THK / Turks-head knot / Turk's head knot : HOW TO ENLARGE
THEM Added 2008
August 24 to 27th
Enlarging allows to go from say a 4L 5B THK you have already
done to the 6L 7B you
want.
There
are TWO types of enlargements :
- one in which you go with the Working End (WEnd)on the RIGHT of the
Standing Part
(SPart)
- the other where you
go on the LEFT
of the SPart-WEnd vector.
LAYING : you do the same type of crossing as the SPart is doing, except
on the BIGHT
rim, at the frontier where something special must be observed.
SPLITTING
: you do (between the two " twin tracks " as in a railway or rail road
tracks )
crossings in such a way
(making the opposite type they are in the tracks) to regain the
Over-Under rhythm or pattern or coding (the so-called Casa
CODING.)
BIGHT
cannot be added on the border
without adding LEADs and with the REGULAR
enlargement you cannot add LEADs without adding BIGHTs on the border.
That is for those THK I am speaking of .
Particular case of
the 3L group
This is a very short and out of context extract
from the THK and MATHEMATICS article
( in Publication
page after validation by mathematicians)
IRREGULAR
ENLARGEMENT
This is coming from a mistake I made when attempting an enlargement and
described
elsewhere in those
pages
You can sometimes add LEADs without adding BIGHTs at the border BUT you
will
add pseudo-BIGHTs inside : so-called INTERNAL BIGHTS
Example
1 adding BIGHT external and internal Example
2 adding internal BIGHT without external BIGHT
THE MAKING OF A 5L 3B THK :
LAYING THE TRACK and
SPLITTING THE
TRACKS / SPLITTING THE PAIRS
This
will show another perspective on " laying the track " :
parallel
progression along the SPart doing the same type of crossing followed by
a second
phase of splitting the tracks (the SPart and its parallel)
building again the
alternating
crossings.
Remember EVEN LEAD
EVEN
BIGHT ARE QUITE IMPOSSIBLE (number of LEADs,
number of BIGHTs must be relatively prime, that is
they admit only 1 as
common divisor )
O - E E - O
E -O
E - O
O - E
E - O O - O
E - O
O= odd (impair) E)even (pair)
COMPARING LEFT AND
RIGHT
ENLARGEMENT ON A THK Added 2008 Sept
10th
A simple drawing of the original
THK and its two enlargements RIGHT
and LEFT.
AN EASY
WAY TO COMPUTE YOUR ENLARGEMENT FROM ANY THK Added 2008
August 30th
I started some lateral thinking.
About ENLARGEMENT people are always obsessed about how many BIGHTs
and
LEADs they start with and how many of each they will add ; many don't
realize ( I was one
of
those once ) that you can enlarge /
expand either from the LEFT
or the RIGHT
of the
SPart WEnd vector (
no
silliness about "that depend on the way you look at it" please ; there
can
be no ambiguity with an oriented vector : if I shot you with an
arrow it will always be the point first
and the notch and feathers last)
, that a given THK can be enlarged
EITHERleftORright.
So I desisted from this obsession and start contemplating "The Void" !
I
was no longer concerned about what already exist and is fixed and
unchangeable but
about what was "coming to existence" and WHERE it can
find room to come to
existence.
In short I ceased to observe the "PINs" in the diagrams and rather
observed the
" empty space in-between"
the PINs.
Of
course there is as much "empty corridor" for a "free run"
than
there are walls (LEADs
going from a BIGHT PIN on a border to another BIGHT PIN on the
other side border ).
I also decided to think "GRAPHS" rather than diagrams, just to keep my
mind off the afore
said obsession .
The
vertices (each vertex is the extremity of one empty corridor)
will be
considered to be
imperatively touched either
twice or zero
(you can
only add pair(s) of BIGHTs with the
laying-splitting method)
and the edges ( what will represents the LEADs
added) will
also
be in "pair".
This drawing illustrates the "circulation
rule in corridors" :
In
the "start" THK, when running in a corridor sided by walls of one
colour (say blue)
you
cannot go through a wall of this
colour (blue) but you
have the magical power to go through
the walls of the colour (yellow)
This is not rocket science so I will only give 2 graphs and let
you figure the rest that interests
you.
Remember
(or be apprised) that with laying a THK on a cylinder with PINs at the
TOP (or RIGHT
if held horizontal instead of vertical)
and at the BOTTOM (or LEFT)
you
have to JUMP or SKIP PINS when going from one
PIN situated on one
border to another pin on the other border.
JUMP can be 0 to n, note (for later use) that JUMP=0
is ONE STEP.
Computation of PIN JUMPING is quite easy.
( excellent diagrams are given there http://www.shurdington.org/Scouts/TurksHead.htm
)
*BUT* PINS JUMPED OVER is not the most rationale nor the nearest to
the
mathematical
reasons that explain it works, that is why I prefer PINS
STEPS ( see
after JUMP)
This gives you the NUMBER od PINs you have to JUMP.
