STANDARD HERRINGBONE KNOTS (SHK) - Charles HAMEL
Contents
1.2. Ke Sa Sak Sa amp gt gt RSS 2 et lt lt ONSE SE SN SE SE ee SNR REESE mp e T3335 OOS Fig 1 A general Herringbone Coding 5555555 522 OP A AR Se Fig 2 A 3 pass Herringbone Coding Copyright Charles HAMEL aka Nautile 2009 Nov PAGE 4 of 33 Schaake remarked that it is impossible to discover general rules if we don t go MODIFIED FROM SCHAAKE amp TURNER beyond this 2 SEQUENCE NUMBER SEQUENCE NUMBER PASS barrier to realize the existence of what he call the fascinating sequence rule GRANT left it at 2 PASS This is probably the reason why there is no SHK over 2 PASS having been published X VV XY 4 V V n This sequence is the sequence of the position of each of the component Turk s head knots forming the SHK relative to the very first laid one the foundation knot oO ANS VV V 1 1 1 1 1 1 5 3 1 9 5 3 IK 2 4 6 8 0 2 4 OOONSAD g
3. SE L even with B such that GDC 1 ODD numbered A SET NUMBER 1 L 2 1 1 L to reference A 1 A 1 A 1 A 1 1 A 1 U O U to L reference A 1 A 1 A 1 A 1 A 1 A 1 1 U U Half period 1 LtoR A 1 A 1 A 1 7 A 1 A 1 A 1 Half period 2 tO L EEUU A 1 Half period 3 0 oaa 1 Half period L to 5 1 28 1 Half period 2B 10 A 1 EVEN numbered A SET NUMBER IM 2 NN L 3 L 2 L 1 L to reference A 1 A 1 A 1 4 7 4 4 1 U U U O R to L reference A 1 A 1 A 7 1 7 4 7 4 1 U O U U O U Hali LtoR A 1 A 1 A 1 A 1 A 1 1 period 1 Half RtoL i eee period 2 Half ie wi dass period 3 Half LioR Si aes period 2B 1 Half 351 ME period 2B Copyright Charles HAMEL aka Nautile 2009 Nov PAGE 14 of 33 Now for an example of how to put a table to use p 5 b 4 Sequence Number odd Set number L Reference R e ga pat m weh R L 4 1 4 1 4 1 1 m m m pod j E bf os rir d Table above and under are quoted directly unmodified from Schaake Let us say A 1 then that read
4. DIFFERENT BIGHT RIM x I 6 en N E 2 en e tI o F o S 2 o 6 MAAN ANN SAA N y ARIMA N st Copyright Charles HAMEL aka Nautile 2009 Nov Copyright Charles HAMEL aka Nautile 2009 Nov PAGE 30 of 33 COMPARE SE 4L 3B the whole with a TYPE SO 3L 2B the smaller area 22 sh ON gt 2 ON This diagram allows to see at one glance Type So AND a Type Se Here are two tables with the coding of those two knots so that you may verify 1 the cordage TYPE 56 4 PASS 3L 2B 121 8B SHK Free run Copyright Charles HAMEL aka Nautile 2009 Nov PAGE 31 of 33 TYPE Sg 4 PASS 41 3B foral6L 12B SHK Free run 1 01 1 01 2 O1 H P2 Ol U1 O1 02 H P 3 U1 H P 3 Ul O1 U1 02 H P 4 UI OI H P 4 Ol U1 02 U2 5 O1 UI H P 5 Ul O1 U2 02 O1 U1 OI O1 U2 02 U2 U2 02 U2 O H P2 02 U2 03 U H P3 U2 02 03 0 H P 4 02 U3 03 U 5 U2 03 0 H P6 03 U3 O3 U U3 03 U3 03 H P 2 03 U3 03 04 H P3 U8 03 U3 O4 H P4 03 U3 O4 04 H P5 UB
5. 1 10 13 13 3 3 3 HP 21 51 03 U4 2 101 133 3134 HP 108 U3 03 U4 3 108 3313 34 HP 4 03 U3 4 U4 4 101 1313 3144 HP 5 108 U3 03 4 U4 HP 5 LtoR 3 3 3 4 4 HP 6 85 03 4 04 U4 HP 6 1011313 14 4 4 HP 7 LtoR U3 U4 O4 U4 HP 7 LtoR 1313141414 HP 8 Rtol 4 04 O4 U4 HP 8 RtoL 3 4 4 4 4 Calculate the value of A and 1 and search the U coding in the second or third row as is the case L to R for odd numbered Half Period and R to L for even numbered H P Copyright Charles HAMEL aka Nautile 2009 Nov PAGE 16 of 33 Using algorithm tables you can make any A PASS STANDARD HERRINGBONE KNOT About the SET NUMBER In a completely finished SHK for each half period it is possible to separate the crossing existing along this half period in SET OF CROSSINGS in such a way that IN A GIVEN SET ALL THE ADJACENT CROSSINGS IN IT ARE OF THE SAME TYPE OVER UNDER IN ALL CASES THE NUMBER OF SETS IS EQUAL TO L the number of LEAD in any one of the component THK components are all of identical dimension BUT we do not give a number to ALL of these SET think about the BLACK CELL either at the beginning or the end of the row of set number The set without number the one with the caviar ed a term from French printed matter censure meaning covered with black ink as if a toast wi
