Nautile
aka Charles Hamel's personal pages

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some so-called THK.

Please, if interested, read :

THK OR NOT THK

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.

Begin as if you were "doubling ", by following the leader , in fact you will be laying a "track"

parallel to the SPart and in a second phase you will " split the tracks ".

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.)

Laying and splitting in 12 photographies

Laying and splitting in 4 drawings ( going from a 4L 3B to a 6L 5B Right enlargement )

See the diagrams they will be clearer than the words.

LEFT enlargement 4L 5B to 6L 7B EVEN LEAD ODD BIGHT

RIGHT enlargement 4L 3B to 6L 5B EVEN LEAD ODD BIGHT

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.

explaining showing by using a diagram

explaining showing by showing the cylinder or mandrel

(in my book explaining is quite different from showing )

now the LEFT SIDE ENLARGEMENT of 5L 3B TO A 9L 5B THK

This one, ODD LEAD ODD BIGHT, is may be easier to understand as I

indicated the nature of the crossings , but first try to study the other diagrams above.

RIGHT SIDE ENLARGEMENT of 5L 3B TO A 11L 7B THK

LEFT SIDE ENLARGEMENT of 3L 5B TO 7L 11B ODD LEAD ODD BIGHT

RIGHT SIDE ENLARGEMENT of 7L 3B TO 11L 5B ODD LEAD ODD BIGHT

LEFT & RIGHT ENLARGEMENT of

3L 2B to respectively 7L 4B and 5L 4B ODD LEAD EVEN BIGHT

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 )

This table show all the "types" that are to be studied and ideally formalized by mathematical

formulas.

A BUNCH OF OTHER ENLARGEMENT DIAGRAMS

Added 2008 August 30th

3L 4B 4L 3B 4L 5B 4L 7B 5L 8B 6L 5B 7L 3B 10L 7B

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 EITHER left OR right.

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.

Added 2011 :: PIN-STEP pdf

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)

NUMBER OF LEADs you want minus TWO is to be DIVIDED by TWO.

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 2 1 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 4 3 2 1 7 6 5 4 3 2 1 7

6 5 4 3 2 1 7 6 5 4 3 2 1 7

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 :

10 9 8 7 6 5 4 3 2 1 10 9 8 7 6 5 4

9 8 7 6 5 4 3 2 1 10 9 8 7 6 5 4 3

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.

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

So now some diagrams (they are graphs !) to illustrate the case of the

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 7B LEFT 10L 17B

then you can observe that

(fR)L + (fL)L = 4 * (s)L 6 + 10 = 4 * 4

(fR)B + (fL)B = 4 * (s)B 11 + 17 = 4 * 7

+implication is that is when you know one enlargement you can get the other immediately.

(fR)L = 4 * (s)L - (fL)L 6 = 4 * 4 - 10 = 16 -10

(fL)L = 4 * (s)L - (fR)L 10 = 4 * 4 - 6 = 16 - 6

(fR)B = 4 * (s)B - (fL)B 11 = 4 * 7 - 17 = 28 - 17

(fL)B = 4 * (s)B - (fR)B 17 = 4 * 7 - 11 = 28 - 11

see again this illustration already given

A little bit more formulas

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)

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