Nautile
aka Charles Hamel's personal pages

PAGE 11

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FOLLOW-ON of PAGE 10

Please be aware that, despite the attention given to proofing and proofing again ,there may persist some

mistakes ( "may" that is like when a bookseller state " books may have reminder marks" : you can be

dead certain the book has them ! )

( ← ) means from RIGHT to LEFT

( → ) means from LEFT to RIGHT

It will probably be best to have in hand colour prints of these diagrams; requirement to be

able to easily "follow" what is coming :

- 7L 5B with crossings

- TOP rim equations (yellow and blue coding for ↖ B.R.T.L and ↙ T.T.B.L are *not*

applied in this)

- BOTTOM rim equations (yellow and blue coding for ↖ B.R.T.L and ↙ T.T.B.L are

*not* applied in this)

On each rim, in direction RIGHT to LEFT ( ←) ( the way the BIGHT were numbered

for ANB) (i) is for the first order Bight and (n) the very next one , the immediately

adjacent one.

recall to mind that :

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

Half-period B.R.T.L ( yellow - ↖ - ODD ) is coming up from a bottom rim BIGHT with

number (i) {bottom rim (i) that is ; there will also be (i), on the top rim}

the half-period number is computed using ( (2 * i) + 3 ).

The Index going hand in hand with the BIGHT is I(i + 1 )

{ (i ) is for (i) on the considered rim and (n) is for the very next -adjacent- one on the

same rim with (n = i + 1 )}

is computed by evaluating :

I(i + 1) = ((i + 1) * L) modulus B

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

This ODD B.R.T.L ↖ will cross an EVEN T.R.B.L ↙ half-period already present and

coming from the top rim.

Already laid half-period ((2 * n ) + 2) related to top BIGHT rim with number (n) by a bight

index number (n + 1) I(n+1) = ((n + 1) * L) modulus B

This I(n+1) is at a distance, measured in columns of bight, left from the bottom rim I(n+1).

This distance( in bight columns) is

(I(n+1) - I(i+1))modulus B == (((n+1)*L) - ((i+1)*L)) modulus B ==

((n - i ) * L) modulus B

We can deduce that half-period ((2*i) + 3) and half-period ((2*n) + 2) cross at a position

some crossing row up from the bottom rim. (easy : there is no crossing on the rim itself so

it is at least 1 row up, hence the "some" ! )

The distance in row is ((n - i) * L) modulus B + (x*B)

x being >=0 and verifying 0<((n-i)*L) mod B + (x*B)<L (you cannot go where there is

no crossing row ! The rim of Bights which are devoid of crossing)

To lay in the real cordage route this ((2*i) + 3) B.R.T.L ↖ half-period we had better be

knowing what sort of crossing it should be doing with the already laid T.R.L.B ↙ half-period.

In order to be able to know if it is Over or Under we have to know on which crossing rows

the crossing to be made is situated.

The already laid T.R.B.L are linked with n-values having a range 0 to (i) [ 0<=n<=i ] so

that (n - i ) has a range of (-i) to 0 [ -i<=n-i<=0] ( look 0 - i <= n - i <= i - i , understand

now ? )

B.R.T.L ↖ makes their crossing with already existing T.R.B.L ↙ on a row of crossing the

distance of which is measured from the bottom rim : this is the justification for numbering

the row of crossings from bottom rim to top rim ( ↑ ) ( ↑ goes with ↖ ) on the

LEFT side of the diagram with the crossings figured

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

Half-period T.R.B.L ( blue - ↙ - EVEN ) is coming down from an top rim BIGHT with

number (i) {top rim (i) that is}

the half-period number is computed using ((2 * i) + 2 ).

The Index going hand in hand with the BIGHT is I(n + 1 ) is computed by

I(n + 1) = ((n + 1) + L) modulus B.

This T.R.B.L ↙ starting on the top rim will cross already laid B.R.T.L ↖ half-period

(2 * ( n - 1) + 3) = (2 * n ) + 1 which is linked with a bottom rim Bight with number (n - 1)

This B.R.T.L ↖ coming from bottom rim point I(n) = (((n - 1) + 1) * L) modulus B

or ( n * L) modulus B will arrive on top rim at point I(n) =(n * L) modulus B

this top rim I(j) is some number bight columns on the left (because ↖) of the top rim

index I(i).

This distance in number of bight is :

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

so in columns of bight it is ( 2 * ((n - i) * L) modulus B)

From Bight 1 to Bight 2 (2=1 + 1) on a given rim there is 2 and not one unit in BIGHT

ROW as) the bight on the other rim are offset. See picture of BIGHT ROW .

Half-period ((2 * i) + 2) cross half-period ((2 * n) + 1) some crossing row under the

top rim.

This distance under the top row is

((n - i) *L) modulus B + (x * B) with x>=0 verifying

0<((n - i ) * L) modulus B + ( x * B) < L

So knowing the crossing row where the crossing is gives you its nature.

