Anyone can measure the length of any track via Google maps = just bring up the visual image of the track, then right click and select "Measure Distance" - and move/click the cursor around the inner periphery of the track.
I did the Meadowlands 10 times and came up with these distances in feet :
5216
5218
5222
5213
5204
5202
5214
5217
5215
5214
for an average of 5214' with a standard deviation of 6'.
But I was measuring exactly along the boundary with the turf course - all horses will be racing at least N feet away from that boundary and therefore will travel 2*pi*N additional feet beyond 5214.
For N=0, D = 5214'
For N=1, D = 5220'
For N=2, D = 5226'
For N=3, D = 5232'
For N=4, D = 5239'
For N=5, D = 5245'
For N=6, D = 5251'
For N=7, D = 5257'
For N=8, D = 5264'
For N=9, D = 5270'
For N=10,D = 5276'
For N=11,D = 5283'
I.e., any horse racing within 11 feet of the turf course boundary will travel slightly less than 1 mile.
These slightly shortened distances can be converted into times in seconds. For a 1:50 pacer
For N=0, t = 0.60
For N=1, t = 0.55
For N=2, t = 0.49
For N=3, t = 0.43
For N=4, t = 0.38
For N=5, t = 0.32
For N=6, t = 0.26
For N=7, t = 0.20
For N=8, t = 0.15
For N=9, t = 0.09
For N=10,t = 0.03
For N=11,t = -0.02
So horses racing about 3-7 feet away from the turf course boundary will have times that are about 1 to 2 fifths of a second too fast.
