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Henry Bushby Transcription pages

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Take S4 6b/18c z = 2
Here it is clear by inspection that
there are 2 exclusions, 4 & 8.
B/z - 1 = 6/2 - 1 = 3 - 1 = 2 [checkmark]
M2n = M x 8 = 8, [16 exceeds 2B]
(M2n = 8, 16, 24, 32, 40 &c.
(M'B = 6, 12, 18, 24, 30, &c.
The rule won't work thus.
M2n - 2B = 8 - 12 [??]
16 - 12 = 4 [checkmark]
24 - 12 (exceeds
equals 2B].
The rule stands without the 3rd
[??]exception, though that seems
to apply when z = 1. But even
then it won't cover the 4 in S6 8b/40c
Test the rule as given on p 67
with S4 9b/27c, z = 1
M2n = M x 8 = 8, 16, [24 exceeds 2B]
M2n - 2B = 6, 24, [22 -------
M2n -2 = 6, 14, [22 ------
B/2 - 1= 8. Exclusions are 6, 8, 14, 16, 22

The rule holds but the numbers
Take S4 11b/33c
n = 4, B = 11, z = 1, 2B = 12
B/2 - 1 = 11/1 - 1 = 10.
M2n = M8 = 8, 16, [24, 32, 40, 48,
M2n - 2B = 2, 12 10, 18 [26 ------
M2n - 2 = 6, 14, [22
[therefore] exclusions are
2, 6, 8, 10, 14, 16 & 18.
4 12 20
In this case they turn out to be all
the even nos, & the rule breaks down.
I believe in every case of 1z,
1 added to an even no. gives a ring.
For there are 4 parts cut, and the
parts are even & odd alternately from
arc to arc. Therefore an odd cut part
will be even in the first bight it
emerges in, odd in the 2nd, even in

[All in pencil]
Let M = any even multiplier & O = an odd one.
Exclusions are O2n
& M2n + 1 [bracketed with] or do. - 2b
which make rings.
Values not to exceed 2B, amend group limited
by B/z - 1
Exclusions are
2 x 2n (even)
(2 x 2n) + 1
3 x 2n
(4 x 2n) + 1
5 x 2n
(6 x 2n) + 1

in S5 So in S6.
Therefore it would seem that
in any S, 1 must not be joined to
M2n, but this must be reduced
where necessary by subtracting 2B.
i.e. values of M2n are to be taken
up to 4B - 2, & 2B subtracted from
those above 2B.
Take S5 12b/48c z = 1
n = 5, b = 12, 2B = 24, B/z - 1 = 11
M2n are 10, 20, 30, 40, 50,
24 24 24
6, 16, 26
giving 6, 10, 16, 20. But this
omits 8 & 18 (p 67. B.)
In point of fact 1 joined to any
even no. in S5 12b/48c gives rings.
Where z = 1, it seems obvious that
as 1 passes into every bight & emerges
as an even no, previous to emerging
as x, if it is joined to any even no.

The part [?] 1 emerges in the nth bight
from itself its own, in the 2nth, in
the 3nth, the 4nth &c.
Does it emerge as an odd or as
an even no?
In S2 4b/4c 1 emerges as 4(=2n) in
the nth bight from its own
In S2 4 5b/5c< as 4(=2n) in the nth bight
& as 8(=2x2n) in the 2nth bight.
Throughout S2 1 emerges always
as an even no.
Therefore in S2 the exclusions are
[Vertical line]M2n. & these only, & the B/Z-1 rule
limits their no.
In S3 1 emerges in the nth bight
from its own, in the 2nth, 3nth &
so on. It emerges as an even no (Ex-
cept in S3 3b/6c where there can be
no prohibitions)
In S4 it emerges as an even no. So

in S5. So in S6.
Therefore it would seem that
in any S, 1 must not be joined to
M2n, but this must be reduced
where necessary by subtracting 2B.
i.e. values of M2n are to be taken
up to 4B - 2, & 2B subtracted from
those above 2B.
Take S5 12b/48c z=1
n=5, b=12, 2B=24, B/z-1=11
M2n are 10, 20, 30, 40, 50,
24 24 24
[horizontal line]
6, 16, 26
giving 6, 10, 16, 20. But this
omits 8 & 18 (p67. B.)
In point of fact 1 joined to any
even no. in S5 12b/48c gives rings.
Where z=1, it seems obvious that
as 1 passes into every bight & emerges
as an even no, previous to emerging
as x, if it is joined to any even no.

