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I need some help on the following problem:

Given that a curve $\mathbf r:I\to \Bbb R ^3$ has constant curvature $k(s)=k$, for all $s$, and constant torsion $\tau(s)=\tau$, for all $s$. Find the curve $\mathbf r$.

I only know that, according to the fundamental theorem, this curve exists and is unique. But, how practically find the parametric equation of the curve?

Thanks.

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    $\begingroup$ One can prove that a curve has constant curvature, $\kappa\neq0$ and constant torsion, $\tau$, iff the curve is a helix; of course if $\kappa = 0$, then talking about torsion doesn't make sense. Surely this helps. $\endgroup$ Feb 2 '14 at 1:08
  • $\begingroup$ What happens if we change the function ? For example lets take $k=h(t) $ and $t=g(t)$ where $h,g$ are continuous functions then how we find that unique curve, is there an algorithm, or we can't find it. $\endgroup$
    – lupus nox
    Jan 17 at 20:50
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Hint: set $r=\kappa/(\kappa^2+\tau^2)$, $h=\tau/(\kappa^2+\tau^2)$. Let $s$ be the arc-length function of $\mathbf r$ with $s(t_0)=0$ and define $\phi(t)=s(t)/\sqrt{r^2+h^2}$. We want to construct an orthonormal basis ($a_1$, $a_2$, $a_3$) and a point $p_0$ such that $$\mathbf r(t)=p_0+r\bigl(\cos(t)a_1+\sin(t)a_2\bigr)+hta_3.$$

First show -- per differentiating twice -- that for $\tilde c=\mathbf r+rN$ there's a point $p_0$ and a vector $v\neq0$ such that $\tilde c(t)=p_0+s(t)\cdot v$. Now chose a suitable orthogonal basis $(a_1,a_2)$ for the orthogonal complement of $\boldsymbol Rv$ such that $\langle -N(t),a_1\rangle=\cos(\phi(t))$ and $\langle -N(t),a_2\rangle=\sin(\phi(t))$.

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  • $\begingroup$ How did you come up with that $r = ...$ ? $\endgroup$
    – Our
    Mar 9 '19 at 9:33
  • $\begingroup$ Simply because with that definition of $r$ all will work well. $\endgroup$ Mar 9 '19 at 11:50
  • $\begingroup$ Don't get me wrong, but I didn't ask who do you choose $r$ in that way; I asked how did you come up with that. $\endgroup$
    – Our
    Mar 9 '19 at 11:52
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    $\begingroup$ If torsion and curvature are constant, the curve should be a helix. And a helix satisfies these equation. $\endgroup$ Mar 10 '19 at 9:25
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To find the arc-parameterization $\mathbf{r}=\mathbf{r}(s)$ you must resolve

$$\frac{\mathbf{r}'\cdot \mathbf{r}''\times \mathbf{r}'''}{\|\mathbf{r}''^2\|}=\tau,$$ $$\|\mathbf{r}'(s)\|=1,$$ $$\|\mathbf{r}''(s)\|=k,$$

for some injective differential map $\mathbf{r}:I\hookrightarrow{\Bbb{R}}^3$ over some open interval $I$.

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  • $\begingroup$ I have a question about the first equation. The LHS is a vector, but the RHS is a scalar? Indeed, I do have the last two equations, but solving them is a trouble for me. Do you have any hints on solving them? $\endgroup$ Feb 2 '14 at 1:59
  • $\begingroup$ you was right, now i did correct. $\endgroup$
    – janmarqz
    Feb 2 '14 at 2:58
  • $\begingroup$ to solve them you gotta begin to unfold $\mathbf{r}(s)=(x(s),y(s),z(s))$ then $\mathbf{r}'=(x',y',z')$, $\mathbf{r}''=(x'',y'',z'')$... etc $\endgroup$
    – janmarqz
    Feb 2 '14 at 3:19
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Central line u = 0 of helicoid surface $ (u \cos t, u \sin t , c t) $ treated as a stand alone space curve has zero curvature and constant torsion.

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