6. SET NUMBER L to R reference R to L reference EVEN numbered A SET NUMBER L to R reference R to L reference Copyright Charles HAMEL aka Nautile 2009 Nov PAGE 12 of 33 Now a pair of whole tables 5 L opp with B such that GDC 1 ODD numbered A SET NUMBER 1 2 3 4 L 2 L 1 L to R reference A 1 1 4 7 1 A 1 A 1 4 7 U O U O U U R to L reference 4 7 4 7 1 A 1 A 1 1 U U U U Half LtoR A 1 A 1 A 1 4 4 7 1 period 1 Half o 1 period 2 Half Lto R I re 1 period 3 Half LioR EXE JW A 1 period 2B 1 Rio L ___ 1 2 EVEN numbered A SET NUMBER MN 2 NES L 3 L 2 L 1 L to R reference A 1 A 1 1 1 A 1 A 1 1 U U U U R to L reference A 1 A 1 A 1 1 A 1 A 1 A 1 U U U U Half LtoR A 1 A 1 A 1 4A 1 A 1 A 1 A 1 period 1 Half xs UE i alll abes period 2 Half LtoR ATI Sy _ 16 period 3 Half LioR tT QA dun period 2B 1 Half 8161 i neun period 2B Copyright Charles HAMEL aka Nautile 2009 Nov PAGE 13 of 33
7. so there will be NO nested bight possible in a SHK as must exist in Herringbone Pineapple MODIFIED from SCHAAKE amp TURNER YEN N The sequence rule implies a rhythm a predictability that allows the construction of tables giving the construction of SHK In a SHK the total number of BIGHT will be A the number of PASS time B number of BIGHT of the component THK Fig 12 The Coding Patterns associated with the upwards braiding direction A PASS SHK with THK components of 4L 5B dimension will have 3 5 15 Bs for component B is for SHK As there are 2 ROW Schaake s frame of reference horizontal mandrel will use all along this article per BIGHT there will be a total of 2 15 30 rows As contrary to what exist in a Herringbone Pineapple all the BIGHT of the SHK are ona unique BIGHT RIM or bight boundary to speak like S amp T there will L 2 L Total number of leads in the SHK In this case 4 3 12 LEAD So L 1 or 12 1 11 columns of crossings alignment the SHK UNIFORM PASS all Aj are have the same value A all the PASS are equal COMPOUND there are several dimensions at least two different of SET of PASS At least two Aj have different numerical value see at the very end of this document Copyright Charles HAMEL aka Nautile 2009 Nov PAGE 6 of 33 Of course the components THK have their L and B complying with the GDC
8. A ODD i 4 68 5 4 14 3 o R to L A 1 U U U 0 UO LioR U U UO 0 5 4 3 2 10 A EVEN 1 3 0 2 4 1 3 0 RtoL UO U O U U A1 O Just one example 3 PASS 3 THK in the SHK 5 Half Period So ODD numbered PASS and L to R ODD numbered H P so upper part above the horizontal line i 5 3 2 1 A 3 1 2 50 U2 O2 U3 O2 3 32 2 3 2 2 3 2 3 2 1 1 A 1 1 1 1 LtoR UO U O U U 0 3 1 4 2 0 3 1 A ODD and so on Copyright Charles HAMEL aka Nautile 2009 Nov PAGE 27 of 33 Last word is left to my source of inspiration SCHAAKE amp TURNER The algorithm diagrams provide a much more compact form of the instructions for the execution of the required steps in the construction of the Standard Herringbone Knots Furthermore the relative position of the component algorithm diagrams associated with the odd and even sequence numbers within the overall algorithm diagram provides a simple mnemonic for the relative bight positions of the interbraided Turk s Head knots in accordance with the sequence rule when the upwards braiding direction is used As am not sure that some persons will not miss some points here are some diagrams of mine in the hope of helping those who can only understand visually As can be seen as a nose in the middle of the face 2 PASS have no discrimina
9. Set number 1 1 2 3 IN Setnumber 1 2 3 4 a HP1 LtoR 0 0 0 0 0 HP 1 Lto R 00 O0 00 00 UO HP2 RtoL 0 0 0 0 0 HP 2 R to L 00 00 00 00 UO HP3 LtoR 0 0 0 110 HP 3 LtoR UO 00 UO 01 UO HP4 RtoL 0 0 1 1 0 HP 4 RtoL 00 O0 UI 01 UO HP5 LtoR 0 0 1 11 0 HP 5 LtoR 00 O0 UI OI UO HP6 RtoL 011111110 HP6 RtoL 00 01 U1 OI UO HP7 LtoR 0 1 1 1 0 7 LtoR 00 OI UI 01 UO HP8 RtoL 1111111110 HP 8 RtoL U1 01 Ul O1 UO Copyright Charles HAMEL aka Nautile 2009 Nov PAGE 15 of 33 S 5 4 Sequence Number even Set number Em e m D Reference u R Reference u m RN rn m D gt 2 m mem m m Dori mM EJ gt D I gt men 59 gt aH a m 1 z PBE d E B ge 2 Pii ed pa gt Let us say that A 4 then that read Set number 12345 Set number 2 13 4 5 HP 1 10 03 U3
