How To Build a Pinewood Derby Car/Physics

This module has some simple physics and, at the end, some chemistry of the pinewood derby. A pinewood derby car converts gravitational potential energy to translational kinetic energy (and resulting velocity), rotational kinetic energy in the spinning wheels, and heat from friction. The fastest car is the one that starts with the most gravitational potential energy (weight in the back), ends up with the least rotational kinetic energy (light wheels with one lifted) and loses the least energy through friction (proper lubrication and alignment). Pinewood derby cars are a five ounce laboratory of practical physics, chemistry and engineering.

The Pinewood derby car problem is treated in two different ways below: 1) using the equations of motion and 2) using energy. Both are equally valid and arrive at the same result.

Equations of motion treatment
We'll build a physical model that considers all of the predominant forces acting on the Pinewood Derby car. First and foremost is the force of gravity, which is the only (allowed) force to accelerate the car down the ramp. Next, we'll consider the frictional forces of the axles on the wheels and the wheel treads on the track. Finally, we'll consider air drag and how it affects the motion of the car.

Throughout this section, we'll consider an idealized car and track. The car has a mass M of 5 ounces (141.7 grams), a length l of 7 inches and wheels each having mass m of 2.6 grams and radius r of 0.5". The track has overall length L and consists of a ramp of length $$L_r$$ at an angle of incline of $$\theta$$ followed by a flat of length $$L_f$$ to the finish. Given this geometry, an expression for the starting height is $$H=L_r\sin\theta$$. To keep things simple, we will not concern ourselves with the transition between the ramp and the flat at this time. Values for a typical Pinewood Derby track are $$L_r = $$ 8' (2.44 m), $$L_f = $$ 24' (7.32 m), $$H = $$ 4' (1.22 m), and $$\theta$$ = 30$$^{\circ}$$.

It will also be helpful to consider the equations of motion for an object with no external forces. This will be the case for the car on the flat part of the track when friction and wheel rotation are neglected. In this case the equations of motion are:


 * $$ a(t) = 0\, ;\, v(t) = v_0\, ;\, d(t) = v_0t + d_0\,$$,

where a(t) is the acceleration, v(t) is the velocity and d(t) is the distance at time t. The initial distance and velocity at time t=0 are given by $$d_0$$ and $$v_0$$, respectively.

Force of gravity on a falling object
First consider the most basic case of a falling object of mass M (an apple?). The force of gravity is given by the following expression:


 * $$ F_g = Mg\,$$,

where g is the acceleration due to gravity (9.807 m/s2 = 32.17 ft/s2 at the earth's surface). This force is directed downward and results in the object accelerating toward earth.

The equations of motion (neglecting frictional forces) for this object are as follows:


 * $$ a(t) = g\, ;\, v(t) = gt + v_0\, ;\, d(t) = \frac{1}{2}gt^2 + v_0t + d_0\,$$.

An object falling the height of 4' (1.22 m) would take 0.499 seconds and reach a maximum velocity of 4.892 m/s. If this object were then instantaneously turned to slide along the 24' flat track with no friction, the time would be an additional 1.496 seconds for a total time of 1.995 seconds.

Force of gravity on a Pinewood derby car
The Pinewood Derby car does not fall straight downward but rather proceeds down a ramp with an angle of incline $$\theta$$. Only the component of the gravitational force along the track surface will result in the acceleration of the car. The resulting accelerating force is:


 * $$ F_a = Mg \, \sin\theta\,$$.

The car only accelerates when on the ramp of angle $$\theta$$. On the flat part of the track where $$\theta=0$$ this accelerating force becomes zero. This can be thought of as acceleration with a reduced gravity $$g_\theta$$:


 * $$ g_\theta = g \, \sin\theta\,$$.

This reduced gravity can be used in the above equations of motion to solve for the car. In a Pinewood Derby on the initial slope, $$v_0=0$$ and $$d_0=0$$ so (neglecting friction and wheel rotation) these equations become:


 * $$ a(t) = g_\theta = g \, \sin\theta\,;\, v(t) = g_\theta t\,;\, d(t) = \cfrac{1}{2}g_\theta t^2\,$$.

