Derby Science - NDR News - Winter, 1998

Can a "big kid be competitive in Derby Racing?

by Dr. Todd Wetzel


 The Soap Box Derby to me is an excellent science classroom. It played such a role in my racing days not so long ago, and the lessons I learned on the track and in the shop have helped me all through my academic and professional endeavors. So that, coupled with a misguided promise I made to the so-called "editor" several years ago is why I'm writing this column.

My goal in this column is to relate hard science to derby racing. All of us love bench racing, and there are dozens of theories out there about different aspects of the send. I will attempt to separate myth and opinion (especially mine) from real science. Perhaps the most interesting thing though, is sometimes "science" doesn't "know" the answer! The topics of this column will be directed primarily by questions from the readers.

Science and math cannot be separated Therefore, inevitably I will use equations in some of my articles I am hoping that many of the Senior drivers will be able to understand and use simple algebra. If nothing else, maybe one or Me kids will get a little more interested in their math classes so they can understand more about their derby cars! I will also present all-important equations with equivalent "forms" that will break down the calculations into simple steps for those less comfortable with equations

Can a kid who weighs more than a feather win in a Derby Car?

A common complaint ever since I started racing in 1981 has been that smaller kids have an unfair advantage over larger ones. Many rule changes have attempted (and often failed) to "even" the playing field. A tiny kid though is no substitute for excellent design, careful construction, and great driving! Larger drivers can be very fast, and this article deals with how to guarantee your car makes weight.

Minimize Extraneous Weight

Whenever possible, make decisions that minimize weight. The worst use of weight in a soapbox derby car is often in the floorboard. A continuous steel plate is not required to be fast! A steel plate, for example, that is 3" wide by 3/4" thick by 71" long will weigh 46 Lbs. Steel plates do not add significant stiffness to a car (perhaps a good topic for a future pride) unless the car is designed poorly. The only function a continuous plate really serves is it is a convenient means for initially aligning your kingpins in the car, and it provides a convenient mount for the axles and all related hardware. None of those benefits makes you fast; they just might make construction a little easier.

Do use individual plates at each axle, and make those plates broad so as to distribute their load to the car as efficiently as possible. Do consider aluminum, which has half the stiffness, but also almost one third the weight, and so is a more efficient structural material. Just be careful with tapped holes in aluminum, as aluminum is much less forgiving to over-torquing bolts than is steel.

Another source of weight is the floorboard itself. A very thick (2 1/4") fir floorboard can weigh over 30 pounds. A well designed car will not be hurt by a thin floorboard, which has the added benefit of lowering the driver's center of gravity Finally, all the other hardware in the car should be as light as possible without compromising safety. Strong axle mounts need not be huge (I once saw a set of sic, mounts that weighed 15 Lbs. each. And some things should go without saying: I knew somebody who raced wearing hiking boots Bad idea!.

Careful weight planning

Careful weight planning is a big key to being competitive, and with a little bookkeeping this can be done with a surprising amount at precision. Truthfully almost all cars are built without explicit calculations of weight balance. Some people do calculate the total weight of the car to make sure that the car and driver will not t, If you have a lot of experience or if you have a very small driver, balancing is not a problem. But if you are building your first car, and if your driver weighs more than 110 lbs., then it is possible to build a car that you can not balance when completed.

Also, it is not always clear where you place the driver in the car. This technique will help you determine drive placement. The technique I will describe is fairly simple but very accurate. It is essentially the same kind of technique used in aircraft design, because an airplane must have a precisely known center of gravity in order to be stable in flight. Basically, you estimate ahead of time the weight and location of every major component. The more components you know the better, and as you actually build them you should update the component weights in your balance estimate.

First, I will show you how to estimate the weight and center of gravity of different components and then we will use the data to determine the center of gravity of your car.

Some General Notes

Everybody knows how to weigh things, but I also want to make sure we are clear on the idea of the "center of gravity", or c.g. The center of gravity of an object is the point in an object through which gravity acts. A simpler way to understand it is by example. One way to find the center of gravity of an object is to balance it on a fulcrum. For example, if you have built your floorboard and you want to find its center of gravity to check it against the estimates you made with the techniques below, you simply need to put the floorboard on a dowel rod and roll it front and back until the floorboard perfectly balances on the rod. Mark the location where the dowel rod is on the floorboard, and that is your center of gravity. It you have a small object, like a brake pedal, then you can estimate the center of gravity as just being roughly where the center of the brake pedal is in the car. And finally, where the kids shoulders are relative to the rear of the car. This may if you have a geometrically symmetric object, like a plate, the center of gravity is the geometric center of the plate.

