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 :
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.)
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 /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 !
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
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
)
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 andTHREE 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 andONE 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:
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 !
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.
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 ¨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 (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)