The time $$T_r$$ to roll down the ramp is given by:


 * $$ d(T_r) = L_r = \cfrac{1}{2}g_\theta T_r^2\,$$,

Solving for $$T_r$$:


 * $$ T_r = \sqrt{\cfrac{2L_r}{g_\theta}} = \sqrt{\cfrac{2L_r}{g \sin\theta}}\,$$.

The maximum speed at the bottom of the ramp is:


 * $$ v(T_r) = v_{max} = g_\theta T_r = \sqrt{2gL_r\sin\theta} = \sqrt{2gH}\,$$.

For the idealized ramp, $$T_r$$ = 0.998 sec and $$v_{max}$$ = 4.892 m/s. This has taken twice the time of the falling case above, but the same maximum velocity was achieved!

The equations of motion for the car on the flat part of the track ($$\theta=0$$) for $$t>T_r$$ are given by the following:


 * $$ a(t) = 0\,;\, v(t) = v(T_r) = v_{max}\,;\, d(t) = L_r + v_{max}(t-T_r)\,;\, t>T_r\,$$.

The time $$T_f$$ to finish the flat portion of the race is given by:


 * $$ d(T_r+T_f) = L_r+L_f = L_r + v_{max}T_f\,$$.

Solving for $$T_f$$:


 * $$ T_f = \cfrac{L_f}{v_{max}}\,$$.

The total time to finish the race for this idealized car is:


 * $$T = T_r+T_f = \cfrac{v_{max}}{g_\theta} + \cfrac{L_f}{v_{max}}\,$$.

The flat-track time is 1.4953 seconds for a total time of 2.4933 seconds. Since the same maximum velocity as the falling case was achieved, the (frictionless) flat time is identical. The total time includes the extra half second it took to move down the ramp.

Energy conservation
Some key understanding of the physics of the Pinewood Derby car can be gained through energy considerations. Through this deliberation, we can consider the potential energy available in the system as well as the linear kinetic energy and the rotational kinetic energy in wheels. Above we considered only the linear motion as the car moved down the ramp and across the track. In fact, the rotation of the wheels requires energy and thereby removes energy from the system that would otherwise be available to accelerate the car. To account for the rotation of the wheels, we will consider the energy of the system.

We'll start out with the potential energy in the system, then move on to the simple linear case, and finally add the rotation of the wheels.

Potential energy
The energy that makes a pinewood derby car roll down the ramp is potential energy (symbol $$U_g$$). At the bottom of the ramp, this energy is converted into kinetic energy (symbol $$E_k\,$$).

The potential energy can be calculated using the formula


 * $$U_g = m g h \,$$.

where m is the mass of the car, $$g$$ is the acceleration due to gravity (9.807 m/s2 = 32.17 ft/s2 at the earth's surface) and h is the height of the car on the starting ramp (about 4 feet = 1.22 meters). The potential energy stored in a 5 ounce pinewood derby car is about 1.7 kg m2/s2= 1.7 J. The quantity 1.7 joule of energy is the same as 1.7 watts of power applied for one second.

Translational kinetic energy
The kinetic energy can be calculated using the formula


 * $$E_k = \frac{1}{2}mv^2$$.

where v is the velocity (speed) of the car, measured in meters per second (in SI units). If all of the potential energy is converted into kinetic energy, then


 * $$U_g = E_k \,$$

and, substituting from above,


 * $$m g h = \frac{1}{2}mv^2 \,$$

since the mass is on both sides of the equation, you can divide both sides by $$m$$ and cancel (remove) the mass from the equation.


 * $$g h = \frac{1}{2}v^2 \,$$

and, rearranging, the velocity is found from


 * $$v = \sqrt{2 g h} \,$$

The velocity doesn't depend on the mass, just as Galileo demonstrated at the Leaning Tower of Pisa and, during the Apollo 15 landing, David Scott demonstrated on the moon. The velocity of any object falling from a height of 1.219 m is 4.890 m/s (11 MPH).

