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Car physics

  1. Aug 5, 2009 #1
    Hi all!
    I've got a problem which I can't solve for almost a week :-/ It's about simulating a car with some physics added. The things is that I cannot calculate the wheel angular velocity right.

    Let's assume that a car has no velocity and is using 7300 N (1st gear, 1000 rpm) of force on the rear wheels at the moment. Now I want to calculate the wheel angular velocity by integrating the wheel angular acceleration over time. Delta time is constant - 0.03s. I only need to calculate the acceleration:
    E=M/I

    M is my force multiplied with the wheel radius (0.34m) - 2482 Nm
    I is the inertia of the wheel - I assume it's about 4 kg*m^2

    So the acceleration of the car is 620 1/s^2. This gives me a wheel angular velocity of 18.6 1/s.

    I think it's much too high, because when I calculate the engine rpm back from the wheel velocity I get:
    engineRpm=wheelAngularVelocity*ratio*60=18.6*6*60=6696 rpm

    It's much too big... What's the problem with my calculations?
    Btw. Sorry for my english :)

    I'll appreciate any help.
     
  2. jcsd
  3. Aug 5, 2009 #2

    berkeman

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    Staff: Mentor

    Welcome to the PF. What are the units you are using "1/s and 1/s^2"?
     
  4. Aug 5, 2009 #3
    's' means second and s^2=s*s.
    velocity - m/s
    force - N=kg*m/s^2
    torque - Nm
    inertia - kg*m^2

    I think that's all for now.
     
  5. Aug 5, 2009 #4

    berkeman

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    Yes, all of that is standard, but things like this are not:
     
  6. Aug 5, 2009 #5
    I checked on the web and the units of angular acceleration and angular velocity is rad/s^2 and rad/s. I thought that I use 1/s^2 and 1/s instead in order not to use too much arithmetics. When I change these things to radians it increases the acceleration and angular velocity So it gives me wrong results too...

    Am I missing the point?
     
  7. Aug 5, 2009 #6

    berkeman

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    The line I quoted above is for the car's linear acceleration, I assumed. In which case the units should be m/s^2
     
  8. Aug 5, 2009 #7

    berkeman

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    Also, it looks like you are just calculating the acceleration of the wheel itself, ignoring that it has to push the car. Don't use torque = I * alpha to calculate the acceleration of the wheel. Figure out how hard it is to linearly accelerate the whole car. The moment of inertia of the wheel is negligible in this problem.
     
  9. Aug 5, 2009 #8
    It was my mistake. I should have written angular acceleration of the wheel.

    Do you have any idea on how to correct this equation? I calculated it like this, because I read it in an article about simulating car physics. It says that I need to calculate the force on the wheel, than calculate the wheel angular velocity, new engine rpm and than look up how much force is transferred by the road on the whole car (slip ratio) which actually gives the car linear acceleration.
     
  10. Aug 5, 2009 #9

    berkeman

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    Where did 7300N come from? The engine produces a torque, not a force. You would use the torque and wheel radius to figure out the driving force, and use F=ma to calculate the car's linear accelertion. What is the mass of the car?
     
  11. Aug 5, 2009 #10
    Some 'constants':
    first gear ratio: 2.66
    differential ratio: 3.42
    efficiency: 70%
    ratio=2.66*3.42*0.7~6.4
    mass: 1500kg
    wheel radius (calculated): 0.34m
    torque=390Nm (1000 rpm)
    torque on the wheel=390Nm*6.4=2496Nm
    force on the contact point=2496Nm/0.34m~7300N

    I approximate the friction forces with speed of the car:
    drag friction=c_drag*speed*speed
    rolling resitance=c_rr*speed

    When the car is not moving they both are 0, so there's only engine force.
     
  12. Aug 5, 2009 #11

    berkeman

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    Oh, so the 7300N is the force on the ground, accelerating the 1500kg car. That's what you plug into F=ma to figure out the linear acceleration.
     
  13. Aug 5, 2009 #12
    Hmpf... That's not my problem :) This force can be the traction force, but it may not. This force just accelerates the wheel to a certain angular velocity. With this velocity I can get the force on the car by using slip ratio and so on... That's not important.

    The problem is that this force gives too high wheel angular acceleration (and than angular velocity). I think that the equation is somehow incorrect or I'm missing some friction force or I should calculate the force with the wheel load... I'm not sure about this and that's why I'm asking :)
     
  14. Aug 5, 2009 #13

    berkeman

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    No, that force does not accelerate the wheel. It accelerates the car via F=ma.

    Using that equation, what is the linear acceleration of the car?

    Using that linear acceleration and the radius of the wheel, what is the angular acceleration of the wheel?
     
  15. Aug 5, 2009 #14
    a=4.8 m/s^2
    E=a/r=4.8/0.34 1/s^2=14 1/s^2

    This approach looks good, but is it really correct?

    In the article I read this was more comlicated (http://regedit.gamedev.pl/Mirror/Car%20Physics%20for%20Games/Car%20Physics%20for%20Games.html [Broken]). You can look at the "Torque on the drive wheels" section only (about 30 lines + an image) in the middle of the article. It's a shortened version of everything before it. It shows a different model. Is it wrong?


    Btw. This E=14 1/s^2 gives the wheel angular velocity of 0.7 1/s and this gives me the engine rpm of 252... It's much too low! We started with 1000 rpm and after pushing the clutch we got only 252?
     
    Last edited by a moderator: May 4, 2017
  16. Aug 5, 2009 #15

    berkeman

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    That's why you slip the clutch as you start rolling, eh?
     
  17. Aug 5, 2009 #16
    You're probably right :) But I can start without a clutch and do a 'burnout' when I use much more torque. Let's use now 4400rpm at start (this is the torque peak point for corvetta). This gives me a total torque 475 Nm of the engine and 2850 Nm on the wheel. This is 8400 N of force; a=5.6m/s^2; E=16 1/s^2; w=0.8 1/s; engine rpm=288.

    Kind of low, and the problem is exactly the same :/ Am I wrong?
     
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