I've now analyzed the square and concluded that the catenary paths as described in the "Baserunner" paper are _not_ optimal for an "orbit" ! [Carozza, et al. Baserunner's Optimal Path. Math. Intelligencer (October, 2009).] The time for a circular path about a square is 99.8% of the time for the catenary path. Very, very close, but not optimal; the circle appears to be optimal. Here's the math. We place a square with vertices at (+-1,0), (0,+-1). (x0,y0)=(1,0), (v0x,v0y)=(0,v0y), (a0x,a0y)=(-1,0). Acceleration. ax=-1/(1+(Bt)^2) ay=-Bt/(1+(Bt)^2) for some B>0 to be determined. At t=tau (tau to be determined), ax=ay. (We assume that acceleration is smooth (no jerk) for the entire orbit.) Solving for tau, we get tau=1/B. Velocity. vx=-asinh(Bt)/B vy=-(sqrt(1+(Bt)^2)-1)/B+v0y At t=tau, vx=-vy. Solving for v0y, we get v0y=(asinh(1)+sqrt(2)-1)/B ~ 1.295587149392638/B. Position. x=(B^2-Bt*asinh(Bt)+sqrt(1+(Bt)^2)-1)/B^2 y=-(asinh(Bt)+Bt(sqrt(1+(Bt)^2)-2*asinh(1)-sqrt(8)))/(2*B^2) At t=tau, x=y. Solving for B, we get B=sqrt((3*asinh(1)+2-sqrt(2))/2) ~ 1.270808246488339. Backsubstituting, we get v0y ~ 1.019498538015292, tau ~ 0.78690077969145. The total time to "orbit" the square is 8*tau ~ 6.295206237531611 ~ 1.001913190486088*2pi. So, the "Baserunner" catenary orbit around a square is approx .2% _longer_ than a circular orbit. --- If we do the same calculations for an equilateral triangle, we get B = sqrt(7*asinh(sqrt(3))/(2*sqrt(3))) ~ 1.631321852686744. v0y = sqrt(2)*(asinh(sqrt(3))+sqrt(3))/sqrt(7*sqrt(3)*asinh(sqrt(3))) ~ 1.079091776648307. tau = sqrt(3)/B=sqrt(6*sqrt(3)/(7*asinh(sqrt(3)))) ~ 1.061746831084396. The total time to "orbit" the triangle is 6*tau ~ 6.370480986506378 ~ 1.013893538875424*2pi. So, the "Baserunner" catenary orbit around a triangle is approx 1.4% _longer_ than a circular orbit. --- For n=5, time = 1.000452457288549*2pi. For n=6, time = 1.000143694223457*2pi. For n=7, time = 1.000055234883237*2pi. For n=8, time = 1.000024299612241*2pi. For n=16, time = 1.000000361898094*2pi. For n=32, time = 1.00000000558828*2pi. For n=64, time = 1.000000000087061*2pi. For n=128, time = 1.000000000001358*2pi. --- So, it appears that the _only_ time the catenary orbit wins is for n=2. My only assumption that may not be warranted is the assumption of "no jerk". Perhaps a very small jerk half-way between vertices could help, but I'm not convinced.