I can follow the math to calculate the radius of curvature of a helix, but what has me confused is there is no $z-$component to the normal vector, so the normal vector points at the $z$ axis at all times. Therefore, I would think the radius of curvature would just be $r$. If the radius of curvature is $(r^2+h^2)/r$, which is larger than $r$, then that means the center of rotation is further away from a moving point than the $z-$axis, and yet the moving point seems to rotate about the $z-$axis, which is only a distance of $r$ away.
Why does a helix have a radius of curvature of $(r^2+h^2)/r$
curvaturecurvesdifferential-geometryorthonormalvector analysis
Best Answer
The radius of curvature is larger than the radius of the cylinder because the helix is climbing as it winds. That means the best matching circle will be larger than the circle of the cylinder.
If you imagine a helix that climbs very steeply you can see that it's locally nearly a straight line, so tangent to a very large circle.