Turksheads

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TURK'S HEAD
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FOLLOW-ON of PAGE 9

Sequential numbers (COLUMN for HALL's, ROW for NAUTILE's frame of reference)
are attributed to ROWs of CROSSINGs.
ROWs of CROSSINGs made with another LEAD encountered by the HALF-PERIOD l
eaving (vector direction) the BIGHT associated with a given (BIGHT) NUMBER.

So to speak the generator of the CROSSING is this HALF-PERIOD. ( before its passage
there is no crossing, only an obstacle that will be ""crossed"" over or under )

Now is time to make a colour printing of this illustration showing all the correspondence
existing between some numbering and indexes that will be useful to make understanding
possible.
This also show ROWs of LEADs CROSSINGs and their numbering in the order the
concerned HALF-PERIOD reads them along encountering them by coming to meet ROW after ROW.

Compare carefully
the UNROLLED knot, unrolled in SPACE *but* showing the TIME distribution,
to the INTACT knot showing the SPACE distribution *but* hiding the TIME
distribution.

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

There are several facts that seem worthy of note:

- the ROW (Column in Hall's frame of reference), figured by 6 green ( of crossings )
horizontal lines, plus one violet line for each BIGHT RIM ( devoid of crossing :
rebound line as in billiard ).
so that makes it
1 + 6 + 1 = 8 ( L + 1) row numbered (↑ ↓) from 0 to L, here 0 to 7.

Finesse :
** the (yellow) ODD HALF-PERIOD going B.R.T.L ↖ read these ROW from bottom
to top (↑) as shown in red on the left side ( this would be read from bottom to top ( ↑) by
the Bottom-Left-Top-Right ↖ ODD numbered HALF-PERIOD for Hall's )

** the (blue) EVEN HALF-PERIOD going T.R.B.L ↙ read these ROW from top to
bottom (↓) as shown in green on the right side ( this would be read from top to bottom (↓)
by the Top-Right-Bottom-Left EVEN numbered HALF-PERIOD for Hall's )

This is what commands the way of writing/reading the numbers sequences.

Note that the (violet) BIGHT NUMBER ( in the angle/corner formed by two half-period
meeting ) can be put in direct correspondence with ANB number for the top rim and
that for the bottom RIM there is an offset of |1|

violet BIGHT NUMBER in the angle/corner minus PNHP ( or value of index) = minus 1

We may have to remember that when writing the equations to find the crossing's ROW

Note that the PNHP can be use to built "tracing numbers" or indexes on both upper
and lower rim. i(PNHP)

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Indexes i(PNHP) can be attributed a computed "value'.
examples

i(3) = 3*L = 3*7=21 21 modulus B = 21 modulus 5 = 1 is in correspondence with
the ANB value for that BIGHT

i(4) = 4*L = 4*7=28 28 modulus B = 28 modulus 5 = 3 is in correspondence with
the ANB value for that BIGHT

on the top RIM (right side for Hall's) and for BIGHT numbered 2 in the angle/corner on
bottom rim ( left side for Hall's)
2 + 1 = 3 or 3 - 1 = 2
but the important fact is that is can be put in relation with the number of a half-period :
The HALF-PERIOD that STARTS FROM that bight

7th   HALF-PERIOD (ODD ↖) starts on bottom RIM (left side for Hall's) at BIGHT
numbered 2 in the angle/corner on the bottom rim
Recall the formula, this number for the BIGHT is given by
(half-period number - 3) / 2    = (7 - 3 ) / 2 = 4 / 2 = 2

6th  HALF-PERIOD (EVEN ↙) starts on the top rim (right side for Hall's) at BIGHT
numbered 2 in the angle/corner on the top rim
(HALF-PERIOD number - 2) / 2    = (6 - 2 ) / 2 = 4 / 2 = 2

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There are other computations if you want to manipulate index i(PNHP)
You can easily find other computations to make, but the main point is not finding
them it is finding a practical use for them.

