In Barrett O'Neil's Elementary Differential Geometry, the following theorem is provided:
Theorem 4.6: A regular curve $\alpha$ with curvature $\kappa > 0$ is a cylindrical helix iff the ratio $\tau / \kappa$ is a constant.
Here, $\tau $ refers to the torsion of $\alpha$. My question is the following: If the torsion of every plane curve is 0, doesn't any plane curve with $\kappa > 0$ (e.g. a circle) satisfy the classification of a cylindrical helix by the above theorem?
My intuition tells me "no," and in fact, Barrett later goes on to provide a list saying that a regular curve with positive curvature is a cylindrical helix iff $\tau / \kappa$ is a non-zero constant, but in his proof of Theorem 4.6 (given below), I don't really see where having a non-zero constant ratio would play a role (or where having a ratio identically 0 would pose a problem). Here's the proof he provides:
Any help/clarification regarding this would be greatly appreciated!
Best Answer
This is a question of definition. Usually an helix is a 3D curve not lying in a plane.
The circle is a degenerate case. But you're right that it satisfies the definition that the tangent has a constant angle with a fixed axis.