A Model of Islamic Rectilinear Interlaced Lattices

Fig.2 A 10 Lattice with some EE

Fig.2. A 10-Lattice with some EE.

LATTICE ORDERS and DISTANCES

ORDER 10. A very interesting and very beautiful case. The angles are the 10 multiples of Theta=360/ 10=36 , 0, 36, 72, 108, 144, 180 and its symmetric by the horizontal. A simple triangle will be the isosceles 36-72-72. Once the base of the triangle is chosen, the distances between parallels which appear now in the triangle are its heights. The ratio of its values are found to be, after little counting, 1.6180.... But this is a prestigious number, older than the civilizations about which we are talking: this is the Golden Proportion, considered as a very harmonic ratio of two distances (like a picture frame, for instance). Its exact value is (1+Square Root 5)/2; let us call it Phi. If now, we take another triangle with the smaller height equal to a linear combination of 1 and Phi, with coefficients a,b, the other one will be (a + b x Phi) x Phi = a x Phi + b x Phi x Phi. But

Phi x Phi =(1 + Square Root 5)/2 x (1 + Square Root 5)/2 
          =(1 + 5 + 2(Square Root 5)) / 4 
          =(3 + Square Root 5) /2 
          = 1 + Phi
therefore

(a + b x Phi) x Phi = a x Phi + b(Phi x Phi) 
                    = (a + b) + b x Phi 
                    = a' + b x Phi
which means that also the other height is a linear combination of the 1 and Phi, and, again in this 10-space, the pair 1, Phi is a base vector. For instance, in Fig.2, all the distances are equal to 1, Phi, 1 - Phi or any other linear combination.

Beautiful and well-known relations can be found for this number Phi; see several developments of this subject in [Coxeter,1973. Ghica, 1977. Rademacher & Toeplitz,1970.].

GENERAL CASE. As we have seen in the former cases, any order which engenders in other radii distances which can be expressed as a linear combination of two, will accept this pair as a base, and any linear combination of them, or what is equal, the distance between any two straights of the same series, will form a legal or permitted triangle (with vertices on some nodes of the lattice), using some others combinations of the base vectors.

If we try all the linear combinations of distances which determine a valid triangle in an N-lattice, we will arrive at the expression:

(ms x n - m x ns) x (sin(Alpha) x mp - sin(Beta) x m) - 
    (mp x n - m x np) x (sin(Alpha) x ms - sin(Gamma) x m) =0
where Alpha, Beta and Gamma are the angles; (m,n), (ms,ns), (mp,np) are the coefficients of the linear combination of the two unknown base vectors for the distances between straights of the three series whose intersection generate this triangle. By trying different combinations with small values (otherwise the job will last a infinite time) we obtain two-vector bases whose ratios allow legal triangles, as we have shown individually for the 6, 8, and 10 cases.

As any elemental figure (EE) of a RIGE can be decomposed into triangles, the found basis also allows legal RIGE, and therefore constitute answers (not all, but some) to the question raised about the distances between straight series. In Table 1 we can see some of these answers. In it are also shown several possible polygons in each lattice, some of which were shown in Fig.1 and 2. The angles are expressed one by one or, when possible, as a multiple, for regular convex and star polygons. In Table 2 are listed base vector ratios for greater values of N.

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05-June-95