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 Post subject: slalom - 1 foot equals .144 seconds lost.
PostPosted: Wed Dec 03, 2003 4:01 pm 
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Joined: Thu Aug 21, 2003 11:00 am
Posts: 1792
Location: Raleigh, NC
For those of you who have doubted my preaching about distance versus speed, this should make you think just a little. You always wanted to know where to get that extra time. From Byron Short:
Quote:
The slalom formula is:

(pi / 8) * sqrt( d / g)

where d is the lateral displacement of the vehicle in feet, and g is the
peak g capability of the vehicle in g's. When we say "lateral
displacement" we mean "how far do you have to move the vehicle from peak
left to peak right to get around the cones. This usually means the
width of the vehicle plus the width of the cone when the cones are in a
straight line, but you can compute it with some offset either positive
or negative if the course design dictates that.

When you solve for the formula you get a number that is typically around
1, which corresponds to the seconds to go from one cone to the next in a
forever, never-ending, consistently spaced slalom. In the real world we
can and do shortcut the first and last cones of slaloms, but the middle
cones are subject to this formula.

Also, you'll note that there's no variable for the distance between
cones because that doesn't matter. Unlike a radius turn, the speed in
this cases increases in direct proportion to the distance between the
cones so unless you get to ridiculously small gates where hand speed,
power steering pump, weird ackerman angles, etc, become factors, the
distance between cones doesn't matter. However, as Rick pointed out,
the distance you miss the cones by increases d and increases the time
considerably.

For instance, let's take an autocross car that pulls 1g and is 5.5 feet
wide, going past a cone that is 1 foot wide at the base. This means

t = (pi / 8) * sqrt( 6.5 / 1 ) = 1.001 sec

If we miss each cone by a foot on each side, then we have increased d
not by 1foot, but by 2 feet (one foot on each side of each cone). So now

t=(pi / 8) * sqrt( 8.5 / 1 ) = 1.145 sec

For a total difference of 0.144 per cone!

This formula was derived with invaluable help from (yep, you guessed it)
Jay Mitchell back around 1994 or so. I've tested it empirically for
years and still find that it holds water, no matter the size of the
vehicle (even dually pickups and Suburbans, down to shifter karts and
kid-karts), no matter the g-loads, no matter the distances (with the
above caveat not withstanding). This simple formula is a lot of fun and
very useful. In conjunction with a G-Cube you can use this formula to
compute how much you missed the cones by, for instance. And you can
certainly predict how fast a street car will slalom if you know it's
width and have a good estimate of it's lateral g limits.

--Byron

_________________
Jim Pastorius
2008 Silverado VortecMax
1992 Camaro CMC#92
2002 BMW R1150R

2009 3rd Place CMC Mid-Atlantic Championship
2009 CMC Hyperfest Winner


Last edited by jimpastorius on Wed Dec 03, 2003 4:18 pm, edited 1 time in total.

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 Post subject:
PostPosted: Wed Dec 03, 2003 4:16 pm 
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Just call me Bo

Joined: Thu Aug 28, 2003 6:39 pm
Posts: 1431
Location: SYPHAJFD
That's pretty cool, no pun intended. :P

I'm guessing that the "cool" emoticon in the formula is really "8 )" ?

That would make a lot more sense... 8)

Jim


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