Note that the center of gravity is a point in three-dimensional space; that is, the true center of gravity of a soap box derby car is located a distance from the tail (longitudinal Center of gravity), a distance up from the road, and a tiny distance to one side or the other (the car can't be perfectly balanced side-to-side) In these analyses we are only concerned with the longitudinal center of gravity. Estimating the, weight and center of gravity of complicated objects, like aerodynamic bodies and shaped floorboards, can be much more difficult. The following sections describe very simple techniques for estimating the weights and c.g. locations of your car. To help simplify the task of weight estimations, I used a computer program to analyze several "good" derby car designs of different width. By "good" it simply means the car must have minimum girth and height, and must be full length and basically streamlined. It turns out that the weights and c.g.'s of the body and floorboard do not very much (< 5%) for cars with different width (or with the location of the widest point either), which may be a little surprising. That is to say, whether you have a 15" round car or a very tiny 13 wide car that are built with the same floorboard and sticks, so long as the car is "properly designed", its body and floorboard, weights are very nearly the same This seems counterintuitive, but the reason is that since the girth is the same for all properly designed cars (49"), then the weight of the body and floorboard is also roughly the same for all good cars, The decisions that do affect car body weight are the body thickness, the floorboard thickness, the type of wood used, and the amount of fiberglass or other cloth that you put on the car. Therefore, you can do a very accurate balance estimate without even knowing the shape of you car, so long as it is "good",

Estimating Driver Center of Gravity (C.G.)

This is the most difficult, and most important, pan of a proper weight balance design. There are two ways to estimate the location of the driver c.g.. The easiest, and less accurate, way of doing it is to place a board on some kind of fulcrum (a piece of pipe or dowel rod will work fine), and have the driver sit on the board in as close to actual order position as possible. Move the board (and driver) back and form on the fulcrum until the driver is balanced. Then measure the distance from the location of the fulcrum to some reference on the driver (I prefer to use the tip of the shoulders).

The better way is to put the driver in an existing car and weigh its balance with and without the driver. The reason I prefer this technique is it more accurately holds the driver in driving position. Also, the weight of the board can introduce errors in the c.g. estimate. So if you have access to another car and a scale, it is easier and more accurate to measure the driver center of gravity. The car does not even have to be balanced or brought up to weight. Just put the kid in with whatever weights are in it. Put his helmet on, insert the helmet into the headrest, and make sure the driver has his knees against the top of the car. Then, you need to make three measurements. First, mark and measure where the kids shoulders are relative to the rear of the car. This may be difficult to do with accuracy, so one way to do it is to mark where the back of the shoulders are actually inside the car, and when the driver gets out of the car, transfer that mark to the center of the floorboard and measure the distance from the rear kingpin to the mark.

By the way, you can use any other body part other than the shoulders; perhaps a good one would be the location of the first helmet rivet, since this will help you locate the helmet in the car The reason I prefer to use the shoulders as my reference is because usually the shoulders are the tightest fit in the car, so I want to be able to check them directly. I will call this location of the drivers shoulders in the car (from the tail) Xshoulders.

Do the Math...

Q. In the figure to the left, Craig, who weights 130 Lbs., first weighs the front axle for the empty car to be 55 Lbs.  Then he gets in the car, and the front axle weighs 03 Lbs.  He measures his shoulders to be 20" from the tail of the car.  Where is his body c.g. in inches from his shoulders?

A. 103 Lbs. - 55 Lbs. = 48 Lbs.

(48 Lbs. / 130 Lbs.) X 64" = 24"

24" + 6" - 20 = 10"

Next, put a scale under the front axle of the car, and weigh the front of the car with and without the driver. We will call the weight on the front axle with the driver in the car WF,driver and the weight of the front axle of the car without the driver in the car WF,empty. Finally, weigh the driver. Then, you can compute the drivers center of gravity, in inches from the drivers shoulders, as:

This formula assumes that the car has a 65" wheelbase, and that the rear axle is the minimum 6" from the rear of the car. Alternatively, I have included the following form you can use to compute the driver center of gravity.

1. Location of drivers shoulders from rear of car, inches (Xshoulders)  
2. Weight on front axle, car empty, Ins. WF,empty)  
3. Weight on front axle, driver in car, Ins. (WF,driver)  
4. Weight of driver, Ins. (Wdriver)  
5. Line 3 - Line 2  
6. Line 5 Line 4  
7. Line 6 x 65"  
8. Line 7 + 6"  
9. Line 8 - Line 1; This is the drivers center of gravity, in inches from the drivers shoulders.  

Estimating car body weight

Since a derby car is a complex shape, there is no easy formula to compute an exact body weight. However, it can be estimated. To compute the weight of something, you first compute its volume and

then multiply by its density. For a "skin" or 'shell", like a derby car body, the volume of wood used to construct it is roughly the surface area times the body thickness. It turns out that since all derby cars have essentially the same girth and length and roughly the same shape, the surface area is very similar from one car to the next, and that value is 3200 in2. The wood weight is then this surface area times the thickness times the wood density. The table to the left lists the density of several woods and metals.

