Bouncing back

The result of an elastic collision is the return of the collision energy to the car. At the median that means a "bounce". Just as a super ball almost returns to the height it is released at, your car may bounce back away from the median at about the same speed it hit it with. It retains most of its initial energy.

A car that is misaligned and shows a tendency for elastic collisions, is characterized by zigzagging down the track banging away at the median as it goes. How much energy might a car lose on this saw-toothed path if all the collisions were perfectly elastic? This and questions like it are addressed by building a simple mathematical model.

Notation:

Important expressions derived for this model include:

Dynamics of zigzagging

First, let's get some information about the path the car might follow in the worst case. The point that we will follow to create the zigzag path is the car's center of mass. That makes the path a trajectory. An interesting point about zigzagging motion is that the car's center of mass may not actually zigzag at all! This happens when the motion is caused by twisting side to side rather than shifting side to side.

Wheels on diagonal opposite sides of the car alternate touching the median in twisting motion. The wheels slide across the track more than spin so that tread friction actually decreases. But energy is lost (Iw2/2 again) as the car opposes the twisting motion via its moment of inertia about the twisting axis. This motion is more likely to occur at top speed on a slippery track.

The distance between collision with the lane median depends on the tread friction since the wheels are not able to direct the course of the car. This tends to be the type of motion caused by putting whiskers too far forward of the center of mass especially on a rear weighted car.

In the more common shifting motion, both wheels on the same side touch the median, then both on the other side shift over and touch the median. Though some sliding may take place, the wheels are largely in control of the direction of the car. The bounce returns most of the collision energy back to the car.

If the car could leave the median at the maximum collision angle, it would immediately touch the median with one of the wheels from the other side of the car. The only way that could happen is if the car twisted as described above. So the car must leave the median at some less dangerous angle. However, the limiting case for this kind of zigzag trajectory is the twisting one described before.

Energy Loss

Likely, the impact angle will be less than the maximum possible, and the impact will be somewhat inelastic. But to get a limit on how much energy can be lost in an elastic collision with the lane median, we assume the worst.

For a car that is shifting back and forth, the only significant energy loss is that associated with the longer trajectory it follows.

In the case of the twisting car, energy is lost in twisting.

The energy loss during a twisting elastic collision would be that required to turn the car from an angle of O to -O, a total of 2*O = 0.1920 radians (2pi5.5/180). A typical value for a car's sideways moment of inertia, I, is about 0.014 ozins2. Then in a collision lasting a hundredth of a second,

E = Iw2/2 = 0.014*(0.1920/0.01)2/2 = 0.014*184.32 = 2.6 ozin

A car has about 200 ozin of energy to "spend" at the beginning of the race. So 2.6 is 1.3% of the total energy, just over 1% lost in just one bad collision.

This means 100-1.3 = 98.7% of the kinetic energy is left after each collision. On a flat track beginning with the maximum linear kinetic energy possible, the amount of forward kinetic energy left after each collision would be 98.7% of the amount before. At the end of the zigzag path, the remaining forward kinetic energy would be 0.987 raised to the 93rd power or about 30% of the starting linear kinetic energy of the car. Granted, this is not a realistic scenario, but it does give a limit we can "play" with.

For example, if a car's initial speed, vi, was 160 in/s, at the end it would be about v = vi\[1-p] = 160\[0.30] = 87.6 in/s. That's a reduction in speed of \[0.30] = 0.55 about one half!

So, from this unlikely example, we can say with certainty that on a well constructed track your car will not be stopped on the track due to bouncing off the lane median alone. It would take a combination of factors. The words "well constructed" were used since tracks that have splinters jutting out of the median might in fact be able to stop a car in a number of ways.

The average force of impact is found by measuring the change in the car's momentum over a short period of time. This is the car's mass, m, times its change in velocity, v-vi, relative to the median divided by the duration of the impact, t-ti. If the collision is totally elastic, the speed before, vi, and after, v, will be nearly the same, say 160 in/s. If nothing falls off the car in the impact, then its mass doesn't change either. So the car's momentum is the same after as before, but going in a different direction. The angle of exit tends to be the same in magnitude, but opposite in direction to the angle of collision. In this model, the sign on the initial speed, vi, is changed to indicate the direction change.

            / t
mv - mvi = | F dt = Fave(t-ti)
          / ti

Only part of the car's momentum is actually at right angles to the lane median, it is the total momentum times the sine of the impact angle, O. Using the worst collision angle, 5.54 degrees, O = 0.09677 in radians.

Fave = m(v-vi)sinO/(t-ti)

Let's evaluate it assuming the collision takes place in a hundredth of a second, that is, t-ti = 0.01. That's enough time for the car to travel over an inch!

Fave = 0.01295(160-(-160))0.09677/0.01 = 40.1 oz

The actual interaction time is probably much less, say 1000 th of a second. Fave is then 400.1 oz! The actual maximum instantaneous force of impact would be much greater still.

Conclusion - Taper hub backs?

This look at elastic collisions on the median tells us that it is important to keep the collision angle low. Each impact is like a hammer blow to the inside of the axle hub as the wheel transmits the energy of the impact from the median.

Tapering the back of the axle hub may help make the collision less elastic by acting as a "spring". When the wheel bore presses into the taper, it lifts the axle up on the taper. If the taper is steep, the lifting will be more difficult and may stop before the wheel bore is "locked up" on the taper. That's when the bore touches all around the hub at the same time. When a wheel is locked up on its axle hub, hub friction is maximized, though it may only be for a short time. The force of the axle pressing down on the taper assures that the wheel will get pushed away from the hub and that some side ways motion will begin after a median collision.

So tapering the backs of the axle hubs may do two things that straight hub backs may not:

1. Absorb some of the shock of a median collision making it less elastic
2. Push the axle (that is, the car) away from the lane median making it behave less inelastic