AW rightly objected to my "foobar" solution. Sorry, my brain malfunctioned. I now reveal the reasoning I had inside the brain... but with error now removed... and considerable more details... View the curve C in the plane as described by curvature=F(arclength). The curve C' of sphere-plane contact drawn on the sphere surface, has the same F in its GeodesicCurvature=F(GeodesicArcelength) description. For polygonal C, we get polygonal (with geodesic edges) C', with the same bend-angles at corners. We want C' to join the N & S poles. We want C to be "closed" which implies integral curvature darclength=2*pi, or for polygonal C the bend angles at corners sum to 2*pi. (The reverse implication would also be true except that you might close a subsegment of C, and there's issues at the two endpoints, and you might need to add straight sections to C...) So anyhow, if C' is smooth in a neighborhood of its start & end, and C is smooth except that there is a corner of interior angle theta>=0 at its start=end point, then C' needs to have integral curvature darclength = (pi+theta). This would be accomplished if C' encloses sphere surface area A and winds round it W times, where W*A=pi+theta and W>0 is integer. (In view of the "angle defect = area" theorem for non-Euclidean geometry.) In view of the isoperimetric theorem on surface of sphere (valid for all areas A less than 2*pi) the shortest closed Jordan curve on sphere surface, which encloses area=pi+theta, is a (non-great) circle with (by Archimedes hat box theorem) circumference = 2*pi*sinR where R is the angular radius of the circle and 2*pi*(1-cosR)=pi+theta=area. Incidentally since theta>=0 we find circle circumference>=sqrt(3)*pi=5.441398... So, consider in the plane, differentiable C of the following form: 1. line segment of length pi/2, then 2. circular arc of arclength Q, subtending angle pi+theta [hence radius = Q/(pi+theta) and curvature=(pi+theta)/Q], then 3. another line segment of length pi/2, closing the curve at its only corner, of interior angle theta<pi. This will yield C' on sphere joining N to S pole, consisting of geodesic segment of length pi/2 to reach equator, then non-great circle of arclength=Q, then finally geodesic pi/2 to reach S pole. The total arclength of C' or of C is pi+Q. And as we saw, Q>=sqrt(3)*pi=5.441398... To make this work: i. The surface area inside the circle on the sphere, needs to equal pi+theta to get the right amount of total curvature. ii. Hence area = pi+theta = 2*pi*(1-cosR), Q = 2*pi*sinR, where R is angular radius of that circle. iii. sin(theta/2)*(pi/2)=cos(theta/2)*Q/(pi+theta) to make C close. Trying to solve these 3 equations for the 3 unknowns Q,R,theta... from iii we have Q = tan(theta/2)*pi*(pi+theta)/2 from iib we have R = arcsin(Q/(2*pi)) from iia we have pi+theta = 2*pi*(1-cosR) so t=pi+theta is expressed as a function of Q, and conversely Q as a function of t, with R eliminated: t=2*pi*(1-Cos[Arcsin[Q/(2*pi)]]), Q=Tan[(t-pi)/2]*pi*t/2 = -Tan[(pi-t)/2]*pi*t/2 = -Cot[t/2]*pi*t/2 hence we can eliminate t: Q+Cot[pi*(1-Cos[Arcsin[Q/(2*pi)]])]*pi*(pi*(1-Cos[Arcsin[Q/(2*pi)]]) = 0 and... there seem to be no solutions for |Q|<=2*pi except for Q=multiple of 2*pi. So... the arclength is Q+pi=3*pi... which is the same solution as Asimov's. This is assuming I have not messed up in all that algebra, and unfortunately it is quite possible I have messed up; little attempt to check. Also, this merely finds an upper bound on the minimum length, not necessarily tight (although it might be; it would be minimum among curves of my class).