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Pinewood derby: What a drag! January 16, 2012

Posted by mareserinitatis in engineering, physics, science, younger son.
Tags: , , , , , , , ,

My husband son competed yesterday in my son’s his scouting group’s pinewood derby race.  For those of you who have never had a kid in cub/boy scouts, they hand out these blocks of wood that you get to make into a car.  The idea isn’t to win the race: it’s supposed to be that dads and their boys spend some time bonding over manly things like woodworking.

(One year, Mike was out of town during this whole thing, so I got to be manly and help the older boy build his car.  All I can say is that Dremel tools are awesome.)

If you look on the web, you’ll find a lot of advice on how to prep pinewood derby cars and make them faster.  One thing that consistently bugs me is that one should pay attention to aerodynamics of the car and give it a low profile.

This makes me nuts.

To understand the following, you might want to know what a track looks like.  So here you go:

Let’s start out with the specifications.  Most pinewood derby tracks have a height of about 4 feet and a length of 32 feet.  I also will note that the ones we’ve raced on were wood, not aluminum.

Most of the pinewood derby cars I watched made it the whole 32 feet, though not all did.  So let’s say that, on average, they travel 32 feet.  The *fastest* ones traveled at an average speed of 10 ft/s (or 3 m/s).  The maximum they can weight 5 oz. or 140 g.

What we’re going to do here is look at how much energy the system puts into overcoming air resistance versus friction.  It’s very hard to figure out exactly how much goes into friction simply using equations, so we’re going to figure out the total energy and the energy lost to drag forces.  Once we have those two quantities, we can subtract the drag forces from the total energy and assume that the difference is equal to the frictional losses.  Finally, we can compare the drag and frictional losses.

First things first: what is our total energy?  It starts out entirely as potential energy as the cars are placed at rest on the top of the ramp.  Potential energy is, fortunately, very easy to calculate.  It is simply the product of the height of the object, its mass, and the gravitational acceleration.  In other words,

We know the mass of the car (0.14 kg), the height is approximately 1.22 m, and the graviational acceleration is 9.8 m/s2.  This results in a total potential energy of approximately 1.67 J.

The potential energy is equal to the total energy in the system since the cars start with no other kind of energy.  In a frictionless and drag-free system, all of this energy would be converted to kinetic energy and the cars would drive forever at the same speed once they reached the bottom of the ramp.  Obviously, however, that’s not what happens.  Eventually, all of the energy is converted to friction and drag, and the cars stop.

Now we need to determine the drag on the cars.  The drag equation is:

The drag force is proportional, therefore, to the density of the fluid (ρ), velocity (v), drag coefficient (CD), and cross-sectional area (A).

The density of the fluid (air) is approximately 1.2754 kg/m3, and the velocity is 3 m/s.  The cross-sectional area of the car, at maximum, is 2.75 x 3 in.  In real *ahem* units, this is 0.00532257 m2.

The drag coefficient for a long cylinder, according to Wikipedia, is 0.82.  Given the cars are all sorts of different shapes, I think this would probably be the closest approximation, although for some cars, this will be high.

All of this put together gives us a force of 0.025 N.  Over a distance of 32 feet (or 9.75 m), this gives an energy of 0.24 J.

If we assume that all of the potential energy is converted to either drag or friction, that means 0.24 J is converted to drag while 1.43 J goes into friction.  This means that roughly 14% of your energy losses are due to drag while the other 86% are due to friction.

My advice, therefore, is to not worry so much about drag and the aerodynamics of the car and to worry far more about the ways to reduce friction.

What should you do to help your pinewood derby car to be fast?  There are three things:

1) You want to maximize your potential energy, so being as close to the 140 gram maximum weight is good.  You can get all sorts of weights commercially to assist with this.  Some people have argued that putting the weights near the back of the vehicle helps even more.  I wouldn’t doubt this as you’re putting the weight at greater height, giving the car more initial energy to start with.

