# Generalized Helixes in 3D

A curve is called a general helix if there exist a direction unit vector $$a$$ such that the tangent of the curve makes a constant angle with $$a$$. I would like to turn a curve $$\gamma = \left(x(t),y(t)\right)$$, a planar curve into $$\hat{\gamma}=\left(x(t),y(t),z(t)\right)$$ a space curve, where $$\hat{\gamma}$$ is a generalized helix over the curve. For convenience's sake, let $$\gamma$$ be a unit speed parameterization of the curve.

What I've tried is letting $$\frac{\tau}{\kappa}=c$$. Then,

$$c = \frac{<\gamma' \times \gamma'',\gamma'''>}{\| \gamma' \times \gamma'' \|^3 }$$

My end goal here is to express $$z(t)$$ in terms of $$x(t)$$ and $$y(t)$$, but I just can't figure out how, and am also wondering if this is the approach I should take.

Note: I have seen this question, and this but the difference is that I need to show existence, rather than proving a generalized helix will have constant curvature/torsion. Thus that is why I am trying to find an expression for $$z(t)$$. Any $$z(t)$$ that will work.

• I think your approach is misguided. $\kappa$ and $\tau$ belong to the space curve. Starting with that, you want to understand the proofs you have linked to recover the unit vector and from that the plane curve. Commented May 23, 2021 at 23:41
• To come to 3d from projection in 2d there is more than one way. A circular helix is just one way. Commented May 23, 2021 at 23:42