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Determination of the Drag Coefficient (Cd)

Many times I have been asked about how to determine the drag coefficient of a car or even a motorcycle. I only know three ways: 1) mathematically using computational fluid dynamics and a very large computer to solve the three dimensional Navier-Stokes equations; 2) a large wind tunnel, one that can fit the full size vehicle in, or a model sized one using a lot of dimensional analyses; and 3) coast down testing. Now, unless you have access to IBM's Deep Blue or have a Cray laying about and have lots of experience in computational fluid mechanics, then method 1 is probably not your first choice. This is probably true as well for method 2. Most of us just don't have a NASA sized wind tunnel in our back yards nor even a smaller one for model testing. Sophisticated instrumentation is generally needed in wind tunnel testing also. So that leaves us with method 3. Fortunately this method can be fairly simple, yet expanded to be really complex to get the real nitty gritty numbers.


What follows is an academic exercise for me. It is not intended to be a definitive answer to anything except my curiosity. To that end, if you find it useful, amusing or down right distasteful,GREAT! You may find errors, wrong assumptions, or other nastiness and I will appreciate you pointing them out to me. Complaints should be backed up by real data, through either analyses of your own or documented information. I am ameniable to changing my thinking iffn there is good reason to. Opinions, by themselves, just don't cut it.

Getting to it, finally.

What is the goal? Well, to me, it is to take empirical data, data obtained on the salt lake (Bonneville), mud lake (El Mirage or Muroc), or concrete strip (Maxton) or even a long straight stretch of hiway in your home town, and then fittiing it to a math model such that we can find Cd.
Well, jeeze, why would we want to do that? Most of us haven't done any math since high school or at least a long time ago, so this could be real agony. Accurately knowing a drag coefficient willpermit us to determine the speed performance of a particular vehicle against horsepower available or needed to meet some goal, like breaking a record.

The "model"

The forces on a vehicle immersed in air on the ground are pretty straight forward. Any number of texts may supply you with it. I started with "Analysis of Vehicle Aerodynamics", SAE SP-1036, Paper 940420, "A Detailed Drag Study Using the Coast Down Method", by M.A. Passmore, Loughborough Univ., and G.M. Le Good of the Rover Group, LTD. It must be stated that the only part of the paper's solution I used was the following model. Solution of the model is purely my effort, no plagarism.

FT = FD + Medv/dt + Mg sin(alpha)

FT = Tractive or motive force
FD = Sum of all resistive forces
Medv/dt = Sum of all inertial forces
Mg sin(alpha) = Gravitational force

(note: here I departed from the text in that I am going to include additional items, later, in the model and analysis)


FT = FD + Medv/dt + Mg sin(alpha)

FT = 0
Mg sin(alpha) = 0

well, FT = 0 because we have slipped the tranny into nuetral and we are coasting!
and, Mg sin(alpha) = 0 because the race track is very flat, no uphill, no downhill.

So, we are left with:

0 = FD + Medv/dt + 0

This model, then is the well spring from which all solutions must arise. We will take the simplified solution set first. Because it is the one that most racers will use. It is particular to the configuration of a given vehicle and imbedded is the sum of all forces. This is likely to make the Cd a tad off, but it will be very useful for determining the vehicle performance when changes to the power plant are made.


FD = - Medv/dt

Plugging in the terms.....

1/2 * rho * V2 * Af * Cd = - Medv/dt

Assumptions I made are that the Cd coefficient contains all drag terms that ever could be and that the Me factor contains all of the inertia terms that could ever be! Big jump here, but, reasonable, I think. Pressing on....

Before getting to far along, let me define the terms used:

rho is the air density in slugs (yipes, what are slugs?)
V is the vehicle velocity in feet per second (ft/sec)
Af is the frontal area, in square feet (ft2) of the vehicle. This needs to be done accurately for your vehicle.
Cd is the dimensionless drag coefficient we are looking for
Me is the total mass (weight/gravity or W/g in slugs) of the car, driver, fuel: in short, everything that is rolling
dv is differential velocity
dt is differential time

Now for some algebra (Aaarrrrggghhhhh....!). Rearranging the terms and separating the variables......

Cd * dt = - (Me *dv) / ((1/2 * rho * Af) * V2)

for simplification lets let

K = - Me/ (1/2 * rho * Af)

now our equation looks like

Cd * dt = K * dv / V2

Now we have a full blown differential equation, which has all necessary conditions for solution. We just integrate and rearrange some terms! We all remember how to do that, right?

