glueguy
glueguy GRM+ Memberand Reader
10/23/12 10:30 p.m.

Reading the STF projection article as the plane was taking off, and one of the comments made me sit and think for most of the flight.

Regarding the Fiat 500 being really narrow - "The rule of thumb says that every 6 inches of extra space a driver leaves between the car and a cone, 0.07 seconds is added to the time."

Who came up with that rule of thumb and how is it calculated? What are the assumptions?

So if width matters, then the course-traveled-distance must be based off of the centerline of the car? If you underdrive the course, say at 2 MPH, a narrow car will still be faster than a wide car because its centerline travels less distance because it's more narrow?

Jaynen
Jaynen HalfDork
10/23/12 10:42 p.m.

What article?

glueguy
glueguy GRM+ Memberand Reader
10/23/12 10:56 p.m.

"Street Fight" by Andy Hollis, October 2012

mad_machine
mad_machine GRM+ Memberand MegaDork
10/24/12 5:45 a.m.

that sounds entirely logical. in that regards, a smaller car WILL travel a smaller arc around the cones, and hence, a shorter journey from start to finish...

foxtrapper
foxtrapper PowerDork
10/24/12 5:53 a.m.

petegossett
petegossett GRM+ Memberand UltraDork
10/24/12 6:21 a.m.

If you take some random values, say a 1-minute time for an autocross run and an average speed of 40mph, then work backwards, you get a course distance of 0.667-miles.

To(completely over) simplify the non-standard nature of autocross courses, we can assume the course is one large constant-radius circle, with a circumference of 0.667-miles. So(if my math is correct) the diameter would = circumference/pi or 0.212-miles, which = 1119'-4.32".

Now lets assume another car which is 6" closer to the apex, thus giving it 1'-less total diameter around the circle, or 1118'-4.32". Working the formula the other direction, we end up with a circumference of 0.6656-miles.

Converting both to feet gives us roughly 3,520' vs 3,514.5', or approximately 5.5' less travelled by the 2nd car. If both cars had the same 40mph average speed, that would translate into a time difference of 0.1-seconds around the same lap.

Of course, autocross isn't that simple. With the amount of transitioning involved, I can see width having an even greater impact on total time.

Not that I provided a direct answer to your question at all.

Andy Hollis
Andy Hollis
10/31/12 5:39 p.m.

That data comes from a theoretical discussion started by Byron Short back in the pre-web days of the Team.net autocross mailing list. He did the math that showed optimal slalom speeds and technique, given the physics of motion and vehicle dynamics. He's the guy who built the Geez cube and associated autocross analysis software. A lot of really good information came out of that period of the sport's history. Mid-to-late 90's.

JohnyHachi6
JohnyHachi6 HalfDork
10/31/12 6:39 p.m.

On the flip side, a wider track width means higher ultimate grip during sustained cornering, which probably translates to faster speeds during and after any long sweepers.

kevlarcorolla
kevlarcorolla Reader
10/31/12 6:44 p.m.
JohnyHachi6 wrote: On the flip side, a wider track width means higher ultimate grip during sustained cornering, which probably translates to faster speeds during and after any long sweepers.

Yes and no wheels up antics either so it must be faster.

You'll need to log in to post.

Our Preferred Partners
tnkcSzWq9kspoacdyGDvQbWkJk5FJnNrUDYb4Z9nhHUwpd3m6BI1PZ9dQOGgCtol