new method for Turk's Head Knot

[Jan van Leiden]

Hi Ruby, nice meeting you too. You seem very interested in THK. Me too.

I have been reading some of the Schaake & Turner and Tom Hall books you mentioned elsewhere. You are right that for cylindrical regular grid diagrams, the greatest common divisor of their dimensions in parts and bights will determine whether they will have one single component. I think that has been known for a long time. You also write about the Fibonacci sequences above. Schaake & Turner show how these sequences turn up when enlarging regular grid diagrams. They coin the term Regular Knot Tree for these things.

The pattern on these regular grid diagrams is a different story. Most people make an O1U1 pattern, but there are many others. I like the U2O2 best. It gives a nice and stable texture.

I do not like ChatGPT, it produces a lot of rubbish when you ask it about knots - as I did above. I am not sure whether it can learn anything at all, but if it makes people happy to run after this new hype, be my guest.

[Ruby]

Schaake's theory is almost 30 years ago, but seems not many people know it. what a pity.

I recently tied may THKs in hand, using 1 meter shoe lace, and find something interested. but after some search I find Schaake has sovlved this long before, just that not many people talking about it. maybe because it is too easy? or too hard?

[Ruby]

some more about THK 9x14:

5x8 to 9x14,   4 parts and 6 bights added,
and split again?
9x14 to 13x20, another 4 parts and 6 bights added,
next? 17x26

it's easy to find the math equation:

p= 5 + 4n
b= 8 + 6n
n= 0,1,2,3......

and obviously this is the  Parametric Equation of a Line.

so point (5,8) , (9,14), (13,20)... etc all lie in a straight line

so this method is a linear enlargement of a THK.


and note that:

4x8  -   6x5  =  2
4x14 -  6x9  =  2
4x20 -  6x13 = 2

this is always true for a right split.

[Ruby]

and what did Ashley say about 9x14?

well, at AboK #1357 Ashley mentioned about from 8x5 to 14x9 to 20x13.

and one interesting thing about THK is the mirror effect.

e.g.
if you apply a RIGHT split to 3x2 and get 5x4
then, the mirror,
you apply a LEFT split to 2x3 and get 4x5
notice that right change to left, part change to bight. bights change to part

so according to Ashley AboK #1357, from 8x5 to 14x9, and I can infer that 9x14 is from 5x8


-------

after I find the fact of the mirror effect, that is to say,
 the method to to tie a PxB thk is the mirror to tie a BxP thk,
 then I often only tie a THK with parts bigger than bights.  so usually  8x5 or 9x7,  not 5x8 or 7x9.

usually I find it's more comfortable to tie THK with more parts(lead) than bights.

[Ruby]

and speaking of a 5x23 THK,
what I'd like to try is a 23x5 THK.

the difference of Parts and Bights is too big,
so the first step is to reduce it somehow,

subtract bights each time,

23x5 --> 18x5  --> 13x5 --> 8x5

23x5  ==>  (8+5*3) x 5

so the method to tie a 23x5 THK is similar to tie a 8x5 THK,
and is the mirror of 5x8.

so if you can tie a 5x8, then 23x5 is almost the same step.
and 5x23 is the mirror of 23x5.


anyway I just tried to tie a 13x5 THK around my finger.
first a 3x1 and double it and do a right split.


maybe I need a more long rope and a more long finger to tie a 23x5 THK.
first tie a 5x1 and double it and do a right split to 23x5

[Ruby]

just tied a THK 23x5, around my finger. not that hard :-)

[Jan van Leiden]

Hi Ruby,

I am not sure why you want to call your 9/14 enlargement 'linear'.
Is it because the regular knot tree positions those knots on a line governed by your two linear equations?

What do you make of a 13/21, 21/34, or 34/55  then?
(or 21/13, 34/21, or 55/34 in the wide analog due to the symmetry of the regular knot tree)
In that type of knot the fundamental enlargements alternate.
The braiding process is messing up the perceived 'linearity'.

I have the impression braiders have only found some simple relations.
Schaake & Turner provide all relations, including a way to find the paths in their tree.
Pity that so few braiders seem to understand this matter, it seems to me.

[Jan van Leiden]

Hi Ruby,

There is one other thing that came to mind about your 'linear' equation.

The set of equations does not really represent a 'linear' relationship, but something that is  'intermittent' at best.

You must have noticed that to retain the U1O1 coding you must skip one node in the regular knot tree.

That is the reason Ashley's Book of Knots speaks of 4 methods for the square THK's.

Schaake and Turner have devoted a pamphlet to some of the effects that patterns have on string runs and vice versa.

[Ruby]

Hi Ruby,

I am not sure why you want to call your 9/14 enlargement 'linear'.
...
The braiding process is messing up the perceived 'linearity'.
...

hi,

you are right, that  something is messing up.
 9/14, ( and 13/20, 17/26  ...) is linear enlargement of 5/8,
but obviously 5/8 is not linear enlargement of any THK.

my point is that, for a speific THK, like 5/8 ,  or any THK you already knows how to tie it,
you can apply a linear enlargement to it.
and it is clear and easy as an instruction.

