windings around an embedded torus This question's not getting anywhere on math.stackexchange.com, so let's see if someone here can say something:
In the $xyz$-space imagine a circle of radius $r>0$ in the $xz$-plane, whose center is on the $x$-axis at a distance $R>r$ from the $z$-axis, and revolve it about the $z$-axis, getting a torus embedded in $\mathbb R^3$.
Its intersection with planes parallel to the $xy$-planes let us call "parallel circles".
A curve winds $m$ times around the long way and $n$ times around the short way (and $m$ and $n$ are not both $0$), and returns to its starting point.  It is situated so that it meets all parallel circles at the same angle.
As a function of $m$ and $n$ and $r$ and $R$ and the position on the curve, what are


*

*the length of the curve;

*the angle at which it meets the parallel circles;

*the curvature;

*the torsion?


When $m=0$ or $n=0$ the answers are obvious.
When $m=n=1$ the answer is surprising: the torsion is everywhere $0$ and the curvature is constant (and equal to $1/R$, so the arc length is $2\pi R$).
Are there other cases where the answer is surprising or elegant or of interest for other reasons?
 A: I think those (1,1) curves are the Villarceau circles.  Remarkably, from knowing that they have length $2\pi R$, you can calculate the length of the $(m,n)$ curves when $n\ne 0$.
There's a conformal map from the embedded torus to the flat torus formed by gluing the edges of a rectangle, and it takes all the parallel circles to lines parallel to one of the edges.  (One way to see this is to draw circles on the torus orthogonal to the parallel circles.  These form a grid, and if they're spaced right, the grid is made of approximate squares.  The conformal map stretches and shrinks so that all the squares have the same size.)  On the flat torus, it's easy to calculate all these curves -- they're just straight lines with slope is determined by $m$, $n$, and the side lengths of the rectangle.
Okay, so what are the side lengths?  Consider the torus formed by gluing the edges of a $a \times b$ rectangle.  There's some $a$ and $b$ and some metric $dg^2=f(x,y)(dx^2+dy^2)$ that makes it isometric to the embedded torus.  In fact, by symmetry, $f(x,y)$ depends only on $y$, so $dg^2=h(y)^2(dx^2+dy^2)$.  Take a closed curve of constant slope that winds $m$ times around the long way and $n$ times the short way.  If $n\ne 0$, this curve spends the same amount of time at each $y$-coordinate, so its length wrt $g$ is its euclidean length times the average value of $h$.  If we call that average value $M$, we get a length of
$$\sqrt{(Mma)^2+(Mnb)^2}.$$
But we already know the lengths of two of these curves: when $m=1,n=1$, it has length $2\pi R$, and when $m=0,n=1$, it has length $2\pi r$.  Therefore, $Ma=2\pi \sqrt{R^2-r^2}$ and $Mb=2\pi r$.  So, the curve that goes around $m$ times the long way and $n$ times the short way has length 
$$2\pi \sqrt{m^2(R^2-r^2)+n^2r^2}$$
and intersects the parallel circles at slope
$$\frac{nr}{m\sqrt{R^2-r^2}}.$$
