The ability to separate the bridge pieces grants us
exactly two additional track layouts which both have the same footprint as the known solution.
To answer this question by hand, we need to understand the two constraints that a configuration of the tiles needs to fulfill to form a closed loop.
The first (and much simpler one) is that we need the orientations of the start and end piece to match up. Since placing a curved tile changes the orientation in increments of 45°, the start and end will match if and only if the numbers of left and right turns differ by a multiple of 8. With twelve such turns, there are evidently only two ways of doing this - either by having a split of 10-2 or an even split of 6-6 (with the intersecting track being the only example among the nine known patterns).
The other constraint of course is that the locations of the first and last piece also need to match up. Getting a handle on this is more demanding. To express the problem precisely, it helps to introduce some notation: I will write $\zeta:=e^{\frac{2\pi i}{8}}$ for this specific primitive eighth root of unity and also write $\lambda:=\zeta-1$. Furthermore, we identify the plane with $\mathbb{C}$ and also assume (wlog) that the start of some piece starts at the origin and parallel to the real line. In that frame of reference, we can handily express the displacement caused by a single piece of track: A straight piece can be placed in any of eight possible directions and therefore causes a displacement of $\zeta^k$ for some $k$. On the other hand, a curved piece is exactly the displacement between two adjacent eighth roots of unity (left and right turn alike) and so corresponds to a number of the form $\zeta^{k+1}-\zeta^k=\zeta^k\lambda$ for some $k$.
Hence, the total displacement of an arrangement of track pieces will be a sum of four terms of the form $\zeta^k$ and twelve terms of the form $\zeta^k\lambda$. The track closes up if and only if this sum is exactly 0. Problem solved! :)
But ... how can we tell which of these sums will be 0 without going mad in the process? There may well be multiple ways forward, but this approach proceeds by leveraging some facts about the arithmetic of the ring $\mathbb{Z}[\zeta]$ in which all of these horrible expressions live.
To bring those guns in, let's first simplify the problem a little bit. Suppose we have some closed track that has at least as many right turns as left turns (wlog). Then, if we cut the track right before some left turn, we can pair off each left turn with the next right turn that exactly counteracts its orientation. After forming these pairs, we are either left over with only the straight pieces (if the split was 6-6), or an additional full set of eight right turns, one in each orientation (if the split was 10-2). In the latter case, these leftover right turns exactly close up and so don't cause any net displacement. Moreover, note that each turn in a matched up pair causes the same displacement. So, the total displacement of the track can be simplified to a sum of four terms of the form $\zeta^k$ and two (or six) terms of the form $2\zeta^k\lambda$.
Here is where knowledge of some algebraic number theory comes in: The number $\lambda$ is a prime in $\mathbb{Z}[\zeta]$. In fact, it is the only prime divisor of 2, which is exactly divisible by $\lambda^4$. Therefore, the contribution of the curved segments is a sum of two (or six) numbers that are each exactly divisible by $\lambda^5$. As such, the sum must even be divisible by at least $\lambda^6$! So, to cancel it out, the sum of the four straight segments must match this, which (finally) gives us a usable restriction:
First, suppose that no two of the straight segments point in the same or opposite direction. Then, their contribution must be of the form $\pm 1\pm \zeta \pm \zeta^2 \pm \zeta^3$. However (as these four summands are a $\mathbb{Z}$-basis of $\mathbb{Z}[\zeta]$), this number is not even divisible by 2, much less $\lambda^6$. So this is impossible. Next, lets suppose that two of the segments point into opposite directions. Then, their joint contribution is 0, so the other two must also cancel out in order for their total sum to be sufficiently divisible. Finally, suppose that no two cancel out but that two of them point in the same direction. Then, their sum is already divisible by exactly $\lambda^4$. Hence, the only way for the remaining two to also be divisible by $\lambda^4$ but not cancel out, is for them to also be parallel. As such, the straight segments will contribute a number of the form $2\zeta^k+2\zeta^\ell$ for some $k,\ell$. In order for this total to be divisible by $\lambda^6$, it is then necessary that $k$ and $\ell$ differ by an even number.
In total, there are therefore three ways the straight segments can behave - they either
(In the provided examples, only numbers 1, 2 and 6 are instances of the third case. All others are of the first kind.)
Now, to see how this restriction reflects back on the curved pieces, we will first look at the case of an uneven 10-2 split between left and right turns.
In this case, there are only two matched off pairs coming from the left turns, so this side of the balance gives a contribution of $2\lambda(\zeta^k+\zeta^\ell)$ for some $k, \ell$. We can compare this to the now known possible displacements from the straight pieces: In the first case (cancelling pairs), the straight pieces have no net displacement. Hence, the curved pieces must also cancel out, which happens if and only of $k$ and $\ell$ differ by 4 (mod 8). In terms of turns, this requires that we have 5 (mod 8) right turns between the two left turns. Since there are only ten right turns available, the only way to do this is by having five on each side.