other than x it must form a ring.
But this is not the case [in?]
when z is greater than 1.
Can this be proved theoretically?
1 emerges in the nth, 2nth, 3nth
bight &c. & where these exceed [?]B in
these bights less B. (The nos. of its
parts are 2n, 4n, 6n, 8n &c. less
2B if necessary.
Nos. are M2n & M2n - 2B.
up to 2B.
Where n=2 & B=3 we have
4, 8, 12, 16, 20, 24, 28
6 6 6 6
[horizontal line]
4 2, 6
Here we have 2,4,6 - all the even
nos up to 2B.
Where 2=2 & B=5
M2n=4, 8, 12, 16, 20, 24
10 10 10
[horizontal line]
2 6 10
and here we have them all.
These are both in Sz.

Take S3. 4b/8c 2B=8
M2n=6, 12, 18, 24, 30, 36
8, 8
[horizontal line]
6, 4, [10
Here we have 4 & 6 only. Yet 2
gives a ring.
It app
Divide [?] by n. If there is no
remainder the exclusions are 2n,3n,
4n, 8n &c. up to 2B. If there is a
remainder it represents a certain
no of bights less than n to the left
of x. Through Under these the line
will pass & emerge in the (nth from x - this
remainder) bight, then in the (2nth
- this remainder) bight & so on. There-
fore to the even no. in that bight it
must not be joined. i.e. to 2 (n-r)
2(2n-r) &c. or M2(n-r.) up to
2B. And if there is still a re-

mainder it will continue to 2(n-r1),
2(2n-r1) &c. until there is no remainder.
Take S6 12b/12c. Divide 12 by 6 =2: no
remainder. Exclusions = 12, 18 24 (=2B.)
Take S6 10b/50c. Divide 10 by 6 =1+4
2B=12. Exclusions are 2n(=12)
It may be said at once that
the first exclusion is 2n, the 2nd
4n, the 3rd 6n, the 4th 8n, but
where these nos. [?] one of these nos.
equals x, there the thing ends, i.e.
if 2B is divisible by 2n (i.e. B by
n). But if it is not then the next
exclusion after the one lower than
2B is the difference from 2B + balance
of 2n, i.e. it is a [sum?] equal to 2n.
Take S6 10b/50c. First exclusions are
12,[24 &c]; but 20/6=3+2 & 2+4=6 20-12=8 & 2z-8=4

(=2n-r, where nr is remainder). This
2n-r=4. Next exclusions are 4,
10, 16, [28 &c). But 16 from 20 = 4
& 2n-4=12-4=8. Next exclusions
are 8, 20 Now 20=2B, & that ends
it. Therefore total exclusions are
12, 4, 16, 8 or 4, 8, 12, 16.
This is correct.
Therefore the Rule of Rings may
be stated thus: Where z=1, an odd
no must not be addes to an even no,
nor an even no to an odd; where z
exceeds 1, if B is divisible by n, then
2n, 4n, 6n &c. up to 2B are excluded;
where B is indivisible by n, then (1)
2n, 4n, 6n &c up to 2B are excluded
& the remainder r difference, d,. (after of whichever
of these nos is next below 2B) subtracted
from 2n, & this no. +2n, 4n &c. are
excluded & so on till a number is reached
which equals 2B.

[Blank Page]

Query: Ist Is it a series of which
the difference is always 2 x the G.C.M.
of B & 2? Thus in S610b/50c the G.C.M of
6 & 10 is 2, & 2x2=4.
In S6 12b/60c, the GCM is 6 & the double
is 12.
In S6 9b/45c, the G.C.M. is 3 & the double
is 6.
In S611b/55c the GCM is 1 & the double
is 2.
In S5 10b/40c the GCM is 5 & the double
is 10
In S5 11b/44c the GCM is 1 & the double is
In S5 5b/20c the GCM is 5 & double is 10.
In S4 6b/18c the GCM is 2 & the double is 4

? Is it true that
y joined to y-1 + Mn,z in any Series n,
gives rings? No -
[Caption: S3 4b/8b]
In S2 7b/7c there are 16 parts 1 + 1 [?:suit]
M2z = M1 = 2, 4, 6, 8, 10, 12, 14
leaving 1 3 5 7 9 11 13
= 6 other cases, & length. This
would give 7 h, but Tait says there are 8
In S2 5b/5c there are 10 pts.
M2z= 2, 4, 6, 8, 10
leaving 3 5 7 9 {Deleted: 4], so 4 other[? rings]
but Tait says there are only 4 knots
Test the knots so formed.

