Car Trajectory

To model the path of a Grand Prix car on a track, it is not really the location of the surface of the track we are interested in. We really want the path of the center of mass (CM) of your car to be modeled. This center of mass is in a sense the "heart" that brings your wooden creation to life. Once this path is known, the acceleration, speed and time of the racing car can be found from the model expressions derived for the ramp and flat. Expressions for the height of the trajectory above the track and the lengths of parts of the trajectory are derived.

Follow your heart

Physically, the center of mass is where your car balances on a pencil. Half its weight is on either side of the pencil. The center of mass is a key to the behavior of each car in a race. Your car rotates about this point when not touching anything else. It is as if the force of gravity acts only on this one point inside your car. Of course that's not true, but we can use this fortuneate behavior to greatly simplify our models.

To properly model the behavior of your car on a track, we must know where your car's center of mass is at all points along the track. This path traveled by the center of mass is called the car's trajectory. In particular, the length of the trajectory paths on the ramp and flat are used with the closed models to determine race speed, time and other dynamics.

In principle, the location of your car's trajectory is just the height of your car's center of gravity above the track surface. This simplicity is hidden somewhat in the way that your car's simple geometry is rotated to the track's ramp angle through trigonometric functions. Finding this extra height on the ramp and flat is straight forward. The extra height and rotation of the car on the ramp changes the distance traveled on the ramp and flat.

Trajectory Geometry

To model your car's trajectory on an official AWANA track, we break it up into ramp and flat parts like the track. In the AWANA track model developed in this manual, the transition section was reduced to zero length for convenience. In general it does have a greater length. But even in the zero length case, the transition produces a different trajectory arc from the ramp or flat. But it can safely be ignored in the zero length transition case.

We will find an expression for the length of a car's trajectory by considering the flat first, then the ramp. This progression allows us to find the easy length first, then the harder ones.

From the time the rear wheels first touch the flat part of the track, your car will travel in a straight line until its nose is over the finish line. Since there is no rotation about its center of mass, the trajectory will be a straight line also. When the front and rear wheel radii are the same, Rf = Rr, the center of mass (CM) is Rr + CMy above the track. CMy is the distance your car's CM is above its rear axle. When the front and rear wheels have different radii, the base of the car is tipped at an angle, *, where * = arctan((Rf-Rr)/B) and B, the wheel base, is the distance between the front and rear axles. Note * is negative when the rear wheels are taller than the front wheels because the car slopes down.

This bit of tilt rotates the center of mass so it is now CMycos* + CMxsin* above the rear axle instead of just CMy. CMx is the distance from the rear axle to the center of mass (a positive number when the center of mass is in front of the axle). So the height of the center of mass above the track is

Rr + CMycos* + CMxsin*, where * = arctan((Rf-Rr)/B).

Where are the ends of the flat part of the track? In this simplified trajectory, the join end is more difficult. Let's do the finish line end first. When your car noses up to the finish line, it's center of mass is behind the line. The distance behind the finish line is the length of the nose plus the wheel base minus the distance to the center of mass from the rear axle. Written in symbols it is,

Finish: N + B - CMx and when * is not zero, (N + B - CMx)cos* + CMysin*

On the initial end of the flat, the intersection of the ramp trajectory and flat trajectory lies along the ramp, flat bisection angle. This is because the trajectory on both the ramp and flat is elevated by the same amount, Rr + CMycos* + CMxsin* or just Rr + CMy when * is zero (Rr = Rf). Projecting the intersection point down to the flat, means that we have a right triangle with the normal to the intersection forming the right angle to the flat (by definition). Call the distance from the normal to the join "x". Call the length of the normal "y". Then y is distance the trajectory is elevated above the flat.

y = Rr + CMycos* + CMxsin* or just Rr + CMy when * is zero (Rr = Rf)

Since the compliment of the ramp angle, O, is pi + O, since O is negative, its bisector is (pi+O)/2. The tangent expression can now be written as:

y/x = -tan((pi+O)/2)

But this can be rexpressed using the identity,

tan(pi/2-(-A)) = 1/tan(-A) for positive A so,

y/x = -1/tan(O/2)

Solving for x gives

x = -ytan(O/2) which is

x = -(Rr + CMycos* + CMxsin*)tan(O/2) after expanding y.

Join: -(Rr + CMy)tan(O/2) and when * is not zero, -(Rr + CMycos* + CMxsin*)tan(O/2)

The length of the flat is Lf. To get your car's trajectory length on this part of the trajectory, L2, we simply subtract the distance from the center of mass to the finish and the distance from the join to the trajectory intersection normal on the flat,

L2 = Lf - N - B + CMx + (Rr + CMy)tan(O/2)

or if * is not zero,

L2 = Lf - (N + B - CMx)cos* - CMysin* + (Rr + CMycos* + CMxsin*)tan(O/2)

The Ramp

On the ramp, the car is already tipped downward by the ramp angle, O. The center of mass, the point directly below it on the track and the point on the track whose normal passes through the center of mass form a right triangle. By definition, the normal line is at right angles to the track surface, so the vertex at the center of mass has an angle of theta. If the front and rear wheels have the same radii, Rf and Rr, the height to the center of mass from the tipped track is (Rr+CMy)/cosO. When Rf and Rr are not equal, the center of mass is rotated about the axle so the height becomes (Rr + CMycos* + CMxsin*)/cosO. This height does not change while the car is on the ramp.

At the starting line, the center of mass is behind it. To determine how far behind the starting line the center of mass is, we can ignore the ramp tilt. The center of mass is behind the line by the length of its nose plus its wheel base minus its distance from the rear axle. This is,

Start: N + B - CMx and when * is not zero, (N+B-CMx)cos* + CMysin*

At the join end, the same considerations apply as on the flat. Only this time, the normal, y, is projected on to the ramp. The logic and results are the same.

Join: -(Rr + CMy)tan(O/2) and when * is not zero, -(Rr + CMycos* + CMxsin*)tan(O/2)

The length of the ramp is Lr. To get your car's trajectory length on this first part of the trajectory, L1, we simply add the distance from the center of mass to the starting line and subtract the distance from the join to the trajectory intersection normal on the ramp,

L1 = Lr + N + B - CMx + (Rr + CMy)tan(O/2)

or if * is not zero,

L1 = Lr + (N+B-CMx)cos* + CMysin* + (Rr + CMycos* + CMxsin*)tan(O/2)

Simplified Model Trajectory On a Grand Prix Track

In summary, the trajectory geometry equations are as follows.