How do we solve for curve coordinates in terms of parameter $t$ in $\mathbb R^3$ having scalar property

$$ \tau \cdot \kappa = const.?$$

The curve would be with infinite twist at straight portions and confined to plane at cusp points.

EDITS (1-3):

I.e., how to find a vector $r(t)$ so that

$$ \tau\, \kappa =\dfrac{r^{'} \times r^{''}\cdot r^{'''} }{||r^{'} \times r^{''}||^2 } \dfrac{|r^{'} \times r^{''}| }{|r^{'} |^3 } =1? $$

The above formulae are of Frenet-Serret vector triad $T,N,B$ and derivatives, textbook derived with respect to parameter $t$.

BTW, is there a word for infinite torsion of a line?( By analogy of infinite normal curvature being a cusp)

The following surface (rotation symmetry assumed) and arc are obtained numerically but not without an extra assumed arbitrary relation of arc angle to meridian and axis of symmetry. Central parts are asymptotic/infinite torsion and the ends have infinite curvature cusp edge/zero torsion just as expected. This singularity terminates numerical computation.

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  • $\begingroup$ Every helix satisfies your condition. $\endgroup$ – Rahul May 20 '17 at 22:58
  • $\begingroup$ Is not the quotient $ \dfrac{\kappa}{ \tau } $ (but not product ) constant for general helices? $\endgroup$ – Narasimham May 21 '17 at 7:35
  • 2
    $\begingroup$ The quotient, product, sum, difference, etc. of $\kappa$ and $\tau$ is constant for any helix because both $\kappa$ and $\tau$ are individually constant. $\endgroup$ – Rahul May 21 '17 at 8:39
  • $\begingroup$ As it is *not * a helix, I placed its name in quotes, Anyhow now changed the title. Proceeding from derivative definition it would not be any helix, I think. $\endgroup$ – Narasimham May 21 '17 at 8:50

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