The Rule of Rings is this in all
cases:^[cannot read pencil notation] 1 must not be joined to
M2 (GCM of B&n)=
M2 (GCM of C/n-1+^&n)
or y must not be joined to y-1
+ M2(GCM of B&n)=
y-1+M2(GCM of C/n-1+n)
But by XIV. (ii.267) the GCM of
B & n = z.
Therefore the Rule of rings is ^?[1m?] S2 ^only 2 bights being cut 1 must
not be joined to M2Z, or, generally,
y must not be joined to y-1+M2Z.
If it is so joined Z becomes Z+1
Problems: Rule for reducing 2 to 1
Rule for further exclusions ^or additions so that C may
remain the same
In S34b/8c join 1 to 4 (=M2Z ie. 2x2x1)
no rings result - Yes they do, & so they
do when joined to 6

S2 6b/6c
Multiples of 4 barred
6c in 1, 2, 3, 6,
5 4 7
3c in 1, 3
2, 4
5c knots [running?]
5c 1, 3-
+ 7
2,0 6,
S2 {1/1}[parenthesis extends to next line]
S6 {1/1}[continuation of parenthesis from line above] = 4

If 1 must not be joined to M2z
[deleted: &z = 1], 1 must not be joined to an even
It may be joined to an odd no. ^Where z = 1
The number of odd nos to which it
may be joined therefore are (2b/2 - 1) where
b = no of bights, or (x/2 - 1), x being
the top left hand number as on p1.
T S2 5b/5c x = 10 x/2 - 1 = 4
but there are only 2 knots (? for each 5c
form in each [seven?]?)
S2 6b/6c x = 12 x/2 - 1 = 5
but there are only 4 knots [But here
S2 7b/7c x = 14 x/2-1= = 6
& there are 8 knots

[blank page]

Experiments in Twisting.
z cords are arranged at equal distances
on the [deleted: rad] circumference of a circle
circular tube at equal distances from
one another, thus: -
[Caption: (1)
2 cords
3 cords
&c -
If the tube is now
simply bent so that the
ends are joined, as
is being done in fig 3,
the ends of each
cord will meet, and
^separate unlinked circles of cord will result. But if
^the end of the tube is twisted through 360[degrees]/z (where
z = no of cords) this will no longer
be the case, & other results will fol-
low which is proposed to investi-
360[degrees]/z is called "[deleted: one] ^an arc", whatever the value

[blank page]

of z, since z being given, the pre-
cise value of the arc can always be
determined if it is of any importance.
For illustration if the tube
is here bent
& one end
twisted through one arc 1 will join 3,
3 join 2, & 2 join 1 & z will be reduced
from 3 to 1. Twisted through two arcs
1 will join 2, 3 join 1, 2 join 3, & z
again is reduced from 3 to 1. But
if it is twisted through 3 arcs (=
z arcs [deleted: )] = 360[degrees]) 1 joins 1, 2 joins 2, 3 joins
3, & z remains 3.
It is of no importance that, twisted
the other way, 1 would join 2, 2 join 3 &c
Since the reverse will be formed & in any
case the same pairs join.

K3 2a/4c
2 arcs 3 cords. Result.
4 Crossings, not alternate,
& [Deleted: not] reducible to the
overhand S2 3b/3c
Crossings over 2 & under 2.
Note: A strip of paper given 2 half twists &
cut along 2 lines (forming 3 bands) gives
[Deleted: S3 3b/[3 corrected to 6]c not S2.] [Deleted: S3 3b/6c]^ 3 linked rings not S2 3b/3c. This is
because in twisting flat paper the central
1/3 part longitudinally must always join itself.
(But in any Sn where n is even the expected
result ought to occur? or might sometimes)

2 cords.
[Revolved?] through Result.
1 S2 1b/1c [tickmark]; tested .& with paper strip
2 S2 2b/2c [tickmark];
3 S2 3b/3c [tickmark]paper strip[tickmark][ bracketed from 1 to 6: everything
] 4 S2 4b/4c [tickmark];
5 S2 5b/5c [tickmark]; tested with paper strip [tickmark];
3 cords.
arcs. result. z
1 S3 1b/2c [tickmark]; circle 1
+2 S2 3b/3c [3 tickmarks];overhand 1
3 S3 3b/6c [tickmark] 3 circles 3
4 anomalous. p 101. z=1.

