiving and building REAL race cars, we couldn't careless about all of the transient effects due to temperature, slip or tire-track conditions because they vary too much for all conditions that will encounter the vehicle. You put your best guess and design your suspension, brakes, aerodynamics and powertrain around it.
     
  10. Apr 5, 2010 #9
    I'm working on an offroad game and in that case, you also must take into account how much gravel your wheels push around, but maybe, that can be added afterwards, not affecting tire friction(?)...

    B.t.w., thank you Jeff for point out an excellent thread and PDF document. I have read the thread but still have to find the energy to read the PDF document. ;-)
     
  11. Apr 5, 2010 #10
    The reason I am asking this I need to roughly estimate the time a car go through different paths. I am not really interested in the value itself, rather in the difference of times between different paths.

    I want to approximate the maximum acceleration of the car in the curve before slipping at time t, knowing the curvature of the road, the power of the car and its speed. The road is flat. Using this acceleration I can now compute its new speed at time t+very small delta(t) and so on. I have already found the maximum speed of the car on the curve so I can bound the speed but I'm having trouble with acceleration. The new acceleration will be the least between the maximum acceleration in function of the power and the maximum acceleration on the curve

    I am having trouble with the maximum acceleration on the curve. How do I do that?
     
    Last edited: Apr 5, 2010
  12. Apr 5, 2010 #11

    jack action

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    [tex]\mu_{max} = \sqrt{\mu^2_{longitudinal}+\mu^2_{lateral}}[/tex]

    [tex]\mu_{max} = \sqrt{\left(\frac{F_{longitudinal}}{F_{vertical}}\right)^2+\left(\frac{F_{lateral}}{F_{vertical}}\right)^2}[/tex]

    And this is true for every tire in each corner.

    Now if you assume your forces are well balance between front and rear, to get an approx of the performance:

    [tex]\mu_{max}g = \sqrt{a^2_{longitudinal}+a^2_{lateral}}[/tex]

    Or in your case:

    [tex]a_{longitudinal \ max} = \sqrt{\left(\mu_{max}g\right)^2 -\left(\frac{V^2}{R}\right)^2}[/tex]
     
  13. Apr 6, 2010 #12
    The car is RWD, and since traction increases as weight on the rear tires increases, and weight on the rear tires increase as acceleration increase, Im thinking about computing the weight on the rear tires at each time [tex]t[/tex] and use that as the normal force under the rear tires. Can I also take that into account? How would it be done?

    Also, I would like to take into account downforce. Is this equation correct?
    [tex]a_{longitudinal \ max} = \sqrt{\left(\mu_{max}\left(mg+downforce\right)\right)^2 -\left(\frac{V^2}{R}\right)^2}[/tex]
     
    Last edited: Apr 6, 2010
  14. Apr 6, 2010 #13
    Also, what does it mean physically when the the interior of the square root is less than 0?
    [tex]
    \sqrt{\left(\mu_{max}g\right)^2 -\left(\frac{V^2}{R}\right)^2}
    [/tex]
     
  15. Apr 7, 2010 #14

    jack action

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    Yes, it is correct if you want to incorporate downforce.

    I don't think you understand the friction circle theory. Let me explain with an example:

    This is the equation you want for any tire on your car:

    [tex]\mu_{max} = \sqrt{\left(\frac{F_{longitudinal}}{F_{vertical}}\right)^2+\left(\frac{F_{lateral}}{F_{vertical}}\right)^2}[/tex]

    Let's say that you have a tire with [tex]\mu_{max}[/tex] = 1.0, a vertical force of 500 lb (due to gravity, weight transfer and downforce) and under an acceleration force of 300 lb. What is the maximum lateral force that the tire can take before loosing traction?

    Rewriting:

    [tex]F_{lateral} = F_{vertical} \sqrt{\mu^2_{max} - \left(\frac{F_{longitudinal}}{F_{vertical}}\right)^2}[/tex]

    [tex]F_{lateral} = 500 \sqrt{1.0^2 - \left(\frac{300 }{500}\right)^2}[/tex]

    [tex]F_{lateral} = 400 lb[/tex]

    Now, if the traction (or braking) force goes to 500 lb, then the maximum lateral force will be zero. This means that if enter a curve in such a condition, the tire will loose traction automatically. Similarly, if the traction force is zero, then you can go up to 500 lb of lateral force; If you go over this, you'll loose traction.
     
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