Upper bound for ropelength of pretzel knots

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1 Upper boud for ropelegth of pretzel kots Safiya Mora August 25, 2006 Abstract A model of the pretzel kot is described. A method for predictig the ropelegth of pretzel kots is give. A upper boud for the miimum ropelegth of a pretzel kot is determied, ad show to improve o existig upper bouds. 1 Itroductio A (p,q,r) pretzel kot is composed of 3 twists which are coected at the tops ad bottoms of each twist. The p, q, ad r are itegers represetig the umber of crossigs i the give twist. However a pretzel kot ca cotai ay umber,, of twists as log as 3, where the kot is deoted by (p 1, p 2, p 3,..., p ), ad defied by a diagram of the form show i figure 1. Fig. 1: A 2-dimesioal represetatio of a pretzel kot. The goal of this paper is to fid a model where the kot is tightest ad the ropelegth is miimized. The ropelegth of a kot is determied by the ratio of the legth of the rope to its radius. Columbia College, Columbia, SC Rop(K) = L r 1

2 It follows from corollary 1 of [1] that the ropelegth of the ukot is 2π. The lowest kow provable upper boud for the trefoil kot is [2], while a lower boud for the ropelegth of the trefoil kot is also kow to be [3]. A upper boud for ropelegth for all types of kots is give i a paper by Catarella, Faber, ad Mulliki [4]. This upper boud is a fuctio of crossig umber, c(l), ad is give by the followig equatio for prime liks: Rop(L) 1.64c(L) c(L) The mai theorem of this paper is a upper boud o ropelegth for ay (p 1, p 2, p 3,..., p ) pretzel kot. Theorem 1. If K is a pretzel kot defied by (p 1, p 2, p 3,..., p ), the Rop(K) π (2 2( 1 p i ) ( p i p i+1 ) + p p 1 ) () A direct corollary of this theorem is a equatio for a upper boud o ropelegth of the (p,q,r) pretzel kot. Corollary 1.1 If k is a pretzel kot defied by (p,q,r), the Rop(k) π (2 2(p + q + r) p q + q r + r p ) It should be oted that oe ca use Theorem 1 to give a upper boud i terms of the crossig umber as well. Oe aive approach follows. For (p 1, p 2, p 3,..., p ) where all p i s are positive, the crossig umber is give by c = p i. If all p i s are egative, the c = p i. Sice ( p i p i+1 ) + p p 1 ( p i + p i+1 ) + p + p 1, ad p i + p i+1 ) + p + p 1 = 2c, we ca simplify the upper boud to Rop(K) c As metioed above, the previously kow upper boud is a quadratic fuctio of the crossig umber. While ours is liear. Of course, the boud give i Theorem 1 is much better tha that of equatio 1. 2 Costructig the Twists I order to prove Theorem 1, we first aalyze the costructio of the twists. Two differet models were cosidered for costructig the twists of the pretzel kot. The first model was a double helix costructed with rope of radius 1. The 2

3 secod model was a sigle helix costructed of rope of radius 1 wrappig aroud a straight rope also of radius 1. For the purpose of this paper, we let the radius of the rope be equal to 1. The secod model was take ito cosideratio due to the aalysis of the (59,2) torus kot i the paper by Baraska, Pieraski, Przybyl, ad Rawde [2]. The costructio of oe helix wrappig aroud a straight rope appeared i the torus kot ad seemed to perhaps miimize the ropelegth ecessary. I the first model, the pitch of the helices must remai 1 to prevet the helices from overlappig. I the secod model, sice we did ot have to worry about oe helix ruig ito aother, the pitch could be reduced to at most The legth of a twist with a sigle crossig for this model is about However, whe compared with the legth of the double helix model for a twist with a sigle crossig, 8.89, this was ot eough to provide a shorter legth tha the first model so the secod model was throw out. Therefore the twists are modeled by the double helix ad give by the parametric equatios: (cost, sit, t) ad (cos(t + π), si(t + π), t). There are a few cocers to esure that o ropes overlap whe miimizig ropelegth. These are show i both [2] ad [4]. The followig are the requiremets as they apply to the costructio of a double helix: 1. The distace betwee the two helices is at least twice the radius. 2. The distace betwee doubly-critical poits must be at least twice the radius. 3. The radius of curvature must be at least the radius of the rope. Lemma 1: The double helix with pitch 1 ad radius 1 satisfies all three of the above requiremets. Proof. To satisfy requiremet 1 we fid the miimal distace betwee oe helix to the other helix. Let the first helix, (cost, sit, t), be H 1 ad the secod helix (cos(t + π), si(t + π), t), H 2. Let t = 0 i H 2, so that H 2 (0) = ( 1, 0, 0). The distace betwee H 1 ad H 2 (0) is d(t) = (cost + 1) 2 + (sit) 2 + t 2. Whe this distace is miimized, t = 0 ad d(0) = 2. Sice the radius of the rope is oe, coditio a is satisfied. Doubly-critical poits occur whe (x, y) K ad (x y) is orthogoal to both the vector taget to x ad the vector taget to y. To show that requiremet 2 is true whe radius is 1, we must fid the vector o H 1 orthogoal to the taget vector of H 1 (0). Thus we solve ( c(t) c(0)) c (0) = 0. I our case: c(t) = (cost, sit, t) c(0) = (1, 0, 0) c (t) = ( sit, cost, 1) c (0) = (0, 1, 1). The vector betwee c(t) ad c(0) is (cost-1, sit, t). If we take the dot product 3

