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Driving car in slope, when does the tire start to slip?

  1. Nov 27, 2007 #1
    1. The problem statement, all variables and given/known data

    I don't know if I am posting in the wrong thread. This is not a question for solving a homework, but a question to help me write some homework students will have to solve.

    I teach algorythmics to mechanics students. The program they are developing is about driving a car along a street. As they are mechanics students, the program will tell them when active driving aids are activated (ABS, traction control and stability control systems). The street has curves and slopes. In curves I have already stated that the maximum speed takes into account the MU friction constant of the surface and the radius of the curve. When braking, if they brake too hard, ABS will start to activate. In fact I would like that the ABS activation were dependent on the speed and surface of the street but well, but I haven't found how. This doubt is related to the one I am asking. I want that when the car accelerates, the traction control activates when the acceleration is too heavy and some tire looses traction, that is, it starts slipping on the surface. I think it is most important if they accelerate in an ascending slope. Let's see if someone can help me.

    2. Relevant equations

    A slope has a % increase, from 5% to 10%. The car has a mass. I have seen in some tutorials that in this case, the sin and cos of the angle play some part. But there is also the surface of the road. If it is wet the MU friction constant is lower and it is easier to slip. When running horizontally, the weight of the car is ignored (simplification) because friction and innertia both depend on the weight. But in slopes there is an angle and then the weight is relevant.

    3. The attempt at a solution

    No idea. I would like a very simplified formula so that I can say, when this happens, the driving simulator will show a message of Traction Control System activation. This condition could be immediate or it could have to take into account what has been done in previous actions.

    Any help? I can add any info you need
  2. jcsd
  3. Nov 28, 2007 #2
    Slippage depends on the static friction coefficient.

    Friction is the force pushing the car up the ramp. Weight is the force pushing the car down the ramp.

    Friction is given as follows:

    [tex]F_f=\mu N=\mu Wcos(\theta)=\mu mgcos(\theta)[/tex]

    On a flat surface, there is no weight. On an angle surface, the force of weight working against friction is...


    So the limit at which slippage occurs is when these forces are equal:

    [tex]\mu mgcos(\theta)=mgsin(\theta)[/tex]


    [tex]\mu cos(\theta)=sin(\theta)[/tex]



    So what this says is that the more traction your tires provide (the better the coefficient of static friction), the steeper the angle you can have before the tires slip.
  4. Nov 28, 2007 #3
    Bill, thanks for your clarifications. Now my question is:

    As you explained, the point where the tire starts to slip is related with the arctangent of mu:
    As a side comment, in fact the car is moving along a street, so the friction to be taken into account is the cinetic one.

    But in any case, this friction calculates the point the car would slip if stopped. If the car is moving, it has a speed forwards. If the car is keeping its speed, it has a constant acceleration forwards. This acceleration causes that the tire is trying to friction with the surface to maintain speed and this could cause that the tire slips. And even more, if the driver decides to accelerate in a slope so the car increases speed, the tire is increasing its work on the surface making it more probably to slip. How can those things be simulated in a simplified formula? Simplified I mean considering a car is a weighted point and the tires are not four points of contact but one, and the tires have an infinite radius. Just some condition that can be easily programmed by a novice first course student not much experienced in programming, where the program can decide that at this moment, the tires slip and traction control must be activated.
  5. Nov 28, 2007 #4
    Actually, it's static friction. Even though the vehicle is moving, the point of contact the tire makes with the road is stationary. It is static friction which drives the vehicle.

    If the wheels were spinning, then there would still be a frictional force (kinetic in this case) driving the vehicle, but it will be less because the coefficient of static friction is greater than the coefficient of kinetic friction. This is why street racers and drag racers try not to spin their tires when they launch. This is why automakers have implemented anti-lock brakes.

    If a car is keeping it speed, that is, maintaining a constant velocity, then its acceleration is 0.

    A car that is driven by its wheels cannot accelerate any faster than that limited by the coefficient of static and kinetic friction:

    [tex]\mu_s g[/tex] (if the wheels are not spinning)
    [tex]\mu_k g[/tex] (if the wheels are spinning)

    What are you asking?

    Do you want to know the maximum acceleration? I just gave it to you:

    [tex]\mu_s g\cos{\theta}[/tex] (if the wheels are not spinning)
    [tex]\mu_k g\cos{\theta}[/tex] (if the wheels are spinning)

    On a flat surface, [tex]\cos{\theta}=1[/tex] so we can just leave that out of the equation.

    But as you can see, maximum acceleration is dependent upon [tex]\mu[/tex] and [tex]\theta[/tex]
  6. Nov 28, 2007 #5
    Oh, I have perhaps explained myself incorrectly. I mean that in the case of my example, as the car is being driven and the wheels are spinning, then the frictional force that applies is kinetic one.

    I wrote it improperly. On a slope up, is the car has to maintain speed, it must accelerate, otherwise it will lose speed. On the contrary, on a slope down the car should brake to maintain speed. In both cases, the tire may slip. If I accelerate on a slope up, depending on the angle and the weight of the vehicle, perhaps the car would loose speed despite the driver trying to accelerate, or it could just maintain speed or even increase it. In a slope down, braking would cause a decrement in speed, a maintainance of the same speed or perhaps an increment if the angle is too high.

    Ok. I was asking how can I detect tires slippage. But I think you already solved it. I am writing now what I understand is the way to express it, correct me if I am wrong.

    First, the simplified simulator calculated changes in speed due to slopes as: slope down added 10km/h to speed, slope up decremented 10km/h. This was done before slope angle was taken into account. Now that slopes have angles, the increment or decrement can be calculated based on the slope angle. The negative acceleration applied to the car in a slope is [tex]\mu_k m g\cos{\theta}[/tex] and the resulting speed can be calculated applying this acceleration to the basic formula [tex]v_{end} - v_{start} = 2*a*distance[/tex] counting a given distance. If the driver accelerates, then he is trying to add 10km/h or at least to maintain the vehicle speed. The simulator can then calculate the maximum acceleration accepted and see if the driver is exceeding it, and then warn the driver that Traction Control has been activated, and then the acceleration applied is the maximum one given the conditions of angle and surface. The final speed then would be calculated on the base of this maximum acceleration.

    This is something students are able to program adequately, I think. But, please, tell me, is there any lie in here?

    I will abuse a bit more of you forum people. Let's take the case of a curve. In curves, cars have the stability control system. When would it activate, when the car exceeds the maximum speed and starts slipping or before? This kind of control usually detects understeer or oversteer, but for simplification in the case of this simple simulator, I will make students detect speed excess. Is it correct that stability control systems get activated when speed makes centrifuge force higher than dynamic friction?

    Thank you very much, you are helping me clear concepts. :) This forum is great.
  7. Nov 30, 2007 #6
    Can someone confirm the solution I propose is correct?
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