Intersection of two congruent spirals Let $S_1$ and $S_2$ be two congruent circular spirals in $\mathbb{R}^3$,
both with their axes passing through the origin. They are congruent in that their
radii are equal, as are their winding frequencies; but aside from being constrained
to have their axes through the origin, one is an arbitrary rigid reconfiguration of the other.
My question is: Do they always intersect,
and if so, where? It appears that they always intersect in two points, as in the example
below (origin: green), but I am not seeing a proof. Any help would be appreciated.
 
 A: My intuition says they usually do not intersect-they are both 1D curves, so a slight perturbation will make them miss.  Let's try two in a special position.  One is along $z$, following $x= \cos (z), y=\sin (z)$, with radius $1$ and pitch $2 \pi $.  The other is along $x$, following $y=\sin (bx+c), z=-\cos (bx+c)$, with radius 1, pitch $2 \pi/b$ and clocking $c$  Looking for intersections, we have $bx+c=z, x=-z=\cos z$.  This has only one solution for $z$, which is about $-0.739$  By selecting $b,c$ correctly, I can make the $x$ part fail easily.  So in this case we get at most one intersection, usually none.
Added:  note that the factor $b$ makes the helices non-congruent.  Following the title, we should set $b=1$
A: No, they don't have to intersect and generally will not.
Each helix lives on a circular cylinder, so if they intersect at all, it must be somewhere on the two orthogonal ellipses where the cylinders intersect. But each helix meets those ellipses only in finitely many points, so if you have a configuration where the helices do intersect, turning one of them by an infinitesimal amount will make it miss the point of intersection without thereby hitting any of the other possible intersection points.

Intuitively, if the winding period is much larger than the radii, then one certainly won't expect any intersection.
