Thanks for that one Malmo.
To those still wondering, I have worked out a couple of examples of weird tracks below. The first one is the "square" track like the one I mentioned. Consider a normal and a square track, each with 100 meter straights (the value is chosen simply because it's nice, other values will also work). For a normal track, that means that you have 200 m of curve, or basically a circle with a 200 m diameter. Again, for ease of calc, assume the outer lane is 10 m beyond. As we have seen, this means that you'll travel 2pi*10m = 62.8 meters further. For the square track, the lane one "turn" is just another 100 meter straight (and let me tell 'ya, if you've never run on one, it's an ugly, ugly thing). Now the outside line is also 100 meters long, but includes two extra 10 meter segments per corner, for a total of 80 extra meters instead of 62.8. This is only true because THESE CORNERS ARE NOT RADIUSED. If they were properly radiused, we'd also get 62.8 meters. (I'm tellin' ya, this was a track from hell).
Example 2. The elliptical track. This is borderline pedantic as I can't imagine why anyone would do this. I mean, the square track is stupid for runners, but I can understand the poor custodian laying out 4 110 yard straightaways. Anyway, the thing to remember is that with a true ellipse, as you add lanes the ellipse becomes less and less eccentric (i.e., more like a circle). This means that the rate of circumferential increase decreases (the lanes don't increase in length in a nice linear fashion). Recall that a circle has the highest area to circumference ratio possible of any 2-D object, and thus by this measure, the more eccentric the ellipse, the less "efficient" it is (similarly a sphere has the least surface area for any given volume contained. This is why you don't see single soap bubbles in the form of cubes, tetrahedrons, and so on, but I digress). I will dispense with the 400 meter requirement and just do a calculation with "nice" numbers to prove the point.
A decent approximation for the circumference of an ellipse is 2pi*sqrt( (a^2+b^2)/2 ) where a and b are the short and long axii lengths. Choose something with moderate eccentricity, say a=1 and b=2 (it's more dramatic with a bigger delta, but you should be able to calc these values in your head). So, the circumference is 2pi*sqrt(2.5). This means that the equivalent circle has a radius of sqrt(2.5) or about 1.581 (yes, I used a calculator to get that to 4 significant digits ;-). Now, consider an outside lane that is (nice number) 1 unit further out. For a circle this yields a circumference of 2pi*2.581, or about 16.2. For the ellipse, this means that a=2 and b=3, so circ=2pi*sqrt( (2^2+3^2)/2 ) or about 16.0. Thus, we see that the outer lane of the ellipse is shorter than the outer lane of the corresponding circle. As they don't agree, it's obvious that the equation derived for a circle is not applicable to an ellipse. Why is it shorter? As I mentioned, the outer lane creates an ellipse that is less eccentric than the inner lane (i.e., it looks more like a circle) and thus will generate smaller length increases. In order to keep the length increase consistant to that of the circle, you'd have to keep the eccentricity of the ellipse constant (basically, keep the ratio of a to b constant). This would necessitate making the lanes wider when they cut the longer "b" axis" than when they cut the shorter "a" axis. Now THAT would truly be a "track from hell"!!
Well, that was fun, but considering what the original poster asked, I'm reminded of a saying about certain professors: "If you ask him for the time, he'll give you the history of clocks."