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Computing how car pass the corner

  1. Apr 15, 2006 #1
    Hi everyone,
    I'm developing onlilne game based on car physics, my model already run on the straight but i need to find out how to compute passing the corners.
    (currently my model calculate high speed that depend on radius and use it until the end of corner)

    any help will be helpfull

    or maybe someone want to participate ?

    thank you in advance
     
  2. jcsd
  3. Apr 15, 2006 #2

    Hootenanny

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    You could model the corner and a circle and then use circular motion to calculate the acceleration of the car.

    Regards,
    -Hoot
     
  4. Apr 15, 2006 #3

    Kurdt

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    You mentioned you wanted to use radius as the computational parameter and as Hootenanny suggests circular motion would be the way to go.

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

    That is about all you should need to know on that web page. I hope this helps!
     
  5. Apr 15, 2006 #4
    Hootenanny, thank you for your answer.
    Its not the end i also need to cary about tyres, drag force and maybe suspension. I'm not a brilliant physic and this sound to me compilicated.
    But I'm good in programming (have created my site www.f1grandprixmnager.com). If you will help me with this equations i will be more than happy
     
  6. Apr 15, 2006 #5

    Hootenanny

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    If you have read the link Kurdt supplied about you will see that all object travelling in a circular path undego centripetal acceleration, the force will be transfered using the tires (friction). The centripetal force experienced by an object is given by;

    [tex]F = \frac{mv^2}{r}[/tex]

    Where m is the mass of the car, v is the tangental / linear velocity and r is the radius. The maximum frictional force that the tires can exert is given by

    [tex]F = \mu mg[/tex]

    Where [itex]\mu[/itex] is the co-efficent of dynamic friction between tires and tarmac (which can reach about 1.7) m is the mass of the car and g is the acceleration due to gravity.

    The air resistance can be modeled using the equation;

    [tex]F_{drag} = -\frac{1}{2}C\rho A v^2[/tex]

    Where C is the drag coefficent (dependant on shape), [itex]\rho[/itex] is the density of air [itex]\approx 1.25 kg\cdot m^3[/itex], A is the cross-sectional area and v is the velocity.

    These three equations should allow you to build up a decent model of the car's motion. However, I do recommend that you read Kurdt's link as circular motion can be quite confusing initally.

    As for the suspension, that is something I cannot help you with. You may want to venture into the engineering forum for that.

    Regards,
    -Hoot
     
    Last edited: Apr 15, 2006
  7. Apr 15, 2006 #6
    Hootenanny thank you for your explanation:
    i have read what i have for cornering (it was creted some times ago by one guy but he missed now and thats a reason why come here :) )
    and tried to combine with what you said

    here is the part of mathcad file:
    % nasty rearranged equation to find max cornering speed with aero downforce
    top_bit = g*(W_br*musf+W_bf*musr)/W_b
    long_bit = 0.5*rho*ClS*(W_br*musf+W_bf*musr)/(M*W_b)
    bottom_bit = 1/R - long_bit % if long_bit > 1/R we end up with sqrt of negative number
    % fraction = top_bit/bottom_bit
    % spd = sqrt(fraction)

    if long_bit > 1/R
    disp('negative square route: flat out corner, treat as straight')
    else
    fraction = top_bit/bottom_bit
    spd = sqrt(fraction)
    if spd>max_v
    disp('potential speed greater than max: flat out corner, treat as straight')
    end
    end

    where:
    W_b - wheelbase
    W_br - distance rear axle - cg [m]
    W_bf - distance front axle - cg [m]
    musf - front lateral friction coefficient
    musr - rear lateral friction coefficient
    R - radius.
    ClS - coefficient of downforce * reference surface

    can you explain me pls why:
    1. long_bit. looks like air resistance but why he use division by (M*W_b)
    2. what is the top_bit and bottom_bit ?

    thank you in advance
     
  8. Apr 16, 2006 #7

    Hootenanny

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    I'm not quite sure on this but it appears the top_bit is for calculating the torque experienced by the tyres, this can be split into two seperate equations (one for front tyres and one for rear) but it seems he has combined the two.