4 of this vector ad the vector taget to the curve at t = c (0) ad set this value equal to 0 we will fid a vector orthogoal to both. I this case the dot product gives us sit + t = 0 which gives us t = 0. Sice this is the same as our origial poit, this tells us that there are o doubly-critical poits whe the pitch is equal to 1. To satisfy requiremet 3 the radius of curvature of a sigle helix must be at least 1. The curvature formula is κ = c (t 0 ) c (t 0 ) c (t 0 ) 3 ad the radius of curvature is 1 κ For the double helix the curvature is κ( c) = 1 2 ad the radius of curvature equals 2. So the curvature of the double helix will ot be a problem. Fig. 2: A double-helix with radius of 1 ad pitch of 1. The legth of oe twist of a sigle helix is foud by l = π 0 ( si(t))2 + (cos(t)) dt = 2π The legth of a double helix with oe crossig is equal to twice the above value. Therefore the legth of a double helix with crossigs is L = 2 2π ad the legth of the double helices i ay pretzel kot is 3 Proof of Theorem 1 l helix = (p + q + r) 2 2 π The followig sectio explais the costructio of a (p,q,r) pretzel kot, or a pretzel kot with three twists. The pretzel kot was costructed by usig partial circles ad straight ropes to coect the helices. The legth of each idividual part is deoted by l helix for the helices, l circ for the partial circles, l vert for vertical straight ropes, ad l hor for horizotal straight ropes. 4

5 3.1 Placig the Helices The issue ow is to place the 3 seperate twists/double helices i such a way that miimizes the ropelegth eeded to coect the helices. Each double helix is cotaied withi a cylider of radius 2. By placig the three double helices i such a way that the cyliders of radius 2 do ot itersect, the double helices are guareteed ot to overlap. Chagig the rotatio of the double helices will miimize the distace betwee the helices to be coected later o. Figure 3 shows the placemet of the helices which will reduce the ropelegth ecessary to coect the helices while esurig that o ropes will overlap. Fig. 3: Placemet of the helices as see lookig dow the z-axis. The equatios for the helices i these locatios are Helix 1: (cos(t + π 2 ) + 4 3, si(t + π 2 ), t) Helix 2: (cos(t + 3π 2 ) + 4 3, si(t + 3π 2 ), t) Helix 3: (cos(t + π 3 ) 2 3, si(t + π 3 ) 2, t) Helix 4: (cos(t + 4π 3 ) 2 3, si(t + 4π 3 ) 2, t) Helix 5: (cos(t π 3 ) 2 3, si(t π 3 ) + 2, t) Helix 6: (cos(t + 2π 3 ) 2 3, si(t + 2π 3 ) + 2, t) 5

6 Fig. 4: The twists of a (2,4,3) pretzel kot These equatios were obtaied by startig with helix 1 ad helix 2 ad rotatig them by 2π 3 aroud the z-axis. 3.2 Coectig the Helices Followig the umberig used i figure 3, helix 1 ad helix 6, helix 2 ad helix 3, ad helix 4 ad helix 5 must be coected o the top ad the bottom of each helix. The first step to be able to coect the helices is to flatte the tops ad bottoms of each helix. At this poit each helix begis ad eds at a 45 agle with the horizotal. I order to flatte these, 1 8th of a circle with radius 1 will be added to the top ad bottom of each helix. These partial circles must satisfy similar coditios as stated above for the helices. First, the distace betwee the ceters of the circles must be at least twice the radius. Sice the ceters of the circles will be located at the same locatio o their helix ad it has already bee show that the helices have a distace of 2, the ceters of the circle must be at least 2 uits apart. I this case there will be o doubly critical poits. Fially, the radius of curvature must be at least 1. The radius of curvature for a circle with radius 1 is 1. So these partial circles satisfy all coditios for the costructio of this kot. The legth of these 1 8 th circles is l = 1 8 (2π). Sice 12 of these 1 8th circles will be eeded the legth that they cotribute to the total ropelegth is 3π, which will cotribute to l circ later. Oce the helices have bee flatteed, it is ecessary to make the helices which are to be coected the same height. So helix 1 must be the same height as 6

