Spiral Tilings with Ccurves


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1 Spirl Tilings with curves Using ombintorics to Augment Trdition hris K. Plmer 19 North Albny Avenue hicgo, Illinois, 0 Abstrct Spirl tilings used by rtisns through the ges using curves re identified. The complete combintoric set of these tiles is determined nd tilings tht cn be mde with them re exmined. 1. Introduction Throughout the history of ornment rtisns from mny cultures hve composed ptterns utilizing spirls. Mny of the compositions mde by these rtisns cn be understood s the substitutions of curves or S curves or combintion of curves nd Scurves for the stright lines of monohedrl or polyhedrl tiling [1]. The subject of this pper is the subset of these tilings composed with curves. Those known nd used by trditionl crftsmen re shown below in Figures 1 nd 2 [2]. Artisns cn pply nd hve pplied n enormous vriety of styles using these three tilings. An exmple of this vriety cn be seen by observing the expression of the sme tiling, (...), in two different styles, simple form in Figure 1 nd eltic style in Figure 2. These underlying tilings re prt of set of tilings with regulr vertices commonly known s Lves tilings []. This pper identifies the complete set of tiles nd exmines some properties of the tilings tht cn be mde with them. This gretly ugments the number of underlying tilings vilble for use by rtisns from three to n infinite number of tilings []. Figure 1 (c) Figure 2 (c)
2 In generl terms, our problem is: 2. The Problem Wht tilings of the plne llow for curves to be substituted for the edges of the tiles We will restrict the problem by imposing the following conditions: i) Spirls must be produced t ll the vertices. ii) The ngles surrounding ny vertex of the tiling must ll be equl. [] As we shll see these tilings re very restricted, but there re nevertheless such lrge number of different kinds tht we cn expect to be ble to clssify them only in some wek sense. As will be described third condition results: iii) Ech tile must hve n even number of sides. Figure ondition ii) shows tht the ngles t vertex tht lies on n edges must be equl to 0 /n. ondition iii) holds becuse the spirl property i) entils tht for ny two curves tht will be substituted for djcent edges of tile one will lie inside the polygon nd one outside. This lterntion cn only be consistently continued round the whole tile if tht tile hs n even number of sides s shown in Figure. The tile in Figure b stisfies condition ii) but not condition iii) cusing contrdiction of condtion i). onversely, however, if we re given ny tiling stisfying conditions ii) nd iii), then fter ssigning clockwise or counterclockwise orienttion rbitrrily to ny one vertex, we cn consistently ssign such orienttions to the other vertices by lterntion. This shows tht in generl there re two wys of orienting the spirls for given tiling.. The Possible Types of Tile We shll sy tht vertex hs type n when it belongs to just n tiles, which will ll hve ngles of 0 /n there. An edge whose two vertices hve types nd b will be sid to hve type (.b), nd tile will be sid to hve type (.b.c.,... k) if its vertices, reding cycliclly (in either direction) hve types,b,c,...,k. Theorem 1. The only possible types of polyhedrl or monohedrl tiles tht cover the plne nd llow curves to be substituted for the lines nd stisfy conditions i), ii) nd iii) re:! (...) Figure (...) (...) (...) (...) (...) (...) (...) Proof. Since the interior ngle of tile t n nfold vertex is 0 /n, the exterior ngle is /n. Now for:
3 n 0/n 180 0/n > < >150 Tble 1 The sum of the exterior ngles of polygon is lwys 0! Since Tble 1 shows tht the smllest vlue for n exterior ngle in these tiles is 0, there cn be t most ngles per tile, nd if there re, tht tile must be of type (...). Otherwise, since the tile must hve n even number of sides, it must in fct be qudrilterl, with n verge exterior ngle of 90, nd smllest one of either 0 or 90. If there re (t lest) two of 0, the sum of the remining two is 20, nd from Tble 1 we hve only two possibilities: nd , corresponding to types (...) nd (...) 