10. 1 rule to be single strand knots For a single strand THK which the component THK are in our UNIFORM A PASS SEMI REGULAR SHK the GDC of L and B is 1 but the SHK will have L A L B Obviously the L amp B do not comply with the GDC 1 rule I hope it is startlingly evident for every one unless you count the foundation THK as a 1 PASS SHK which would be ridiculous in the real 3D world if acceptable in the theory So L and B of SHK have a GDC suk gt 1 for the SHK with A gt 2 PASS This GDC sux Either IS divisible by A number of PASS Or IS NOT divisible by A That makes for two different sub families see first diagram illustration given If GDC sux IS divisible by Then that means that n n being a positive integer 1 to as great a number as you want that is obvious 3 PASS 24L 158 GDC 3 1 2 PASS 301 248 GDC 6 n 3 2 PASS 24L 20B GDC 4 n 2 A SHK made with n A strands with n A gt 1 of course will be a UNIFORM A PASS SEMI REGULAR n A STRANDS HERRINGBONE KNOTS BUT there is still a dichotomy in those knot to be observed Those where n is an ODD number they will have n A TURK S HEAD KNOT COMPONENT and Those where n is and EVEN number they will have MATTHEW WALKER KNOT COMPONENT see illustration next page SCHAAKE insists on this point We emphasize that the interwoven knots of a Semi Regular Herringbone Knot play an important
11. COMPOUND PASS PAGE 32 of 33 Copyright Charles HAMEL aka Nautile 2009 Nov 200 AVA AN OS NAAT AA Abad AAT AAA bss NT AAA AAV A PANNI IAA INI HIN FANS NNI AN WY NNN SNN ANNI ANN N AIL 8 EWN ANNOS NAE NOSE NOSE SOSA NS u u 61 65 Y iD V A af 3 7 Pass PAGE 33 of 33 STUDY THE Se and Se BIGHT RIM AND THEIR CHARACTERISTICS Copyright Charles HAMEL aka Nautile 2009 Nov y N S f 6 e n Ay 4 ENING A NY EA NOI EANST EAN VS UA A V M CY COL GUY CO LA CY
12. Nov PAGE 25 of 33 SCHAAKE s reading rules all the unspecified number of crossing belonging to a coding type and corresponding with a bight number is equal 10 A 1 unless this bight number is smaller than or equal to the bight number associated with the half cycle under consideration in which case the number of crossings is equal to A For this 7 half period have explained elsewhere why prefer period to cycle We get U4 O4 U4 05 U5 05 U4 Now the 10 half period in this 5 PASS EVEN numbered H P so R to L with bight index 10 2 224 We get 0 1 2 3 4 5 0 RtoL O U U O U 4 5 5 8 5 5 4 Or 1 A A A A 1 U4 05 U5 05 U5 O5 U4 It is the same thing for an EVEN numbered PASS and its ODD or EVEN H P only you then use the part UNDER the horizontal line It is pretty much the way to use the THK algorithm we have seen in my other web pages Turkshead and in my HP48GX programs for knot made along a THK cordage route ALGORITHM DIAGRAM FOR Se A real example from which you will infer the general procedure L 8 B 5 8 mod 5 2 or use m 1 and DELTA 3 after calculating them with the formulas Complementary cyclic bight number sequence is 03 14 2 MIND THAT the reference values for L to R and R to L are different from the Type So Copyright Charles HAMEL aka Nautile 2009 Nov PAGE 26 of 33 LtoR UO U U A1 O 0 3 1 4 2 6 3 1
13. Schaake made a distinction between horizontal mandrel frame of reference two braiding how hate that word in this context direction One laying direction is UPWARD he other is DOWNWARD After this very short introduction let us dig just a bit deeper Copyright Charles HAMEL aka Nautile 2009 Nov PAGE 3 of 33 Lifted straight fromSCHAAKE amp TURNER in fair quote as this cannot be told better and illustrated better A GENERAL HERRINGBONE CODING In the Introduction we have already mentioned that a herringbone coding is a row coding It consists of sets of 2A adjacent rows with i equal to 1 2 9 example of such a coding is illustrated in Fig 1 where reading top to bottom 41 is equal to 3 hence 24 2A 6 43 is equal to 2 hence 2A 243 4 is equal to 5 hence 24 243 10 is equal to 2 hence 24 244 4 and 45 is equal to 4 hence 2A 2 45 8 Each set of 24 adjacent rows consists of two sub sets and Aj each with adjacent rows The rows within a sub set all have the same coding The coding in all the sub sets Ai is the same and opposite to the coding in the sub sets A When the value of A is the same say A for all 2A sets the coding is an A pass herringbone coding such a herringbone coding is illustrated in Fig 2 a 3 pass herringbone codine gt gt gt gt gt gt gt
