I've computed a generalization of the orbit around the regular n-gon inscribed in the unit circle, n=2,3,... This version simply increments angle of the constant absolute value (=1) acceleration vector discretely, allowing the orbiting body to hop over the next vertex, just grazing it with its parabolic path. Thus, a 12-gon would be like a clock, with a vertex at each number, and the acceleration vector would change angle discretely half-way between each pair of adjacent numbers. Note that n=2 simply moves along the line segment from (1,0) to (-1,0) and back again, but constantly accelerating either towards (-1,0) or (1,0). The cool thing about the parabolic segments is that one can compute their length _in closed form_. n = number of vertices in n-gon. tau = time to go from vertex to next acceleration shift; T = 2*n*tau = total time for the orbit around the n-gon. T = 4*n*sin(pi/n)/sqrt(3+cos(2pi/n)) L = total length of orbit. L = 4*n*(sin(pi/n)+cos(pi/n)^2*asinh(tan(pi/n)))/(3+cos(2pi/n)) r = radius of orbit at acceleration shift (perigee?). r = 4*cos(pi/n)/(3+cos(2pi/n)) R = radius of orbit at vertex (apogee?). R = 1. v0 = velocity at vertex/apogee. |v0| = 2*cos(pi/n)/sqrt(3+cos(2pi/n)) v1 = velocity at perigee. |v1| = 4/(3+cos(2pi/n)) The total time T has the following series approximation: T(n) ~ 2*pi*[1+(1/12)*(pi/n)^2-(11/480)*(pi/n)^4-(491/40320)*(pi/n)^6-(11159/5806080)*(pi/n)^8...] T(1)=0 (I don't know what this means!). T(2)=sqrt(2^5) ~ 90% of 2pi T(3)=sqrt(6^3/5) ~105% of 2pi T(4)=sqrt(2^7/3) ~104% of 2pi T(5) messy ~103% of 2pi T(6) ~102% of 2pi T(7) ~101.6% of 2pi T(8) ~101.2% of 2pi T(16) ~100.3% of 2pi T(32) ~100.1% of 2pi So T(n)>2*pi, except for n=1,2. I.e., this discrete "clock" form of acceleration shifting only betters a circular orbit for the line segment case. Total length L has the approximation: L(n) ~ 2*pi*[1-(7/120)*(pi/n)^4-(41/2520)(pi/n)^6+(83/120960)*(pi/n)^8+...] (Note missing (pi/n)^2 term.) L(n)<2*pi. L(1)=0 L(2)=4 ~ 64% of 2pi L(3)=(6/5)(asinh(sqrt(3))+2*sqrt(3)) ~ 91% of 2pi L(4)=(8*asinh(1)+sqrt(2^7))/3 ~ 97% of 2pi L(5) ~ 99% of 2pi L(6) ~ 99.5% of 2pi L(7) ~ 99.7% of 2pi L(8) ~ 99.85% of 2pi L(16) ~ 99.99% of 2pi L(32) ~ 99.9995% of 2pi L(n)/T(n) is the mean linear speed of the orbiting object over a complete orbit, compared with a circular orbit. L(n)/T(n)=(1+cos(pi/n)*asinh(tan(pi/n))/tan(pi/n))/(3+cos(2pi/n)) L(1)/T(1)=? L(2)/T(2)=1/sqrt(2) ~ 70.7% L(3)/T(3)=asinh(sqrt(3))/sqrt(30)+sqrt(2/5) ~ 87.3% L(4)/T(4)=asinh(1)/sqrt(6)+1/sqrt(3) ~ 93.7% L(5)/T(5) ~ 96.2% L(6)/T(6) ~ 97.5% L(7)/T(7) ~ 98.2% L(8)/T(8) ~ 98.6% L(16)/T(16) ~ 99.7% L(32)/T(32) ~ 99.9%
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Henry Baker