Can we show that geodesics with a rational radius $ r_{mid-equator} =a q/p \,(p>q ) $ at mid-equator on a spindle type surface of revolution of constant Gauss curvature $ (K=1/a^2 \, $in $\, \mathbb R^3 ) $ form closed loops around the symmetery axis in terms of $ (p, q)$? Is the arc length of a loop/orbit expressible as being related to $ (2 \pi a, p,q)?$

If a Riemann sphere radius $a$ is isometrically mapped/deformed without twist (roll squeezed) along equator will its geodesics deform this way due to isometry?

The figure shown below is drawn for $ q=2,p=3 $ respecting geodesy/constant $K$ in the present investigation. In all such cases numerical work established/ confirmed existence of such closed loops.

What existing related literature can be found on the topic?

Thanks in advance of your response.

EDIT1:

At first $k_g=$ for constant geodesic curvature closed loops were expected (by me) but not the entire geodesic line closing in as a single loop line as that would be counter-intuitive. In the picture shown numerical value $a k_g=1.4$ is assumed.

EDIT2:

The following geometrical results of radius and arc length are now verified in this investigation and reported here:

Radius variation with respect to arc length $s$, $ \alpha$ being angle between arc projection and symmetry axis at mid-equator start:

$$ \left(\frac{r}{r_{mid-equator}}\right)^2 + \left(\cos \alpha \cdot \sin \frac{s}{a}\right)^2=1 \tag1 $$

which can be also written $ r_o= r_{cuspidal-equator}$

$$ \cos \dfrac{s}{a} =\sqrt{\dfrac{r^2-r_0^2}{r_1^2-r_0^2}} \tag2 $$

Arc length of a single circuit $ L_{circuit}$ from right cusp equator to right cusp equator back via left cusp equator is found to be:

$$ L_{circuit}= 2 \pi a \tag 3$$

as can be expected in full isometric mappings.

momentaround symmetry axis $F\cdot ( r \sin \psi) $ is constant, $\psi$ is the angle thread makes to meridian. 2) A physical action of winding of string on a cylinder should result in a thick bead / lump at the ends but not at middle of cylinder after several cycles of geodesics are gone through, as mentioned in his reply. ( contd).. $\endgroup$ – Narasimham May 28 '18 at 9:49