Drag Numbers
 

I made the mistake of thinking yesterday afternoon and it wasn't long before I realized my mistake and quit, but not before I wondered:

Is there a figure (pounds, foot-pounds, whatever) that represents the wind force acting against a vehicle's forward motion at a particular speed? And are these numbers linear when force is plotted against speed? And how do they change from slippery cars like ours, to Mack trucks?

It takes force to accelerate a vehicle, but the force necessary to overcome the vehicles existing inertia is not what I'm thinking about. Once a vehicle is moving slowly at a constant speed of, say, two mph, the "maintenance" power needed has to fight only a teeny tiny bit of wind, and some wheel bearing resistance (minor). At a hundred mph, wheel bearing resistance is increased but still not great...and wind speed is much more ferocious. In fact, a few dozen MPH over that and wind resistance is about all a powerful engine can handle. So, we must be talking about quite a lot of resistance. How much? Where are these numbers?

I hope to get a grip on this "thinking" problem I have from time to time.

it's not linear it's more like an exponential ratio. I believe it is something like (not exactly, but very simplified)

Cd*f a*v^2

I'm not sure of the exact equation, but the important factors are

Coefficient of Drag
Frontal Area
and the speed

actually cars like ours aren't slippery at all
they are closer to mac trucks than say, a new audi. The bellows bumpers cars have a Cd of well over .4 while many new cars are closer to the low to mid .3's. The reason our cars do so well is that they have a relatively tiny frontal area. That's why Ruf did so much aero work when he did the yellowbird. When they changed to the 964 I think the Cd went down to the low to mid .3's.

Hey, here's tons more info for you

http://www.centennialofflight.gov/essay/Theories_of_Flight/coefficients/TH12.htm

Drag = Cd*1/2*rho*v^2*s

rho is the density of the fluid

s is the frontal area

v is velocity

Note that if you want to calculate the force used to change speed, you have to integrate the above equation between the two speeds. Make sure your units are consistent, too -- if you use square meters for frontal area, use meters per second for velocity.

Oh, except the above equation isn't quite right. It's actually:
Fd = 1/2 * rho * Cd * Af * v^2
Fd -- Force of drag
rho -- density of the medium; IIRC 1.2kg/m^3 works for air at sea level
Cd -- drag coefficient. Masraum's got good numbers for those -- see Frere's book on the development history of the 911 for the exact numbers for any car.
Af -- frontal area -- should be a little less than 2 square meters.
V -- velocity.

I don't remember the exact numbers now, but because of that v^2 business, a little more speed hurts a lot more. Worse, after you integrate to find the force required to go from one speed to another, you get a v^3, so acceleration at higher speeds is more difficult by a function of the cube. Or more simply, the faster you go, the harder it is to go faster faster. Or for example, 60-90 might take you 4.2 seconds, but 90-120 is going to take a _whole_lot_longer_.

Does that help even a little bit?

Dan

Not offended, but just out of curiousity what am I missing that is in your formula?

I also have

density -- rho
frontal area -- s
velocity -- v
Coef of drag -- Cd

Mine are in a different order, but since it's all multiplication that doesn't matter.

You guys are making my head hurt. I'm not so smart as you think. How much force is acting on my car at 120 mph? How much is acting on the Peterbilt truck that's trying to pass me at that speed? How 'bout the Audi?

The problem is I don't have many of the numbers.

For bellows bumper 911 if I use 1 for the density, .45 for the Cd of 911, 120mph (53.67 m/s) for v, and 2 for frontal area I get

~1300 for a 911 (again, this number is pretty bogus since I fudged several things, but it should show you somewhat the relationship)

lets say audi, Cd.32, speed - same, frontal area 2.3
~1060

for mack truck Cd.65, speed same, frontal area 5
~4320

Again, I have no idea what the frontal area of any of these vehicles are, and I have no idea what the Cd of a mac truck is, and I just left out the density of air, but you should get a general picture.

So, our cars can hit the top speeds that they do based on the fact that they are small, and the engine and transmission were designed as a unit for going fast, not getting the best gas mileage.

New cars, Mercedes, audi, bmw, etc have better Cd in many cases, so they can more easily push through the air despite the fact that they are bigger cars.

If you want to make a 911 a top speed machine get a 964 or newer because you'll need much less power than in an older 911

same thing, but for 964, so Cd.32
~920 or 30% less power.

There used to be a web page with a database of Cd and frontal areas of common cars, with a calculator for determining how much power you would need to get to a given speed.

