## Determination of airdrag coefficient

So, we're trying to simulate an electric car running with a DC-engine. To get the calculations right we must determine the rolling resistance coefficient and airdrag coefficient.

(We've already determined the rolling resistance to about 0.014, doing some linear calculations at low speeds < 2m/s, where the airresistance is ~ 0)

We've gathered data of the deacceleration of the car (doing a runout test) using GPS data. This means that we have the speed as a function of time. However, no matter what we do, it seems that we get the calculations wrong, as we expect the rolling resistance coefficient to be somewhere between 0.010 - 0.020 and the air drag might be 0.3 - 0.9.

Here is a graph of the speed as a function of time.

The graph is generated with MATLAB using this data: http://dl.dropbox.com/u/202411/p2/v.txt , notice that this is just the speed, however, we sample 5hz, which means Δt = 0.2s between each datapoint.

So, how do we calculate the correct airdrag- and rolling resistance coefficient, preferably with the lowest possible error?

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 Oh yea, so, there are no force from engine or anything else, just the wheels and airdrag. The rolling resistance is; $F_{rr} = C_{rr} \cdot m \cdot g$ $C_{rr}$ is the coefficent of rolling resistance, m the mass and g the gravitational constant (9,82 m/s^2). and the air resistance is $F_{air} = \frac{1}{2} \cdot \rho \cdot A \cdot C_d \cdot v^2$ and likewise.. $\rho$ the density of air, A the front area of car, $C_d$ the air drag coefficient and v the speed of the car. In this case, the deacceleration should be calculated like this: $F_{res} = F_{rr} + F_{air}$ $m \cdot a = F_{rr} + F_{air}$
 If you have velocity versus time just determine the acceleration and then if you know the rolling resistance you can easily solve for the air drag assuming that is the only other force acting on you car. What exactly are you having trouble with?

## Determination of airdrag coefficient

We seem to get a function a(v), where as the other function is v(t), and something here messes up. However, when we do the calculations manually with just Δa at the time Δt, we seem to get approx values that sound plausible. But we want to use it mathematicly right in MATLAB.

 Why don't you supply us with the vehicle mass and frontal cross section? With your very low speeds (under 5 m/sec), I am surprised that air drag is important.

 Quote by Bob S Why don't you supply us with the vehicle mass and frontal cross section? With your very low speeds (under 5 m/sec), I am surprised that air drag is important.
Frontal area is:
$A = 1.425m \cdot 1.462m = 2.08 m^2$

We use the density of air at 17 degree celsius:
$ρ = 1.2250 \frac{kg}{m^3}$

The mass of the car is 740 kg, and weight of our 2 passengers approx 75kg each:
$m = 740kg + 2 \cdot 75kg = 890kg$

 There is an empirical method to find those coefficients in the Bosch Automotive Handbook (4th ed) and it is reproduced here.
 Here is a plot of your car deceleration using a Crr of 0.017. There is no curvature. If I add air drag, I see no effect until the drag coefficient is over 1. Stokes (linear) air drag may apply at these speeds. How level is your track? A change in slope of 0.1% makes a difference. Attached Thumbnails

 Quote by jack action There is an empirical method to find those coefficients in the Bosch Automotive Handbook (4th ed) and it is reproduced here.
OK, We will take a look there.

 Quote by Bob S Here is a plot of your car deceleration using a Crr of 0.017. There is no curvature. If I add air drag, I see no effect until the drag coefficient is over 1. Stokes (linear) air drag may apply at these speeds. How level is your track? A change in slope of 0.1% makes a difference.
I can't deny that there might be a slope of 0.1%, but it should be more or less as level as it can be... So you might recommend us doing calculations at higher speeds?

 Quote by Bob S Here is a plot of your car deceleration using a Crr of 0.017. There is no curvature. If I add air drag, I see no effect until the drag coefficient is over 1. Stokes (linear) air drag may apply at these speeds. How level is your track? A change in slope of 0.1% makes a difference.
Yes, if you just use Crr there won't be any curvature. Can you tell me how you made the calculations/graphs using both rolling and air resistance, and tried changing the drag coefficient?

 I used The program shown in the attachment (written in True Basic). A slope of 0.1% (1.2 inches per 100 feet) in your track is equivalent to changing your Crr by 0.001 ( i.e., from 0.016 to 0.017. You might run the test in the opposite direction, or using a different section of track. Higher speeds will make the turbulent air drag larger, and use a larger section of track, but running test in both directions will decrease the dependence on slope. Attached Thumbnails
 OK, Thanks for the help everybody. Just to close this topic, here's what we did. (I hate when there are no ending answers when looking around internet forums, so here we go :D) Here is an example of data we gathered using GPS (something similar is already presented in the topic). We quit trying to solve the differential equation that we started out with, and solved it using average change in deacceleration during the time. $C_d = \frac{m \cdot \frac{\Delta v}{\Delta t} - m \cdot g \cdot {\mu}_{rr}}{\frac{1}{2} \cdot \rho \cdot A \cdot (\frac{\Delta v}{\Delta t})^2 \cdot \Delta t^2}$ This gave us some possible values, here's an example, using data from the image i posted above. $C_d = \frac{890kg \cdot \frac{ 4,85 m/s - 0 m/s}{ 38,5s - 6,5 s} - 890kg \cdot 9,82 m/s^2 \cdot 0,014}{\frac{1}{2} \cdot 1.293 kg/m^3 \cdot 2,08 m^2 \cdot (\frac{4,85 m/s - 0 m/s}{38,5s - 6,5s})^2 \cdot (38,5s - 6,5s)^2} = 0,3962$ To prove that this was right, I created a program in MATLAB which calculates the run with different drag coefficients, rolling resistances and slope of the road. (This is done by making a for loop calculating the deacceleration, calculating the new speed ... and so on until you hit 0m/s) .. Playing around with different values proved that the calculated values were possible with a slope of 0.1 % (0.1 meter rise over a distance of 100m, and we can't really neglect this being a possibility)... Also, airforce and windspeeds (with or against the car) were not measured either, so these could have influence.
 Although it gives out the same answer, I think the correct formulation for your equation would be: $C_d = \frac{m \cdot \frac{\Delta v}{\Delta t} - m \cdot g \cdot {\mu}_{rr}}{\rho \cdot A \cdot v \cdot \Delta v}$ Where $v$ is the average speed. That being said, I'm not sure you have enough data to determine $C_d$. You need more data at a higher speed. To prove that, if we used your equation with the points {6.5; 4.85} and {35; 0.15} instead of {6.5; 4.85} and {38.5; 0}, we get $C_d = 0.822$, which is really different from 0.3962. With points {6.5; 4.85} and {25; 1.5}, it is even worst with $C_d = 2.57$.

 Quote by jack action To prove that, if we used your equation with the points {6.5; 4.85} and {35; 0.15} instead of {6.5; 4.85} and {38.5; 0}, we get $C_d = 0.822$, which is really different from 0.3962. With points {6.5; 4.85} and {25; 1.5}, it is even worst with $C_d = 2.57$.
Yes, I agree, there are big changes with just a little change in the data used. To determine it more accurately we need measurements at much higher speeds (70-100 km/h)... However, we are modelling and simulating an electric car, which means that our calculations will only need to satisfy speeds at 20-40 km/h.