14. again on that well trodden ground Just go back to the necessary pages and PDF in Publication and in Turkshead pages Study the books and THE BRAIDER to be fully enlightened METHOD amp METHOD B They differ from one another by the method used to write the complementary cyclic METHOD A use the slow slogging method of L modulo B METHOD B use the DELTA and DELTA where This formula gives DELTA m 1 p p b with m being of such value integer value negative zero or positive that the quantity is the smallest possible m blah 1 p li Let us say that we were able to write the complementary cyclic forL 9 B z4 using either the programs gave or the equations that SCHAAKE gave in THE BRAIDER and other books 0 3 2 1 Then we write 0 1 2 3 4 5 6 7 8 0 3 2 1 0 1 2 3 4 5 6 7 8 Then 0 3 2 1 0 3 2 1 0 0 1 2 3 4 5 6 7 8 This last is called the SET DIAGRAM It is all that is necessary to be able to get the algorithm tables entries HAVE TO BE INCREASED FROM A 1 to A We address that a bit later this set diagram can be extended to algorithm diagrams which make the algorithm tables discard able as it is possible to make the SHK directly from them In SCHAAKE s own words However this method is generally not recommended for the novice he should first master all the procedures outline
15. 03 04 04 H P6 03 4 O4 04 A curiosity stolen from SCHAAKE amp TURNER with the high respect have for those men You will get dozens of grids and a lot of drawings by Tom HALL illustrating those grids The grids on the influence of colours on pattern are mesmerizing will give just sample to entice you to buy the CD mind you the pdf book on SHK has a few printing mistakes and a handful of pages illustration not text are missing NN RS o gt ee EEE CEE NC QS CS 383 373 2 493 2 3 2 333323 COFCO CELE OLAS 49 049 99 094 RE RT EER EES SECADI ENSE SSS SS SESS SS SSS Sb SELLE LEL 49 99 49 4 99 gt S NU CS NS gt gt 27372732348 3233333333 429749 E SUE 666665 S SE E SSS 27327 3 Ke KER SSS 2 ESS S S SES 25S X S lt 2 Ke ANN ENSE NENNEN AEN M Vi Nj NES NES ANE SS 22 2 Fig 9 2 3 pass Semi Regular Herringbone Knot 25 parts 20 bights Note the MIXED PASS
16. Copyright Charles HAMEL aka Nautile 2009 Nov PAGE 1 of 33 STANDARD HERRINGBONE KNOTS SHK It was difficult to summarize in less than 3 dozens of pages hundreds of pages so pray forgive me if this is a bit of a shamble to read Sources my own observations and reflection THE BRAIDER Schaake amp Turner BRAIDING STANDARD HERRINGBONE KNOTS Schaake amp Turner BRAIDING STANDARD HERRINGBONE PINEAPPLE KNOTS Schaake amp Turner made this diagram from the way Schaake dismembered the SHK CLASS of REGULAR CYLINDRICAL BRAID LOOKING knots with aHerringbone pattern CLASS of CLASS of REGULAR SINGLE STRAND knots SEMI REGULAR more than 1 STRAND knots SUB CLASS SUB CLASS REGULAR or SEMI q PERFECT REGULAR Sub HERRINGBONE HERRINGBONE elass knots knots Z FAMILLY FAMILLY FAMILLY FAMILLY UNIFORM COMPOUND UNIFORM COMPOUND PASS PASS PASS PASS SUB FAMILLY of knots SUB FAMILLY of knots whose whose GDC of L amp B is GDC of L amp B is DIVISIBLE by the NOT DIVISIBLE by the number of PASS A number of PASS A Copyright Charles HAMEL aka Nautile 2009 Nov PAGE 2 of 33 The knots we are speaking about here the STANDARD HERRINGBONE KNOTS SHK are assemblies of Turk s Head knots just as Standard Herringbone Pineapple knots SHPAK are assemblies of Turk s Head knots and just as are many interweaves of THK that are neither SHK or SHPAK BUT a big a very BIG BUT in the SHK the components