Also, the weight of the fiberglass or carbon fiber that you put on the body adds significant body weight. When you buy fiberglass or carbon fiber, its "weight" will be the primary specification. For example, a very common fiberglass cloth is "10 oz. cloth", which means the cloth weighs 10 oz per square yard (1 sq. yard. equals 1296 sq inches).

Do the Math...

Q. Estimate the weight of a carbody using 1/4" thick spruce sticks.

A. Density of spruce is 0. 021 lbs./in.3, so:

3200 in 2 X 1/4" x 0.021 lbs./in.3 = 16.8 lbs.


When saturating cloth by hand, typically it takes 2 ounces of resin for every ounce of cloth; therefore, the total weight per square yard is roughly 3 time the "ounce" rating of the cloth. Therefore, If Wcloth is the "weight" of the cloth in ounces, you can estimate the weight of each layer of fiberglass or carbon fiber plus the resin to be:

For example, each layer of 10 oz. cloth you use on your car will add 4.6 tbs to the car body weight, and two layers of 10 oz cloth will add 9.2 lbs. to the body weight. layers

In all cases, the center of gravity of the body is roughly at 41" (slightly behind the middle) from the tail of the car.

Do the Math...

Q. Estimate the weight of 2 layers of 8 oz. fiberglass cloth plus resin on the car body.

A. 2 x .046 x 8 = 7.36 Lbs.


Estimating Floorboard Weight

Estimating the weight of the floorboard is also an inexact science because the floorboard typically has an irregular shape. Also, in the method presented above for computing the wood weight of the car body, some eight of the floorboard was implicitly included because the surface area of 3200 in2 includes the surface area all the way around the car, including the bottom of the car.

Floorboard Thickness, inches



.75 350
1.5 850
2.25 1600

Taking this into consideration the following table lists some rough estimates for typical floorboards of varying thicknesses. These estimates are based on the same computer calculations I did for the body shapes in the previous section. In all cases, the center of gravity of the floorboard is roughly at 41" (slightly behind the middle) from the tail of the car.

Do the Math...

Q. Estimate the weight of a 1.5" thick maple floorboard.

A. 850 inch3 x 0.025 Lbs./inch3 = 21.25 Lbs.


Axles and Wheels

One axle with two plastic wheels and wood airfoils weighs 10 lbs, They are located at 6" (rear) and 71" (front).

Plates and Weights

Again, the weights of plates are very easy to compute. Compute their volume and multiply by the density of the type of metal used Since many plates are just rectangular, multiply their width, length, and thickness to get the volume.

Do the Math...

Q. Estimate the weight of a steel plate measuring 71" long, 3 inches wide and 3/4 inch thick.

A. 71" x 3" x 3/4" x 0.284 Lbs./inch3 = 45.4 Lbs.


The center of gravity of the plate is of course the middle of the plate. But we need to always make sure that all center's of gravity are referenced from the tail of the car. To keep track of all of the locations of the different components, it is good to make a simple sketch of the entire car.

Beware of Extraneous Weight

Note that we have only analyzed the "major" items In your car However these are only estimates, and while they can be surprisingly accurate, they will not predict every ounce weight of you, car Items such as body filler, paint, headrests, hatch latches, etc., can add up. So never design your car right at maximum weight. Instead, aim for at least 30 Lbs. of lead in your car so you have a buffer.

Putting it All Together

So now we are going to use these techniques to find whether or not the car will balance. For review, the information we need for each component is its weight and the location of its center of gravity in the car. Make a list of all of these components. Then, for each component, multiply the weight (W) by its location IS) to get its "moment" (M=W*X). The units of a moment are inch-pounds, so a moment is the same as a torque. Do this for all of the components, and add the moments up to get the total moment, Mtotal Also, add up the weights of all of the objects to get Wtotal (which better be less than 250 lbs to race at Akron and 260 pounds to race in NDR).

Remember, when entering the driver's c. g. the number we computed above is the distance of the driver c. g. from the drivers shoulders. The value required for this calculation is the location of the driver's c.g from the tail of the car, so add on the distance from the tail to the drivers shoulders. This is how you can determine where the drive needs to be in the car.

To finish our analysis what we'll calculate is the amount of lead in the car and where it must go to balance the car out. You first need to specify the maximum weight of the car, Wmax, which is 250 Lbs. for All-American rules and 260 Lbs. for NDR rules. I will assume we are trying to achieve a balanced car. The weight of the lead will be Wlead = Wmax Wtotal. This should be at least 20 or 30 Lbs. Then the location of this weight required to balance the car is calculated from:

Note that this location is where you can place all of the weight, or you can spread the weight out in other locations such that the center of gravity of all of the weights combined equals this location. Whether you try to center all of your weight in one location, or put it at the axles, or spread it throughout the car is a complete topic in and of itself.

