(*30Oct13*: Now solved; see **Addendum**.)

Define a curve, a

*ratchet spiral*, $S(r_0,\epsilon)$ as follows, where $r_0 > 0$ and $\epsilon < 1$.

$S(r_0,\epsilon)$ begins with the arc of an origin-centered circle of radius $r_0$, $C(r_0)$, starting from the point $p_0=(r_0,0)$. The arc follows $C(r_0)$ until it can jump radially by less than $\epsilon$ to a lattice point $p_1$ exterior to $C(r_0)$. Let $r_1=||p_1||$. The process continues, always jumping to a larger radius $r_{i+1}=||p_{i+1}||$ from an arc of $C(r_i)$ that enters a disk of radius $\epsilon$ around a lattice point $p_{i+1}$.

In the above example, $p_1=(3,1)$, $p_2=(3,2)$, and so on. Notice that the curve does not jump to $(3,4)$ because the distance is $5-3\sqrt{2} > \frac{3}{4}$.

For sufficiently small $\epsilon$, the curve remains a bounded distance from the
origin. E.g., $S(3,\frac{1}{8})$ is just a circle:

(*Added*.) Answering a question raised by Sam Hopkins in the comments, $S(3,\frac{1}{2})$
is not just a circle but remains bounded when the radius reaches $2\sqrt{5}$:

My questions are:

Q1. For which $r_0$ and $\epsilon$ (if any) does $S(r_0,\epsilon)$ spiral to infinity?

Q2. If any $S(r_0,\epsilon)$ spirals to infinity, how fast does $r_i$ grow as a function of its winding number?

Continuing the same example, after $59$ steps, the spiral has
wound once around the origin, and the radius exceeds $30$:

(*Added*.) And here is $S(10,\frac{1}{2})$,
an interesting contrast to $S(3,\frac{1}{2})$ above:

This question occurred to me while thinking about the Gauss circle problem
and Rational points on circular spirals.

**Addendum**. The main question (

**Q1**) is now solved by Abhinav Kumar's answer to the related question, Lattice-point-free buffers around circles: For every $\epsilon > 0$, there is an $r_0$ such that the rachet spiral $S(r_0,\epsilon)$ is unbounded. The reason is that, for sufficiently large $r_0$, for all $r > r_0$, the "buffer" between $C(r)$ and the next exterior lattice point is smaller than $\epsilon$. Thus the spiral will always jump radially, since the jump is smaller than $\epsilon$. Conversely, for any $\epsilon > 0$, $S(r_0,\epsilon)$ might only be bounded (terminating in a circle) for "small" $r_0$, $r_0$ less than some finite bound. See Abhinav's answer for details.