Tubular neighbourhood of a smooth curve In plane for a smooth non self intersecting curve $C$ the tubular neighbourhood can be constructed with non intersecting starigh line segments normal to the curve from points of the curve.
If we use the fact that the $\varepsilon$-tubular neighbourhood is constructed diffeomorphically by extending the normals we could arrive at a necessary condition that $\varepsilon < 1/K$, where $K=\sup_{p\in C} k(p)$ and $k(p)$ is the curvature of curve at $p$. 
Suppose the curve $C$ is in a surface embedded in $R^3$.  The tubular neighbourhood can be constructed similarly along the normal geodesics to the curve.  Is it true that for 
$K=\sup k_g(p)$ where $k_g(p)$ is geodesic curvature of curve at $p$, the necessary condition is $\varepsilon < 1/K$?
 A: No. 
Consider the example of a right circular cylinder as the surface in question and let $C$ be a "generator" (i.e., a line on the cylinder parallel to the axis). Then the geodesic curvature is everywhere zero, but if epsilon is greater than half the circumference there is no epsilon-tubular neighborhood.
A: As shown in Richard Palais' answer, the radius $\epsilon$ of the larger tubular neighborhood can be much smaller than $1/K$.
On the other hand, at least for smooth embeddings of $\mathbb{S}^1$, the best uniform radius $\epsilon$ is also never larger that $1/K$. So in this sense yours is a necessary condition (with weak inequality though). 
Consider a simple closed curve in, say, a Hilbert space $H$, with a $C^2$ arc-length parametrization $\gamma:\mathbb{S}^1\to H$.  Thus, $1/K=\inf_s |\ddot  \gamma(s)| $. 
Let $r > 1 / K $. So there is $s _0 $ such that  $ r > |\ddot\gamma (s_0) |$. Consider the point $ x _ 0 \in H $ at abscissa $r$ on the normal at $\gamma (s _0 )$:
$$ x_0 : = \gamma(s_ 0) + r \frac {\ddot \gamma(s _0) } {| \ddot \gamma(s_0)|}\,.$$
Computing the second derivative of $|\gamma(s)-x_0|^2$ at $s = s_0$ we find that the distance from $x  _ 0$ to $\gamma(s)$  has a strict local maximum at $s _ 0$. Thus, in particular $s_0$ is different form the global minimum, which is also a point whose normal meets $x_0$.  Therefore $x_0$ belongs to at least two normal disks of radius $r$, and in conclusion $r$ is larger than the radius of any tubular nbd. 
(All that with minor changes also holds for Riemann or Hilbert manifolds; and for non-closed curves too, with some care at the end-points).
