Many runners are intensely interested in their milage. The same goes
for walkers, and this sometimes leads to conflicts between runners and
walkers at the local track. Runners doing speedwork like
to use lane 1 so that they know how far they have run, and walkers
like lane 1 for the same reason!
In fact, as nearly everybody knows, for most modern outdoor tracks
the classic 4 laps in lane 1 is equal to 1600 Meters, which is about 9 meters
short of a mile. If you're willing to call 4 laps a mile then you should be
willing to run or walk several other lane/lap combinations which are as close
or closer to an even-mile distance. Below are your best bets on a standard
outdoor track ( 1 lap in lane 1 is 400 M and each lane is 42 inches wide.)
The first figure is the number of miles you run if you run in the lane
indicated by the second number for a number of laps given by the 3rd number:
.9941939076, 1, 4
22.99692551, 2, 91
6.003999297, 4, 23
12.99521346, 5, 49
7.003709132, 6, 26
Thus, for example, you could do that Sunday long run of 23 Miles
by going to the local high school track and running 91 laps in lane 2.
( The accuracy indicated here is absurd, of course. The milage figures
quoted assume that the track has been measured exactly, and uses the
exact conversion from metric to English. In practice, the accuracy of
track surveying falls well short of this ideal.) Note that if you want
to do a 6 mile run it is _much_ more accurate to run 23 laps in lane
4 than it is to run 24 laps in lane 1 -- where the roughly 9 meter
discrepency gets multiplied by 6.
Here are some similar figures for a 200 Meter indoor track with 36 inch
lanes:
.9941939076, 1, 8
10.99460379, 2, 86
4.994415896, 4, 37
.9948693772, 6, 7
In general, if you run in lane L on a track with lane width W then
the additional distance you run per lap over the distance you would run
in lane 1 is 2*Pi*W*(L-1). (Don't forget that you must measure W in the
same units you want the answer.) To see why this holds, note that most tracks
consist of two semicircular endpieces attached to parallel straightaways. Since
all lanes run the same distance along the straightaways, for the purpose of
computing the excess distance in Lane L we can cut out the straightaways and
assume we are running around a circle. The result follows by comparing the
circumference of two circles: Lane 1 with radius r (say), and Lane L with
radius r + (L-1)W. The excess is the difference of the circumferences:
2*Pi*(r + (L-1)W) - 2*Pi*r = 2*Pi*(L-1)*W.
What happens when the track has a different shape? Some tracks, for example
those in health clubs, may have odd shapes, and the derivation above does not
apply. Surprisingly, the same result always holds. (Well, almost always.)
Here is a simple way to see it. Suppose the inside edge of the track is
marked by a simple closed curve. Imagine taking a rod A-------B of length
W and carrying it around the track so that end A traces out the inside
edge curve. Then, in the course of this motion, end B will trace out the
inside edge of "Lane 2." The question then is, "How much further does B
move than A?" To answer this we can subtract out the motion of A. We are
left with A fixed and B tracing out a circle. Thus, the extra distance B
moves is just the circumference of this circle: 2*Pi*W.
If the shape of the track is everywhere convex (i.e., any two points in
the "infield" can be connected by a line segment which does not leave
the infield ) then the argument just given is absolutely correct. If
the track is allowed to curve back in on itself ( imagine, e.g., the shape
of the outline of a kidney bean ), then there can be a problem if the lane
width is too large. The difficulty is that end B of the rod described above
can actually retrograde, i.e., traverse some portion of its circle of
motion more than once. The technical condition which must be met is
|W/R| < 1, where R is the minimum radius of curvature of the inside curve.
It is a nice calculus exercise to prove, subject to this restriction, that
the excess distance in lane 2 is indeed independent of the shape of the
track. It all boils down to the "Gauss-Bonnet Theorem" which says
that the integral of the curvature around a simple closed plane curve is
equal to 2*Pi.