    I'm not sure what he has done with the long_bit or bottom_bit. His name tags are a little unhelpful :confused:

    ~Hoot
     
  9. Apr 16, 2006 #8
    maybe full version will be more usefull:

    clear all


    % VEHICLE DATA
    M = 610; % weight [kg]
    CdS = 1.035; % coefficient of drag * reference surface [m2]
    ClS = 2.325; % coefficient of downforce * reference surface [m2]
    h_cg = 0.25; % cg height [m]
    W_d = 0.43; % percentage of weight at the front
    W_b = 3; % wheelbase [m]
    W_bf = W_b *(1-W_d); % distance front axle - cg [m]
    W_br = W_b - W_bf; % distance rear axle - cg [m]
    r = 0.33; % wheel radius [m]
    f = 0.01; % rolling resistance coefficient (assuming drag force = M * f)
    mu_lat_r = 1; % rear lateral friction coefficient
    mu_lat_f = 1; % front laterl friction coefficient

    % CONSTANTS
    g=9.81;
    rho = 1.22; % air density

    % engine power curve X = rpm, Y = power [kW]
    X = [13500,14500:500:18500];
    Y = [452,485,517,536,548,558,570,580,588,593];
    max_v = (2*max(Y)*1000/(rho*CdS))^(1/3); % theoretical max speed considering only aero drag

    R = 500; % m radius



    % nasty rearranged equation to find max cornering speed with aero downforce
    top_bit = g*(W_br*musf+W_bf*musr)/W_b
    long_bit = 0.5*rho*ClS*(W_br*musf+W_bf*musr)/(M*W_b)
    bottom_bit = 1/R - long_bit % if long_bit > 1/R we end up with sqrt of negative number
    % fraction = top_bit/bottom_bit
    % spd = sqrt(fraction)

    if long_bit > 1/R
    disp('negative square route: flat out corner, treat as straight')
    else
    fraction = top_bit/bottom_bit
    spd = sqrt(fraction)
    if spd>max_v
    disp('potential speed greater than max: flat out corner, treat as straight')
    end
    end

    %
    % if we have a flat out corner, we can treat it as a straight,
    % so recalculate that sector as straight with length = original corner length
    %
    % if this means we have a straight followed by a straight, they can be combined to make one long straight
    %
    % if it is not a flat out corner then we can use the spd value to calculate time for that corner as before
    % time = distance/speed
    % for two consecutive corners we will still have to put up with a step change in vehicle speed.
    %
    % with corner speeds set we have entry & exit speeds for the straights.
     
  10. Apr 16, 2006 #9

    Hootenanny

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    Ahh, I think I've got it. top_bit refers to the top of the fraction and bottom_bit refers to the bottom of the fraction. Obvious, I know. I can't belive I didn't see it before! top_bit is using an application of [itex]F = \mu mg[/itex] to calculate the maximum centripetal force that the tyres can hold, as I said before the equation calculates the difference in force for the front and rear tyres and the torque produced.

    I'm still working on the other bits. Unfortunatly, I've got to do some work now :frown:, but if I get chance I'll try and take a closer look at it tonight.

    Regards,
    ~Hoot
     
  11. Apr 16, 2006 #10
    Thank you Hootenanny !
    I have strong experience with applications for web.
    So if you need somehelp (scripts, hosting and so on), it will be pleasure for me to help you.
     
  12. Apr 17, 2006 #11

    Hootenanny

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    I think the long_bit is the additional downforce added by the aerodynamical properties of the car. He has divided by the mass of the car because he divided the top_bit by the mass. The top_bit is effectively [itex]F = \mu g[/itex] but the equation for firction is given by [itex]F = \mu mg[/itex].

    You are very generous.

    Regards,
    ~Hoot
     
  13. Apr 17, 2006 #12
    Hootenanny thank you for explanations.
    Drag force is here and now all that i need is to add suspension and tyre stuffs.

    Hootenanny, have you enough time to help me with this stuff ?
    Its going to be complicated enough for me.
    my knowledge is enough to calculate air density regarding weather, but not enough for this.

    And i as said before please do not hesitate to ask me for web stuff.
     
  14. Apr 17, 2006 #13

    Hootenanny

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    I will help you as much as I can, but I do have some work to do, so you will understand if I do not reply immediatly. In addition, there are others here at PF who will be more than willing to help you if I can't or am too busy. If you post your questions I'm sure someone will answer them given enough time.

    As I said before the workings of the suspension are beyond me, you are probably better asking in the engineering forum for that. As for the tyres, it depends how complex you wish to go, I may have to read up on some bits. I will do my best to answer your questions, but if I can't there will be someone here who can :smile:

    What would you like to know bout tyres?

    Regards,
    ~Hoot
     
    Last edited: Apr 17, 2006
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