17. NER you will have them and their full User Manual by buying the CD with the 5 PDF of the books Calculation table for additional set number intersections for Herringbone PINEAPPLE knots Calculation of the SE7 WUMBEAS where an ADDITIONAL intersection occurs additional to reference quantity For Standard Herringbone PINEAPPLE KNOTS 13 SaD t where 1 2 3 discard eet numbers the range 1 2 3 8 1 rs w ere m 1 2 3 and 0 1 2 1 as for ha Fcyc e 2n Copyright Charles HAMEL aka Nautile 2009 Nov PAGE 21 of 33 Calculation table for REGULAR KNOTS those made on a THK cordage route Calculation of the COLUMW NUMBEHRS where an intersection occurs For REGULAR KNOTS knots made 3 i on a Turk s Head cordage route 12 2 where m 1 2 3 discard column numbers outside the range 1 2 3 mb rp where m 1 2 3 and r 0 1 2 n 1 as for ha Fcyc e 2n With any of those methods A B and C you should be able to draw the algorithm tables for any STANDARD HERRINGBONE KNOT you care to make SCHAAKE amp TURNER provided a LOT of tables pairs in their book on those knots and Tom HALL provided the hand drawings helping the untrained brains and eyes with interpreting the isometric grid diagrams Copyright Charles HAMEL aka Nautile 2009 Nov PAGE 22 of 33 Now let us see how to get the algorithm tables entries THA
18. T HAVE TO BE INCREASED FROM 1 to A Reminder The reference algorithm for any ODD numbered half periods is L 1 U A 1 O A 1 U A 1 O A 1 U A 1 O A 1 U A 1 O R 1 U The reference algorithm for any EVEN numbered half periods is R 1 U A 1 O A 1 U A 1 O A 1 U A 1 O A 1 U A 1 O L 1 U The algorithm for Half Period ONE is always identical to the reference algorithm for an ODD half period NO CHANGE OCCURS So half period 1 has each of its cells with the value of A 1 As for the rest of the half periods the reference values L 1 and 1 may have increased by ONE for some cells in set number 1 and the A 1 may have increased by ONE for some cells in some set number So L 1 becomes L and R 1 becomes while 1 becomes In the last column in the reading direction L to R and R to L respectively no change occurs Look at the table in page 19 the first table given for METHOD C which allows to calculate in which cells set number those changes have to be done This is what my HP48 program does Let us let go of the algorithm tables and attempt to look at the algorithm diagrams using a set diagram that were overflown quite swiftly in METHOD A and METHOD B Copyright Charles HAMEL aka Nautile 2009 Nov PAGE 23 of 33 ALGORITHM DIAGRAMS FOR THE SHK Remember We need a PAIR of tables one for ODD numbered PASS one for E
19. THK are ALL IDENTICAL plus those THK are not using several staged BIGHT RIM as Pineapple do SHK use only one rim as THK do The immediate implication is that there is NO NESTED BIGHT in a SHK while those nested bights are a must have for a SHPAK The first step in the making of an A PASS SHK A being a positive integer number 1 to N is the laying of a foundation knot in other word the making of a first THK When this first laid knot is finished then a second THK of identical dimensions in Lead and Bight is interwoven with it following a particular coding This make 2 THK assembled so it is a 2 PASS SHK That is the biggest PASS dimension was able to find published before Schaake amp Turner If a third THK is put in this will gives a 3 PASS SHK and so on 2 PASS 3 PASS 4 PASS 5 PASS 6 PASS 7 PASS 15 even PASS odd PASS even PASS odd PASS 55 odd PASS 1 2 3 7 8 n 3 2 n 1 n is called by Schaake the sequence numbers in the making of the knot As we shall see when using a pair of universal algorithm tables to make a SHK one of the pair of tables is for the even PASS and the other for the odd PASS THE INTERWOVEN THK HAVE TO BE POSITIONED RELATIVE TO THE FOUNDATION THK ACCORDING TO A CERTAIN ORDER This order is given by the SEQUENCE RULE Of course as regular as possible a physical interval positioning of the start points of the component THK around the rim of the SHK is the best to do
