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ABOUT LONG ( LEAD - BIGHT >= 2 )

TURK'S HEAD KNOT / TURKSHEAD / TURKS-HEAD KNOTS

( HELP TO TIE TURK'S HEAD KNOTS )

After being exposed to Don WRIGHT Mammoth PineApple Knot and Jimbo remarks I wrote to them:

[open self-quote]

you guys do long THK without any existential angst, as if peeling a banana, I never made one long THK but I know about them a lot I don't really need ;-D )

I was quite dissatisfied at my yesterday formulation of things so...

I burned the midnight candle , err...electricity till 01:15 last night and made more long THK that probably you will ever make.

All were made as the electrons speed !

I ran a number of them on my HP48GX using my programs and those many simulations led me to this provisional knowledge :

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first remark

I only found in my books (a lot) long THK with EVEN number of BIGHT ( nB ),

probably because they are easier than those with an ODD nB that stay "possible"

yet I found one ODD nB on the Net (look at it ONLY after reading this COMPLETE

TREATISE : http://www.youtube.com/watch?v=Zy_1afLE_WI

How am I doing with hyperbola ?

Will I find a publisher ? (yes I can tell you in confidence tell you that I am negotiating

with one small publishing business : Nautile's, but the guy there is a bit particular.)

--------------------------------------------------

second remark :

about LEAD / BIGHT we can have :

ODD number of LEADs ( nL ) / ODD number of BIGHTs (nB)

EVEN number / ODD number

ODD number / EVEN number

of course there is no true THK with BOTH EVEN number of LEADs , EVEN number of BIGHTs (true = standard O1 -U1 SINGLE STRAND cylindrical knot using the fabled THK cordage route ( cordage shadow for the abstract minded, cordage route is for down to Earth practical people with a wee bent toward the theoretical ) due to Greatest Common Divisor and the geometrical nature of THK.)

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ODD number of LEADs is compulsory if the long THK is the PRIMARY KNOT of a

'true' PINEAPPLE KNOT (Herringbone PineApple Knot)

----------------------------------------------------

ODD /or EVEN number of BIGHTs depends on need/want and calculation of "ideal"

number of BIGHTs according to circumference of the finished knot. (we will address

that in a Post Scriptum ; better that than in a Post-Morten ! )

Still EVEN nB are faster to make (more OVERs than UNDERs) the "ideal" depend on

diameter of lace or cordage, circumference of knot/support, number of PASSes, tightening !

----------------------------------------------------

*** LEADs

*** BIGHTs and PINs

BUT first some words about the frame of reference for the sake of having a "common

ground" for this mail.

Knot is made on a vertical cylinder.

First Half-Period ( HP), the First going with the cordage from one BIGHT RIM to the other BIGHT RIM is from BOTTOM RIGHT ( as each odd numbered HP will be ) to TOP LEFT ( second HP and each even numbered HP will go from TOP RIGHT to

BOTTOM LEFT

. Note that you have as many HP than nB*2 ' B HP are EVEN ; B HP are

ODD)

First PIN USED IS N°1 (PINs are numbered 1 to nB in CW: in the same

direction the knot is laid )

Whether nL is ODD or EVEN the HP1 (the first going from BOTTOM RIM to TOP

(RIM ) will be a FREE RUN (NO CROSSING) and consist only of 0 (not complete)

to N WRAP ( wrap = 360° turn around the cylinder )

You *may* escape calculating the coding of each half-period ( unless you use on the THK cordage route another coding than the O1 - U1 ( U1 - O1 ) in which case you must compute the coding ( for example using the paper & pencil method I wrote using Schaake work )

-------------

EVEN number of BIGHTs ( yes BIGHTs ):

HP1 : the very first going UP from BOTTOM to TOP is always a FREE RUN with no

crossing and only WRAP to do (see how to compute number of WRAP depending on

the number of LEADs you want on this number of BIGHTs ).

HP2 : the first going DOWN from TOP to BOTTOM is all OVER crossing.

HP3 : the second going UP from BOTTOM to TOP is all OVER crossing.

After that you let yourself be guided by the crossings already there to make the new ones.

circuits (or 8 steps ) and 3 steps and 12== 3 complete round (or 12 steps ) and no

additional step )

going DOWN

going UP while being parallel to HP2

HP4 = U1 - O1 - U1 - O1 - U1 - O1 - U1 - O1 - U1 - O1 - U1 ( your are sort of

splitting pairs )

HP5 = U1 - O1 - U1 - O1 - U1 - O1 - U1 - O1 - U1 - O1 - U1

HP 6 = U1 - O1 - O1 - U1 - O1 - O1 - U1 - O1 - O1 -U1 - O1 - O1 - U1 - O1 - O1 -

U1 - O1 ( you are making new pairs = the O1- O1)

HP 7 = U1 - O1 - O1 - U1 - O1 - O1 - U1 - O1 - O1 -U1 - O1 - O1 - U1 - O1 - O1 -

U1 - O1

HP 8 = U1 - O1 - U1 - O1 - U1 - O1 - U1 - O1 - U1 - O1 - U1 - O1 - U1 - O1 - U1-

O1 - U1 - O1 - U1 - O1 - U1 - O1

if top get 12 STEP and bottom 11 ( remember to use modulo nB to get the Pin's number )

BOTTOM 1 4 3 2 you see that it is NOT " the next pin" !