Note that the Right to Left ( ← ) reading of PNHP ( the value attributed to i(x) is,
starting from :0 ( the start of the cordage run ) on the unrolled diagram :
0 2 4 1 3 ( return to 0 or 5 on modulus 5)

We know that on the unrolled in space diagram we have the temporal sequence of
the creation of the BIGHT, the order of BIGHT to immediately next BIGHT in the
temporal sequence ( do no confuse that with the immediately next BIGHTin the
spatial sequence that will be read from Right to Left ( ← ) 0 3 1 4 2 ,
the PNHP order)

From one BIGHT to its immediate neighbour in SPACE sequence there was
L bight ran over in the TIME sequence.

-----------------------------
so from 0 to 2 = 1 "step" of L BIGHT
(1*L) modulus B == (1 * 7) modulus 5 = 2 this is indeed where you are arrive after
step of L BIGHT from 0
-----------------------------
from 2 to 4 also L BIGHT - that is not interesting as each 'interval will be 'L'
-----------------------------
from 0 to 4 there is 2 "step" of L BIGHT
2 * 7 modulus 5 = 4
-----------------------------
from 0 to 1 there is 3 "step" of L BIGHT
3 * 7 modulus 5 = 1
-----------------------------
from 0 to 3 there is 4 "step" of L BIGHT
4 * 7 modulus 5 = 0
from 0 to the closing on 0 ( ready to "double" or to "enlarge" there is 5 'step" of
L BIGHT
5 * 7 modulus 5 = 0 you have come back on the starting point
hence the correspondence

i(PNHP) = i (number of "step") = (number of "step" * L) modulus B

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The sequential algorithm of ROW (column for Hall) the complementary sequence
is written along "something" that stands for the ROW of crossing ( # or '' or @ )
( # does not stand here for BIGHT per se as it did previously)

There are( L-1) ROW of crossings.

(L+1) ROW grand total but..
there is one ROW for each crossless BIGHT border ( total 2 )
so (L+ 1) - 2 =( L-1) ROW (column for HALL's) between the rim of BIGHT

so of the ( L+1) , 8 positions, only (L-1), that is 7, are used by each sequence of digits.

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

We must somehow put into relation the BIGHT from where the cordage route
( half-period, odd and even ) comes of or arrives to with the LEAD that are
already laid and will be crossed so forming a crossings that will themselves be
forming the ROWs of crossing we have numbered.

as there are (L + 1) ROW ( here 7 + 1 = 8 ) to be represented with '#'

BUT 2 of those '#' are not ROWs of crossings : they are the 2 RIMs ROWs OF
BIGHTs , one at each extremity of our line of '#'.
They will not be "coded" for the nature of the crossing as there are none there.
We sort of discard them "in action" , but keep them "virtual" as place holding items.

must be verified thorough fully from here consider that possible garbage pending new consultation of my old notes

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

upper row of digits is read from left to right ( → ) (that correspond for me to the
bottom RIM as the left side of Hall's ref has been turn CW Pi/2 radian ( 90°) to be
put on an horizontal line)

lower row of digits is read from right to left ( ←)

( → )
0   2 4 1 3 0 2
@ @ @ @ @ @ @   @ (ROW)
2 0 3 1 4 2 0
( ←)

The '#' are replaced by the coding of the ROW concerned. ( COLUMN for HALL)

0   2 4 1 3   0 2
  @ \ /     \ / \    /    @
2 0 3 1 4 2    0
# holding place for the row of BIGHT on the rim

Note that upper line of digit and lower line of digit have the same relation as

B D A C X
X C A D B this is a reverse ;-) ( a mirror would be
   mirror of BDACX

( Ashley did enough mistaken usage of "reverse" for me to refrain from letting
the occasion go unused ! )

This should be useful to make "short cuts" later on, I think.

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

The algorithm just above is "just data", we need to transform that "into information"
that can in its turn be made " into knowledge ( three very distinct logical levels
that are often mistook one for the other : there is no knowledge in book, no
information, only data. With this data your brain can get some 'information'
that will be transformed into knowledge)

To decipher the "algorithm" data into information we need some tools :

This is where the formula computing BIGHT NUMBER from HALF-PERIOD
NUMBER will come handy.
That and the acquired knowledge that ODD are B.R.T.L ↖ and EVEN are T.R.B.L ↙.

To lay the cordage of an ODD HALF-PERIOD then it is immediate that it is a B.R.T.L
route or an↖ going upward, hence we start somewhere on the upper line
(correspond to bottom rim) ( in HALL it correspond to left side).