0 + 0, corresponding to types (...) nd (...). If there is precisely one such 0 exterior ngle, the sum of the other three must be 00, which forces the lest to be 90, forcing nother 90 ngle nd leving precisely 0 for the lst one, corresponding to: (...) nd (...). Finlly, if there is no 0 ngle, ll four ngles must be 90, since this is now both the smllest possible vlue nd lso their verge vlue, nd we hve the only reming type: (...). nd this completes the proof. c We remrk tht the ngles of these tiles do not suffice to determine their shpes: for instnce tile of type (...) might be ny shpe of rectngle. However, we shll regrd tiling s wellenough determined when we hve specified ll the ngles of its tiles. There is then cnonicl tiling with these ngles, obtined by using tiles with the prticulr dimensions shown in Figure. From now on we shll lwys drw only these cnonicl tilings.. Tilings With Type Vertex. Theorem 2. No tiling involves tile of type (...), nd there is combintorilly only one tiling tht involves one of type (...). F D E A B Figure 5 Figure Proof. We exmine the type vertex of tile A of type (...) shown in Figure 5. If both the djcent tiles B nd t this vertex hve tye (...) then the tiles D nd E of Figure re forced to hve types (...), nd the tile F between them hs n impossible type (...). So we cn suppose tht B hs type (...) s in Figure, nd this forces the tile of tht Figure to hve type (...). The numbers p nd q of tht Figure must be nd in some order. But if q, then r, mking tile F to hve the impossible shpe (..,). So we must hve p, q, nd so r, nd gin F hs n impossible type, (...). A p B q r
4 A B Figure 7 Figure 8 Now tht we hve dismissed type (...), we see tht ll the tiles with ny type vertex of type (...) must hve tht type, forming rosette s in Figure 7 nd the tile of tht Figure hs type (...) which cn only be (...). This in turn forces tiles D nd E to hve type (...), yielding two more rosettes round vertices Y nd Z. Now it is cler tht the tiling continues uniquely s in Figure 8 thus concluding the proof. c 5. Tilings with Edges of Type (.) or (.) Theorem. If tiling hs n edge of type (.) or (.), it must be one of the tilings H, H1, H2,.., Hh, or Hh* of Figures 9. H Figure 9 H1 Figure 10 H2 Hh Figure 11 Figure Proof. The only types tht contin two djcent s re (...) nd (...). The ltter is lso the only type contining two djcent s. So it follows tht if one edge of tile is of type (.), then the opposite edge is either (.) or (.), while the edge opposite to one of type (.) is necessrily of type (.). Hh* Figure 1 The tiles contining ll these opposite pirs of edges therefore form ccordionrows like tht of Figure 1, ech such row consisting of n infinity of hexgons ech of which is either single tile of type (...) or is split into two tiles of type (...). We suppose first tht some ccordionrow does contin two unsplit hexgons, nd select one for which these hexgons (A1 nd A2) re seprted by the smllest possible number of tiles (B) tht hppens in tht
5 A A E``` ``` E```` E```` ```` ```` F``` F``` D``` E``` ``` E`` F`` F`` F`` F`` `` D`` D`` E`` `` E` F` F` F` F` F` F` E` ` D` D` D` ` Figure 1 E F F F F F F F F E D D D D A1 B B B B B B B B B B A2 tiling (Figure 1). Then the two tiles mrked in the Figure hve type (... ) tht cn only be (...), nd we cn lso determine the type of the tiles mrked D, since (... ) cn only be (...). We cn proceed upwrds in this wy, forcing ll the types of tiles (nd so the shpe of the Figure) until we conclude triumphntly tht there must be third hexgonl tile t A. In the third row, for exmple, we cn see tht the end tiles E hve type (...) (...), nd this forces type of F1 to be (...) (...), which in turn forces F2 to to hve type (...) (...), nd so on. The rgument then repets, with row of tiles ' nd D' behving like nd D followed by one of tiles E' nd F' behving like E nd the Fi, nd so on, until just below the summit we hve row of just two tiles ''''... surmounted by one of just two tiles E''''.... The tile A t the summit hs type (......) which cn only be hexgon (...). We hve found n equilterl tringle of tiles whose vertices re A1, A2, nd A. But we could repet the rgument to show tht butting this tringle long its edge A2 A there is nother one, with vertices A2, A, A. In the cse illustrted in Figure 1, the centers of ny two nerest unsplit hexgons re seprted by units, nd so we cll this the tiling H. But the seprtion could be ny positive integer. The rgument lso hndles the cse when n ccordionrow contins only one unsplit hexgon which is nturl to cll Hh (Figure 11), nd tht in which it hs no unsplit hexgon, Hh* (Figure ). This concludes the proof. c. The Tiling With No Vertex of Type At ny time from now on, the tilings we hve lredy found will be clled the old tilings. Any new tiling cn hve tiles of only the types: (...), (...), (...), (...). Theorem. The only new tiling with no vertex of type is the rhombic tiling R of Figure 15. Figure 15 Proof. Plinly ll its tiles must hve type (...), nd moment s thought suffices to show tht these cn only but s in Figure 15. c