7 helix 6, helix 2 the same height as helix 3, ad helix 4 the same height as helix 5. This will be accomplished by simply coectig a straight rope to the top of the shorter of the two helices. This straight rope will ot itersect itself or ay other sectio because it will exted vertically from the attached 1 8th circles which are of distace of at least 2 apart. The radius of curvature ad doublycritical poits will ot be a issue sice this is a straight rope. The legth of this vertical straight rope will be equal to the differece i height betwee the two helices. Sice each twist has height of π the legth cotributed to the total ropelegth by these vertical straight ropes is l vert = ( p q + q r + r p )π Fig. 5: Costructio of two double-helices. Now that the helices to be coected are at the same height, the helices ca be coected. The coectios will be costructed by a horizotal straight rope with a 1 4 circle of radius 1 o either ed. It must be show that both the horizotal straight rope ad quarter circles satisfy the coditios give above for the costructio of a kot. The quarter circle will be similar to the 1 8th circle described above. The ceters of the circles will be o their respective helices, which are at least distace 2 away from each other, i the directio of the helix that they are to coect to. Because of the placemet of the helices o two quarter circles will overlap. Agai, there are o doubly-critical poits ad the 7

8 radius of curvature is 1. The horizotal straight ropes will be similar to the vertical straight ropes described above. The 1 4 circle coects the top or bottom of the twist with the horizotal coectig rope. The 1 4 circles will have legth of 1 4 (2π). Oce agai 12 of these 1 4 circles are eeded. So the total legth that the 1 4 circles alog with the 1 8 th circles cotribute to the total ropelegth is l circ = 9π Fig. 6: Coectio at the bottom of two helices created by a straight horizotal rope ad two 1 4 circles. The legth of each horizotal rope is the distace betwee the two 1 4 circles to be coected. I our costructio this distace is for a (p, q, r) pretzel kot. Sice 6 of these horizotal straight ropes are eeded, the legth these cotribute to the total ropelegth is l hor =

9 Fig. 7: Fial costructio of a (2,4,3) pretzel kot. 3.3 Total Ropelegth Usig this costructio a upperboud for (p,q,r) pretzel kots ca be foud. Whe all of the idividual parts are added a upper boud for ropelegth is give. Rop(L) π (2 2(p + q + r) p q + q r + r p ) It was foud that this costructio ca be applied to all (p1, p2, p3,..., p ) pretzel kots. We ca geeralize all parts of the above equatio to give a upper boud o ropelegth for all pretzel kots with twists. The helices are foud by rotatig helix 1 ad helix 2 by θ = 2π aroud the z-axis where helix 1 ad helix 2 are placed a distace of si2 π alog the x-axis. The distace betwee the two quarter circles to be coected for a (p1, p2, p3,..., p ) pretzel kot is r si2 π 1 + q 1 2 cos 2π ( 2π si +1 q 1 ) 2 si 2π si 2π ( 2 si π +1 q 1 ) 2 So the legth of the horizotal straight pieces is 2 less tha the above distace. However, sice this legth decreases as icreases, we ca simply use the value foud for the pretzel kot with three twists, , i the upper boud. The followig are the geeral equatios for all parts of the ropelegth: Lhelix = 2 2( X pi ) π Lcirc = 3 π 1 X Lvert = (( pi pi+1 ) + p p1 ) π Lhor = ( ) cos 2π 2

10 Combiig these parts gives us Theorem 1. 4 Coclusio Whe this upperboud is applied to the (2,4,3) pretzel kot we are give a upper boud of The previously kow upper boud for kots gives a upper boud of Compared to the previous upper boud, this upper boud greatly improves upo the upper boud of ropelegth for (p,q,r) pretzel kots. This ca also be see i the equatios by the fact that both are fuctios of crossig umbers. However, as metioed before the previously kow upper boud is a quadratic fuctio while the upperboud give i this paper is a liear fuctio, resultig i a lower value. However, it should be oted that there is room for improvemet o this upper boud. It may be possible to brig the helices eve closer together, which would slightly reduce the upper boud o ropelegth. There could also be a better costructio that would miimize the ope space i the middle of the kot, which would also hopefully reduce the upper boud o ropelegth. Ackowledgemets This research ad paper would ot have bee possible without the advisemet ad assistace of Dr. Rollad Trapp, Califoria State Uiversity at Sa Berardio. Refereces [1] Literlad, R.A.; Simo, J.; Durumeric, O.; Rawde, E. Thickess of kots. Topology ad its Applicatios, 91: , [2] Baraska, Justya; Pieraski, Piotr; Przybyl,Sylwester; ad Rawde, Eric. Legth of the tightest trefoil kot. arxiv:math.gt/ v2 16 Oct [3] Dee, Elizabeth; Diao, Yuaa; Sulliva, Joh. Quadrisecats give ew lower bouds for ropelegth of a kot. Geometry ad Topology, 10:1-26(electroic), [4] Catarella, Jaso; Faber, X.W.; ad Mulliki, Chad. Upper bouds for ropelegth as a fuctio of crossig umber. Topology ad its Applicatios, 135: ,

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