Herringbone partitions of regions and surfaces Let $R \subset \mathbb{R}^2$ be a region of the plane bounded
by a Jordan curve.  The boundary $\partial R$ could be a polygon,
or a smooth curve—there are variations depending upon boundary assumptions.
I would like to partition $R$ into regions $R_i$ that can be striped by
parallel lines with the property that each stripe meets the boundary
$\partial R_i$ at an angle that excludes the open 
interval $(\frac{1}{4}\pi,\frac{3}{4}\pi)$.
In other words, the stripes meet the boundary of $R_i$ at $45^\circ$ or more sharply; they cannot meet the boundary nearly orthogonally. 
(Here the boundary is the boundary of $R_i$, which might include portions of 
the boundary of $R$.)
And the ultimate goal is to partition $R$ into the minimum number of such regions.
For example, a rectangle can be partitioned into one region,
but it seems a circle may need four regions(?):

          


A number of questions suggest themselves:

Q1. For $R$ a polygon of $n$ vertices, what is the largest number
  of herringbone regions needed for a fixed $n$, and how many regions always suffice for
  a fixed $n$?
Q2. For $R$ a circle in $\mathbb{R}^2$, or the surface of a sphere in
    $\mathbb{R}^3$, what is the optimal (fewest regions) herringbone partition?
    Can the circle be herringbone-partitioned into fewer than four regions?
Q3. What is the optimal herringbone partition of a torus?

I'll stop here, as you can spin off these questions as easily as can I.
My original focus was on compact surfaces in $\mathbb{R}^3$ (the sphere and torus above),
when the pattern is drawn by parallels to a geodesic,
but it already
seems interesting in $\mathbb{R}^2$.
Thanks for any insights and/or pointers to relevant literature!

          
          


Update. Sergei's idea, as articulated by Cristi, leads to a 3-region herringbone partition
of a circle.  Here his Cristi's illustration from his comment:

          
          
 A: Three regions is the minimum for any convex figure with a smooth boundary. To construct a partition, let $AB$ be the diameter of $R$. The segment $[AB]$ splits $R$ into two pieces $R_1$ and $R_2$. Let $C_i$ be a point at maximal distance from $[AB]$ in $R_i$, $i=1,2$. Draw perpendiculars $C_iD_i$ from $C_i$  to $AB$. The segments $[AB]$, $[C_1D_1]$ and $[C_2D_2]$ split $R$ into 4 regions which can be filled by parallel lines just like the 4 quarters of the circle. Then you can merge two regions by removing the segment $[D_1D_2]$ or using a zig-zag if $D_1=D_2$.
Two regions are impossible. Indeed, parametrize $S^1$ by tangent directions of $\partial R$ (say, oriented counter-clockwise). For each region $R_i$, consider the set $R_i\cap\partial R$. The tangent directions of this set form angles at most $\pi/4$ with a fixed line, hence they belong to the union of two arcs of length $\pi/2$ in the circle of tangent directions. It is possible to divide $S^1$ into 2 pair of arcs like this, but then the two corresponding pairs of arcs in $\partial R$ cannot be connected by disjoint paths (I assume you don't want connected but not path-connected examples).
The same arguments, modulo minor details, works for any $n$-gon with angles sufficiently close to $\pi$. For example, for a regular $n$-gon with $n\ge 10$.
A: We can extend Sergei's answer to non-convex shapes.
If $n$ is the minimum number of arcs it takes to partition the boundary of the shape in a way such that the tangent angles for each arc lie in a union of two arcs of length $\pi/2$ in the circle of tangent directions, then it takes at least $n/2+1$ herringbone regions to partition the shape.
Proof: Consider the herringbone partition as a graph with vertices, edges, and faces. Its Euler characteristic must be $1$. Remove all leaves, then the Euler characteristic is
$|F|-\sum_{v\in V} \frac{d(v)-2}{2}$
where $d(v)$ is the degree of a vertex, and all summands are nonnegative.
All $n$ vertices on the boundary have degree $3$, so the number of faces is at least $n/2+1$.
Thus, a smooth curve that has a lot of Thurston-style corrugations can have an arbitrarily high lower bound on the number of regions needed.
A: 
Remarks:


*

*This works as well for smooth figures, and for figures with corners, and polygons.

*Different orientations of the diagonal lines may yield different number of pieces, so we should supplement this with a method of choosing the optimal direction.