20. VEN numbered PASS for EACH type So Soaa and SE Seven of SHK depending on the parity of L the component THK number of LEAD do hope that Schaake was not mistaken given my experience with persons saying they are knot tyer fear he was when he wrote what looks like wishful thinking Most braiders will eventually use the a gorithm diagram method exclusively due to its overall compactness and simplicity Especially with regard to the Standard Herringbone Knot of Type S a very compact and simple algorithm diagram can be constructed This type of Standard Herringbone Knot is the most universally applicable one since it always create a symmetrical colour pattern when colour strings are used in its construction We must stress that there are a number of applications where the aesthetic quality of the braided artifact would be considerably enhanced by using a Type Se instead of a type So Standard Herringbone Knot For this reason we shall not discuss the most simple and compact form of the algorithm diagram that can be constructed for the Type So Standard Herringbone Knot but rather discuss the form which is similar to the one used for the Type Se Standard Herringbone Knot As we have seen previously in this document we need the complementary cyclic bight numbers of the THK components that are interwoven to build a SHK treated that complementary cyclic at length in other web pages on Turk s head knots and gave persona
21. d in all the other Chapters Copyright Charles HAMEL aka Nautile 2009 Nov PAGE 19 of 33 Given that most persons encountered in the flesh by mail or on Net forum don t seem to be easily able to grasp how to use the simpler methods then this is a very dire warning indeed Brains will over fry for sure ROTFLOL First this is also given as an HP48GX program of mine and explained in some Turks head pages it is necessary to call back to mind the relationship between half periods and bight numbers We need to calculate the bight number associated with the half period have pity on you recall heven 2 2 and i 3 2 where his the number attributed to the half period METHOD C use Calculation of the SE amp 7 WUMBERS where an ADDITIONAL intersection occurs 5 2 72 to reference quantity mb 3p 2ZoooNuNDpURBON gt vs ws os ut os ot ms me wr 0 we m where 1 2 3 discard set numbers eufs amp fe the range 1 2 3 mb where 1 2 3 and r 0 1 2 n 1 as for halfcycle 2n This table looks a lot like the table in use for STANDARD REGULAR KNOT knot on a THK cordage route For the Standard Herringbone PINEAPPLE knots a table built along those lines is used Copyright Charles HAMEL aka Nautile 2009 Nov PAGE 20 of 33 The 3 TABLES GIVEN HERE ARE QUOTED FROM SCHAAKE amp TUR
22. ir frame is concerned but the value to be entered in each of the yellow cells depends on the So or Se nature Copyright Charles HAMEL aka Nautile 2009 Nov PAGE 10 of 33 So or SE L opp orL even with B such that GDC 1 ODD numbered A SET NUMBER T u L 2 L 1 AN Lto R reference RtoL reference Half L to R MESSI period 1 Rio L period 2 Half L to R SO period 3 Half LtoR period 2B 1 Half R to L BEES period 2B EVEN numbered A SET NUMBER MEE 1 2 3 L 3 L 2 L 1 L to R reference R to L reference Half Lio R period 1 Half R to L BEES period 2 Half Lio R NN period 3 Half L to R period 2B 1 Half R to L period 2B Copyright Charles HAMEL aka Nautile 2009 Nov PAGE 11 of 33 Let us be slow in our progress READ VERY ATTENTIVELY DECIPHER COMPLETELY BEFORE PROCEEDING ALONG DON T RUSH THE VERY SOFT LEARNING CURVE I AM TRYING TO BUILD FOR YOU S L opp with B such that GDC 1 ODD numbered A SET NUMBER L to R reference R to L reference EVEN numbered A SET NUMBER L to R reference R to L reference Se L even with B such that GDC 1 ODD numbered A