K3 3a/6c
3 arcs 3 cords: result.
[Caption: non-alternate.
crossings over 2 under 2.]
[Caption: =]
[Caption: K3 6a/8c]
best form

K3 -- 101

4 arcs, 3 cords : result .

not a Carrick.

This knot is undoubtedly
the result, having been
tried many times, and
appears to be an ir-
regular irreducible knot
of 8 crossings. There are 2 pairs of non-
alternative crossings.

K3 -- 103
5 arcs 3 cords. results.
[Caption: Crossings not alternate.]

K3 --- 105
6 arcs 3 cords. Results 2 = 3
[Caption: non-alternate.]

K3 --- 107

76 arcs 3 cords-
[Caption: Drawn from rule.]

K3 --- 109

8 arcs 3 cords
[Caption: rule.]

 (Empty page)

K3 9a 111
[Caption: Drawn from rule]

    (Empty page)

K3 10a 113
10 arcs 3 cords

  (Empty page)

K3 11a 115
11 arcs 3cords

 (Empty page)

K3 12a 117
12 arcs 3 cords

 (Empty page)

K4 1a 119
1 arc 4 cords
a twist.

(empty page)

K4 2a/6c
2 arcs 4 cords
[Caption: Drawn from rule.]

[Editor: this page contains manuscript in pencil which has been overwritten in ink. The text below reflects the manuscript written in ink. The manuscript in pencil is illegible in places, and has not been transcribed. The manuscript in pencil appears to be a fragmentary earlier version of the manuscript in ink. The content of this page needs to be reviewed.]
[Editor: First four lines have a curly brace to their right with the text "Tested" to the right of the curly brace.]
K2 1a/c = K1 2a/c
K3 2a/4c = K2 3a/3c
K4 3a/9c = K3 4a/8c
K54a/16c = K4 5a/15c
|| Does Kn (n-1)a/c = K(n-1) na/c? Yes.
If so K5 4a/16c = K4 5a/15c Yes. [checkmark]
(Note that the crossings are 1 less & the arcs
1 more.) & that the reducible knots have
{ Does Kn (n-2)a/c = K(n-2) (n+1)a/c? No.
If so K4 2a/c = K2 5a/c. No.

[Editor: mixed pencil and ink manuscript again. I have tried to sort out the mixed sentence and figure that starts with "Reduces to..." in the way that seems most faithful to the manuscript and also makes mathematical sense, as the mathematical expressions and the figures are two representations of the same concept. The content of this page needs to be reviewed.]
K4 3a/9c
3 arcs 4 cords
[Caption: Drawn from rule.]
Reduces to
K43a/9c reduces to = [FIGURE]
by experiment
= K3 4a/8c.
Full formula:4a/8c
Kn (n-1)a/(n-1)2c = K(n-1) na/[(n-1)2-1]c

This series may be expressed, instead of
e.g. S3 4b/8c as K3 4a/8c = 3 cords, re-
volved through 4 arcs, give a knot of
the form S3 4b/8c with 8 crossings in alter-
nate sequences of (3-1) = 2 (q. v. p 101[alpha]) z
in the result being the same in both series.

4 arcs 4 cords
[Caption : b w 9 2]
[Editor, I'm unsure about the "b" and "w"]
Crossings not alternate but over 3 &
under 3.
Rule of Twists.
If in the Plait series we read
arc for bight, & no. of original cords
for series number then for [Editor: illegible text crossed out] n cords
& x arcs we shall get a form
similar to Sn xb/c, c being the same
BUT, instead of being alternate they
are in alternate sequences of (n-1) over
& (n-1) under. Z is also the same in
the result.

