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I'm trying to come up with an example of a curve that has constant curvature and torsion, both exactly 1.

Let $\vec{r(t)}$ be a parametrization for the curve $\gamma$. By calculating with formulas for curvature $\kappa$ and torsion $\tau$ I got that the length of the second derivative $\dot{\vec{r}} $ must be the square of the length of the first derivative. Similarly, I got that the length of the third derivative must be a cube of the first derivative. But I couldn't get this any further.
I'm guessing it could be something like a spiral... that would give us a constant curvature. And this spiral should bend...

What would be such an example?

(For definition of the curvature and torsion: https://en.m.wikipedia.org/wiki/Frenet%E2%80%93Serret_formulas)

cmk
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Coupeau
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1 Answers1

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Your intuition is correct. Consider the unit-speed helix $$\beta(s)=\left(a\cos\frac{s}{c}, a\sin\frac{s}{c},\frac{bs}{c}\right),$$ where $c=\sqrt{a^2+b^2}$ and $a>0.$ You can check that $$\kappa(s)=\frac{a}{a^2+b^2},$$ and $$\tau(s)=\frac{b}{a^2+b^2}.$$ In particular, if $a=b=1/2,$ then both $\kappa$ and $\tau$ are equal to $1$.

cmk
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  • This is great. But is there any systematic way of finding such curves? Or can we prove that this is the only such curve with both curvature and torsion exactly 1? – Coupeau Sep 07 '19 at 13:05
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    You can prove that regular curves in $\mathbb{R}^3$ with constant, non-zero curvature and torsion are circular helices! – cmk Sep 07 '19 at 13:12
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    There is a unique curve (up to rigid motion) once you specify curvature and torsion. This is called the Fundamental Theorem of Curve Theory. See, for example, my differential geometry text. – Ted Shifrin Sep 07 '19 at 16:43