23. l HP48 programs that deal with all that is necessary The algorithm diagram being a PRACTICAL TOOL it will be made in such a manner that the L to R ODD numbered half periods that are read L to R will be indeed read L to and that the R to L EVEN numbered half periods that must be read R to L will indeed be read R to L This is a very important point to keep in mind Many persons find that difficult even with the simple Regular Knots THK for example Mind to not forget to comply with the sequence rule Copyright Charles HAMEL aka Nautile 2009 Nov PAGE 24 of 33 ALGORITHM DIAGRAM FOR So A real example from which you will infer the general procedure L 7 B 6 First we get the complementary cyclic bight number sequence 7 mod 6 5 also usable are m and DELTA SD s qw o as many as B 0 0 1 0 2 1 05 4 2 1 We have now to put that complementary cyclic in as many than L 1 and one at each extremity LtoR UO U O U 0 5 4 3 2 1 O0 A ODD 0 1 2 3 4 5 0 R to L U O U U LioR A1 U U O U 0 U 0 5 4 3 2 1 0 A EVEN 0 1 2 3 4 5 0 RtoL UO U O U A1t U You are set to go PASS 5 5 THK component 4 after the foundation knot so ODD PASS so UPPER part above the line We want the half period 7 ODD numbered half period so L to R Compute the bight number i2 7 3 222 Copyright Charles HAMEL aka Nautile 2009
24. opyright Charles HAMEL aka Nautile 2009 Nov PAGE 17 of 33 The second and third rows give the REFERENCE QUANTITY A 1 and the TYPE OVER UNDER of the CROSSING SET NUMBER L to R reference R to L reference SET NUMBER L to reference R to L reference The algorithm tables previously given were entered with the NON CHANGING values As for the remainder of the cells they may stay at A 1 value or may have to be push up to A value Which cells are affected by the change depend on L and B We have to address the calculation of the SET NUMBERS which are to be increased HOW TO CALCULATE THE ALGORITHM TABLES SCHAAKE offers THREE METHODS A do not mix that with A for PASS B and C METHOD A and METHOD quote are essentially the same A does not use mathematics whereas B does Method C also uses some mathematics quote METHOD A and METHOD B use the complementary cyclic bight number scheme of the interwoven THK component That scheme we have seen else where Combining this complementary cyclic with the set numbers we can determine for a given half period which of its set number are increased from 1 to A MODULAR ARITHMETICS are in use of course Copyright Charles HAMEL aka Nautile 2009 Nov PAGE 18 of 33 BE REMINDED that in publications 3 are my HP48 programs They calculate all that will be used here so will not go
25. role in its classification Let us stay with SCHAAKE In this book we shall limit our discussion to the A pass nA string Herringbone Knots for which the value of n is equal to 1 hence the value of A is greater than 1 these knots are called the Standard Herringbone Knots Since 1 is an odd positive integer an A pass Standard Herringbone Knot with p parts and b bights consists of A interwoven Turk s head knots each of which has p p A parts and b b A bights PAGE 7 of 33 Copyright Charles HAMEL aka Nautile 2009 Nov BRAIDING STANDARD HERRINGBONE KNOTS does not treat of the COMPOUND PASS semi regular herringbone knots or the UNIFORM PASS semi regular herringbone knots The Authors treat differently the UPWARD and the A ot 30 parts 24 bights n 3 GDC 30 24 ARAN UNO s HEAD KNOTS COMPONENTS VAAN TURK Fig 6 2 pass Semi Regular Herringbone Kn 3 E MATTHEW WALKER KNOTS COMPONENTS Fig 7 2 pass Semi Regular Herringbone Knot 24 parts 20 bights n VAS AAA INNEN AAA AN DOWNWARD ways to make these knots will keep only the UPWARD mandrel frame 2 GDC 24 20 PAGE 8 of 33 Copyright Charles HAMEL aka Nautile 2009 Nov 17 WANWAN AN AAN ASN WANN Fig 14 The regular positioning of the starting points upward
