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We've probably all had measurements where there are cars or other objects on a course that cause you to ride farther out from the curb than 20cm. On a straight section, no big deal, just do an offset until past the obstacle.

But on a curved section, riding wide has the effect of making the measurement longer and thus shortening the actual course.

I'm sure there's a mathematical trick to figuring out what this extra distance is, no doubt involving Pi (Pete Riegel was great at this). Anyone have a quick and dirty formula for this?

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The difference in arc length is


where W is the offset (the width of the car), and theta is the angle you turned through in radians.  If you are going around a 90-degree turn and stay (1+W) feet from the curb the whole time, then the formula works with theta replaced by pi/2. But for less turn you'd have to estimate theta, and I would be concerned about people, including myself, trying to do that.

BTW, the formula works whether the curvature of the turn is constant or not. The difference between adjacent lanes on a track is 2*pi*W, where W is the width of the lane, even if the turns are not a constant curvature, i.e., circular.

Instead, what I do when faced with this situation is to hop up on the curb and walk my bike past the car on the inside. It's usually much less than 90 degrees, so it's less than a foot shorter than following the true SPR since the W in that case is about 1 foot.

Jim, you mentioned staying 20 cm. from a curb. The rule is 30 cm. (1 foot) from a curb. The 20 cm. is the distance from a painted line on a track.

I still have a letter from Ted Corbitt which starts out, "The old one-meter rule is dead." He goes on to say the rule is now 30 cm. Anyone else still around from the days of one-meter?

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