All the already laid B.R.T.L ↖ half-period are linked with n values ranging from 0 to (i)

( 0 <=n<=i ) so the value of ( n - i) ranges from(-)i to 0 ( -i)<= (n - 1) <= 0

the distance under the top rim is counted from this top rim which justify that the row

crossing on the RIGHT side were numbered from 0 to L (↓ goes with ↙ starting from the

top rim

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

A "happy coincidence" : computation of ROW distance ( from the rim they come from)

of the crossing made by half-periods (( 2 * n) + 2) and ((2 * i) + 3) are identical in row

of arrival.

ROW distance from upper rim to crossing of half-period ((2 * n) + 2) is equal to the

ROW distance from the lower rim to crossing in half-period ((2 * i) + 3)

so from bottom

or from top

you arrive in the same row of crossing ( but this ONE row of crossing has a label number

that is different in the left "scale" ↑ and on the right "scale" ↓ )

I mean third for bottom is 3 on the left sequence corresponding to 7 - 3 = 4 on the

RIGHT sequence, the one measured from the top. ( related to the (+ 2 )and (+ 3),

discrepancy is 1)

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

Let us consider 2 crossing half-period verifying the relation (n - i) = minus 1

( remember : circular numbering of BIGHT, here we have come around , and are on the

right side of 0)

( minus 1 modulus B is equivalent to ( B)-(1 modulus B)

(-1) mod 5 == 4 (5) -(1 mod 5) == 5 - 1 == 4

Their BIGHT ( of the 2 crossing half-period) attached on the rim from which the considered

half-period starts are BIGHT(i) and BIGHT(n) following one another immediately,

adjacent (spatial) in time on the given rim, consecutively( temporal)in the cordage run.

adjacent/consecutive in time does not necessarily means adjacent in space along a rim.

Their first crossing point (must be verified pointedly) is at ((minus L) modulus B) ROWS from

the rim on which BIGHT (i) is coming from..

Next crossing ( the second one ) is yet farther along by B ROW crossing ( obliquity, slant

oblige ) from this rim, and on and on it goes

(minus L) modulus B is also in the relation of intersections rows ; those intersection row are

climbed up or down along the slant, as in a staircase from one BIGHT to the next .

(must be verified pointedly)

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

We should have by now enough to discuss directly on the algorithm diagram where

(hopefully) we will "read" the consecutive ROW of crossings ordering. In other words the

crossing met along the run of the considered half-period.

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

Put in line B points (#), with one of them with "0", the leftmost in the usual way of writing

counting step distribute from left right the digit 0 to (b-1) or 0 to 4

7L 5B

7/5 = (5*1) + 2 5-2 = 3

You may also compute ( I prefer it ? What with the error prone sign "minus"put

just before the lone L dispensed with )

(B - L ) modulus B (5 - 7) modulus 5 = (-2) modulus 5 = 3

( the 'pendant is B - [(B - L ) modulus B ] 5 - 3 = 2

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

↓ →

0 1

# # # # #

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

↓→

0 2 1

# # # # #

.....

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

0 2 4 1 3

# # # # #

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

Let us label the 'digit written' at a given distance from starting point 0 ( # of included in

the distance) : DW (too late to modify but strictly it should have been Number Written

instead of Digit Written)

DW = n - i ???????????????VERIFY different sorts of cases, beware of "flukes"

it is placed at (minus DW * L) modulus B '#' from point 0

(-DW) * L modulus B == B - (DW * L) mod B

(-4 * 7) modulus 5 == 5 - (4 * 7) modulus 5

-28 modulus 5 == 5 - 28 modulus 5

== -23 modulus 5 == 2 second place for number 4 (DW)

example

2

2=2-0

(-2 * 7 ) modulus 5 (-14) modulus 5 = 1 2 is one step after 0

4

4=4-0

(-4 * 7) modulus 5 (-28) modulus 5 = 2 4 is two steps after 0

ROW distance from the rim of the ROW of crossings belonging with the first crossing made

by the half-period ((2 * i) + 2) and (( 2 * n) + 1 ( with n - i = DW)

TO the rim where the starting point of ((2 * i) + 2 is placed is computed by

( minus DW * L) modulus B (already seen a few lines above )

ROW of crossing distance for the first crossing on half-period ((2 * i) + 2) implies for

(minus DW) a range of values - i <= (minus DW) <= 0 which can also be written as

0 <= DW <= i ( slant oblige )

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

Let us look at the works with 7L 5B

(minus 7 ) modulus 5 = 3

(B - L ) modulus B (5 - 7 ) modulus 5 2 modulus 5 ==3

five # ( one for each BIGHT) and numbers 0 to 4 to be put ABOVE the # line

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

→

0

# # # # #

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

0 1

# # # # #

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

0 2 1

# # # # #

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

→

0 2 4 1 3 COMPLEMENTARY PERIODIC SEQUENCE

# # # # #

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

The sharpest readers will have remarked that those are the BIGHT NUMBER ( but not

those "belonging" to it ). Remember differentiate SPATIAL from TEMPORAL.