26. s braiding method THERE IS ALSO AN ILLUSTRATION FOR THE DOWNWARDS BRAIDING METHOD Get yourself a copy of the book over 200 pages Tables will be for L and B values of the component THK even odd and L We will have to distinguish L In the book there are 32 PAIRS of READY MADE tables looking like the one down For each pair of L even amp B values there will be a pair of Sg tables here in page 14 For each pair of L opp 8 values there will be a pair of So tables One for the ODD numbered PASS One for the EVEN numbered PASS One for the ODD numbered PASS One for the EVEN numbered PASS Copyright Charles HAMEL aka Nautile 2009 Nov PAGE 9 of 33 How does one draw the LAYOUT of the cells of the ALGORITHM TABLES Here is some tips SET NUMBER Lto R reference RtoL reference Hali LtoR period 1 Half RtoL aa period 2 Half LtR a period 3 Half LtR as period 2B 1 Half aa period 2B ALL the tables will be drawn like this one just above Now for the particular TYPES TYPE So and TYPE Sg So THK component that is NOT the whole SHK with ODD number of LEAD Se component with EVEN number of LEAD The following template applies to both TYPE So and TYPE Seg as far as drawing the
27. t Om MO Obviously SHK has a HERRINGBONE PATTERN one that is INTER Fig 11 The Sequence Rule for the upwards and the downwards braiding direction BIGHT CODED ROW CODED for SCHAAKE s frame of reference horizontal mandrel but is COLUMN CODED on the vertical cylinder that is why dislike these appellations that are much too frame dependent What propose is not INTER BIGHT and PARALLEL BIGHT but have expounded on that point elsewhere in my web pages You can see that sequence rule in the pattern when using colours The above and under illustration are modifications in a fair quote of SCHAAKE amp TURNER colorized them not to make them prettier or different but to make them clearer To quote the authors themselves of BRAIDING STANDARD HERRINGBONE KNOTS We do know that the presentation of the various grid diagrams would be considerably improved my emphasis with italics by the use of colour printing however the very small demand for these books and hence the small number of copies printed a maximum of 50 unfortunately prohibits this Copyright Charles HAMEL aka Nautile 2009 Nov PAGE 5 of 33 SHK are an assembly an interweave of 2 to n that is what define the number of PASS A Turk s head knots that are ALL of the same dimension in LEAD and BIGHT contrary to Herringbone Pineapple who have TWO SET of THK even if one of the SET can be empty
28. th caviar cell is THE ONE WHERE THE NUMBER OF CROSSING WILL NOT CHANGE WHEN MAKING THE KNOT THE CELL SET WHICH WILL NOT BE NUMBERED AND WILL GET THE BLACK CELL IS N THE TABLES THE VERY LAST ONE in the case of the NUMBER SEQUENCE of A being ODD THE VERY FIRST ONE in the case of the NUMBER SEQUENCE of being EVEN So only L 1 sets will get a number but there will be L cells in the SET NUMBER row the discrepancy being the black cell The same SET NUMBER say 3 figures in both and ODD numbered HP L to R and in an EVEN numbered HP R to L BUT that set number DOES NOT GO with the same crossings so NOT to the same SET of crossing belongs to the FIFTH SET on the ie SER le SSR Qr 7 29 24 29 2 22 RIE 2 222 RED ODD SBE e SSS Se SOS numbered PIV gt gt gt gt d S323 half period going 222257625247 Left to Right and to SSS the FOURTH SET SF gt on the GREEN EVEN numbered NEA AN NS S AQ A half period going gt S aS SS gt BEES Right to Left e 2 0 CN A RO SET is N SSS SSS SLP Mb SSS b gt both H P Fig 5 3 pass Semi Regular Herringbone Knot 24 parts 15 bights n 1 C
29. tion 1 1 are similar BUT IF 5 3 1 2 Aissimilarto 4 135 BOTH ARE WILDLY AND WIDELY DIFFERENT FROM 12345 or 5 4 3 2 1thatdo not comply with the SEQUENCE RULE 1 2 2 PASS Type So 2 PASS are not near enough to show the SEQUENCE RULE PAGE 28 of 33 Copyright Charles HAMEL aka Nautile 2009 Nov V 17 9 y V 9 9 1 Y V V 7 V 9 7 9 4 Type So 4 PASS 3 2B 12L 8B SHK With these two diagrams it is easy to compare two 4 PASS SHK one complying with the SEQUENCE RULE and the other NOT v4 AN AUAM 4 NANNAN NS ANA x V 4 Type So 4 PASS NOT COMPLYING WITH SEQUENCE RULE lt x lt a o o ES lt PAGE 29 of 33 Se Type 4 PASS 41 3B fora 161 128 SHK S 47 4 at _ A S p 2 ue DN VT 8 NUN Y d v f M M A e PN m 4 PTOI NOVA 4 e Q P
Download Pdf Manuals
Related Search