I thought it was on Mulsanne's corner, but I can't find it (lots of other cool stuff there though).

Anyone remember what the page was?

Tom

What units are those numbers expressed in? I understand the guesswork regarding the variables and that these numbers are not reliable or accurate. But what units is this hypothetical "1300" expressed in? Could this be expressed in something like "torque" numbers. Like lb/ft?

My engine makes somewhere around 200 lb/ft of torque, peak. So, if my car were geared high enough, I might reach a speed where the power generated by the engine exactly cancels out these various forces (wind resistance, bearing resistance and tire deflection). With this information, it would be possible to isolate these forces enough to estimate their values. From this, one could estimate the speeds at which they can outperform another specific vehicle, and at what speeds that other vehicle might be able to outperform us, heaven forbid.

well density is kg/m^3, v^2 would give m^2/s^2, and frontal area is m^2, so this would all cancel out to kg m/s^2 which I believe is Newtons or the metric equivalent of Force (lbs for us US guys)

So, where a jet engine is often rated for lbs of thrust you would also have to do the same thing for a car.

I'm not sure if hp has an easier or more direct route to pounds than torque does, but torque shouldn't be too far since you only have to get rid of one unit for them to be the same. ftlbs, if you can get rid of the "ft" then you are set. ;)

Porsche Cd

924 - .31-.33
944T - .33-.34
911 Carrera -. 38-.39
911 Cabriolet - .40-.41
Boxster .31

Mercedes c class .27
Lexus LS430 .25
G-35 .26

Jeremy, for what years and with what aids, tails, etc...

Now you see why billions a students rejoiced whenever the physics problem stated "Two bodies in a vacuum..."
-Chris

Even things like ride height and <FONT FACE="SYMBOL">a</FONT> will affect C_d #s

Here's a better list - with dates

Vehicle Year and Model Cd
'89 911 Cabriolet 0.32
'86 911 Carrera 0.32
'95 911 Carrera 0.33
'92 911 RS America 0.32
'90 911 Speedster 0.32
'94 911 Speedster 0.31
'80 911 SC 0.34
'80 924 0.35
'80 924 Turbo 0.35
'86 924 S 0.35
'80 928 0.34
'90 928 0.34
'93 928 GT 0.34
'93 928 GTS Automatic 0.34
'88 928 S4 0.34
'86 944 0.35
'89 944 0.35
'91 944 S2 0.33
'86 944 Turbo 0.33
'88 944 S 0.35
'90 959 0.32
'92 968 0.34
'92 968 Automatic 0.34
'90 Carrera 2 0.32
'91 Carrera 2 Tiptronic 0.32
'89 Carrera 4 0.32
'88 Turbo 0.37
'89 Turbo 0.37
'91 Turbo 0.36
'94 Turbo 3.6 0.35
'93 Turbo S2 0.36

Jeremy, where did you get those numbers. I don't have Frere's book in front of me, but they seem off from what I remember Porsche 911 Story publishing.

Not saying you're wrong, just that I could have sworn that the bellows bumper years were pretty bad, and that after that they improved by approx 33%.

remember that power = force * velocity, meaning that the power required to propel your car goes as the velocity to the power three.

This explains why my 70 hp Opel Kadett could get me going close to 100 mph back in the days. To make it go 200 mph, it would have needed roughly 70*2^3 = 560 hp. (It wasn't a very aerodynamic car.)

/ Johan

Damn Jeremy, I just did the calculations with 0.38 for a carrera and you now tell me it is 0.32!?! @#$^#$ give me a second....

ok
Fd=0.5*Cd*A*rho*v^2
rho=1.225kg/m^3 at STP


Carrera

assumptions
Cd =0.32
Standard Pressure and Temp (STP)
A is approximated by a trapaziod 55" at base 50" at top and 50" tall

Fd @ 60mph = 66lbf (pounds force)
Fd @ 100mph=156lbf
Fd @ 150mph=315lbf

theoretical Mack Truck
Cd=0.8 (Mark's handbook for Mechanical Engineers)
A=6m^2

Fd @ 60mph = 959lbf
Fd @ 100mph= 2650lbf !!


Superman, the earlier statements are correct, drag force increases with the speed squared. power to overcome the drahg increases with the speed cubed!

drag due to rolling resistance increases linearly with speed.

here is a graph for a truck of the forces involved

http://forums.pelicanparts.com/uploa...1076088565.jpg

Jeremy, Interesting Find