So: Do you know, geniuses
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F1 cars are not designed for minimal drag. If they were, we wouldn't see any wings on the cars. It's a compromise between drag, the needed downforce and the power with which the engine can pull on the straights. Therefore you would not expect to see a perfect drag coefficient in an f1 car.Ciro Pabón wrote:Are you crazy? Or am I? What kind of units are you using? Imperial gallons of drag? I may be only a highway engineer, but I am PRETTY SURE Cd is around 0.3 for a well designed car. And I mean Cd like in:
F = 1/2 * rho * Cd * A * V^2
Where, of course, F is drag force, rho is air density, A is the frontal area and V^2 is velocity squared. I HAVE measured Cd between 0.5 (for a sphere) and 1 (for a box).
So, please, explain to me how could an F-1 car have a Cd of 1 point something? Do you, F-1 aerodynamicists, use another equation?[/url]
Bentley EXP Speed 8 LMGTP, '01
Lift-to-drag ratio: 3.70:1
Coefficient of drag: .5038
Coefficient of lift: -1.87
1999 Ferrari 360 Modena
Lift-to-drag ratio: .73:1
Coefficient of drag: .34 (factory claim)
Coefficient of lift: -.25 (factory claim)
2003 Lister Storm LMP
Lift-to-drag ratio: 4.81:1 (peak L/D better than 5.0:1)
Coefficient of drag: .5499
Coefficient of lift: -2.65
I don't know exectly situation now but in 2000 and 2001 in soe track not so fast and slow maclaren has Cd that used in F = 1/2 * rho * Cd * A * V^2 formula near 0.67 and 0.65F = 1/2 * rho * Cd * A * V^2
Yes, correct, they are. You still end up with the same final CD in the end, which in f1's case is hightened due to the induced drag from the airfoils on the car. As I mentioned, it's a compromise, and a higher drag coefficient to a certain limit doesn't matter as much as grip when you have a strong engine to chew through it.Ciro Pabón wrote:Thanks, Bernard. Very reasonable. But, again: are not Cd and Cdi (coefficient of drag, induced) measured independently?
Seems ok, this is pretty much kindergarden stuff.wothLike in an airplane. Cdi equation is a different one, as you surely know.
Cdi = (Cl^2) / (pi * AR * e)
where: Cl = coefficient of lift, pi=pi, e = efficiency factor and AR = s^2 / A (s = span, A = wing area).
And, of course, Cd total = Cd + Cdi. And I would presume you want both coefficients as low as possible and a Cl as high as possible. Could you clarify it for me, please? [-o<
Using the equations we don't have much choice on how to reduce drag and maintain a high Cl. Either increase the AR or the efficiency. I know that deep endplates effectively increase the AR so that is one way.Cdi = (Cl^2) / (pi * AR * e)
where: Cl = coefficient of lift, pi=pi, e = efficiency factor and AR = s^2 / A (s = span, A = wing area).
And, of course, Cd total = Cd + Cdi. And I would presume you want both coefficients as low as possible and a Cl as high as possible.
Does this account for 'spoon-shaped' rear wing profiles?Lifting line theory shows that the optimum (lowest) induced drag occurs for an elliptic distribution of lift from tip to tip. The efficiency factor e is equal to 1.0 for an elliptic distribution and is some value less than 1.0 for any other lift distribution. The outstanding aerodynamic performance of the British Spitfire of World War II is partially attributable to its elliptic shaped wing which gave the aircraft a very low amount of induced drag. A more typical value of e = .7 for a rectangular wing.
Actually this slipped past my eyes, so I take it was a mistake from you too, but that's actually Cdo + cdi=cd total, as in cd at zero lift. Otherwise it's correct.Ciro Pabón wrote:And, of course, Cd total = Cd + Cdi. And I would presume you want both coefficients as low as possible and a Cl as high as possible. Could you clarify it for me, please? [-o<
I take it that means this type of elliptical distribution of lift?Does this account for 'spoon-shaped' rear wing profiles?