Curve with constant torsion and no curvature When curvature and torsion are given a curve is fully defined (upto Euclidean motions) in 3-space.
$ k=const , \tau = 0 $  represents a circle in a plane ;
But what does the space curve
$$ k =0 , \tau= const,$$
represent?
The center line $ (u=0) $ of a right handed twisted helicoid with parametrization  $( u \cos v, u \sin v, c \;v ) $  is a good example.
Curvature/Torsion of $u=0$ line of helicoid.
Clearly u=0 is a straight line at the helicoid mid with zero curvatures ( both normal (asymptotic $ k_n=0$) and geodesic  $k_g=0$ ) as valid for a full straight line.
Using Enneper-Beltrami theorem  torsion of the central parametric line at  $\; u=0$ is found constant:
Evaluating Gauss curvature K
$$ K= \dfrac{-c^2}{(c^2+u^2)^2}, \tau = \sqrt{-K}= \pm \dfrac {1}{c}$$
The sign for the torsion of right helicoid is positive and, for the left handed helicoid it is negative.
A physical example is of a long human hair that can be twisted right or left with constant torsion even if the twist is not clearly visible. Other examples include long straight portions of DNA and other polymer molecules which inhabit such a surface.
EDIT1:
In another example the straight line parameterized by
$$(x,y,z)= (a, b t, c t) $$
has zero curvature and non-zero torsion in this example when it becomes asymptotic on certain (arbitrary?) surfaces surfaces of negative Gauss curvature.

EDIT2:
What I meant by torsion without curvature is shown in the first figure Twist of Helicoid's straight/geodesic Spine. The special asymptotic line intrinsically characterizes how twist occurs during parallel transport in tangent spaces.
 A: The whole theory of curvature and torsion of "curved lines" is based on the implicit assumption that the curvature is not zero except for isolated points. It is zero on a segment if and only if the curve segment is a straight line. 
As the Wikipedia article Torsion of a curve states:

Let $\bf{C}$ be a space curve parametrized by arc length
   $s$ and with the unit tangent vector $\bf{t}$.
   If the curvature $\kappa$ of $\bf{C}$ at a certain point is
  not zero then the principal normal vector and the binormal
  vector at that point are the unit vectors
  $$ \bf{n}=\frac{\bf{t}'}{\kappa},
\quad \bf{b}=\bf{t}\times\bf{n}, $$
  where the prime denotes the derivative of the vector with
  respect to the parameter $s$. The torsion $\tau$
  measures the speed of rotation of the binormal vector
  at the given point. If is found from the equation
  $$ \bf{b}' = \tau\,\bf{n}. $$

Note carefully that the definition of $\bf{n}$ involves dividing by the curvature. Hence, if the curvature is $0$, $\bf{n}$ is not defined.
Because this vector is not defined, this implies that the
torsion can not be defined either since its definition uses
$\bf{n}$.
Of course, you could just pick any fixed unit
vector $\bf{n}$ which is perpendicular to the tangent
$\bf{t}$ and then define $\bf{b}$ the usual way
which leads to a constant binormal and hence by the
definition of torsion we find $\,\tau=0.\,$ Obviously,
the normal vector $\,\bf{n}\,$ is not unique, but
in any case torsion is forced to be $\,0.\,$
A: If a 'curve' $\gamma:\>s\mapsto{\bf r}(s)$ is parametrized with respect to arc length then by definition $\kappa(s):=|\ddot{\bf r}(s)|$. The assumption $\kappa(s)\equiv0$ then implies $\ddot{\bf r}(s)\equiv{\bf 0}$, and this leads to ${\bf r}(s)={\bf a}+s {\bf u}$, where ${\bf u}$ is a unit vector.
Now, Torsion measures the failure of a curve to be planar. If $\gamma$ has zero torsion, it lies in a plane. Hence for $\kappa=0 \implies \tau = 0$ corresponding to a line. Lines look very much like lines, and they are certainly planar. 
