Maximum crossings of curvature-constrained curve

Question

Let $C$ be a curve in the plane whose curvature is everywhere $\le 1$. If $C$ has length $L$, what is the largest number of proper self-crossings of $C$ as a function of $L$?

For example, the curve below has length $L=2 \pi (n + \epsilon)$ and has $n(n-1)$ proper crossings, where $n=5$ in the figure. So this pattern achieves $L (L-2\pi) /(4 \pi^2))$ crossings as $\epsilon \to 0$, and so grows quadratically in $L$.

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Q1. Can anyone see a pattern that improves on the above curve? E.g., can more than $20$ intersections be achieved with $L \approx 10\pi$?

Q2. Can the number of crossings grow faster than $L^2$?

Answer 1

There is no upper bound. Consider the curve which runs along the circle of radius $1+\epsilon$ and turns around this circle $n$ times. The length is $2\pi n(1+\epsilon)$, and the curve has infinitely many self-crossings, and curvature less than $1$. Now perturb the curve slightly, so that the curvature is still less than $1$, but it has finitely many self-crossings. It is clear that the number of self-crossing can be made as large as you want, for every given $n$.