Rotational kinetic energy
Moving forward isn't the only kind of kinetic energy; it also takes energy to spin the wheels. The rotational kinetic energy of each wheel is given by


 * $$E_r = \frac{1}{2}I\omega^2$$,

where I is the moment of inertia of a wheel and ω is the angular velocity. Since we consider only a wheel that does not slip on the track, the angular velocity is related to the linear velocity through
 * $$\omega = v / r\,$$,

where r is the radius of a wheel. The moment of inertia for a ring of radius r and mass $$m'$$ is


 * $$I_{ring} = m'r^2\,$$

and for a disk, it is


 * $$I_{disk} = \frac{1}{2}m'r^2\,$$

For a Pinewood Derby wheel, the moment of inertia can be approximated by


 * $$I = 0.58\,m r^2\,$$,

where m is the mass of one wheel and r is the radius of the wheel.

The rotational kinetic energy for N wheels is therefore


 * $$E_{r} = \frac{N}{2} (0.58\,m r^2)\omega^2 = \frac{N}{2} (0.58\,m)v\,^2\,$$.

This turns out to be 0.094 J of energy stored in four wheels at the car's top speed of 4.75 m/s (see the table below). Compare that to the 1.7 J gravitational potential energy from above. Almost 6 % of that energy is stored in the spinning wheels and about 94 % is available for the car's forward velocity.

If we again convert all potential energy into kinetic energy, we can find the maximum kinetic energy and velocity:


 * $$U_g = E_k + E_r \, \Longrightarrow \, M g H = \frac{1}{2} \left( Mv_{max}^2 +  0.58N\,mv_{max}\,^2 \right)\,$$.

Solving for the velocity,


 * $$v_{max} = \sqrt {\cfrac {2gH} {1 + 0.58N \cfrac {m}{M} } } \, = \sqrt {\cfrac{2g H}{\alpha_R}} \, $$

Note that this result is again independent of the track geometry. The rotational factor, which has the effect of reducing the gravity, is expressed as follows:


 * $$\alpha_R = 1+0.58N \cfrac{m}{M} \, $$.

This factor can be used with the gravity in the above equations of motion. It must be combined with the reduced gravity due to the slope of the track. The resultant effective gravity is


 * $$ g'_\theta = \cfrac{g\sin\theta}{\alpha_R} \, $$.

This reduced gravity with wheel rotation can be substituted into the equations of motion to calculate the car's distance, velocity and acceleration at any point. the equations of motion become:


 * $$ a(t) = g'_\theta = \cfrac{g\sin\theta}{\alpha_R} \,;\, v(t) = g'_\theta t\,;\, d(t) = \cfrac{1}{2}g'_\theta t^2\,$$.

Now it is clear why the heaviest possible car is the fastest: for large M, $$\alpha_R$$ approaches 1 and the velocity is the same as the above case for the frictionless block: 4.9 m/s. With standard 3.6 g wheels, a 141.7 g (5 ounce) car has a velocity of 4.75 m/s at the bottom of the track. A 113.4 g (4 ounce) car has a velocity of 4.72 m/s.

The advantage of light wheels is also clear. The slowest car is one with the most mass in the wheels. If $$m=M$$ (four 35 g wheels rolling down the track!) the velocity is 2.7 m/s. With 3.6 g wheels, the velocity is 4.75 m/s; with 1 g light wheels, the velocity is 4.85 m/s. Another way to look at it is that a light wheel car has less than 2 % of its kinetic energy in the wheels while the standard wheel car has almost 6 %.

Raising one wheel reduces the rotational energy as long as that wheel doesn't touch the ground. The velocity of a three wheel (3.6 g) car is 4.78 m/s, while a raised light wheel car can reach 4.86 m/s.

