Nautile aka Charles Hamel's personal pages
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(elementary but quite important, compulsory to absorb and digest if you want to
continue reading my delirium without trouble)

Tip of the day : your reading of the text below will be made easier with 
a printing of this diagram 

This is a very ordinary 5L 3B THK but what will be said about it is valid for
other true THK.

On the drawing the cordage (in red) is laid from 1( lower)' to 3(upper) to 3 (lower).
This is a FULL PERIOD.

1(lower) to 3(upper) is the first half-period  ↖ 3(upper) to 3(lower) is the second
-period.  ↙

Notice that in this case they are represented by being drawn of different length. and the
slant of the slopes are very different, asymmetric.

This is not always so visible and this is not always so : drawing artifact. ( see the THK
drawn as a curve). (added 2010 : this is due to *not* usingthe only safe tracing paper :

Note also than whether you distribute L/2 = 5/2 = 2.5  so 2 and 3 between TOP and  
BOTTOM RIMs, in 2 , 3 or 3, 2 there will always be 5 between the 2 pins at the foot of
 the period on the lower rim )

Compared with the "anatomical" numbering order (structural  numbering order) of 
BIGHTs which is '1' with yellow halo, '2' with green halo, '3' with pink halo , their
"embryological" numbering, their process numbering makes it that '3' is the first
laid, '2' is the second one and '1' is the third one to be completed.

see   this picturethat picture

How the hell can we get, in a reproducible and reliable manner, the ordering of their
appearance in the making without tracing a diagram or without  making the knot
with cordage ?
Answer with a bit of observation and with mathematics,  with modulus being among the

Important anatomical fact about the true THK  [please I do hope you will
have by now realized why it is very important and not simply my whims to put
all the so-called THK in the correct slot or in a *possible* correct slot, at least
 a *distinguishing and separating* slot, using a sieve ! That was done months
ago in THK OR NOT THK.]
the STEP  [the 'why' I do prefer L/2 or STEP  to (L-2)/2  or JUMP] or rather
the number of STEPs made BIGHT-wise, along the LOWER RIM,  in ONE
FULL PERIOD is equal to the Number of LEADs.
But you will also have 5 STEPs from
1(lower) to pin 1(upper)
and 5 STEPs from 1(upper) to pin 3(lower)  running along the period zig-zaging
between the 2 RIMs.

Here 5 LEADs  hence 5 STEPs  from pin 1(lower)  to pin 3(lower).  ( 1 2 3 1 2 3 :
six digits or "pickets" but only 5 intervals or steps  :........................................ 0 1 2 3 4 5

How come ?  between '1' and '3' there is only '2' !

At this point you must remember that the curve re-enters, that it closes on itself so that to get
to '3' from '1' using 5 step you must go over '1' and '2'.
You just reset the counter to "zero" so that the next step will be on 'one' and not on 'four'.

Instead of using '1', '2'... 'n' up to the number of BIGHTs sometime it will be easier to go
from '0', '1' to 'n-1' (one less than the number of BIGHTs)

That is where MODULUS is great :
5 LEADs leading to 5 STEPs in the PINs using as modulus the number of BIGHTs (here 3),
result of 5 modulus 3 is 2 so you go the whole 1, 2, 3 ( a full 'tour', revolution) and you
count again 1, 2 steps and there you are : on PIN 3.

Another important anatomical fact about the true THK :
going from PIN 1(lower) to PIN 3(upper)   you will encounter ( not always 'cross' mind
you - as the PIN
3(upper) is the starting point of the fifth LEAD in our particular
reckoning here
) the number of LEADs there are in the THK.
If you want a justification for 5 LEADs to be crossed look at this drawing.

When there is no artifact produced by the  representation used you see things as different ;
such as  in this picture :
- equality of the 2 HALF-PERIODs
- LEADs crossed
- the counting '1' to '2' , '2' to '3' then  '3' to '1' ,  '1' to '3' , and final fifth step  '2' to '3'

Four LEADs are crossed before going on the fifth after PIN 3(upper)  () which is its start.
Same thing from PIN 3(upper) to PIN 3(lower)  .

In a FULL PERIOD, running on it up and down the two half-period will encounter
twice the number of LEADs.

