*some spirolaterals (graph here)*

In the formula below, a represents the angle that you turn by at each step, and m represents the maximum length of the sides that you count up to before repeating. If m = 1, all sides are length 1, if m = 2, then the sides alternate between length 1 and length 2, for m = 3, the lengths of the sides form the repeating squence 1, 2, 3, 1, 2, 3... etc.

If the angle that you turn by is a rational multiple of 2

*pi*, then the first spirolateral (

*m*= 1) will trace out either a regular polygon or a regular star polygon - it is always adding a length of 1, and since the angle is a multiple of 2

*pi*you will close the loop and end up back where you started.

Different story if the angle not a rational multiple of 2

*pi*: you will never get back to the starting point, and the first spirolateral is going to look like an annulus after enough iterations (below is

*a*= 2, zoomed in on the left, zoomed out on the right):

*spirolateral that does not (soon) meet up - graph here*

For higher values of

And here are a few star-isogonals:

What about the beasties that spiral off to infinity? One set of these springs occur when

And then there are the tangled stars... not sure where to begin with these.

*m*, although the shapes traced out vary widely they are always equiangular (we are always turning by the same angle, after all). So for*m*= 2 we end up with isogonal figures: every vertex is the same - it always has the same angle and always has sides of lengths 1 and 2 on either side of it. Here are some isogonal polygons - the first, the rectangle, is pretty familiar; the second, @solvemymaths tells us, may be a*ditrigon*.And here are a few star-isogonals:

What about the beasties that spiral off to infinity? One set of these springs occur when

*a*=*pi/k*for some positive integer*k*and*m*= 2*k*.And then there are the tangled stars... not sure where to begin with these.

Trying classify even a few of these strange creatures reminds me of Borges' Book of Imaginary Beings.

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