2) The biggest issue is reducing the friction between the wheel and the axle, and there are a few easy ways to do that: sanding and polishing the axles as well as using graphite or teflon as a lubricant.  (Our troupe doesn’t allow graphite because it marks the floors, but teflon is allowed.)

3) Reduce the friction between the car and track.  The best way to do this is to make sure your wheel alignment is as straight as possible as this will keep it from rubbing against the center of the track.

The winner of our pinewood derby actually had a very blocky car designed to look like a platypus.  I’m guessing the winner didn’t spend much time focusing on aerodynamics and instead chose to minimize their frictional losses.



1. scott - January 21, 2012

i have read that ‘lifting’ one of the front wheels off of the track ( 1/16th inch ) will decrease the friction…any comments


mareserinitatis - January 21, 2012

It hypothetically shouldn’t help at all. The friction will be proportional to the weight of the car, not the area in contact with the surface. Therefore, by making only three of the wheels load-bearing, you simply increase the amount of friction on each wheel but do not change the overall amount of friction.


Deej - February 7, 2013

You failed to factor in rotational inertia. If the 4th wheel never spins ( very important assumption), then there is a significant difference. The 3 wheel car will reach its top speed sooner, and cross the finish line ahead of the 4 wheel car. If the track is long enough, then no difference as the 4 wheel car can convert the stored rotational energy of the fourth wheel back into forward motion.

You are correct that friction matters most. But all things being equal, the margin of victory can easily be found in the 14 % loss from aerodynamics.


2. Ed Ciereszynski - January 23, 2012

Interesting and very accurate scientific formulas, reminds me of my college physics course. I failed it miserably.

We, my son and I, just did our first pinewood derby. He took first place in his den, first place in the pack, and fastest time and speed in the entire field. His car won every heat entered. Most of the cars and scouts had done this before. These were experienced boys and dads.

I did not spend a great deal of time on the project. I didn’t use any sophisticated tools or jigs that are available on the web. I focused on a very aerodynamic outline with significant weight behind and over the rear wheels. The final paint was a mirror finish with a clear sealer and a buffed out paste wax finish. Not all the weight was in the rear. I proportionately distributed weight to the front wheels. We weighed in at .2 oz under 5 ounces and chose to stay with it.

I really believe that the way the weight is distributed is critical. the shape as well as the finish are also key contributors.

Don’t bother with the weight forward teardrop shape. Not a single one of these was a winner.

I did no more than the basics when it came to the axles and wheels. I started out with the parts from the basic kit. I smoothed out the axles, took out the visible defects and smoothed out the tire treads with sandpaper. I made no big effort to place the axles, or square them up. If anything, when I tried out the car on my hardwood floor, I was shocked by what I thought was the poor performance. I really thought we didn’t have a chance.

I lubed the axles with some commercial grade graphite I bought at Walmart for $1.00

So now we go on to the regionals. I have no idea what were up against. I probably need to do some dissection and examination
of our car.


Paul - August 30, 2016

One possible caveat: if the car is still moving at the end of the track then it has some kinetic energy remaining so not all of the potential energy goes into drag and friction. Some is left as kinetic energy.


3. Farhad Mirbod - February 19, 2017

Your post provided some helpful information in thinking through my 6th graders science project, thank you. However, I’m in agreement with Paul regarding your conclusion summarized in this statement:

If we assume all of the potential energy is converted to either drag or friction, that means 0.24 J is converted to drag while 1.43 J goes into friction. This means that roughly 14% of your energy losses are due to drag while the other 86% are due to friction.

Since the car has velocity at the finish, all the energy is not only drag and friction but also remaining kinetic energy. I think your derivation of the drag energy is good but you would need to estimate energy lost due to friction (i.e. the wheels) to be able to make a final conclusion. Perhaps calculating the actual/instantaneous kinetic energy at the finish using KE=1/2*m*v^2 (would need to estimate velocity at the finish somehow, radar gun?) then subtracting it and your drag energy from the total potential energy would result in energy lost due to friction. Then I think you could compare friction energy loss to drag energy loss.


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