The following form can be used to carry out these computations. When doing these calculations, if the weight of your lead is negative, that means your car is going to be overweight! Likewise, if the location of your lead is negative, that means your car is going to be too nose‑heavy to be balanced; and, if the location of your is greater than 84", then your car is going to be too tail‑heavy to balance Play around with driver location and component weight until you find a combination that works.

Component Weight, Lbs (W) c.g. location inches (X) Moment in, - Lbs. (W x X)
1 Driver
2 Car body 41
3 Floorboard 41
4 Rear axle, wheels 10 6 60
5 Front axle, wheels 10 71 710
6 Plate (if full length)
7 Rear axle mounts (+ plate optional)
8 Front axle mounts (+ plate optional)
9 Steering and brake
10 other
11 Total weight (Wtotal) add lines 1      
through 10 in the "Weight" column through 10 in the "Weight" column
12 Specify maximum weight (Wmax, 250 Lbs. for AA, 260 Lbs. for NDR)      
13 Weight of your lead (Wlead) Line 12 Line 11      
14 Total moment (Mtotal), add lines 1 through 10 in the "Moment" column      
15 Multiply Line 12 by 38.5
16 Line 15 - Line 14
17 Line 16 / Line 13 (This is where you need to place your weight)

Examples: David and Todd

To help understand these techniques, I have two examples for you David and Todd both are designing cars for themselves according to Akron rules I'll just give the answers, but go through the calculations yourself to make sure you get the same answers. Here is the pertinent data for both:

David is building a 14 wide car, using 1/4" thick sticks of Douglas Fir covered with 2 layers of 10oz. fiberglass cloth. His Douglas Fir floorboard is going to be 1.5" thick with a 3" wide, 71" long, 1/2" thick steel plate running through the length of the car. His axle mounts weigh 7 Lbs. for the front and 5 Lbs. for the rear. He has a 1 Ib. brake and steering assembly located 50" from the rear of the car. David weighs 140 Lbs. and his center of gravity is 20" in front of his shoulders. He is going to put his shoulders 20" from the rear of the car.

Todd, on the other hand, is building a 14" wide car, using 1/4" thick sticks of basswood covered with 3 layers of 6 oz. fiberglass cloth. His basswood floorboard is going to be thick with two aluminum plates, 3" wide, 6" long, and 1" thick located at each axle. His axle mounts weigh 3 Lbs. for the front and 2 Lbs. for the rear. He has a 1 lb. brake and steering assembly located 50" from the rear of the car. Todd weighs 150 Its. and his center of gravity is 20" in front of his shoulders. He is going to put his shoulders 20" from the rear of the car.


Car body weighs 25.2 lbs.

Floorboard weighs 17 lbs

Plate weighs 30 lbs c.g. is locatedat 38.5

Wtotal = 245.2 lbs. Not much weight to work with!

Mtotal=9832 in.- lbs.

If you calculate where to place Davids 4.8 lbs. of lead you will get an answer of minus 43, or 43 behind the car You obviously cant do that so Davids car is hopelessly noseheavy. You will find that if you move David only 2 back in the car, the car can be balanced but only barely.



Car body weighs 21.5 lbs.

Floorboard weighs 6 lbs.

Plates weigh 1.8 lbs. each and are located at 6 (rear plate) and 71 (front plate)

Wtotal= 207 lbs. So Wlead = 43 lbs.

Mtotal = 8311 in. lbs.

Place weight at 30.6 from rear of car

Todd's car is not over weight (almost 40 lbs of lead!) and can be balanced.


 NDR News Editors Note: Although Todd did a much better job of planning out the weight in his car, when these two race, David will always win because he is a much better driver than Todd.


The most difficult part of this whole analysis is locating the driver's center of gravity. However, if we had enough data from enough different drivers of different weights and heights, it should be possible to develop a chart or a technique of some kind to allow a driver to estimate his or her center of gravity without stepping foot in a derby car. If you already have a driver and a car, take the driver and go through the steps outlined in the "Estimating Driver Center of Gravity" and make the following measurements for me (of course, all data will be held in confidence):

1. Driver's weight

2. Driver's height

3. Driver's gender

4. Location of driver's shoulders from the rear of the car

5. Weight on front axle of car without driver

6. Weight on front axle of car with driver

Send those measurements and any questions you may have for future articles to the NDR Newsletter via mail or email. If we get enough responses, I will use the estate develop a driver c.g. estimation technique in a future article.

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