TITLE: Curves with constant curvature and constant torsion
QUESTION [9 upvotes]: Describe all curves in $\mathbb{R}^3$ which have constant curvature $κ > 0$ and constant torsion $τ$. 
Any ideas what we can do to describe all such curves? 
Do we have to use the formulas of the curvature and of the torsion?

REPLY [2 votes]: From the Fresnet-Serret formulas, we can derive
$$\ddddot\gamma(s)=\kappa\ddot n(s)=-\kappa(\kappa^2+\tau^2)n(s)=-a^2\ddot\gamma(s),$$ with $a=\sqrt{\kappa^2+\tau^2}$.
The general solution for this fourh order linear equation is
$$\color{green}{\gamma(s)=C\cos(as)+S\sin(as)+C_1s+C_0},$$ 
which describes an elliptic helix.
Anyway, computing the curvature for any $s$ (such as $s=0,s=\dfrac\pi{2a}$),
$$\kappa=\|\dot t(s)\|=\|\ddot\gamma(s)\|=a^2\|C\cos(as)+S\sin(as)\|,$$
we must have
$$\|C\|=\|S\|=\frac\kappa{a^2}\text{, and }C\cdot S=0$$ for the curvature to be constant (the normal vector describes a circle).
Similarly,
$$\kappa\tau=\|\kappa\dot n(s)+\kappa^2t(s)\|=\|\ddot t(s)+\kappa^2t(s)\|=\|-a\tau^2(C\cos(as)+S\sin(as))+\kappa^2C_1\|$$ requires $$C\cdot C_1=S\cdot C_1=0\text{, and }\|C_1\|=\frac{\tau\kappa}a$$ (the tangent vectors describes a circular cone).
Hence the curve is a circular helix.