TOP 1 4 3 2

or if top get 11 STEP and bottom 12

BOTTOM 1 4 3 2

TOP 4 3 2 1

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ODD number of BIGHTs : here it is a bit less easy

HP1 : the very first going from BOTTOM to TOP is always a FREE RUN with no crossing and only WRAP to do ( see how to compute number of WRAP depending on the number of LEADs you want on this number of BIGHTs )

HP2 : the first going DOWN from TOP to BOTTOM is

BUT

HP3 : the second going UP from BOTTOM to TOP

After that you let yourself be guide by the crossings already there to make the new ones.

circuits ( or 10 steps ) and 1 step and 12==2 complete circuits (or 10 steps ) plus

2 steps)

HP2 = U1 - O1 - U1 - O1

HP3 =

HP4 = O1 - U1 - U1 O1 - O1 - U1 - U1 - O1 - O1

HP5 = O1 - U1 - U1 O1 - O1 - U1 - U1 - O1 - O1

HP 6 = O1 - O1 - U1 - U1 - U1 - O1 - O1 - O1 - U1 - U1 - U1 - O1 - O1

HP 7 = O1 - O1 - U1 - U1 - U1 - O1 - O1 - O1 - U1 - U1 - U1 - O1 - O1

HP 8 = U1 - O1 - O1 - U1 - O1 - U1 - U1 - O1 - U1 - O1 - O1 - U1 - O1 - U1 -

U1 - O1 - U1 - O1

HP9 = U1 - O1 - O1 - U1 - O1 - U1 - U1 - O1 - U1 - O1 - O1 - U1 - O1 - U1 -

U1 - O1 - U1 - O1

HP10 = U1 - O1 - U1 - O1 - U1 - O1 - U1 - O1 - U1 - O1 - U1 - O1 - U1 - O1 -

U1 - O1 - U1 - O1 - U1 - O1 - U1 - O1

if top get 12 and bottom 11

BOTTOM 1 4 2 5 3 not the "next pin" obviously !

TOP 3 1 4 2 5

or if top get 11 and bottom 12

BOTTOM 1 4 2 5 3 not the "next pin" obviously !

TOP 2 5 3 1 4

TWO cases :

That is OK

during the laying of the knot if the tyer is meticulous and attentive.

if you use PIN

need to think PIN SKIPPED or PIN STEPPED ON

The rational explanation follows :

just for the sake of example ( but will work with another nB ODD or EVEN ) we will

consider a 4 BIGHT

you will remember that PIN SKIPPED = (nL - 2) / 2 and what has my personal preference PIN STEPPED ON nL/2

case nL= 73 73 modulo 4 ( the nB ) == 1

case nL= 74 74modulo 4 ( the nB ) == 2

case nL= 75 75 modulo 4 ( the nB ) == 3

case nL = 76 76 modulo 4 ( the nB ) == 0 or 4

case nL= 77 there as 73 modulo 4 ( the nB , remember ? ) is == 1 and 73 modulo 4 is also 1 we get back in MODULAR fashion ( think CLOCK )

It is immediate : the pin skipped or stepped on will change depending on the nL just because of nL modulo nB

A word about modulo and the first HP and WRAP and division with only INTEGER

PART and NO DECIMAL PART

---------------------

case nL= 73 73 / 4 = 18.25 if we use the decimal part which is a no-no for us so

73 = (4 * 18 ) + 1 (this '1' is the remainder and it is also the result of the modulo operation)

It means that

on BOTTOM and on TOP ! because ODD nL )

case nL= 74 74 / 4 = 18.50 if we use the decimal part which is a no-no for us so

73 = (4 * 18 ) + 2 (this '2' is the remainder and it is also the result of the modulo operation)

It means that

---------------------

case nL= 75 75 / 4 = 18.75 if we use the decimal part which is a no-no for us so

73 = (4 * 18 ) + 3 (this '3' is the remainder and it is also the result of the modulo operation)

It means that

different on BOTTOM and on TOP ! ( because ODD nL ) and different from the

above cases

for nB=4 it will be pin skipped =

---------------------

case nL = 76 76 / 4 = 19.00 if we use the decimal part which is a no-no for us so

73= ( 4 * 18 ) + 4 or 73 = (4 * 19.00 ) + 0 (this '4' or '0' is the remainder and it is also the result of the modulo operation.)