Example for the dumbfounded ones :
- you want 7 LEAD
- so 7 minus 2 gives 5
- 5 divide by 2 give 2.5
AH ! how do I jump "half of a pin" ?
well easy 2.5
is between 2
and 3 so on
one BIGHT border you will
always skip 2
pins
and on the other 3.
so for a 3 BIGHTs to get 7 LEADs you JUMP (2 TOP
and 3 BOTTOM )
3 2 1 3 2
1 3 2 1 3 2
1
3 2 1 3 2
1
2 1 3 21
3 2 1 3
2 1
3 2 1 3 2
1
read as in Arabic : from RIGHT
to LEFT
You see a line but it stands for the perimeter of the cylinder and the
PINs are numbered
in clockwise direction.
for a 8 BIGHTs to get 7 LEADs you JUMP (2 TOP and 3 BOTTOM )
7 6 5
43
2
1 7 6 5 4
3 2 1 7 6
5 4 3 2
1
7 6 5 4 3 2
17
So if 2 PINs are to be skipped,
the one from which the cord comes being
"zero", you
"alight" your cord on the" fourth" ,
1 at the BOTTOM go to 1 at the TOP, skipping TOP PINs N° 3 and 3
1 at the TOP go to 2 at the BOTTOM, skipping BOTTOM PINs 2
3 1 in the first "round"
before arrival.
My preference goes to L/2 or STEP
as it is more immediately derived
from the
anatomical
and mathematical structure of the THK than the above JUMP (we
will see the math
foundation later on)
But PINs JUMPs and PINs STEPs are like the old pickets and intervals
problem :
either you count the pickets or you count the interval between *two*
pickets. In that case
number of intervals = number of pickets (NP) minus one.
JUMPs over 2 pins : start is A then JUMPed PINs are
say B and
C, alighting PIN is D.
STEP 3 times = start is A then first STEP
is B, second STEP is C,
arrival STEP is the
third or D
To get the "count" of the next PIN
in a direct fashion (
instead of the number of skipped
pins between the 2 extremity pins )
: L / 2
ex 7 / 2 =
3. 5 so 3 and 4
-----------------------------------------------
Here is a fully worked example for a 17L 10B
17 / 2 = 8.5
8.5 so it will be 9 and 8 , let us say 9 at the TOP , 8 for the BOTTOM
DO NOT forget that the pins go a merry go round so use MODULUS(B), here
MODULUS(B)=10
Start is at the BOTTOM, ODD -numbered HP go from BOTTOM-RIGHT to
TOP-LEFT
(cylinder) and EVEN-numbered HP go from TOP-RIGHT to
BOTTOM-LEFT.
Note that you have the choice to number your PINS either 1 to 10 or 0
to 9.
(no importance for "in the cord" but highly important when using
formulas). So it is :
start at BOTTOM PIN
1
go to TOP PIN 10 ( 1 + 9 = 10 ; 10 modulo(10) is
'0')
go to
BOTTOM PIN 8 (
10 +8 =18 ; 18 modulo(10) == 8)
go top TOP PIN 7 ( 8 + 9 = 17 ; 17 modulo(10) ==
7)
go to BOTTOM PIN
5
go top TOP PIN 4
go to BOTTOM PIN
2
go top TOP PIN 1
go to BOTTOM PIN
9
go top TOP PIN 8
go to BOTTOM PIN
6
go top TOP PIN 5
go to BOTTOM PIN 3
go top TOP PIN 2
go to BOTTOM PIN
10
go top TOP PIN 9
go to BOTTOM PIN
7
go top TOP PIN 6
go to BOTTOM PIN
4
go top TOP PIN 3
go to BOTTOM PIN
1 curve has closed on itself,
the Working End meets with the Standing Part.
5L 8B RIGHT enlarged to 11L 18B +6L
+10B
or 3
* 2
5 *2
5L 8B LEFT enlarged to
9L 14B +4L
+
6B or 2
* 2
3 *2
That stated I must add that this is interesting
mainly from a theoretical
point of view.
From a practical
point of "good sense" view and daily practice make your start
with
basic
THK as "seed"
or starting THK, (for those basic THK I added in 2009
August some
topics elsewhere.)
HAVE YOU PERCEIVED THE INTERESTING POINT verified on all the
other
enlargements) in the GRAPHS illustration ?
Let
us denote by L(s) and B(s) the start knot L & B
and by L(fR) (
respectively L(fL) )
and B(fR) ( respectively B(fL)
) the
final knot for RIGHT
and LEFT enlargement
4L 7B
RIGHT
6L 11B 4L 7BLEFT
10L 17B
then you can observe that
starting THK is denoted a (x) L (y) B then addition of L and B
in the final THK will be
Added L (RIGHT)
= n * 2 or 1
*2
n
= 1 2 leads
added
Added L (LEFT)
= m * 2 or 3
* 2
m
= 3
6 leads added
Added B (RIGHT)
= k *2 or 2
* 2
k
= 2 4 bights
added
Added B (LEFT)
= j * 2 or 5
* 2
j = 5
10 bights added
n + m = x or 1 + 3 = 4
k + j = y or 2 + 5 = 7
(n) is the number to add to the JUMP count in the starting
THK to get the JUMP
count in the final THK for the RIGHT
enlargement ( respectively (m) for the LEFT
enlargement )
Tip
: for deep mathematical reasons (more of that in my MATHEMATICS AND
THK PART 1 and PART 2 ) for a given span of time
you will always get ONE MORE or ONE
LEAD LESS laid that there were BIGHT laid.
If (z) BIGHT laid then you
will have either (z + 1) or ( z-1) LEAD laid. ( it is one of
the
explanation why you can NEVER have B = L for the single strand THK, the
only TRUE to
type one as far as I am concerned. ( please confer to the
two publications linked at the top
of this page for more
argumentation ) Now it "just" remains to find the rule for deciding
-- which is the predominant enlargement : RIGHT
or (n) or LEFT
or (m)
-- and the way to find (n) and (m) knowing (x) and
to find (k) and (j) knowing (y)
Copyright 2005 Sept - Charles
Hamel / Nautile -
Overall rewriting in August 2006 .
Copyright renewed. 2007-2012 -(each year of existence)