An idealized track
Now that we know how fast an idealized car can go, let's let it run down an idealized track. A typical pinewood derby track is 32 feet long with 4 feet for the starting gate and arresting gear, leaving 8.534 m for travel. The first part of the track has a slope of about 20°, a curved transition section, followed by a flat section to the finish. The track can be approximated by a straight ramp section with length $$d_1$$ followed by a flat section with length $$d_2$$. Times calculated using the straight track approximation differ by less than 0.001 s compared to those obtained using a curved track and numerical integration. With the straight track and 20° slope, $$d_1$$ = 3.564 m and $$d_2$$ = 4.970 m. The car travels down the sloped section $$d_1$$ starting at zero velocity and accelerating to a velocity $$v\,$$ with an average speed of $$v/2\,$$. It then travels the distance $$d_2$$ at a velocity $$v\,$$. The total time is then


 * $$t= 2\frac{d_1}{v} + \frac{d_2}{v} $$

or 12.098/v seconds, if the velocity is specified in meters per second. Run times under different conditions are given in the table below.

!Conditions ||$$h\,(m)$$ || $$m\,(g)$$ || $$m'\,(g)$$ || $$v\,(m/s)$$ || $$t\,(s)$$
 * + Car Speeds and Times
 * Frictionless Block ||  1.219  ||  any || 0 || 4.89 || 2.47
 * Stock Wheels ||  1.219  ||  141.7 || 3.6 || 4.75 || 2.55
 * Light (4 oz) Car ||  1.219  ||  113.4 || 3.6 || 4.72 || 2.56
 * Light Wheels ||  1.219  ||  141.7 || 1.0 || 4.85 || 2.49
 * One Raised Wheel ||  1.219  ||  141.7 || 2.7 || 4.78 || 2.53
 * Light Wheels, One Raised ||  1.219  ||  141.7 || 0.75 || 4.86 || 2.49
 * Rear Bias (Stock Wheels) ||  1.243  ||  141.7 || 3.6 || 4.80 || 2.52
 * }
 * One Raised Wheel ||  1.219  ||  141.7 || 2.7 || 4.78 || 2.53
 * Light Wheels, One Raised ||  1.219  ||  141.7 || 0.75 || 4.86 || 2.49
 * Rear Bias (Stock Wheels) ||  1.243  ||  141.7 || 3.6 || 4.80 || 2.52
 * }
 * Rear Bias (Stock Wheels) ||  1.243  ||  141.7 || 3.6 || 4.80 || 2.52
 * }
 * }

Center of mass
Up to this point, we have approximated the car as a point mass that starts 1.219 m above the finish and winds up at 0 m. As anyone familiar with pinewood derby knows, it is usually best to keep the center of mass as far back as possible, in part to increase the gravitational potential energy. How much faster is a car with the center of mass shifted to the back? Consider two cars, Car A with the center of mass at the center of the block and Car B with the center of mass 25.4 mm (one inch) in front of the rear axle in an extended wheelbase configuration with the axle 15.9 mm (5/8") from the rear of the block. The center of mass of the Car B is 47.6 mm behind Car A and 23.8 mm higher on a 30° slope. Car A starts at 1.219 m and (as shown above) has a maximum velocity of 4.75 m/s and 2.55 s time; Car B starts at 1.243 m and has a maximum velocity of 4.80 m/s and a 2.52 s time.

Friction forces
Friction is the force that opposes the motion of two objects in contact. In general, there are two types of frictional forces - static and kinetic. Static friction is a force that resists motion. Kinetic friction, also called sliding friction, is a force exerted between two objects in motion relative to one another. The primary types of friction associated with a Pinewood Derby car are the kinetic friction of the axles on the wheels and the static friction of the wheels rolling on the track. Both of these will be discussed in further detail below. Air Drag is considered a special topic and is discussed in detail below.

Other types of friction, including contact between the wheel inner hub and car body, the wheel outer hub and the axle head, the wheel and the track guide rail, can be greatly minimized or completely eliminated through proper car design and tuning.