So in a COMPLETE PERIOD, for the BIGHTs RIM line at the foot of the "PERIOD"
there are L STEP between one and the other (you now know why JUMPs are much less

This imply that for a HALF-PERIOD, specifically, the first half here, the one from
PIN 1(lower) to PIN 3(upper)  there is a correspondence with the STEPs
( BIGHT offset ) to be calculated then complied with to get the exact number of LEADs
 you do want in your THK.

This number of STEPs for BIGHT offset is independent of the number of BIGHTs (LEADs
meddle in the life of BIGHTs - as rightly LEADs should since BIGHTs should not exist.
BIGHts area rather undifferentiated part of LEADs - , as already explained,  see first page

How can our new knowledge leads us to understand.
To understand well beyond mere handed down recipes, transmitted with first rate craft
abilities but no real intelligence of what is behind all that ?
That will be the object of future topics.

Already some things should have been guessed by the logical ones among you.
No need to be a mathematics lion. I am not one as it is plain to see, "as the nose in the
middle of the face".



Have another look at this image  (a printing of it, will make things easier for you while

Now, at this point, we know that the order of the "structural" numbering of the BIGHT is not
identical to the "procedural" numbering of the order in the making of the BIGHTs.

We are presently going to compute in which order the anatomical structural, BIGHT
'1' to 'n' or '0' to (n-1)  were made with the cordage traveling its route on the cylinder.

Modulus : The Return!

Still staying with our numbered example THK 5L  3B

RF/RC  (Radius Fixed /Radius Rolling Circle) is 3/2, in other words that is 3 rotations and
2 revolutions. (reference : trochoid)

Number of BIGHTs is RF  
Number of LEADs is RF+RC  using signed values)

Each rotation lay a BIGHT so 3/2= 1.5 BIGHT laid per revolution
For each revolution ( look back on the formulas  )  there is one more of one less LEAD
laid than there were BIGHTs laid.

In our case it is one more.  [  (RF + RC)/ RC or (RF/RC) + 1  ]
1.5 + 1 = 2.5 LEAD laid per revolution.
2.5/1.5 = 5/3 or L/B laid per rotation.( 1.666 ) meanwhile in this rotation there was only

sum total of STEPs for the full 3B is 5 * 3 = 15 ( that is L*B)
For each B at the "inflexion" point of a PERIOD there is one up-going branch and one
down-going branch (hence 2*B branches all told   branch ==half-period)

To get one branch offset (the offset of BIGHTs to get the wanted number of LEADs will be
[L /2] measured in PINs    -  we meet L/2 again...
Now for computation of the order in the making (start position on the lower rim will *not*
be '1' but '0' and last will not be (n) but (n-1) ).

Start on the LOWER RIM then go on to the UPPER RIM to make the first BIGHT
there : 1(upper) then go back to LOWER RIM to do the 1(lower) and alternatively
down-up-down  till  the circuit is closed by coming back to starting point
completing the last BIGHT last(lower) there.

To get a better understanding of the words to come : see this picture.

If on the cylindrical diagram we denote the first BIGHT completed or fully done (it will be
on each rim)  by CB(x)(y)  x being the processing order, y being the anatomical order

We have for a 4L 5B

UPPER RIM     CB(0)(III)     CB(1)(II)      CB(2)(I)        CB(3)(V)      CB(4)(IV)
LOWER RIM      CB(0)(V)      CB(1)(IV)     CB(2)(III)      CB(3)(II)       CB(4)(I)

If  read from RIGHT to LEFT ( ) (see picture for clarification ) from 'I' to 'V' in the
anatomical order then
--- UPPER RIM reads as  2 1 0 4 3 (2)
--- LOWER RIM reads as  0 1 2 3 4 (0)  

We have also for each successive BIGHT from the first to the last which is the fifth :

(L * 1) modulus B =(4 * 1) mod 5 = 4 mod 5 = 4    with  1 added = 5  =   V

(L * 2) modulus B = (4 * 2) mod 5 = 8 mod 5 = 3    with  1 added = 4  =  IV

(L * 3) modulus B = (4 * )3 mod 5 = 12 mod 5 = 2  with  1 added = 3III

(L * 4) modulus B = (4 * 4) mod 5 =16 mod 5 = 1  with  1 added =  2  =  II

(L * 5) modulus B = (4 * 5) mod 5 = 20 mod 5 = 0  with  1 added = 1 =   I
Look at this picture

