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The problem is to find the trajectory of the center of mass of a car when it is rotating and translating (after an off center collision, for instance)

What i know as input is the trajectory of the wheels (skidd marks deposited on the surface of the road) I also am given the geometry of the vehicle, so the distance of all four wheels to the center of mass is know, as are the Wheelbase ($WB$) and Track lenghts ($T$).

Let $y = f_i(x), i=1,..,4$ be the function describing the trajectory of the $i$th wheel, and $r_i$ be the distance of the $i$th wheel to the center of mass.

How would i go about finding a function $y = f_{CM}(x)$ that describes the trajectory of the center of mass? Assume planar motion.

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  • $\begingroup$ Is this a homework problem or an exam question? $\endgroup$
    – Solar Mike
    May 1 at 17:13
  • $\begingroup$ Shouldn't the center of mass be used as the reference point for the equation of Y with respect to the displacement in X at time t? $\endgroup$
    – r13
    May 1 at 17:14
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This requires a few assumptions, and its not the most optimal/efficient algorithm, but its relatively easy to program.

Assumptions:

  • you have(or you can create) a list of points for the coordinates of each tire $FR(i),FL(i),RR(i),RL(i)$ (Front Right, Front Left, Rear Right, Rear Left). So for $FR(1)$ you would get something like (3,2.5), meaning 3 m on the axis and 2.5 m on the y-axis
  • the thickness of the tires is really small (so orientation does not have an effect.
  • The car and the tires do not deform

The main idea is that you will start with one track -lets assume FR()- and you find which points from the other curves correspond to that point, based on geometric constraints.

Algorithm:

  1. You start by the first point in the list FR(i=1).
  2. With center that point you draw 3 circles with radius ($WB$, $T$ and $\sqrt{WB^2+T^2}$ (if you want you can generate circles with $\pm$ a few cm)
  3. if the circles
  • coincide with only one point on each other curve, then you find the index of that curve.
  • intersect more than one points on the other curves then you use the orientation of the vehicle to select the best fit.
  1. Assuming you arrive at indexes $FR(i), FL(k),RR(l),RL(m)$, you know that the center of geometry is exactly between the coordinate $FR(i)$ and $RL(m)$.
  2. you proceed to the next point ($FR(i+1)$) and go back step 2. The main difference, now is that assuming points $FR(i)$ and $FR(i+1)$ are close you have a good starting point.
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