Enigmatic patterns in Archimedean spirals Distributing the natural numbers as circles evenly along the Archimedean spiral yields surprising patterns when changing the radius of the circles: they cover more and more of the plane, finally covering it completely. But shortly before this happens, more or less obvious and intricate patterns shortly show up, and I'd like to understand them.
The numbers are arranged by these formulas with $\hat k = \frac{\sqrt{k}}{2}$
$$x_\alpha(k) = -\hat k\cos(\alpha\cdot 2 \pi\cdot \hat k)$$
$$y_\alpha(k) = -\hat k\sin(\alpha\cdot 2 \pi\cdot \hat k)$$
– the distance between consecutive numbers along the spiral being controlled by the parameter $\alpha$.
This is how the number spirals look like for $\alpha = \frac{\sqrt{2}}{2}, 1, \sqrt{2}, 2,2 \sqrt{2},4, \ldots = \sqrt{2}^k$:

[Click image to enlarge.]
When the radius of the circles is enlarged until they almost cover the plane, one observes different spot patterns: a spiral, a cross with $8$-fold rotation symmetry, another cross with $4$-fold rotation symmetry and a series of horizontal stripes.

[Click image to enlarge.]
When we compare $\alpha = 0.99, 1, 1.01$ we see that the straight cross for $\alpha = 1$ is the limit of two bundles of eight spirals running in opposite directions.

[Click image to enlarge.]
Here is my question:

How can the cross with the 8-fold rotational symmetry (for $\alpha=1$)
  be explained? ("For $\alpha=1$ there is a cross with a 8-fold
  rotational symmetry, because ....")

Note that for $\alpha=1$ the square numbers are aligned along the horizontal axis:


One final remark: The observed patterns come as a surprise (somehow) because "from the distance" the spiral looks almost like a set of concentric circles thus having seemingly total rotational symmetry. But in fact the spiral has no rotational symmetry at all (only for 360°). And the ever changing patterns are due to this fact: the concentric circles do show only patterns reflecting the density of numbers along them. 
In this example the density is $10$ per $2\pi$. Note how the triangle numbers (compared to the square numbers in the case of the spiral) are arranged along the horizontal line. Let $\triangle(k) = \frac{k(k+1)}{2}$ take the values $0,1,3,6,10,15,\dots$ We observe along the horizontal line $n(k) = 10\triangle(k) + 1$ (to the right) and $m(k) = n(k) + 5k$ (to the left):



To see how the spiral patterns continue here for $\alpha = 4\sqrt{2}, 8, 8\sqrt{2}$:

There's a little Youtube video where you can see the number spiral continuously change for $\alpha$ going from 1 to 4.
 A: Let me try. We see a pattern where there is particularly much space between the circles. This happens when the circles are lined up with each other, so four neighbouring circles make an approximate rectangle.
In the $n$'th revolution starting at the positive $x$-axis, there are $(2n+1)^2-(2n-1)^2 = 8n$ circles. This means that we can divide the revolution into 8 parts with $n$ circles each, and the circles must line up with each other at each division (and nowhere else). This explains the pattern.
Have you tried making a circular pattern with $8n$ numbers in each circle? I suspect the same pattern would emerge (if the distance between the concentric circles are chosen appropriately). 
This also explains the other symmetric patterns. We can work out that if $\alpha = \sqrt 2^c$ for $c\le 3$, then the circles that lie just before the positive $x$-axis will be $k=2^{-c}\cdot((2n+1)^2-1) = 2^{2-c}\cdot n\cdot(n+1)$, which implies that there will be $2^{3-c}\cdot n$ circles in the $n$'th revolution. This means that the circles are lined up just below the positive $x$-axis, and we have $2^{3-c}$-fold rotational symmetry. This matches what we observe on your plots, which correspond to $-1 \le c \le 4$. ($c=4$ is not explained by my calculation). I believe that you can see the 16-fold symmetry in the first plot, if you squint a little. 
A: To give some visual sugar to user Milten's nice answer find here highlighted those numbers that make approximate rectangles. You'll easily see the number pattern and see the role that the number $8$ plays, giving typical triangle number like sequences, e.g. $a = 2,12,30,56,90,\ldots$ with $a_k = 8\triangle(k) + 2(k+1)$ or $b = 3,13,31,57,91,\ldots$ with $b_k = a_k +1$, $k=0,1,2,\ldots$.

For the sake of comparison the circular pattern with $8n$ numbers in each circle. The similar numbers make approximate rectangles:


What needs to be shown analytically is that for example for the numbers $a_{k},b_k, a_{k+1}, b_{k+1}$ we have
$$|p(a_k) - p(b_k)| \approx |p(a_{k+1}) - p(b_{k+1})|$$
and 
$$|p(a_k) - p(a_{k+1})| \approx |p(b_k) - p(b_{k+1})|$$
with $p(k) = \langle x(k), y(k)\rangle$ and $x(k) = -\hat k\cos(2 \pi\cdot \hat k)$, 
$y(k) = -\hat k\sin(2 \pi\cdot \hat k)$ and $\hat k = \frac{\sqrt{k}}{2}$.
A: This answer complements the other one by Milten; it explains the linear patterns in the diagrams. First consider the case $\alpha=2$, which corresponds to the function $z_k:=\sqrt{k}e^{2\pi i\sqrt{k}}$ (it is different from OP's function by a half). Take different cases of $k$. 
(i) $k=n^2+m$. $$z_n=x_n+iy_n=\sqrt{n^2+m}\,e^{2\pi i\sqrt{n^2+m}}=\sqrt{n^2+m}\,e^{2\pi i n(1+\frac{m}{n^2})^{1/2}}\approx ne^{\pi im/n}$$ so $x_n\approx n$, $y_n\approx n(\frac{\pi m}{n}-\frac{m^3\pi^3}{6n^3})\approx\pi m-\frac{m^3\pi^3}{6n^2}$, that is $$y_n\approx \pi m-\frac{A}{x_n^2}\to \pi m$$ This explains the right-hand lines in OP's diagrams with $m=\ldots,-1,0,1,2,\ldots$.
(ii) $k=n^2+n+m$. $$z_n=\sqrt{n^2+n+m}\,e^{2\pi i\sqrt{n^2+n+m}}\approx (n+\tfrac{1}{2})e^{2\pi in}e^{\pi i}e^{\pi i m/n}\approx-(n+\tfrac{1}{2})e^{\pi im/n}$$ This is similar to case (i) but reflected, so the lines appear on the left side.
Now consider $\alpha=2\sqrt{2}$ which corresponds to $z_k:=\sqrt{k}e^{2\pi i\sqrt{2k}}$. The same type of analysis gives $$z_n\approx n\,e^{2\pi im/n}$$
