Parameterized hexagonal spiral equation(s) I need to design a fully parameterized hexagonal spiral, where I can choose/change the distance between turns, the inner and outer radii etc. I understand that it will be somehow related to the Archimedean spiral, whose equations are $x=rcos(\theta)$ and $y=rsin(\theta)$, and I understand that it will also be linked to the $\pi/3$ value, since each internal angle of a hexagon measures 60 degrees, but I can't pull this out.
Can anyone help me with this? It would be something like the picture below. Thanks in advance!

 A: This is a parametric equation for a $n$-gon spiral (thus set $n=6$ in your case): 
$$
x=\cos\left({2\pi\lfloor t\rfloor\over n}\right)k\lfloor t\rfloor(1-(t\bmod 1))
+\cos\left({2\pi(\lfloor t\rfloor+1)\over n}\right)k(\lfloor t\rfloor+1)(t\bmod 1)\\
{\ }\\
y=\sin\left({2\pi\lfloor t\rfloor\over n}\right)k\lfloor t\rfloor(1-(t\bmod 1))
+\sin\left({2\pi(\lfloor t\rfloor+1)\over n}\right)k(\lfloor t\rfloor+1)(t\bmod 1)\\
$$
As usual, $\lfloor t\rfloor$ is the floor of $t$ (greatest integer lower than $t$) and $(t\bmod 1)$ is the fractional part of $t$ (that is $t-\lfloor t\rfloor$).
Variable $k$ controls the space between turns and every unit of parameter $t$ corresponds to a side of the spiral. The logic should be clear: that's nothing else than a sequence of segments, joining points 
$P_i=ki\big(\cos{2\pi i\over n},\sin{2\pi i\over n}\big)$,
that is those points on an Archimedean spiral which are separated by a central angle of $2\pi/n$.
The spirals in figure below were obtained for $k=1$ and $0\le t\le12$ (small spiral) or $5\le t\le42$ (large spiral).

EDIT.
If one needs the spiral to have parallel sides, then one can construct it starting with a first side of length $1$ and at each step construct the following side by rotating the previous side by $\pi/3$ and prolonging then the result by a fixed amount, which we can choose as equal to $1$. 
If we use the complex representation of points in the plane, then the vertices of this spiral are given by
$$
P_n=\sum_{k=0}^n kz^{k-1}=\frac{n z^{n+1}-(n+1)z^n+1}{(z-1)^2},
$$
where $z=e^{i\pi/3}$ (set $z=e^{2i\pi/n}$ for a more general $n$-gonal spiral). Switching to real coordinates $(x_n,y_n)$ this can be written as
$$
x_{n}=
+\frac{\sqrt{3}}{2} (n+1) \sin \left(\frac{\pi n}{3}\right)
-\frac{1}{2} (n-1) \cos \left(\frac{\pi n}{3}\right)
-\frac{1}{2},\\
y_{n}=
-\frac{\sqrt{3}}{2} (n+1) \cos \left(\frac{\pi  n}{3}\right)
-\frac{1}{2} (n-1) \sin \left(\frac{\pi n}{3}\right)
+\frac{\sqrt{3}}{2}.
$$
As in the previous case we can then write the parametric equation for the spiral as a sequence of line segments joining those points:
$$
x=k x_{\lfloor t\rfloor}(1-(t\bmod 1))+k x_{\lfloor t\rfloor+1}(t\bmod 1)\\
y=k y_{\lfloor t\rfloor}(1-(t\bmod 1))+k y_{\lfloor t\rfloor+1}(t\bmod 1)\\
$$
As before, parameter $k$ controls the spacing: parallel sides are distant $3k\sqrt3$ between them. The following figure was obtained with $k=1$ and $0\le t\le20$:

