Curvature and Osculating Circle of a spiral

Question

I want to clear out a misunderstanting in an excercise i came across recently. Let $r(t)=(Rcost, Rsint, vt)$ be a curve (the usual spiral). Then with a bit of calculations you find that the curvature is $ k=R/(R^2+v^2)$ and torsion is $ \sigma = v/(R^2+v^2)$. The excercise says that if $k=\sigma =1$ then find R,v. Solving the system we easilly find R=v=1/2. My question is this : The spiral is just a circle who is by some sense going upwards inside a cylinder if you like, so shouldn't the osculating (adjacent? i m not sure what the right word is) circle be the base of the cylinder? The radius of an osculating circle is $r=1/k$ where k is the curvature, which here gives us $r=1$. On the other hand, the circle of the spiral (the base of the cylinder) has R=1/2. So clearly the two circles are different. Shouldn't those two circles be the same?!

Answer 1

The osculating circle is not in the horizontal plane here. It's in the osculating plane, which contains in particular the upward tangent vector to the helix.