<BLOCKQUOTE><font size="1" face="Verdana, Arial">quote:</font><HR>Originally posted by CobraCam:
So ten years down the line, when a Corvette C6 Z06, with incredible aero can beat a 550hp Viper down the track with 400hp, and can go faster in a straight line, what will you say? Like it or not, the GTS is no aerodynamic wonder, it rates a .36. The Corvette C5 is a .29! And yes, aero does come into play on the drag strip, after about 80-90mph.
<HR></BLOCKQUOTE>
I decided to put my physics degree to work tonight to see whether it is really true that .36 vs. .29 Cd makes a difference in the 1/4 mile. I decided to make a simplifying assumption that acceleration from the motor is constant over the 1/4 mile, which obviously it's not, but I don't expect the results would vary dramatically with a more sophisticated assumption. Say 0-60 in 4.5 sec (just to pick a number); that equates to accleration of about 6.0 m/sec^2 (conversion to metric). Call this acceleration a0 = 6.0 m/sec^2.
Now drag results in an acceleration of aD = -1/2*d*Cd*A/m*v^2 = -k*v^2, where
d = atmospheric density = 1.2 kg/m^3 on the earth's surface
Cd = 0.36 or 0.29 in the two cases to be studied
A = cross-sectional area into the air; i.e., front-view area, approx .85*width*height = 1.83 m^2 based on 76" width, 44" height
m = mass = 3445 lbs curb + 250 lbs driver and gas = 1678 kg
v = velocity (m/sec)
k = 1/2*d*Cd*A/m to simplify the notation
So, net acceleration a = a0 + aD. Doing a bunch of calculus and algebra results in the conclusion that, starting from a dead stop, the time T required for the vehicle to travel a distance X is given by
T = ((a0*k)^(-0.5))*arccosh(exp(k*X)), in seconds
where arccosh = inverse hyperbolic cosine, and exp(y) is 2.718281^y (i.e., the number e raised to the power y).
The 1/4 mile = 402.34 m, since I'm doing everything in metric, and plugging into the above equation results in 1/4 mile times as follows for the two cases of different drag coefficients:
With Cd = 0.36, T = 11.77 sec to the 1/4 mile
With Cd = 0.29, T = 11.80 sec to the 1/4 mile
The difference is .03 sec, which is a very small percent (a quarter of 1%). For what it's worth, if you plug in an RT/10 Cd of 0.5, you get T = 11.88 sec; still not a very big difference.
Bottom line, Cd doesn't matter a whole lot in the 1/4 mile. Of course Cd does affect terminal velocity (vT) in a bigger way; where vT = sqrt(aT/k), where aT = top-end acceleration. So, the ratio of terminal velocities for two different Cd's (Cd1 and Cd2) is sqrt(Cd2/Cd1), which is about 11.5% for Cd1 = 0.29 and Cd2 = 0.36. Meaning if my GTS has a top speed of 185 mi/hr, and if I could magically decrease Cd from 0.36 to 0.29, my top speed would go to 1.115*185 mi/hr = 206 mi/hr.
However, I happen to agree with Mark O. I will never drive anywhere near the top end of the car, so in the straight line, all I really care about is 1/4 mile time since I'm never going to be driving much above 120-130 mi/hr anyway. And in the 1/4, Cd doesn't matter a whole lot.
Again, this means I'd take more outrageous styling at the expense of Cd going from <.3 to somewhere above .3.... Just another man's opinion!