You can go to the Golen Engine site and they have free drag racing calculators you can use to figure out unkown values like HP from know stuff on your time slips. What I'm looking for is a formula to figure out how much a head wind takes off your top end in the 1/4. I ran yesterday and there was a 20 to 30 knot wind blowining down the track. I was down several MPH and ET and was wondering if there was a way to figure the loss.

 

I assume you would use dynamic air pressure and coefficient of drag... something like F=C*A*rho*g*V where C is the coefficient of drag (something like .4?), A is frontal area of the car (height X width), rho is the density of the air, g is the acceleration of gravity, and V is the velocity of the wind. This gives a force, and I bet you can use the radius of your rear wheels to calculate torque (T = r X F). Since you have ~20% losses through the drivetrain, multiply by 1.2 or so to get approximately the amount of extra torque you would need from the crank to compensate for the headwind.

You could also use newton's law (F=ma) and kinematics (x = x0 + v0*t + .5*a*t^2) to get the extra ET.

Seem reasonable?

 

This would have to be done using differential equations. It wouldn't be too hard but it would involve a ton of things as torque to the track varies the entire time, force done by drag goes up exponentially etc.

 

Compare 60 ft 330 ft and 660 ft times to your previous (low wind) runs. Obviously 60 ft times won't be affected by the wind. You might plot no wind times and windy times graphically using seconds and mph as the axes. I would expect to see the curves overlay at 60 ft, and gradually diverge at each (higher speed) point.

If the 60-330 times didn't agree, shift the windy curve to match at the 60 ft times and see what you get. Unless you had wind tunnel data for your car, and very accurate wind vector, calculation would probably be less accurate.

Of course you could have run the strip with the wind as well as against it and average...but getting stopped might be a problem.

 

Force by drag should be fairly constant since we are only looking at the headwinds effect. If you wanted to, you could try to get a better estimate by taking your final speed and ET and form a simple equation for v(t). Add the 20-30mph headwind, then go with my analysis about to see how different it is from using your original v(t).

This won't give you a 100% perfect number, but I assume you're looking for ballpark, right?

 

The extra drag from the headwind is not fairly constant. The added drag from the headwind is about 10x more at the end of the quarter mile than at the beginning depending on the exact trap speed and headwind.

 

well, it seems to me that since drag is proportional to V^2, then let's do a quick calculation.

Assuming final trap speed is 100mph and headwind 30mph, then at the beginning, we have wind's contribution as a constant times 30^2 = 27000 and at the end, its contribution is 130^2 - 100^2 = 6900. This means that wind resistance has a REDUCTION in effect by about 4x at the end as compared with the beginning.

 

30^2 = 27000?

 

New Math folks are reaching adulthood.

 

Id take an old timeslip do a quadratic interpolation and do the same for the new one with the headwind. Graph the curves as has been suggested. The difference will be get you pretty damn close to the force of the headwind.

 

I hate math.

 

bah, just add 20-30 knots to your trap speed...

If nothing else, you have a new best!

 

****, no idea how that slipped by... i'll concede.

 

That's just a typo, he meant 30^3.

 

i didn't "mean" 30^3, but i gave 30^3... silly mistake that made me completely wrong. anyway, the force is quadratic, as stated earlier..