EDIT: Here are pictures of these two projections to help clarify the situation:
https://docs.google.com/open?id=0B5BVGcL23IkoSTlNNGZKZnZtT1E
With regards to Dylan's question about how this projection arose, I was considering the parametric curve $S(t) = (t,t^2,t^3) \subset \mathbb{R}^3$. In a comment or answer on a past MO post which for the love of me I cannot find now, someone had shown in a very simple way that no two chords of $S$ intersect one another anywhere in $\mathbb{R}^3$ (unless they share an endpoint). Thus, for each $n$, $K_n$ can be embedded in $\mathbb{R}^3$ as the set of chords connecting the points $(1...n)$. Now for a fixed tame knot or link $K$, the Robertson-Seymour theorem implies there is a finite set of forbidden minors for graphs which are $K$-lessly embeddable. Hence, every embedding of $K_n$ for $n$ sufficiently large contains $K$, so one may ask (like I did) "what $n$ is sufficient for a given $K$, and what order must I connect the points $(1...n)$ in order to realize $K$?" The crossing condition comes from looking at the chords projected onto the $yz$-plane as viewed from the $-x$ direction (which I think is the direction I did my crossing calculations from). Finally, I moved the $yz$-projections of the integral $t$-valued points around so that they lay on the unit circle to make the pictures easier/clearer.
