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TURK'S HEAD
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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 :

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

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.)

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

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 !

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

We now have to address two different points

*** 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 )


******   First  point the LEADs-------------------------------------------------------

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.

Example for a 23L 4B      PIN STEPS   11 & 12   ( if modulo 4   11==   2 complete
circuits  (or 8 steps ) and 3 steps  and 12== 3 complete round (or 12 steps ) and  no
additional step  )

HP1 =     FREE RUN    FIVE WRAP (  23 / 4 = 5 plus a decimal part not used )

HP2 =    O1 - O1 - O1 - O1 - O1   corresponding to crossing the 5 WRAPS when
going DOWN


HP3 =    O1 - O1 - O1 - O1 - O1   corresponding to crossing the 5 WRAPS when
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

pin circuit ( CW ) will be
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

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

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 NOT all  OVER crossings
BUT 
alternating OVER / UNDER

HP3 : the second going UP from BOTTOM to TOP   NOT is all OVER crossings 
BUT    alternating OVER / UNDER

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

Example for a 23L 5B   PIN STEPS   11 & 12   ( if modulo 5   11==2 complete
circuits ( or 10 steps ) and 1 step  and  12==2 complete circuits (or 10 steps )  plus
2 steps)


HP1 =     FREE RUN   4 WRAP   ( 23 / 5 = 4 and some decimal part not used )

HP2 =     U1 - O1 - U1 - O1

HP3 =    
U1 - O1 - U1 - O1  end of the easy part

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

pin circuit ( CW ) will be

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     


******  Second point  BIGHTs and PINs----------------------------------------------

TWO cases :

you don't use PIN because nB is at most 4. 
That is OK because as the BIGHT are "floating" they can readjust themselves
during the laying of the knot if the tyer is meticulous and attentive.

BUT
if you use PIN
then BIGHTs are FIXED and unless you get awfully lucky  you will
need to think PIN SKIPPED or PIN STEPPED ON because it will not always be "the next pin" but that will always depend on nL and nB  as seen previously but here is another hammering on the nail.

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 number of WRAPs has to be 18 BUT PIN SKIPPED will be different
on BOTTOM and on TOP !  because ODD nL )

for nB=4  it will be pin skipped = ZERO or Four ( which is equivalent ) on one rim and THREE on the other

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

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 number of WRAPs has to be 18  BUT PINs skipped will be identical on the 2 RIMs but different from the above example and from the below example !

for nB=4  it will be pin skipped = ZERO or Four ( which is equivalent ) on both RIM

---------------------
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 number of WRAPs has to be 18  BUT  BUT PINs SKIPPED will be
different on BOTTOM and on TOP ! ( because ODD nL ) and different from the
above cases
.

for nB=4  it will be pin skipped = ZERO or Four ( which is equivalent ) on one rim and ONE on the other.

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

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 number of WRAPs has to be 19  BUT PINs skipped will be identical on the 2 RIMs but different from the above and from the under !

for nB=4  it will be pin skipped = ONE on both RIM.

---------------------
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 number of WRAPs has to be 19 BUT   BUT PINs SKIPPED will be
different on BOTTOM and on TOP ! ( because ODD nL )

for nB=4  it will be pin skipped = ONE ( which is equivalent ) on one rim and TWO
on the other

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

IT WILL BE "the next pin" ONLY if PIN SKIPPED = ZERO ( or 4 )


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 !

[end self-quote]


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(odd)
If EVEN it is N(even) 
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 * Nodd      ( use nL=B*Neven if you want )

Observe that |B*Nodd/even| modulo B = B or 0

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 ĘPIN 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 (half as many needed as the number of bights, since they go all the way through

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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