Sliding (axle) friction
Sliding friction occurs whenever two surfaces are in contact and are in motion relative to one another. For the Pinewood Derby car, the primary source of contact friction is axle on wheel. This friction occurs within the car "system" and always occurs with the metal axle on the plastic wheel. The friction force on the axle and wheel ($$F'_f$$) is as follows:


 * $$F'_f = \mu N_f \,$$,

where $$N_f$$ is the normal force (perpendicular to the contact surface) exerted between the surfaces and $$\mu$$ is the coefficient of friction. The coefficient of friction is a property of the two materials in contact. For a Pinewood Derby car, the value of $$\mu$$ is about 0.24 for smooth plastic on metal and can be reduced to 0.10 or less if the axle is polished and a lubricant, such as graphite, is used.

The normal force is the force of the body mass on the axles. On a track with slope of $$\theta$$, the total normal force of all load-bearing axles is


 * $$N_f = (M-Nm)g \cos\theta \,$$,

where M is the mass of the car, m is the mass of a wheel, and N is the number of load-bearing wheels. $$M-Nm$$ is 127.3 g for a car at maximum weight with four 3.6 g load-bearing wheels.

The force of friction relative to the track is reduced by the mechanical advantage, which is the ratio of the wheel radius to the wheel bore hole radius. For every rotation of the wheel, the car travels one wheel circumference: 95.0 mm. The axle travels only 7.19 mm around the wheel bore hole circumference per rotation. The wheel-to-axle friction is then


 * $$F_f = \mu \left( \cfrac{r}{r_B} \right) (M-Nm)g  \cos\theta = \mu' (M-Nm) g \, $$,

where $$r_B$$ is the radius of the wheel bore hole and the adjusted coefficient of friction $$\mu'$$ is:


 * $$\mu' = \mu \left( \frac{r}{r_B} \right) \, $$.

The linear acceleration due to this force, including the effect of wheel rotation, is opposite the direction of motion and is


 * $$a'_f = \cfrac{\mu' g \left( 1-\cfrac{Nm} {M} \right) \cos\theta}{\alpha_R} = \cfrac{\mu' g \alpha_m \cos\theta}{\alpha_R}\,$$,

where the factor for wheel mass $$\alpha_w$$ has been defined:


 * $$ \alpha_w = 1- \frac{Nm}{M} \,$$.

Because this is a constant acceleration, it can be written as an effective reduced gravity in the equations of motion. The reduced gravity for all geometric, rotational and sliding frictional effects thus far is


 * $$g'_{\theta f} = g \cfrac{\sin\theta - \mu' \alpha_w \cos\theta}{\alpha_R} \, $$.

Using the above values, the wheel-to-axle frictional force is 0.00946 N.

Note that the wheel-to-axle friction for a light wheel car is 8% larger than for the stock wheel car due to the greater mass of the body (137.7 g body plus four 1.0 g wheels). However, the frictional force can be cut in half using needle axles, which are allowed in some open class races. A 1 mm needle axle has a mechanical advantage of 30 and a frictional force of 0.00453 N (assuming light wheels).

Another result is that a raised wheel does nothing to reduce wheel-to-axle friction. A three wheel car simply distributes the weight among three axles rather than four and the frictional force is the same. All of the raised wheel advantage stems from rotational kinetic energy.

Rolling friction
There are two kinds of friction: static and kinetic. Static friction is the resistance of an object at rest, for example, a heavy box at rest on concrete floor. Rolling resistance is the friction between a rolling wheel and surface. It is a static friction because, even though the wheel is moving, it is not sliding against the surface. Kinetic friction is the force that opposes the motion of a moving object. Using the box example, this would be the force that must be overcome as you slide the box across the floor. In the wheel example, rolling resistance is static, while skidding or doing a burnout involves kinetic friction. In this sense, rolling resistance is for the most part a "good" friction for a pinewood derby car because it does very little work and because it helps to convert what would otherwise be skidding friction into the significantly lower friction between the wheel and axle.

The force that results from rolling resistance is given by:


 * $$F_{r} = C_{rr} N \, $$

where Crr is the coefficient of rolling friction (CRF), and $$N$$ is the normal force (perpendicular to the track). The CRF value for a low rolling resistance tire is about 0.002 (see the rolling resistance article).

On a track with a slope of $$\theta$$, the normal force is perpendicular to the track and is given by


 * $$ N_\theta = Mg \cos\theta \,$$.

