Take $1$ step forward, turn $90$ degrees to the left, take $1$ step forward, turn $90$ degrees to the left ... and keep going, alternating a step forward and a $90$-degree turn to the left.
Where do you end up walking? It's very easy to see that you end up walking on the perimeter of the same square, with side equal to $1$ step (every $4$ steps you are back at the point where you started), if you are walking on a plane. But what happens if you walk on the surface of a sphere?
The answer obviously depends on how long is your step in relationship to the sphere's radius $R$. If your step length is $\pi R$, for example, you end up walking back and forth over the same lune, formed by two half circumferences at $90$ degrees angle. If your step length is $\pi R/2$, you end up walking on a triangle, formed by quarter circumferences at $90$ degrees angle. I believe one can prove the following:
a) For any $x\geq 0$ and any arbitrarily small $\epsilon>0$, there is a step length $y\in(x,x+\epsilon)$ such that the process above takes you back to the starting point in a finite number of steps.
b) No step length that brings you back to the starting point in a finite number of steps is a rational multiple of the length of the circumference, $2\pi R$, except for integer multiples of $\pi R/2$ (which keep you cycling over the same $1$,$2$, or $3$ points).
c) Any step length that does not bring you back to the starting point in a finite number of steps eventually brings you all over the surface of the sphere, in the sense that for any point $P$ on the sphere's surface and any arbitrarily small $\epsilon>0$, you eventually end up within distance $\epsilon$ of $P$. In fact, for any point $P$ and any angle $\alpha$, you eventually end up within distance $\epsilon$ of $P$ and with an angle between $\alpha$ and $\alpha+\epsilon$ of your starting direction (measured as the angle between the respective great circles)!
Can you prove any or all of the above? Note that a proof of b) would automatically yield an answer to this related question.