They are in relation to the BIGHT number on the LOWER RIM WHICH HAVE THE

SEQUENCE 0 3 1 4 2 which is the PERIODIC SEQUENCE

We can get this last sequence by using ( no, not (minus L) modulus N )

( +L) modulus B here (+7)modulus 5 = 2 note 3 + 2 = complement to 5

I do prefer the look of B - [(B - L ) modulus B ]

5 - ((5-7)mod5 == 5 - 2mod5 == 5 - 3 == 2

The formula I found and that I prefer B - [(B - L ) modulus B ] IMMEDIATELY show

that is is COMPLEMENTARY - complementary to THE PERIOD in other words to

NUMBER OF BIGHT ( see MATHEMATICS AND THK )

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

→

# # # # #

0 1

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

# # # # #

0 1 2

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

→

# # # # #

0 3 1 4 2

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

xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

My note in passing :

0 2 4 1 3 COMPLEMENTARY # # # # #

# # # # # 0 3 1 4 2 PERIODIC

to get directly the periodic leave 0 on the first # and write LEFT to RIGHT the

complementary read RIGHT to LEFT.

xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

Those numbers are consecutive or proximate neighbour BIGHT

PERIODIC = TOP RIM SEQUENCE FOR THE CORDAGE RUN so the first BIGHT is

given "0", then BIGHT "1" for the second made, then...

COMPLEMENTARY PERIODIC = BOTTOM RIM , will GIVES SEQUENTIAL

ROWS or the row of crossing met by the half-period starting from the BIGHT with a

certain number.

Just to make READING of the sequence ROW more immediate we write the

complementary above the ROW of crossing represented by a file of @ ( a sort

of matrix B x row of crossing )

So (L + 1 ) ( here 7 + 1 = 8 )

ABOVE and UNDER that are representing :the TWO BIGHT RIM

PLUS between them (L-1) ( or here 7 - 1 = 6 ) ROW of crossings..

The leftmost point represent the BOTTOM RIM of BIGHT and the rightmost is for the

TOP RIM of BIGHT ( will be @ ) °°°

@ @ @ @ @ @ @ @

bottom row of top

rim crossings rim

°°° imagine it is like that ( Pi/2 radian , 90° CCW rotation )

@ top rim

@

@

@

@

@

@

@ bottom rim

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

BOTTOM rim sequence written ( ← ) remember the reverse / mirror with the letters in

page 10 B D A C X X C A D B

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

0 2 4 1 3 0 2

@ @ @ @ @ @ @ @

2 0 3 1 4 2 0

TOP rim sequence written ( → ) but having been read ( ← ) in order to reverse it.

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

the

0 2 4 1 3 0 2 and 2 0 3 1 4 2 0

have the same colour coding as in the drawing

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

0 2 4 1 3 written here LEFT to RIGHT ( → ) in the usual manner

what was read RIGHT to LEFT ( ← ) on the BOTTOM rim so this is the existing

BOTTOM sequence written here RIGHT to LEFT ( ← )

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

0 3 1 4 2 written here LEFT to RIGHT ( → ) in the usual manner

what was read LEFT to RIGHT ( → ) on the TOP rim so this is the existing TOP

sequence written here LEFT to RIGHT ( → )

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

the red 7 numbers line stands for the ROW of crossings numbered BOTTOM to TOP ( ↑ )

for the use of ODD yellow B.R.T.L ↖ half-period ( using 5 numbers for 7 places we

had to do a "carriage return")

the green 7 numbers line stands for the ROW of crossings numbered TOP to BOTTOM ( ↓ )

for the use of EVEN blue L.R.B.L ↙ half-period( using 5 numbers for 7 places we had

to do a "carriage return")

Hence the justification for writing the line of numbers in opposite directions for red ↑ and

green ↓.

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

Nothings at the @ on extreme left and extreme right : there is only crossless

BIGHT RIM hence their absence of numbers

In fact we could as well ( but much less rigorous and much less clear ) have written

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

BOTTOM rim sequence written ( ← )

2 4 1 3 0 2

@ @ @ @ @ @

2 0 3 1 4 2

TOP rim sequence written ( → )

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

Next we will have to replace the @ with the coding of the corresponding ROW of

crossings.

example / \ / \ / \

Look again at the 7L 5B with crossings

For the left side numbering of rows of crossings bottom to top ↑

0 and 7 are without any crossing so are not really put to use

even 2 - 4 - 6 are OVER for the yellow B.R.T.L half-period ↖ upward going

even 1 - 3 - 5 are UNDER for the yellow B.R.T.L half-period ↖ upward going

For the right side numbering of rows of crossings top to bottom ↓

0 and 7 are without any crossing so are not really put to use

even 2 - 4 - 6 are OVER for the blue T.R.B.L half-period ↙downward going

even 1 - 3 - 5 are UNDER for the blue T.R.B.L half-period ↙ downward going

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/