The frictional force from rolling resistance is then


 * $$ F_{\theta r} = C_{rr} Mg \cos\theta \,$$.

This force acts in the opposite direction of the motion of the car. The linear acceleration due to this force is


 * $$ a'_{\theta rr} = g' C_{rr} \cos\theta = \cfrac{g C_{rr} \cos\theta}{\alpha_R} \,$$,

resulting in an effective rolling resistance reduced gravity of


 * $$ g'_{\theta rr} = g \cfrac{\sin\theta - C_{rr} \cos\theta }{\alpha_R} \,$$,

The rolling resistance reduced gravity can be substituted into the equations of motion to solve for the car distance, velocity and acceleration at any point in time.

On a flat section of track, $$\theta=0$$ and


 * $$F_{r} = C_{rr} Mg = 0.002 \times 0.1417\,kg \times 9.807\,m/s^2 = 0.0028\,N \ $$.

The wheel-to-axle friction and rolling resistance are of comparable magnitude. From the above section, the rolling resistance was estimated to be about 0.003 N compared to the wheel-to-axle friction estimated above to be about three times larger.

Car modifications that can reduce the CRF can reduce the rolling resistance by the same factor. For example, a "V" or "H" tread pattern limits the contact area between the wheel and track and thereby reduce the CRF and rolling resistance. Note that a raised wheel does not reduce the overall rolling resistance since the full weight of the car is distributed over the remaining three wheels and increases their rolling resistance. However, the rolling resistance is greater for the rear wheels in a rear biased car since they bear a greater fraction of the weight.

Motion with friction
The "all-in" reduced gravity that includes the effects of track geometry, wheel rotation, sliding friction and rolling friction is:


 * $$g'_{\theta fr} = \cfrac{g_{\theta fr}}{\alpha_R} = g \cfrac{\sin\theta- \mu' \alpha_w \cos\theta - C_{rr} \cos\theta }{\alpha_R}  \, $$.

Note that the reduced gravity can be negative. In particular, it is negative on the flat track when $$\theta=0$$. Also of interest is that there will be some minimum value for $$\theta$$, below which the car will never overcome frictional forces and begin rolling - even on the sloped track!

The equations of motion become:


 * $$ a(t) = g'_{\theta fr} \,;\, v(t) = g'_{\theta fr}t+v_0\,;\, d(t) = \cfrac{1}{2}g'_{\theta fr}t^2+v_0t+d_0\,$$.

At the start of the race, $$v_0=0$$ and $$d_0=0$$. At the start of the flat section, simply reset the time to zero and use $$\theta=0$$, $$v_0=v_{max}$$ (you must calculate $$v_{max}$$ from these equations) and $$d_0=L_r$$.

Measuring friction
The static and dynamic friction can be measured using the apparatus in Figure 1. From Newton's Second Law of Motion we know that the forces balance, therefore the normal force balances the gravitational force on the car




 * $$ N_f = mg \,$$.

At the point where the mass M just balances the static frictional force


 * $$ \mu N_f = mg \,$$

and


 * $$ \mu = M/m\,$$.



The reader can verify that, in the inclined case of Figure 2, the coefficient of friction is given by


 * $$\mu = \frac{ \left ( \frac{M}{m} - \sin(\theta) \right ) } {\cos(\theta)} \,$$.

The kinetic friction can be determined using either of the above apparatus, but the procedure takes a bit more patience. Add or subtract from the mass M until the car moves at a constant velocity when pushed. At this point, the gravitational force is equal to the kinetic frictional force. The trick is that the car can neither accelerate nor decelerate; we want F=ma=0, but this time when the car is in motion.

