Originally Posted by
Jon Kerr
I'd like to know what kind of g forces these cars pull under acceleration. At Reedy, the invite guys are getting from maybe 10 mph coming through the esses to 50+ at the end of the straight less than 120 feet away. Anyone know the math to figure that out? It's been 13 years since I was in physics class and can't remember that stuff.
John,
You can compute average acceleration using the relationship:
(ending velocity – starting velocity) / (time interval)
The guestimating begins by making an assumption about when the cars hit their top speed of 50 mph. If we assume they do it half way down the 120 foot straightaway, this would yield:
Initial velocity = 10 mph = 4.47 meters / second
Final velocity @ end of 120 feet = 50 mph = 22.35 meters / second
Average velocity over first 60 feet = (10 mph + 50 mph) / 2 = 30 mph = 13.41 meters / second
Distance of half straightaway = 60 feet = 18.29 meters
Time to cover first 60 feet = 18.29 meters / (13.41 meters / second) = 1.36 seconds
Average acceleration = (22.35 m/s – 4.47 m/s) / 1.36 seconds = 13.12 meters / second^2
Dividing by 9.81 gives an average straight line acceleration of ~1.33g The peak, instantaneous acceleration is obviously more, but I'm not sure how much more.
Andy (I guess I’m having a slow day at work too - ha!)