If we do not impose any symmetrical rule, any distance is possible. But if we do, as Tradition demands, once we choose one, the gyration for central symmetry will create 'alias' in another radius, thus creating new distances as a necessity. Let us see some cases from the simplest to some more complicated ones.
ORDER 6. This is a very simple case. The angles are the 6 multiples of q = 360/6 = 60 that is, 0º, 60º, 120º, 180º, 240º and 300º. If we consider a triangle, it must be equilateral, as shown before, and the angles will be 60º, 60º, 60º. If we choose any distance on the horizontal as a side, the same distance appears by rotating it on the oblique side, for equilateral. Thus we have only one distance necessary in the case N=6. We can use more, of course, but the order does not impose it, one is enough.
ORDER 8. A simple, but already interesting case. The angles are the 7 multiples of q = 360/8 = 45 , that is, 45º, 90º, 135º, 180º, 225º, 270º and 315º. If we consider a triangle with angles 45º-45º-90º, once we gyrate the hypotenuse on the horizontal side, we obtain a new distance as a consequence of the first one. The ratio between them is the Square Root of 2. So the lattice order implies a specific distance ratio. See this ratio in Fig.1
Now the problem is to know how many different distances we need. If we raise another perpendicular to the horizontal at point Ö2, the new distance on the diagonal will be Ö2 x Ö2 = 2. But this distance is 2 times 1, that is, a second distance equal to the first, 1: we do not need another distance in this case. In general, any gyration of a linear combination of these two distances, 1 and Ö2, will provide us with another combination of them: they form what is known as a Vector Space, with 1 and Ö2 as a Base of the Space:
(a + b(Ö2) x Ö2 = 2b + a(Ö2) = b' + a(Ö2)
Back to Elementary Figures
Forward to Lattice Orders and Distances - part 2