6 7. Tilings With Line of Type (...,...,...) We sy tht stright line hs type (...,...,...) if it consists entirely of type (.) edges of the tiling. (c) (d) Figure 1 (Some exmples of mny possible rrngements of prllel tilings ) Theorem 5. The only tilings contining such line re the prllel tilings illustrted in Figure 1. Proof. We consider the tiles tht but the given line on one side. Any chosen one of these hs type (...) which cn only be (...) s in Figure 17 or (...) s in Figure 18. In the first cse the tiles neighboring the given one hve shpes (...) which mkes them lso (...), nd inductively we see tht in fct ll the tiles on tht side hve tht type, nd re bounded bove by nother of type (...,...,...). B A A A A A A B In Figure 18 the neighboring types re (...) or (...), nd so re necessrily (...), nd we see gin tht ll the tiles (A) on tht side of the line hve this type. But this time they re bounded bove by zigzg line whose vertices re lterntely of types nd. The tiles (B) tht but them from bove hve types (...) (...), nd we obtin Figure 18b. The tiles () bove these hve types (...) which my be either (...) or (...), nd it is gin esy to see tht in fct ll of them must hve the sme one of these types. If this is (...) they re bounded Figure 18 Figure 17
7 bove by nother line of type (...,...,...) (hinted t by the dotted line in Figure 18b). If not, we hve nother row (D) of horizontl rhombs (...), nd so on. c 8. Ptches nd the Prity Rule From now on, we shll seprte the edges of our tilings into two clsses, indicted by thick nd thin lines. We mke this decision rbitrrily for one edge, nd then continue by using the rule tht ny pir of djcent edges of tile re to be the sme thickness if they but t type or vertex, but of different thickness if they but type vertex. This rule is consistent round ny tile becuse ll possible types hve n even number of type vertices, nd round ny vertex becuse we cn mke its edges ll hve the sme thickness if it hs type or, or lternting thicknesses if it hs type. The thick edges now divide the tiling into regions which we shll cll ptches, consisting of vrious numbers of tiles. The edges of ptch will consist of chins of edges of the underlying tiling, with possibly some internl verticies which must ll be of type nd end vertices which must be of type or. Plinly ny vertex in the interior of ptch must hve type or, nd two djcent such vertices must be one of ech type (in new tiling). An exmple is shown in Figure 19. Figure 19 We shll now prove wht we cll The Prity Rule : Theorem. The two ends of n edge of ptch tht hs ny internl vertices re of the sme type ( or ) if tht edge hs n odd number of internl vertices; of opposite types (one, one ) if it involves n even number. Figure 20 Proof. As in the proof of Theorem 5, the tiles tht but this edge inside the ptch must ll hve type (...) except for the two t the ends. Figure 20 shows three possibilities: we see tht these tiles re bounded on the side wy from the edge of the ptch by zigzg line whose ends hve type, while the others hve lterntely types nd. Since the only possible type for the two end tiles is (...) the ends of the edge must hve the types opposite to those of the outermost internl vertices of this zigzg, which estblishes the theorem. c
8 Figure 21 There cn exist ptches tht violte the prity rule, but they my contin no internl vertices. It is esy to see tht ny ptch contining no internl vertices is of one of the shpes shown in Figure 21. (Not ll of these violte the prity rule.) Theorem 7. Any tiling with ptch tht violtes the prity rule is one of the Prllel Tilings found in Theorem 5. Proof. If one edge of ptch violtes the Prity Rule, then we see from Figure 21 tht so does the opposite edge. The ptches djcent to the given one long these edges must lso be in Figure 21, since they eqully violte the rule. We cn continue in this wy to obtin n infinite concertin of such ptches, which we see contins some lines of type (...,...,...), nd so must be covered by Theorem 5. So no new Tiling cn contin ny edge tht violtes the Prity Rule. This concludes the proof. c 9. The Difference Rule, nd the Shpes of Ptches Since ech vertex ngle of ptch is either 0 or 0, ptch cn be t most hexgonl. We shll stte nd prove the Difference Rule first for hexgonl ptches, since this is in wy the most generl cse. b 9 5 A 7 5 c Figure 22 B 11 Theorem 8. For hexgonl ptch, the numbers A,b,,,B,c of lines tht cross the successive edges of the ptch (see Figure 22) obey the Difference Rule : A B b c. b g Figure 2 b g Proof. We introduce three scores, b, g, tht re ttched to the ends of the thin edges of tiles ccording to the direction in which they point, s in Figure 2. Since the two ends of ny edge cncel, the totl scores of ALL the edgeends in our ptch must ll be zero.