Somehow this reads like the LOWER RIM sequence as it is in
CB(0)(V)      CB(1)(IV)     CB(2)(III)      CB(3)(II)       CB(4)(I)

If  starting on the third line {L * 3 modulus B} and going to circular permutation then it is
the upper rim. (offset between the two rim seems to be 2 , well that is just 4/2 in other word
the  STEP  L/2) CB(0)(III)     CB(1)(II)      CB(2)(I)        CB(3)(V)      CB(4)(IV)

Now how can we get, in a reasoned fashion rather than by foraging about, the fact that
starting at the third line the UPPER RIM sequence is obtained.
Easy :  the offset between I on the lower rim and I on the upper rim is the STEP  L/2  so
 I plus  L/2 is 3   III  hence start at the third line to get the upper rim


SOME MISCELLANY MUSING  or thinking aloud :


Starting from 1 (lower)  [I] it will be  1+ 1*STEP ( upper) [III]
then 1+(1 * STEP) +( 1 * STEP)    (lower)[V]
then 1+(1 * STEP) + (1 * STEP) +(1 * STEP)    ( upper) [II]
and so one till back to staring point.

e.g  the 1+(1 * STEP) + (1 * STEP) + (1 * STEP)  has to be expressed by its modulus (the
 number of BIGHT).


[1 + 1 * (L/2)] mod B    3  mod 5 = 3
[1 + 2 * (L/2)] mod B    5 mod 5 =  0
[1 + 3 * (L/2)] mod B    7 mod 5 =  2
[1 + 4 * (L/2)] mod B    9 mod 5 =  4
(1 + 5 * (L/2]) mod B   11 mod 5 = 1

general formulation is 1 + (x * (L/2) ) with x <= B
where can we use that ?


--- concerning the numbering of BIGHTs at the moment  I can only imagine :

*first : low level anatomical order on each rim ( with just the offset included in the drawing of
the pins positions and going CW on the cylinder as I do my knots so from Right to Left
(could be the other way around; only thing is to stay consistent and congruent.)
On both rim it will read the same. This is "dead data" with no life in it

**second : the embryological order or the order in which BIGHTs are completed during the
laying process of course, ha! this data is a living one.

***third somehow making an order which ( with using permutation when written in a linear
manner on paper to reproduce what happen on the cylinder where you turn around ) will
read on lower rim as it read on the upper rim.

Immediate idea is using the SPart-WEnd vector : lower position 1 ( will be 0)  on the lower
rim is where the SPart enter the would be knot and go the the pin whose offset for number
of LEADs has been computed.

This first attained PIN on the UPPER RIM will be  upper position 1 (will be zero if we count
 not from 1 to n but from 0 to n-1 ), then numbering from RIGHT to LEFT ( ).

That will ensure that all BOTTOM-RIGHT to TOP-LEFT  get the same digit on each
extremity and that the TOP-RIGHT-to-BOTTOM-LEFT get digits with an offset of  'L'
(use modulus B when necessary ).
All very coherent and with a logical and anatomical basis : very first portion  of SPart-WEnd

****random staring point , but random and reproducible computation in such a case does
not really agree well.


Why use mathematics, what are they to me ?

It is  a matter of finding relations and finding a way of representing these relations in a
computable manner being transmissible without the ambiguity that seems to cling to

We already know how to compute "holes" or crossings in a true THK : (L-1) * B but it
would be nice (would be highly useful for the computing of ENLARGEMENT) to know
how to compute existing crossings between up and down LEADs just as computing the
order of BIGHTs making compared to BIGHTs ordering in the finished structure will
be nice.

- corridors should be a good tool I expect

- An  "enlarged" THK inherits the PERIODICITY /STEP  *while in the actual process* of
being enlarged and going to another periodicity.
This, it seems to me is a bit involved (look at all the diagrams I offered and at the
Excel worksheet propose for your own searching

Copyright 2005 Sept - Charles Hamel / Nautile -
Overall rewriting in August 2006 . Copyright renewed. 2007-2012 -(each year of existence)

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