TITBITS
(UK) or TIDBITS (US) of THK ANATOMY AND THK MATH (elementary but quite important, compulsory to
absorb and
digest if you want to
continue reading my delirium without trouble)
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
half-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 :
ISOMETRIC grid)
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
completely laid, '2'
is the second
one
and '1' is the
third one to be completed.
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
tools.
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 from1(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
suited).
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
of MATHEMATICS
AND
THK PART TWO )
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".
HOW TO GET THE ORDER IN WHICH THE BIGHTs ARE COMPLETED
IN THE PROCESS OF MAKING A TRUE THK:
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
1 BIGHT
laid.
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
number.
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 = 3
= III
(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
laid.
****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
words.
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)