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here's a first guess
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Posted by nt moore (more from nt moore) on Thu, 14 Feb 2002 11:47:37 Share Post by Email
In Reply to: coefficent of drag, mike, Wed, 13 Feb 2002 20:03:29
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first find a classical mechanics text at your local library. My copy of "Classical Dyanmics of particles and systems" by Marion and Thorton is a good book with ~15 pages on air resistance.

I think the hard part of this problem will be estimating how much energy you lose in combustion and internal losses in the engine.

In general if you are moving with a constant speed (no acceleration) then the sum of the forces on the car will be zero. This means that the engine is pushing the car forward just hard enough to compensate for air resistance and internal losses. Instead of thinking about forces though it might be better to calculate the energy you'll putting into and losing from your system.

(Energy in) = (waste energy: out the tailpipe, and internal friction) + (work you're doing)

The work you're doing is pushing the car through the air. One formula for this is the "Prandtl Expression" where work done (in units of energy) is:
(0.5)(Cw)(density_of_air)(cross-sectional-area)(velocity^2)*(distance_you_push)

Cw is the drag coefficient, sometimes carmakers will publish this. Your cross-sectional area is easy, just estimate the width and height of the car's profile into the wind (you can do fancy things with sloped surfaces, but I'll bet that's beyond the accuracy of your calculation).

To get a number for the power (energy per unit time) dissipated by air resistance, first replace the "distance_you_push" with "velocity*time." Now work/time = power, so the power dissipated by air resistance is then

(0.5)(Cw)(density_of_air)(cross-sectional-area)(velocity^3)

Now you need to figure out how much energy is going into the system (the car) per unit of time. A car gets energy from burning gasoline, Gasoline has a certain energy associated with combustion. You should be able to figure out what this quantity is from a chemistry text. With this number you can make a definitive statement about how much energy you're putting into the system (car) by referencing your Mileage. For example if you're burning gas at a rate of (MPG) and you're traveling at a speed V, your energy input per unit time is V/MPG.

Of course not all the energy you burn in your car goes into combating air resistance. Some goes into heating up the engine block and exhaust pipe, some of the energy goes into running the alternator and powersteering pump... There is a ratio that you're going to have to calculate that will tell you how much of the energy you put in will be used to turn the tires. You'll want to read about the "Otto" cycle as well as send an email to Saab asking them about the efeciency of their automobile (I assume this will be rather low, ~5-15%)

I hope this helps a little,

NT Moore

ps I foind the combustion reaction for octane,
(fuel rich mixture, octane + oxygen = carbon monoxide, carbon dioxide, water)

1(C8H18) (liquid) + 12(O2)(gas) -> 7(CO2)(gas) + CO(gas) + 9(H2O)(liquid)


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