K4 5a/c
[Caption: rule.]
Does this = K5 4a/c? Yes [checkmark]

[Caption: 5 arcs 4 cords]
[Caption: fill in this]

[blank page]

6 arcs 4 cords
7 arcs 4 cords

[blank page]

8 arcs 4 cords
9 arcs 4 cords

[Page left blank]

10 arcs 4 cords
11 arcs 4 cords
12 arcs 4 cords.

[Page left blank]

K5 1a/c
K5 2a/c

K5 4a/c
Does this = K4 5a/c ? Yes [Tick/check mark].

K5 3a/c
K5 4a/c

[Page left blank]

General Principles.
(1) A r.k. is a knot expressed in
one of the figures of which ^ the charac-
ter is shown in vol ii. ad fin.
(2) Every complication can be expressed
as a r.k. but in this form it
may or may not show its least
no. of crossings.
(3) Every group (G) of ^ r. knots such ^ e.g.
as Plaits (P) or ^ and Torsions (K) has
alternate equal sequences of crossings
over & under.
(4) The sequence defines the group
& must be a (& may be any) measure
of (n - 1) from n - (n - 1) [=1] to
n - 1. Thus the seq. in P is 1 or
n - (n - 1); in K it is (n - 1). These
two groups represent the two ex-
tremes. Between them lie :-

Groups will be in Series with n = 1 + [M  ?]

n - (n - 2) = 2
n - (n - 3) = 3
& so on up to n - 1.
Therefore any factor measure of
(n - 1) may be a sequence number,
but and form a Group, but only
those figures will be rated in
which the given
values of n will
supply figures, in which the sequence
no. selected is a factor measure
of (n - 1). Thus the P group with
seq. of 1 will embrace every figure; so
will the K group with seq. (n - 1) for
both these seq. nos. are measures of
(n - 1). But a group "II" with seq.
2 will only embrace the series
in which n is odd (since then
(n - 1) is divisible by 2. Group III
will embrace only S4, S4, S7, S10 &
so on; Group IV will be in S5, S9
& so on. And in some cases the
groups ^ or series of different groups will coincide. Thus S2, being

{1 = n - (n - 1)
{n/n n - 1/n - 1
{ n - 1 = n - (n + 1) = n2 - n/n = n - n/n
{= n2 - 2n + 1/ n - 1 = (n - 1)2/n - 1
n Seq. = n - (n +/- 1) covers K & P
3 odd n - 1 = II 3 - 1 = II
4 - 2 = III
5 odd n - 5 - 3 = IV
6 - 4 = II ( [ ? ] )
3 - 1 = II
5 - 3 = II
4 - 3 = I

"I"(=P) appears in S[subscript]2 as with
a sequence n-(n-1)=P[subscript]1, or as
with a seq. (n-1)=K[subscript]1 for both
are equal to 1. In S[subscript]3 (factors
of n-1=1,2,)we have "I"(P) &
2"II" coinciding with K.(n-1). and
no others. Thus
Series no (n) (n-1) Factors Sequences
1 0 0 0
2 1=(n-1) 1 I(P,K.)
[curly bracket]3 2 1 ? I(P)
4 2=(n-1) II(K.)
[curly bracket]4 3 1 I(P)
3=(n-1) III(K.)
[curly bracket]5 4 1 I(P)
2 [Circled:II]
4 IV (K.)
[curly bracket]6 5 1 I(P)
5 V(K.)
[curly bracket]7 6 21 II(P)
32 [Circled:II]
63 [Circled:III]
6 VI(K.)

?shld reduce to [vertical line]K4 9B/27C-4 then to less
Parallel Case wld be III10 3B/27C reducing to 1K4 9B/27C & then
Parallel Knot III10---
to less
So I2 1B/1C reducing to I2 1B/1C is also parallel.
I2 1b/1c parallel[Vertical:Cutting line +-+--+-+
II5 2b/8c(before ?rings)
reduces to
1K3 4b/8c(after ?rings)
reduces to
which is "A" in Vol ii pp 100-1.

n (n-1) Measures Sequences
8 7 1,7 I (P)
7 VII (K).)
9 8 1 I (P)
2 [Circled:II]
4 [Circled:IV]
8 VIII (K.)
10 9 1 I (P)
3 [Circled:III]
9 IX (K)
and so on.
It seems probable that in
every [?] group except I there is
a law of reduction for knots before
rings. But observe P always = I,
but K=I,II,III,IV ... in succession
& the sequences vary. The law of
reduction for K is stated in vol vi.
It was found that II5 2b/8c reduced
to I.K.3 4b/8c, & that this reduced
to informal knot of 6c, as shown
on opposite page.


