Just chattin with my bud about 1/4 mile times and we were talkin about race weight and he said that for every 100 pounds you knock off a car you gain a .1 in the 1/4 mile.

I was like "really? isn't it for every .1 (tenth of a second) its 4 car lengths?"

can someone shed some light on this debate? There was a theread on this before but jsut try and find it with the search engine, its not really a drect topic you can search

thanks guys.

leo

 

Well, it's not as simple as just saying 4 car lengths per tenth at the end of a 1/4 mile. There are just too many variables. I'm sure if someone really wanted to spend the time, they could come up with some type of complex equation that took several of the variables into account. These variables include traction off the line, acceleration rate(impacted by torque/horsepower curve and such), velocity equation, and things of that nature. However, I'll give a simple little estimation...

**This is by no means a very effective manner to approach this problem, but it'll work for this application.**
Let's assume that a car is running a constant velocity of 100mph since several f-bodies are somewhere in that neighborhood when crossing through the traps in the 1/4 mile. This would mean that the car is traveling roughly 147 feet per second which would be 14.7 feet per .1 second. This is just shy of one f-body length, so by no means would I say a 4 car lengths per tenth. If you want some type of rule of thumb, I would say 1 car per tenth(at most) with a car trapping somewhere in the neighborhood of 100mph.