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Originally Posted by Storm Cadet
Bill we spoke about this 2 years ago with CJ and the track super at Sar, remember? Does your equation figure in for the whole trip, one turn turf, two turns? Is your equation for a circle...which a race track is NOT. The straight aways can't figure in to the added distance, ONLY the turns.
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Picture a circle split into 2 semi-circles. Now, separate the 2 semi-circles by a distance of 2 furlongs and connect them by straight lines, which is the approximate configuration of a race track. The equation holds in either case and the straight segment does not add anything (a 2 furlong straight segment is 2 furlongs no matter where it is relative to the hedge). As ArlJim78 mentions, the equation is 3.14159 (Pi) times the "rail out" distance for each turn.
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Also, doesn't the distance out differ from inner and outer turf courses on your equation? The added distance HAS to be greater on the outer Belmont course than the inner with it's huge area and sweeping turns?
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Nope - the change in circumference is linear. Also note that the same equations apply for a horse running off the rail.