It means that

for nB=4 it will be pin skipped =

---------------------

case nL = 77 77/4 = 19.50 if we use the decimal par which is a no-no for us

so 73 = (4 * 19 ) + 1 ( this '1' is the remainder and it is also the result of the modulo

operation.)

It means that

different on BOTTOM and on TOP ! ( because ODD nL )

for nB=4 it will be pin skipped =

on the other

---------------------

Little reminder :

ODD and EVEN nL don't give the same process :

look:

nL=73---------------------------------------------------------------------------------------

PINs TO BE SKIPPED (73-2)/2= 71/2= 35.5 How do we skip half a pin? Simple!

You take 0.5 and add it or take it off and then you have 35 and 36 ( just as 35.5*2 = 71 you get 35+36=71)

You consistently use "this" one on this BIGHT RIM and "that" one on that BIGHT RIM.

nL=76---------------------------------------------------------------------------------------

PINs TO BE SKIPPED (76-2)/2= 71/2= 37 . here it is easy 37 on each rim !

You see you cannot escape the PIN count : either you get it right by happenstance/serendipity/ angels watching over you.... or you get it right by calculation or it will be hopelessly wrong !

As response Jimbo sent his find : a mail with subject :

Say What You Will About McDonalds...

Photo 1 Photo 2 ( they are in the .pdf file linked just above )

I sent a response giving the maths in the topic preceding this one ; that got in

return

[open quote]

First, please feel free to forward your last msg., or at least the math, to Don. ....

As to the #B, {many}L THKs, I always thought the <= 4B meant those who'd attempt such a project had sense enough to know that 4B on a painfully large #L could take many days to finish, and each additional B increases the workload exponentially...

That's what I thought anyway.

Or maybe they're "CHEAT"ing (<G>) and using ABoOK #790's layout plan...

[end quote]

THANKS YOU JIMBO !

got my EUREKA ( no no running in the nude in the streets! ) just at the end of your mail.

It worked for you and Don to go "random" as far as the number of WRAPs is concerned.

That is the state of affairs ; So me-think either they are extraordinary lucky OR some law is helping them to be lucky.

see

PINs SKIPPED/JUMPED = (nL-2 ) / 2 PINS STEPPED ON = nL/2

Now let us pose that nL is a "number of time" nB . "Number of time" is denoted N ,

if ODD it is N

If EVEN it is N

N being an INTEGER ( a NATURAL DIGIT )

In fact as I will be using MODULO B it is of no import that it is an EVEN or an ODD

number of WRAP.

nL = B * N

Observe that |B*N

So PIN SKIPPED = (B-2)/2 ( PIN STEP = B/2 )

*** Case B= 2 pin skipped (2-2)/2 = 0 so NO PIN SKIPPED == it is the PIN

IMMEDIATELY NEXT that is used.

*** Case B = 4 pin skipped (4-2)/2 = 1 so ONE PIN SKIPPED = it is the PIN

IMMEDIATELY OPPOSITE that is to be used ( pin 1 and pin 3 are opposite just as

pin 2 and pin 4 are opposite on the rim)

*** Case B= 3 pin skipped (3-2)/2 = 0.5 so 0 on one BIGHTs RIM and 1 on the other

so NO PIN SKIPPED ON ONE BORDER = it is the immediately NEXT

and it is ONE ¨PIN SKIPPED ON THE OTHER.

As it is 3 PIN one skipped make it THE PIN IMMEDIATELY BEFORE ( If you skip a pin on a three pins circle then you are immediately before you starting pin ) in all 3 cases above a "immediate visual correction" is possible

*** Case B = 5 pin skipped (5-2)/2= 1.5 so 1 and 2

It is getting complicated and offer no "easy rule" in absence of actual calculation.

This explain enough why LONG THK are often 2 / 3 / 4 BIGHT and seldom above if I am to believe what I have seen.

WOOW THE GENIUS ! ( just hope that I got that all right ! )

to that Jimbo answered

[open quote]

Glad you liked it! Obviously the "hard part" is just beginning, as fairing and tightening these is a real chore.

The straw idea "just came to me". It's the "right" size for ~2mm cordage, and the

high-quality plastic holds pins well (

Ashley "specifies" a 45-degree angle, but I lost my protractor! :^D

[end quote]

Copyright 2005 Sept - Charles Hamel / Nautile -

Overall rewriting in August 2006 . Copyright renewed. 2007-2014 -(each year of existence)

Url : http://charles.hamel.free.fr/knots-and-cordages/

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