Air drag
The frictional force of a gas or liquid on a body is called drag and the drag force Fd can be calculated from


 * $$ \mathbf{F}_d= {1 \over 2} \rho A C_d v^2 = b v^2$$

where ρ is the air density (about 1.2 kg/m3), v is the velocity of the air relative to the body, A is the frontal area of the body, and Cd is the drag coefficient, which depends on the shape and surface of the vehicle (but about 0.4 for a typical car). The constant b is defined above as a convention. For a Pinewood Derby car, A is dependent on the car design - an uncut block is 0.0014 m2. Using the 4.75 m/s for a 5 ounce car with standard wheels gives a drag force of 0.0081 N. Using $$F=ma$$, we get an acceleration of -0.057 kgm/$$s^2$$, about 0.6% of the gravitational force, but still enough to slow the car from 4.75 m/s to 4.69 m/s on the flat section of the track alone. A numerical integration with a similar track resulted in an a run that was 1.8% slower. In practice, all cars experience some air drag, but a low and sleek design may have an advantage of a few hundredths of a second against a less aerodynamic car.

At the sizes and speeds of pinewood cars, the viscosity of air does not matter at all; the only important air resistance force is the drag force, proportional to the density of the air and the speed squared. The drag force is a kind of ram pressure.

To get to the equations of motion for a vehicle subject to ram-pressure drag, we consider the car on a sloped track with rotating wheels and friction forces and add to this model the air drag force. The acceleration of the car based on the balance of forces is the following:


 * $$ \frac{\mathbf{F}_d}{M}= a(t) = \cfrac{g_{\theta fr} - \frac{b}{M} v^2}{\alpha_R} $$,

Recall that the force is related to acceleration through Newton's second law of motion and the acceleration is the derivative of velocity. Thus, an equation for the velocity of motion can be written:

This is a non-linear differential equation for the velocity. It's solution is not straightforward but is tractable. The equations of motion are:


 * $$ a(t) = g'_{\theta fr} \left[ 1 - \tanh^2 \left( t \sqrt{\cfrac{g'_{\theta fr} b}{\alpha_R M}} \right) \right] \, $$,


 * $$ v(t) = \sqrt{\frac{g'_{\theta fr} \alpha_R M}{b}} \tanh \left( t \sqrt{\cfrac{g'_{\theta fr} b}{\alpha_R M}} \right) \, $$,


 * $$ d(t) = \frac{\alpha_R M}{b} \ln \left[ \cosh \left( t \sqrt{\cfrac{g'_{\theta fr} b}{\alpha_R M}} \right) \right] \, $$.

Tribology
Tribology is the study of friction, lubrication and wear. Since friction is the force that slows a Pinewood Derby car and lubrication reduces friction, tribology is an essential science of the sport. Wear is not an important consideration since, as we have seen above, the forces are low. The lubricant must reduce the friction between the wheel and axle and also be compatible with the polystyrene wheels. Expanded polystyrene is known as styrofoam and makes a great cup for hot cocoa, but is a poor container for an organic solvent such as gasoline or naptha (ask your Dad about the old auto shop trick). Pure polystyrene can be crosslinked to form a copolymer with polybutadiene that is called high-impact polystyrene (HIPS) or high-impact plastic. HIPS is excellent for injection molding, so it is no coincidence that it is the material used for Pinewood Derby car wheels. Nonetheless, care must be taken to keep inappropriate solvents from lubricants, paints, or cleaners away from the wheels. Another consideration for lubrication is the possibility of fouling the track with excess lubricant, especially liquids. Some race rules permit only dry powder lubricants for this reason.

Graphite

Graphite, like diamond is one of the several forms of carbon. Unlike diamond, graphite is a poor gemstone, but it is an excellent lubricant. This is because the chemical bonds in diamond form a three dimensional structure, while the bonds in graphite form a two dimensional structure. The two dimensional sheets of graphite slide easily over each other, giving graphite powder its lubricating properties.

Molybdenum Disulfide

Molybdenum disulfide has the chemical formula MoS2 and is similar in chemical structure to graphite in that it forms two-dimensional layers that slide upon each other. In pinewood derby applications, molybdenum disulfide powder is typically mixed with graphite powder.

Teflon

Teflon is the DuPont brand name for polytetrafluoroethylene (PTFE), which is a synthetic fluoropolymer (a fluorine containing polymer). Teflon has the lowest coefficient of friction of any known solid material. It is used in powdered form as a Pinewood Derby lubricant.