Straight and narrow is the road to the finish line

As your car travels down the track, it probably won't roll perfectly straight. It may slide side to side, or "park" against one edge of the track median dragging the inner part of the wheels on that side all the way to the finish line. The path your car follows is an indicator of other things that are happening in the race.

But before we can hope to model the real causes, we need to know some basics about drift. Just how straight must a car be to avoid the median totally? What is the effect of hitting the median on something you care about like speed? Later we find out the amount of speed lost depends on the collision angle. So here we find out what the greatest angle can be.

All of this discussion centers about the kinematics of drift. We're just going to stick to the geometry and motion of a drifting car. On other pages, the dynamics of drift is discussed. That's where you'll find out about the origin of forces that cause the motion.

• First, how straight at the starting gate?
• A look at lost speed is next.
• What happens when the car runs into the median?
• Then how many times can a car hit the median?
• If it hits that many times, what is its path?

Notation

Important expressions derived for this model include:

Straight at the gate

The effect of placing your car crooked at the starting line is difficult to measure. If your car does not have perfect wheel alignment, the track is not perfectly flat and smooth or a number of other things, placing your car crooked on the track may not be noticeable at all because those other causes of drift will take over very soon after the race begins. Likewise, placing your car perfectly straight won't help either if your car's wheels and track are not reasonably perfect already. However, we can establish how straight your car should be for the best runs if it is a perfect car and track otherwise!

How much can the car wander?

The lane strip or track median is 1.375 inches wide. From the inner edge of one wheel to the inner edge of its compliment is about 1.75 inches not accounting for the short inner wheel standoffs. In total that makes 0.375 or 3/8 inches of room to wobble down the track side to side between the lane median and the inner edges of the wheels.

Ideally, when your car is placed on the track, there will be 3/16 inches clearance on the inside of each wheel. Let's call this clearance "c". c = 3/16 for a typical AWANA car. For a car to not have its wheels touch the lane median it cannot vary more than this 3/16 inch either way for the entire 30 foot track length. If we call the track length "L", we can model this "no drift" limit as a straight line with an angle "D" (though it's probably more of an arc). When we do this we can answer the question, "How much longer is the car's path with this minimum amount of drift?".

Using a little trigonometry, we know that c/L = tanD since the wheel to median clearance and the track length are at right angles to each other. Solving for D, we get D = arctan(c/L) where arctan is the inverse of the tangent function.

For c = 3/16 inches at 30 feet ( L = 360 inches), D = arctan((3/16)/360) = 0.03 degrees!

To get the extra distance traveled on this slightly wayward path, we can use the angle D in the expression L/cosD - L or the pathagorus theorem. The pathagorus theorem applies to right triangles formed by lines like those measured by c and L. It tells us that the longest line in the triangle, the hypotenuse, h, has a length related to the others by h2 = c2 + L2. That means h = \[c2+L2]. In our case, then, the extra length of the path, h, over the track length, L, is \[c2+L2] - L.

Either way, you get the same answer. The length of the minimal drift line is 0.00005 inches longer than the track! So according this simple straight line model of drift, if your car's drift angle is zero or up to 0.03 degrees, you aren't going to lose a race from the extra time it takes to traverse it; it's straight for all practical purposes.

However, there are many possible road hazards that can cause your car to deviate more than 0.03 degrees even with perfect starting line placement! Above 0.03 degrees, the inside edge of one or two wheels of your car will hit the lane median at least once somewhere on the track. Also, chances are, that the same wheels have been rubbing their axle hubs and the opposite wheels have been rubbing the car body. These areas can be lubricated to reduce the effects, but it is still added friction.

Lost speed

When the inside edge of a wheel hits the median, it rubs the median. If the angle is a small grazing angle, O, there may be little transfer of the car's energy to the median. If it is larger the wheel can lose rotation speed and the car can lose the component of kinetic energy in the cross-track direction.

Linear kinetic energy can be expressed as mv2/2 where m is the mass of the moving object and v is its speed. For a collision, we need to express the energy before and after the collision. If nothing falls off the car, its mass won't change, so we need a variable, vi, to hold the value for the initial speed.

Now we can set the initial kinetic energy of the car at mvi2/2. This initial kinetic energy is equal to the final kinetic energy of the car, mv2/2, plus that absorbed by the median. To make the model simple, a loss factor, (1-p), can be multiplied by the initial kinetic energy to get the final. The relation looks like,

(1-p)mvi2/2 = mv2/2 for p less than or equal to 1.