Cd * t = K / V

Now, plug in the limits of the integration and you get:

Cd * (tf - ti) = K * (1/Vf - 1/Vi)

The terms here are simple:
ti  is the time, in seconds, when you entered the measured mile or trap and or "0" if using a stop watch
tf  is the time, in seconds, when you exited the measured mile or trap and or elapsed time if using a stop watch
Vi  is the speed when you entered the measured mile or trap and corresponding to time ti
Vf  is the speed when you exited the measured mile or trap and corresponding to elapsed time tf

For simplification let's let (tf - ti) be ET for elapsed time, in seconds.


Cd = K * (1/Vf - 1/Vi) / ET

ET, Vi, and Vf are just the elapsed time to go from the start coasting speed to the end coasting speed. In the case of Bonneville it would be the time to coast through say a one mile measured trap and the speeds would be the entry speed and the exit speed. Pretty simple, huh? We still have to plug the K stuff back in and that happens next. Oh, you don't have to do this for a mile, it can be for any distance as long at the start time corresponds to the entry speed measurement and the final time corresponds to the exit speed measurement. I'll run trhough a couple of examples in a bit.

Cd= - Me * (1/Vf - 1/Vi) / ((1/2 * rho * Af) * ET

Remembering that Mass = Weight divided by gravity, then Me = Wt / g where Wt is the total vehicle weight including driver. Plugging this back into our result and moving some numbers about gives us:

Cd= 2 * Wt * (1/Vf - 1/Vi) / ((g * rho * Af) * ET

Only one more thing to do and that is to get rid of the term "rho". How, you ask? Well, P / rho = g * R * T, with P in pounds per square foot, rho in slugs, g is 32.174 ft. per second per second, and T is temperature in degrees Rankine (459 + To Fahrenheit). Cross multiplying and dividing we see that rho * g = P / (R*T). Now to get P in lbs / ft2, we need to multiply by 144 inches2. R, the universal gas constant for air is around 54. Also remember that V has to be in feet per second, so a correction factor to permit using mph is 1.4667

So here is what it looks like:

Cd= - 2 * Wt * (1/Vf - 1/Vi) * R * (459 + To F) /((P) * 1.4667 * Af * ET))

Here it is once again with all the numbers in it:

Cd= - 2 * Wt * (1/Vf - 1/Vi) * 54 * (459 + To F) /(P * 144) * 1.4667 * Af) * (ET)

Reducing the numbers and we find:

Cd= - 0.51135 * Wt * (459 + To F) * (1/Vf - 1/Vi) / (P * Af * ET)

An example. I hope it works!!

Lets pretend we have a car that we are just about ready to run on the salt. We just weighed it right before we left home and have added some fuel. Withthe driver, fuel, and everything else, it weighs 2000 pounds. We stop by the ERC fuel trailer and find that the current barometer is at 25 inches of Mercury (12.3 psia) and the ambient temprature is 95 deg F. We had previously determined that our car has a frontal area of 20 square feet. Whats left? Only the coast time and the associated change is speed. We make our run and coast through the mile: it takes 30 seconds to coast through. The entry speed for the mile was 200 mph and the exit speed was 100 mph. Plugging the numbers in:

Cd= - 0.51135 * 2000 * (459 + 95) * (1/100 - 1/200) / (12.3 * 20 * 30 )

Cd= - 0.51135 * 2000 * 554 * ( 0.010 - 0.005) / (12.3 * 20 * 30)

And thusly:

Cd= 0.3839

So where did the minus sign go? I know, but, do you?

A final note. I hope this works out for you and it was fun for me. I tried to put in all the steps so everyone can follow my reasoning and math so that errors can be caught. I should mention again, that you do not need to coast for a mile. All you have to do, is get up to some speed above where you want to start you coastdown, put the vehicle in nuetral, start the stop watch, note speed at the same time, coast for a while, stop the watch and note the speed. Use these numbers in the formula and it should work. If there is sufficient interest, I will try my hand at including the other components such as rolling resistance, tire drag, rotating inertia of the wheels and tires. To use that formula will however, require that we determine the moment of inertia of each wheel and tire with rotating parts attached, such as brake drums and rotors and spindles. Gear drag would have to be measures as would the moment of inertias of the rotating parts of the gears, drive shaft, axles, etc.

Copyright (C) 1998 - 2004, all dates inclusive, L.E. Mayfield - All Rights Reserved
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