 Schaake's theory is not that easy to follow.
 though it actually can guide me to reach any THK.
 
 
 and
I think this linear enlargement is part of Schaake's theory. they do not conflict.

a  tiny small part of it.

by walking at that so called RKT tree,
    or actually  a Stern-Brocot Tree,
at  any THK you have infinite ways to enlarge it to any THK belongs to current branch,
by changing direction randomly left and right.
 
 and among all these ways, the simplest ways are  the 2 linear enlargements:
RightSplit and LeftSplit, like from (3,2) to (5,4), or from (3,2) to (7,4),
by not changing direction,  and walk only left 2 steps, or only right 2 steps.


---

so, I am not going to say that I can tie a 13/21， or 21/34 , with a linear enlargement method,
but,
if I can already tie a specified THK, say 13/21， 21/34， or any p/b THK,
or if somebody else just give to me a finished THK like 13/21， 21/34， or any p/b THK,
then, without knowing how he tied it beforewards,
I can apply a linear enlargement method to it, and build a series of enlarged THK.
(also infinite, actually the same size of all THKs , mathematically)

[Jan van Leiden]

Hi Ruby,

Viewing cylindrical regular grids as numbers (p/b), you can indeed speak of the Stern-Brocot Tree.

The two fundamental enlargements provide you with binary options to branch one way or the other.

The 'linear enlargements' you speak of are forced to emerge when you oblige yourself to follow the U1O1 pattern.

If you ignore any pattern and just consider the bare-bone string-run skeleton, you will notice that the sequences I mentioned above will alternate between the fundamental enlargement methods.

Schaake's theory is not that easy to follow. Though it actually can guide me to reach any THK and I think this linear enlargement is part of Schaake's theory. They do not conflict.

Indeed, I think you got it.

If you want to find the paths in the regular knot tree you need to play around with Euclid's algorithm.

That will tell you which turn to take while making the grid.

Your mind will have to figure out a way to keep track of which pattern you are aiming at. 

[Ruby]

Hi Ruby,

There is one other thing that came to mind about your linear equation.

The set of equations does not really represent a linear relationship, but something that is  intermittent at best.
...

well, I think the set of equations can represent any linear relationship.

look at the relation of RKT tree and  a Stern-Brocot Tree,
I see that:
   a THK represent any  co-prime pair of number,
   a THK represent any rational number
   a THK represent any fractional number

according to the  Bezou s lemma
for any co-prime number pair, you can write a  Diophantine equation  to get any number.

so the size of equations is the same of the size of all THKs.
 or the size of all integer numbers,
 or the size of all rational numbers
 
 for any THK p/b, you can write  a  Diophantine equation like mp - nb = 1, or  mp - nb = 2,
 and this is the linear equation of a linear enlargement.
 

[Ruby]

...
If you want to find the paths in the regular knot tree you need to play around with Euclid's algorithm.
...

yes Euclid algorithm is easy. just grade school algorithm for finding gcd.

and later I saw Schaake gave detailed method to get every OUOU or O2U2... steps for any THK.
not that hard, but thanks to here : http://freakinsweetapps.com/knots/knotgrid/advanced.html
so we do not need to calculate it any more

so theoretically this help you to make any THK, with the help of a woggle , and many pins


---

if you want a method to tie pratically sized THK by hand, then it is another story.

to make a THK , to braid a THK, is much different with to tie a THK.

no need to tie any THK,
since the number is same with all natual number, why bother tie any of them?
we need to find some speial method to tie some special THK,
which is very easy to tie, without  the help of any woggle or pins

find some squar one, p=b+-1 or p=b+-2，
or some long one,   p / b > 2
or some wide one,   b / p > 2

then it is almost enough.
no need to count parts or bights, just run your rope with your finger,
and build a  pratically sized THK by hand easily.

[Jan van Leiden]

Using computers in braiding is like using chatGPT to get a question answered.

• If your knot is too big, your machine will suffocate when it runs out of time and space.
  
• If the answer to your question is not (pre-)programmed somehow, your machine will provide you with suboptimal information.

I think you may want to read up on Bezout's Lemma with respect to the uniqueness of its coefficients.

The path in the regular knot tree is quite unique and this reflects in the Diophantine Equation you will be needing in order to produce your desired pattern.

[Ruby]

maybe people now cannot live without computer? or can not live without  a smart phone?

[Ruby]

THK[13/8]?   = first THK[3,2] and then double it, and then apply a LeftSplit
THK[21/13]?  = first THK[5,3] and then double it, and then apply a RightSplit.
THK[34/21]? = first THK[8,5] and then double it, and then apply a LeftSplit
THK[55/34]? = first THK[13,8] and then double it, and then apply a RightSplit
using Euclid algorithm I can see that it is always 1111111111.....
so the direction is always changing,  left right left right.....