Hence, the track with all straight segments taken out must look exactly like this:
The full track then differs from this by inserting straight segments in the gaps in such a way that, for each of the letters, the number of segments in the upper case gap(s) matches the number of segments in the lowercase ones. Most of the given examples in the OP fall in this category. However, as they all must put the bridge in only one gap, the other opposite straight segments must go in gaps of the same letter. If we separate the pieces, we can mix and match the letters as we like. However, clearly none of these tracks will form an intersecting path, so these are of no actual use to the question.
On to the second case: In this option, the straight segments are all parallel, which nets a displacement of $4$ units in some direction. As it turns out, this is actually impossible to match (the curved displacement cannot be divisible by $\lambda^8$ without cancelling out completely). So, the only remaining case is that the straight segments come in two parallel pairs of orthogonal lines. The only way to match this is for $k$ and $\ell$ to be consecutive (mod 8). Here, there are two cases to distribute the turns - we can either have two and eight right turns in the gaps or ten and none. For both of these possibilities, we can draw another picture as above. Sadly, it turns out that in each case, there is only one gap to fit in each parallel pair. So the ability to separate the pieces is of no additional use here as well, and we are left only with the already known patterns (numbers 1 and 6 in the OP).
This exhausts all the tracks with an imbalance of turns, so it "only" remains to tackle the case of an even split. Here, there are six matched pairs, so the situation looks a lot more complicated at first glance. It pays to first take a step back and consider the forming of matching pairs again. Now, if we step through the track tile by tile, it is clear that the orientation changes only one step at a time as well. As such, the set of orientations must form an unbroken segment of the circle. Now, since each of the left turns is exactly matched by a right turn, this shows that also the (multi)set of orientations of left turns occurring on the track must form an interval of the circle all by itself. In other words, we can guarantee that the exponents $k$ in the terms $2\zeta^k\lambda$ in the total displacement must cover some interval. By turning the track, if necessary, we can assume that this interval is $\{1,...,\ell\}$ for some $\ell$.
Now, if $\ell=6$, every orientation of a left turn occurs only once. This is clearly only possible if all the right turns (and all the left turns) occur in a row. This is the case in example 2, the only (vanilla) self-intersecting track. In that example, the straight segments are paired off in orthogonal parallel pairs. From our previous analysis, we then conclude that this is in fact the only possible way to account for this exact displacement of straight pieces. Unfortunately, we can see by inspection that there (again) is only one gap in which we can possibly fit each orientation of the pairs. So, the only way of obtaining a new solution here is by permuting the straight pieces around. If the two bridge pieces are aligned, there are two cases. If the elevated parts match right up, we have exactly the original solution. Otherwise, we can flip both pieces around and match them at ground level. This yields a track that is mostly elevated, with the bridge pieces forming a tunnel under the other straight segments. The only other possibility is to split the bridges up. Then, in order to not intersect at the same level, one of the pieces must elevate towards the crossing and the other one away from it. (This is exactly the layout in Lezzup's answer.)
What about a shorter interval though? Well, if the interval is of length $\ell=4$ or shorter, each of the left turns fits into the upper half circle and each right turn into the lower half circle. Thus, there will be no turn after which the track ever points into a direction with a positive real part. As a consequence, no tiles whatsoever could then point in a direction with a positive real part, so there is no possibility to form a closed track.
Lastly, suppose that the interval is of length $\ell=5$. In that case, there would be exactly one repeating orientation, giving a term $2\zeta^k\lambda$ for some $k\in\{0,...,4\}$, so the total displacement of the curved track would check out to be $2(\zeta^5-1+\zeta^{k+1}-\zeta^k)$ after some cancellation. The five values obtained in this way are
| $k$ |
|
| $0$ |
$-4$ |
| $1$ |
$2(\zeta^2-2\zeta-1)$ |
| $2$ |
$2(\zeta^3-\zeta^2-\zeta-1)$ |
| $3$ |
$2(-\zeta^3-\zeta-2)$ |
| $4$ |
$4\zeta^5$ |
Out of these, the middle three are useless, but the two extreme ones could in principle exactly pair off with four straight segments that are all aligned. However, if we actually draw out the track layouts corresponding to these values (a succession of left/right turns in groups of 5-1-1-5), we must sadly find out that there is not a single point at which the track would point in the direction needed to insert all the straight pieces. So, after all this, there is unfortunately nothing new to report, it seems.
Solving this was a lot of fun, even though this approach was (clearly!) quite complicated and took some work. Did you solve this by hand, Pranay? If so, I would be very interested to learn about a slicker way through this! :)
