This is a spin-off of my previous question. I will take a moment to reintroduce the notation. The problem concerns elastic "ropes" in $\mathbb{R}^3$, which are modeled as sequences of points $\gamma=(x_1,x_2,\dots,x_m)$. A rope is supported on the union of line segments $$ S(\gamma) := \bigcup_{j=1}^{m-1} \overline{x_jx_{j+1}} \subset\mathbb{R}^3. $$ The $j$-th segment of the rope has direction $$ \tau_j := \frac{x_{j+1} - x_j}{|x_{j+1}-x_j|}, $$ and the force $F_\gamma$ associated to the rope $\gamma$ is described by the vector-valued measure $$ F_\gamma := \sum_{j=1}^{m-1} \tau_j (\delta_{x_{j+1}} - \delta_{x_j}). $$

This time I am curious about "generic" ropes $\gamma$. We say that $\gamma$ is generic if the points $\{x_1,x_2,\dots,x_m\}$ are in general position. That is, the points are all distinct, no three are collinear, and no four are coplanar. In particular, no two of the line segments $\overline{x_jx_{j+1}}$ and $\overline{x_kx_{k+1}}$ intersect except when $j=k+1$ or vice versa (so there are no loops in $S_\gamma$).

Question: Suppose that $\gamma$ and $\gamma'$ are two ropes with the same force, so that $F_\gamma = F_{\gamma'}$. If $\gamma$ is generic, does it follow that $\gamma'$ is also generic?

What I believe I can show is that, if $\gamma'$ is not generic, it must have at least three loops. I would also be curious if there was a reasonable strengthening of the "generic rope" condition that satisfied this conjecture. So what I am really after is something like the question below.

Question'Is there a set $S$ of "good ropes" that can be described explicitly with the following properties:

- For two ropes $\gamma,\gamma'$ with $\gamma\in S$ and $F_\gamma=F_{\gamma'}$, $\gamma'$ is guaranteed not to have any self-intersections (that is, $S(\gamma')$ has no loops).
- The set $S$ is "generic" or at least "dense".

For example, I am happy to consider only ropes that do not have any corners with angle $2\pi/3$, if that were to help at all.