It is unsafe to def deduce a
law from this ^[?G]K because :K
may be any number, but here
we deal with II only, though K
also is II; hence what is uni-
versal for K ought to be universal
for all groups.
[On the analogy of the K
formula we should have
II5 2/
IIn xb/x(n-1)c=I.K.x+1(n-1)b/x(x-1)c
Observe, here the reduction is to an
1.K, & even that is not ultimate.
For x(n-1)c we may write simply
yC as usual.
Although it is convenient to
have P=Group I, & equally convenient to
have K=Groups I,II &c. whenever
equal to (n-1) yet it seems probable
that one must choose between the

  • Fallacy. Because a rule applies to Torsions
    in every Numeral Group (I,II &c.) it does
    not therefore apply to every [??] individual
    of each group. vi.268.
    We have
    IV52b/8c reducing to I25a/5b
    II52b/8c [Horizontal line] 1K34b/8c

systems & decide that I shall be
single & complete as a Group; II a
complete group & so on. In that
case K would enter into every
group instead of every group entering
into K.
Since K enters into every group
any rule universal for all Ks
is presumably applicable to each
& every of I,II, III &c. But it
is found that the Law of Reduction
needs modification to embrace the
case of K II5 2b/8c reducing to an
1K3 4b/8c. According to the K. law it
should reduce to S2 5[?]/5c, which
it certainly does not, being an ir-
[Vertical Line ? for emphasis?]reducible i.k.of [?6c] ultimately. Of
course the Torsion S52b/8c is IV52b/8c
& the other is II52b/8c

1/3 [?] 1=1 s.2
Combinations of n things r at a time are
[Fomula: |n / |r |(n-r)] [In modern mathematical notation this would be
n!/r!(n-r)! where n! = nx(n-1)x(n-2).... e.g. 5! = 5x4x3x2x1 = 120]
[Bracket spanning 3 lines:] Permutations n things r at a time are
n(n-1)(n-2) . . . (n-r+1)
[No of Permutations = n!/(n-r)!]
Permutations of n things all together are [n?] [n!]
Possibilities of y crossings 2^y. [2 to the power y, 2y]
Notes on Knots. Vol 4.
Ap. 1902.


Irregular Knots,
with alternate crossings.

Preliminary Investigations as to the con-
nection (if any) between these and the
Regular Knots.

For purposes of clear reference in any
diagram of a regular knot (2.230 ff) the
^parts of the bights, if cut, are supposed to be numbered
from the top by the right from 1 to x.
Thus in the case of S25b/5c, the numbers
would be as follows:--

[Diagram] The same plan

can be adopted
in the case of
any r.k., and
by quoting 2
nos, e.g. 4,5,
any particular

bight can be indicated. In forming i.ks
from r.ks. since more than 1 bt. must
always be cut, the top bt. 2-1 is as-

1z 2z 3z 4z
2c (1B knot) I2 2b/2c [checkmark]
3c None
4c I3 2b/4c [checkmark] [pencil hook? knot] I2 4b/4c [checkmark)
6c "A" (6c twist.) [checkmark] [pencil 14 2b/6c I failed?]
8c I5 2b/8c [checkmark]
12c Twist Knot. [checkmark]

[?] to be one of the cut bts. in every case.
Provisional Rules.
Let o = an odd no & E, an even no. in
the above system. Let [symbol] = joined
(1) o [symbol] o +1 or e[symbol]e -1} makes a twist & annuls 1c.
(2) o [symbol] o -1 e [symbol] e, +1} restores original bight.
(3) o [symbol] o + 2x? -1 e [symbol] e [symbol e -2x? +1 } rings (x? = S width.)
(4) o [symbol] o+2 e [symbol] e +2} knotsencloses?/loose end involosing?
/ new C & then cutting of 2 adjacent B outside enclosure.
(5) o [symbol] [symbol?]o +3 either encloses / uncut B
or 2 l. ends involosing? 2 new C.
[library number 39891]

To reverse a twist knot: Kew round
[Oro?] 180 [degrees]; & put bight across from right
to left or [v.v.?] The only twist knots
I could reverse were I3 2b/4c = Figure
of 8), "A" & the 12c knot, and ? is this
last [egnel?] to I5 3b/12c ?
It seems probable that twists knots of MGc
are reversible.