9 b Figure 2 b b g b b  g g b  b b  g g g b g b  b g b  g g b b  g g b  b g b  g  g 0 b  bg 1  b 1 g Now s seen in Figure 2 the three edgeends t type vertex hve either: b g 1 or b g 1, while the totls for the six edgeends t type vertex re b g 0. It follows tht if we sum over ll edgeends t ll internl vertices of ptch, the three scores stisfy b g, nd therefore the sme must be true for the scores of the edgeends t boundry vertices of the ptch. But for these the scores re A, b B b, g c, proving the rule. The possible shpes of ptch re shown in Figure 25. The second nd third rows of this figure show the smllest ptches possible which do not contin the rhombic tiling R. The finite ones cn be regrded s hexgons in which some edges hve length zero (nd so contribute zero to the pproprite score), nd so the Difference Rule nd its proof continues to hold for them. The infinite ptches shown in Figure 2 cn lso be regrded s hexgons in which some edges hve infinite length, while others hve receded to infinity. The proof of the Difference Rule fils, becuse the sums involved become infinite; nd in fct those of the numbers A,b,,,B,c tht survive re only restricted by the Prity Rule. This concludes the proof. c 10. All Remining Tilings All the tilings not lredy discussed re obtined from tesselltions of the plne by pieces of the shpes indicted in Figure 25 in the following wy. We ttch to ech finite edge number, which is the number of thin edges which tht edge is to cross. These numbers must stisfy the Prity Rule, nd for finite ptches lso the Difference Rule. If they do so, they determine unique tiling, in which ech ptch is filled by portion of the rhombic tiling R. These ptches my be combined with the ptches shown in Figure 21. Figure 25
10 Figure 2 One exmple of composition using trditionl eltic style to express one of the new underlying tilings presented in this pper cn be seen t the front of the Bridges Section. Below re the five sphericl tilings with regulr vertices tht llow curves. Acknowledgment I thnk Professor John onwy of the University of Princeton for spending hours of his time to help put my rguments into mthemticl lnguge nd dedicte this pper to him. References [1] The complete set of tiles tht includes Scurves, curves nd combintions of Scurves nd curves, study of those compositions used by trditionl rtisns nd mny tht ugment trdition utilizing them is presented in Scurls nd Whirl Spools by hris K. Plmer, VisMth Volume 10. [2] Some interesting exmples of trditionl compositions tht use tilings with irregulr underlying tilings not ddressed by this pper cn be seen in eltic Art the methods of onstruction, George Bin, Dover ISBN , pge pltes 11. Pges 5 Pltes 710 contin exmples of frieze ptterns nd rosettes utilizings spirls lso not ddressed in this pper. [] After the fmous crytllogrpher Fritz Lves. See Tilings nd Ptterns, Grunbum nd Sheprd, ISBN , pges [] Recursive methods developed by the uthor from 1997 to the present show tht within the restrictions imposed by the rules described in this pper n infinite vriety of spirl tilings cn be composed. These will be presented in forthcoming ppers nd s prt of doctorl thesis. The principles presented in this pper pply for tilings on spheres: Hyperbolic tilings cn lso be composed with these tiles nd the ddition of other even sided tiles.
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