# Trajectory of the center of mass of a vehicle

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.

• Is this a homework problem or an exam question? May 1 at 17:13
• 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?
– r13
May 1 at 17:14

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.