To lay an EVEN HALF-PERIOD then it is immediate that it is a T.R.B.L or
an ↙ going downward so we start somewhere on the lower line (correspond to top
rim).

To get the "somewhere" we have to start from we must compute the position or the
BIGHT NUMBER making use of the number of the HALF-PERIOD we want to lay.

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

Note that the Number of the BIGHT to which a given HALF-PERIOD is "attached"
is given by :

--- if ODD numbered half-period or " ↖" cordage route going from
Bottom-Right to Top-Left

PNHP(lower) = (half-period - 3) / 2    is the "origin" Bight of the half-period
first half-period (1 - 3) / 2 = -1 something amiss there ? or is it just indication "start" or 0
think modulus ( -1) modulus 5 == 4
third half-period (3 - 3) / 2 = 0
fifth half-period (5 - 3) / 2 = 1
seventh half-period (7 - 3) / 2 = 2
ninth half-period (9 - 3) / 2 = 3

--- if EVEN numbered half-period or " ↙" cordage route going from
Top-Right to Bottom-Left
PNB or PNHP(upper) = (half-period - 2) / 2 is the "origin" Bight of the half-period
second half-period ( 2 - 2) / 2 = 0
fourth half-period ( 4 - 2) / 2 = 1
sixth half-period ( 6 - 2) / 2 = 2
eighth half-period ( 8 - 2) / 2 = 3
tenth half-period ( 10 - 2) / 2 = 4

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

Now looking at this diagram with ROWs and CROSSINGs :

We plainly see that for the yellow ↖upward ODD HALF-PERIOD vector reading the
RED ROWs NUMBERs on the left, the crossing met on an ODD numbered ROW of
CROSSINGS ( 1 - 3 - 5 ) is an UNDER crossing the crossing met on an EVEN
numbered ROW of CROSSINGS ( 2 - 4 - 6 ) is OVER

For the blue downward ↙EVEN HALF-PERIOD vector reading the GREEN
ROWs NUMBERs on the right the crossing met on an ODD ROW of
CROSSINGS ( 1 - 3 - 5) is an UNDER crossing the crossing met on an EVEN ROW
of CROSSINGS ( 2 - 4 - 6 ) is OVER

So if we can somehow know (data transformed into info, info transformed into
knowledge) for sure what half period we are on and at what distance from the BIGHT
ROWwe are then we can know for sure the TYPE of the CROSSING we must make
from knowing the ROW of CROSSINGS we are on.

We know how to compute BIGHT NUMBER from HALF-PERIOD NUMBER ( and
vice versa ), so how are we going to proceed after that ?

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

We will have to do both cases :
--- the TOP RIM BIGHT and its associated HALF-PERIOD
--- the BOTTOM RIM BIGHT and its associated HALF-PERIOD.

We know that for the HALF-PERIOD associated with BIGHT (i) the HALF-PERIOD
is known by computing :

((2 * i) + 2) if the vector is coming from an UPPER RIM BIGHT ( blue) it has
direction T.L.B.R or the downward going '↙" and it is EVEN numbered.

((2 * i) = 3) if the vector is coming from a lLOWER RIM BIGHT ( yellow) it has
direction B.R.T.L or upward going "↖" and it is ODD numbered.

The associated BIGHT Index I(i) will be respectively

I(i) = (i * L) modulus B

I(i+1 ) = ((i+1) * L) modulo B (offset of PNHB between two rims for a given column
of BIGHT)

plus you will recall : between the 2 'feet' on a rim of the given period there are
'L' BIGHT

So from the point of departure -one given BIGHT - to the point of arrival on the same
rim -immediate neighbour of the departure BIGHT- we have stepped :
I(i+1) - I(i) BIGHT
that is 2 BIGHT (look at the drawing ) in SPACE for ONE BIGHT IN TIME.

Remember 1 BIGHT spans 2 BIGHT COLUMNS
2 BIGHT== 4 COLUMN

One BIGHT has a 'feet span' of 'L' BIGHT on the rim.

So 2 BIGHT is 2 * L here
2 * 7 = 14
14 measured with modulus 5 is congruent with 4.
The 4 BIGHT COLUMNS of translation
((i+1) * L ) modulus B - (i * L) modulus B
((i*L) + L) modulus B - (i * L) modulus B

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/