Divide through by m/2 and take the square root of both sides and take vi2 out of the square root,

v = vi\[1-p]

This simple result shows that the speed of the car after the collision varies directly with the initial speed, but decays with the squareroot of the proportion of energy kept in the collision. So if the car loses 5% of its energy, it retains \[1-0.05] = \[0.95] = 0.97 or 97% of its initial speed.

As the car drifts, its kinetic energy is not aligned with direction of the track anymore. A model can be made by splitting the velocity, v, into two parts at right angles. One part is along the track, vt, and the other across the track, vc. The initial linear kinetic energy can then be written m(vt2+vc2)/2 using the pathagorus theorem for vi2 = vt2 + vc2. The cross-track part mvc2/2 is wasted motion.

From trigonometry and the right triangle formed by the two directions of the speeds, we can write,

vt = vicosO

vc = visinO

Using the small angle approximation (sinO=O), vc = viO. In order for this to work, the collision or drift angle O must be measured in radians and it must be less than 0.34 radians (about 20 degrees). So the wasted energy, not accounting for the collision of the tire and median, is mvi2O2/2.

To get p, the proportion of lost energy, we divide the lost energy by the initial energy to get,

p = (mvi2O2/2)/(mvi2/2) = O2, the square of the collision angle in radians.

If the car continues to rub the median, total friction is increased. Rubbing is sliding friction and depends on the force between the wheel and median NOT how fast the wheel is spinning.

If the car ricochets back across the track, the track has returned some of the car's cross-track energy but the wheels that hit the median will likely rub the car body and the opposite wheels their axle hubs. Because the car rebounded, it may settle in the lane or drift again to repeat grazing the median. The drift angle, O, may change with each collision. If it does not, one may estimate the increased distance the car travels as L/cosO - L in inches.

What is the greatest angle your car can make with the median?

In pinewood racing, the angle at which your car hits the lane median and how fast it is going at that moment are the two most important factors in guessing how bad the collision will be. A good start is to find out what the greatest collision angle can be. This greatest collision angle will tell us how much energy can be transferred to the lane median (remember p = O2?). Any loss of energy means a slower race.

Assume one wheel touches the median in front and one in the rear on opposite sides of the car. For an AWANA kit, the axles are B = 3.875 inches apart (B stands for the wheel base). When we looked at wheel alignment, we found that a Grand Prix car has only about c = 0.375 inches of "play" across the median. These two distances give us this greatest collision angle, O, that's possible.

From the right triangle formed by the lines measured to get c and B we get,

sinO = c/B = 0.375/3.875 = 0.09677 so O = arcsin(0.09677) = 5.55 degrees.

Of course, if the wheel base, B, was shorter, the angle would be greater and the bad effects of drift would increase. On the other hand, lengthening the wheel base to B = 5.8 inches, the maximum for an AWANA kit car, O is,

O = arcsin(P/B) = arcsin(0.375/5.8) = arcsin(0.065) = 3.71 degrees

That's a third smaller than the standard kit. You can work out how much of an advantage a longer wheel base might be from the other results on this page.

How many median collisions are possible?

Suppose the car ricochets back and forth down the track with two wheels touching opposite sides of the median at this angle at every bounce. It's "drift" would be the worst it could be. The wheels would follow a saw-tooth course and the car would travel forward BcosO inches with each impact. The most impacts over a 30 foot (360 inches) coarse using the kit wheel base would be

360/BcosO = 360/(3.875x0.9953) = 93 impacts!

For the longest wheel base, 5.8 inches, and its maximum collision angle, 3.71 degrees,

360/BcosO = 360/(5.8x0.9979) = 62 impacts, a third fewer.

What is the longest trajectory possible?

Suppose a car veers from one side to the other at the maximum angle, O = 5.5 degrees, bumping both wheels on the same side of the car momentarily against the median. Repeating this behavior would bring the center of mass back and forth across the track 0.75 inches each cycle.

There would still be 93 median impacts possible for this twisting car. As above, the car's center of mass travels BcosO (=3.857) inches forward, but now 0.75 inches sideways as well. The total distance is

\[(0.75)2+(BcosO)2] = 3.93 inches.

Over 93 cycles, that's 365.5 inches. 5.5 inches longer than a 360 inch trajectory for a straight path. Considering that the possible advantage of other factors like lifting wheels, being under weight and weight placement can be much less than 5.5 inches, this distance is significant. At a goodly speed of 160 inches per second, the zigzagging alone (without accounting for the energy loss of impacts) would add about 0.034 seconds to the car's time.