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 :
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 / \ / \ / \
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)
