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Automotive Tracing the Path followed by a car

  1. Mar 17, 2014 #1
    I would like to know the calculations regarding tracing of the path followed by a car during turning process. i have the following details , kindly do tell what other information is required and how the equations are formed.

    front wheel track - 1559 mm
    rear wheel track - 1605 mm
    overall height - 1468 mm
    wheel base - 2909 mm
    overall length- 4961 mm
    turning circle - 11.5 m
    kerb weight - 1735 kg
    gross vehicle wt - 2320 kg
    steering ratio - 16:1

    By calculating the radius of curvature and the path followed by it , will it aid in finding m,y solution.Thanks in advance.
  2. jcsd
  3. Mar 19, 2014 #2
    The interaction of tyres with a road surface is an extremely complex subject. If you are willing to make the assumption that there will be no lateral movement of the tyre across the road (ie cornering well below the grip limits of the tyres) then it becomes a relatively straightforward geometry problem. You can assume that each wheel only moves in the direction it is facing, thus to work out the centre of rotation of the car you can draw a line out perpendicular to each wheel (when looking from above) and find where they intersect. Each wheel will follow an arc about this centre point.
  4. Mar 19, 2014 #3

    Ranger Mike

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    how high is the sky?
    seriously, welcome to the forum
    there is not enough critical information ..like speed of vehicle entering the corner, degree of banking of the turn, radius of the turn,....Looking for Gs here.. what is the track? asphalt, concrete, what is % of weight front to rear, left to right?

    Its tuff enough just trying to keep a purpose built car inside the guard rails at speed...to try to predict where a stock production door slammer will track with dubious tires and rolling around like a big old whale is going to be more of a challenge. Add to this the mushy shocks ( dampers) and built in understeer most mcphereson strut cars have just add zest to this endeavor.

    bottom line is -you can not fit a static model into a dynamic environment.
    Last edited: Mar 19, 2014
  5. Mar 20, 2014 #4
    Thanks sgb27 & ranger mike for the reply ...
    i shall add some more information on the matter.consider there is no slip...i am just calculating the theoretical value of the path followed by the car when the steering wheel is locked to one side , so it should produce a circle right?

    consider the weight distribution to be 50:50
    lock to lock steering turns : 2.5 (900 degrees) so towards right 450 degrees and vice versa
    turning radius : 5.75m
    velocity = 20 km/hr

    i have tried it and i got a epicycloid path ...
  6. Mar 21, 2014 #5


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    Assuming that ideal Ackerman steering is implemented on the front wheels, and no slippage or contact patch deformation occurs (slow speed, low g turn), then when the front tires are turned inwards, the extended axis lines from the rear wheel axle and both front wheel axles will all cross at the same point, and this point will be the center of a circle that the car travels around.
  7. Mar 21, 2014 #6
    What exactly did you try that gave an epicycloid path? If you post your method someone will be able to tell where you went wrong with your working.
  8. Mar 23, 2014 #7
    @sgb27 ..it was my mistake ...i did get a circle ....it was the scaling on the xy plotter which made me see it as an elipse. As of now i have got the 2D version of plotting the traced path ...but now i have to implement the 3D version of the traced path ....(when climbing a slope ) ....Is it done with the help of using longitudinal acceleration ?

    these are the equation i used

    inner wheel angle = steering wheel angle/steering ratio

    outer wheel angle= tan inverse(1/cot(inner wheel angle) + (wheel track/wheel base))

    using Pythagoras theorem and finding the respective data in the ackerman diagram ..i was able to find the value of the radius of curvature as 5.1124m and the angle of orientation as 20.85 degrees.

    X(c)=Xo - Rc sin(angle of orientation)
    Y(c)=Yo + Rc cos(angle of orientation)

    X(t)= X(c) + Rc sin(w(t1-t0) + angle of orientation)
    Y(t)=Y(c) - Rc cos(w(t1-t0) + angle of orientation)

    where X(c),Y(c) are the coordinates of the centre of curvature
    X(t) ,Y(t) are the coordinates of the vehicle traced path
    w= angular velocity = constant velocity/radius of curvature
    t1-t0 =time stamp

    w(t1-t0) = traced path in angles or radians

    now i need to find the position of the vehicle while going up a slope , so the height is also to be considered...kindly do tell me which all equations is required for this. Thanks in advance
  9. Mar 26, 2014 #8
    So long as the road slope is not to great to cause slipping of the tyres, the path will still be the circle, just at an incline to match the road. You can calculate the new equations for X,Y,Z based on this inclined circle.
  10. Mar 26, 2014 #9

    can you please tell me those equations to calculate the path traced while climbing a slope?? how do i find the slope traced by the car??
    Last edited: Mar 27, 2014
  11. Mar 27, 2014 #10

    Ranger Mike

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    suggest you read Race car suspension Class in this forum
    see the last page..
    it discusses why the path of a car in a turn will not be circular.
  12. Mar 27, 2014 #11
    @Ranger Mike thanks a lot for the input...but mate i am just trying to calculate the path traced on the basis of theoretical knowledge using ackerman's theorem....its by using simple mathematical equations (sin , cos and all) so far i had been able to trace the path on a level road but when the car ascends a slope , we dont have the required data to find the height at which the car climbs or the slope .. i am just calculating the path traced with a normal passenger car, race cars dynamics and application is on another level i guess.....thanks mate...sincerely hope you might be able to find those equation and help with the rest of the calculations. have a good day
  13. Mar 28, 2014 #12
    Your equations are still correct for going uphill, it's just your Y coordinate now points up hill rather than horizontal. To find the actual horizontal and vertical distance you can draw a right-angled triangle to represent the slope, with your "Y" distance going up the hill. You can use the trig relations to find a formula for the horizontal and vertical distance in terms of Y(t) (edit: and the slope angle).
  14. Mar 28, 2014 #13
    @sgb27 i do get your point but i dont have the arc length or the slope angle for using the trigonometric equations.
  15. Mar 28, 2014 #14
    You need to know the slope angle of the road, the vertical and horizontal position of the car will depend on the slope.

    If your car has travelled Y(t) up the slope (given by your equations), then that is Y(t)cos(a) horizontally and Y(t)sin(a) vertically, where a is the slope of the road. The height of the vehicle is then just given by your equation for Y(t) above multiplied by sin(a), plus perhaps some constant if you want a certain point of the turn to be zero height.
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