Given two different points (point $A$ and $B$) in 3D space ($\Bbb R^3$) and a given perimeter length, $X$, how could I go about a generating the complete (if possible) series of smooth functions which cross both points $A$ and $B$, with a length of $X$? Is there a viable mathematical functional to be able to do this?
Any source/book would be helpful.
1. A good first-order approximation is to rotate the line segment from $A$ to $B$ in space to be the $x$ axis, and parametrize the curves you want as $\vec p(t)=<t, y(t), z(t)>$. Then the exact arc length is $L=\int_a^b \sqrt{ 1+ (\dot y) ^2 + (\dot z)^2} dt$. A Taylor expansion shows this is approximately $$\int_a^b 1+ \frac{1}{2}[(\dot y) ^2 + (\dot z)^2 ]dt$$ Thus the class of functions of constant total length is approximately that class of functions for which the Total Squared Energy Integral $$I= \frac{1}{2} \int_a^b [(\dot y) ^2 + (\dot z)^2 $$ is constant. One convenient way to describe that latter family is in terms of Fourier series. Assume for simplicity that units have been chosen so that the parametrizing interval is $0<t<2\pi$. Then expand the vector $$ \vec r(t)=<y(t), z(t)>= \sum_{k\geq 1} \vec A_k \frac{\sin (kt)}{k} $$ This has been constructed so that $\vec r(0) =\vec 0$ and so that $$\vec {\dot r} =\sum_{k\geq 1} \vec A_k \cos (k t)$$ Recall that all these cosine functions have average value zero on the specified interval. Thus the total integral $\int_0^{2\pi} \vec {\dot r} = \vec r(2\pi)- \vec r(0) =\vec 0$, which ensures that the curve $\vec p(t)$ that starts on the $x$ axis returns there at the end.
Observe that $\vec {\dot r} \cdot \vec {\dot r }= \sum_j \sum_j (\vec A_j \cos (j t)\cdot (\vec A_k \cos ( kt))$
You can use the mutual orthogonality of the various oscillating trig terms to confirm that the Total Squared Energy is $ I=\pi \sum_k \vec A_k \cdot \vec A_k )=\pi \sum_k ||\vec A_k||^2 $.
Conclusion: you have complete freedom to specify the numerical values of all the Fourier coefficient vectors $\vec A_k$, as long as the sum of all their squared lengths is held constant.
2. A more precise mathematical solution that does not use the Taylor approximation is probably too difficult to actually implement. FWIW, here is another attempt at a solution. (It is probably technological overkill for your applications however.)
Given any reasonably smooth initial curve from $A$ to $B$, parametrized by arc length $s$, one can treat it as the edge of a narrow ribbon of paper that twists in space. (Imagine a narrow track on a twisty amusement park roller coaster.) If the two opposite end segments of the ribbon happen to be oriented parallel to one another in space, then when you draw an adjacent interior stripe on the ribbon, its ends will be close to being parallel copies of $A$ and $B$. That is, you can regard that parallel curve as another approximate solution of the problem of connecting A to B using the same length.
Note: if you can construct a family of equal-length paths that connect given points on the flat piece of paper, you can transfer that family directly onto the twisted paper version.
In more detail, the curvature and torsion of the Frenet frame for the original ribbon edge determines the shape of the paper ribbon surface (The ribbon surface is called a developable surface because it is isometric to a portion of the plane.) The construction of the paper surface is described in this classic paper:
On the Geodesics and Geodesic Circles on a Developable Surface William Caspar Graustein Annals of Mathematics, Second Series, Vol. 18, No. 3 (Mar., 1917), pp. 132-138
Note that every new way you twist the ribbon generates a new paper shape that creates a new class of nearby solution curves. Thus there are lots of solutions!
As I recollect, there is also a nice article about twisted paper surfaces written by S. Tabichnikov. It is reference [7] in the following preprint:
Tabichnikov