Ok, after rotating the n-gon, I finally got the catenary curves orbiting the n-gon to behave as advertised. Here are the results for the n-gon inscribed in the unit circle "orbit" problem: The acceleration vector a, |a|=1: a = [-1/(1+(Bt)^2), -Bt/(1+(Bt)^2)] (B>0) We focus on the path segment from z=[z0x,0] to z=[cos(pi/n),sin(pi/n)], where t goes from t=0 to t=tau. The velocity vector starts vertically v=[0,v0y] when t=0, and becomes parallel to [-sin(pi/n),cos(pi/n)] at t=tau. Let D = 1+cos(pi/n)*asinh(tan(pi/n))/tan(pi/n). B^2 = D/2/cos(pi/n)^2 B = sqrt(D/2)/cos(pi/n) v0y = sqrt(2)(sqrt(D)-cos(pi/n)/sqrt(D)) z0x = 2*(cos(pi/n)^3-1)/cos(pi/n)/D+(2-cos(pi/n)^2)/cos(pi/n) tau = sin(pi/n)*sqrt(2/D) Time = 2*n*tau Percent2pi = Time/2/pi n=2,B=1.1548311487678426E+16, v0y=1.414213562373095, z0x=4.595339800345407E-15, Time=5.656854249492381, Percent2pi=0.90031631615711 n=3,B=1.661428901969911, v0y=1.059537362214989, z0x=0.96408602060014, Time=6.255040365008328, Percent2pi=0.99552059333041 n=4,B=1.274058570137272, v0y=1.016897637016049, z0x=0.99490419407429, Time=6.279146177038037, Percent2pi=0.99935715247218 n=5,B=1.156505513430619, v0y=1.006591443623277, z0x=0.99878961735182, Time=6.282222778602842, Percent2pi=0.99984680881914 n=6,B=1.102711571990029, v0y=1.003097017726408, z0x=0.99961516043849, Time=6.282878865388524, Percent2pi=0.99995122827418 n=7,B=1.073049873498318, v0y=1.00164592643468, z0x=0.9998520082811, Time=6.28306738560551, Percent2pi=0.99998123219859 n=8,B=1.054794889128636, v0y=1.000955201356099, z0x=0.99993488083036, Time=6.283133399939409, Percent2pi=0.99999173870614 n=16,B=1.013052493113118, v0y=1.000058267649409, z0x=0.99999902987078, Time=6.283184533299711, Percent2pi=0.99999987683319 n=32,B=1.003225198474834, v0y=1.000003619879664, z0x=0.99999998501889, Time=6.283185295227248, Percent2pi=0.99999999809773 n=64,B=1.000803965385648, v0y=1.000000225903149, z0x=0.9999999997666, Time=6.283185306993373, Percent2pi=0.99999999997036 Note that the main vertex does not lie on the X axis, so z0x /= 1, but slightly smaller, as the curve bows in slightly inward from a circular path. It is this slight bowing that produces the advantage over the circular orbit. So the catenaries always beat the circle (just barely). It's amazing how close the circle is to being optimal even at n=3. Note that the formula even "works" for n=2! (Sorta. The 2-gon line segment goes from [0,-1] to [0,+1]. B = oo, z0x=0.) This formula may be the most obscure formula for pi ever invented! I apologize for bothering everyone with my bad math. I thoroughly enjoyed the exercise, tho. It was fun getting Macsyma to produce closed form solns for this problem.