I could reverse no T.H. except I3 2b/4c
nor the Weaver, nor I2 6b/6c<be>nor I4 2b/6c
Probably I7 2b/12c is reversible, in which case
a series, [riff?] 4c [wld?] be indicated, x=2
& is always odd.
See I. p. 152-3. Every other twist knot of this
series 3, 6, 9, 12 [etc?] (viz 6, 12, 18 [etc?] and of
the series 4, 7, 10, 13 (viz 4, 10 [etc?] are reversible.
? of the 2z series 5, 8, 11, 14. Probably 8 & 14 are
reversible. Anyhow 15 2b/8c is. Is this a twist knot.
The 2z 8c twist is reversible.

(6) e [symbol] e + 3 encloses 2 l. ends involving
a twist (and [onff?]) or 2 new C.
(7) o [symbol] o +4 } 1 loose end involving 1 new C
e [symbol] e + 4 } & an uncut B, or 3 l. ends
of [?.] 2 may make a twist
counteracting the 1 new C
made by the 3rd, or 3 [?][?]
involving 3 new C.
(8) o [symbol] + 5 encloses 2B.
(9) e [symbol] + 5 " 1 B, 2 l.ends.
(10) o [symbol] + 6 " 2B, 1 l.end
(11) e [symbol] + 6 " 2 B, 1 l.end
(12) o {symbol] to any other o=o' encloses (o'-1)B
(13) e [symbol] higher o=o'-2) B & 2 l.ends.
(14) o [symbol] to any higher e = e' encloses E'-2/2 B4 & 1 l.end

[diagram] [diagram] 5c [diagram] 8c [diagram] 7c
{[diagram] 5c [diagram] 4c [diagram] 3c
[diagram] 1 [diagram] 6c
[diagram] 2c

(15) e [symbol] to any other higher e=e' encloses also
e'-2/2 B + 1 l.end.
(16) There is always a loose end when o is
joined to a higher o'.
(17) Always 2 l.ends (which may be joined
subject to o rules) when e [symbol] e'.
(18) For every single enclosed l.end a twist
must be supplied by cutting 2 B adja-
cent to one another.
Note: A join, for purposes of Description
is supposed to be made by the right (as
the hands of a clock.) But in effect
it makes no difference how it is made.
Crossings must be kept alternate.

Plait or Twist Knots as on p152 [ft?].
of vol 1.
Their [double underline] reversibiltiy.
The law of reversibilty for these knots
is as follows:-
Let [sigma] = any odd no. of strands >1
Then [sigma]-1 is the difference of an
Arithmetical series of which it is also
the first [terin?], and in this series every
[terin?] represents a reversible Plait (Twist)
Knot with crossing equal to the [terin?].
But except in the case of [sigma]=3 the first
[terum] duplicates a knot in n ^(not the) preceding
series, and therefore, when all the series
are being considered from 3([sigma]) onwards,
it is best, except with [sigma]=3, to make
2([sigma]-1) the 1st [terin?].
Thus: Required reversible knots: [sigma]=5.
[sigma] = 5, [sigma] - 1 = 4 = 1 st terin (since the series
alone is being considered.
Are = 4c, 8c, 12c, 16c, 20c . . . . . knots.
(p g follows)

Required: Reversibles of 12c. 16c.
[sigma] diffces are 2, 4, 6, 8 etc.
16 is divisible by 2, 4, 8 only.
Values of [sigma] are 3, 5, 9. _ 3 knots.
Therefore, except in the case of 2c, of
which there is 1 knot, given y (the
coefficient of c) there are as many
knots as there are measures equal
to 2 or multiples of 2 ^ less than y which will
divide y, & values of [sigma] = these +1.
Example : Required, reversibles of 24.
24 is divisible by 2 _ [sigma] = 3
[symbol] 2 x 2 _ [sigma] = 5
[symbol] 3 x 2 _ [sigma] = 7
[symbol] 4 x 2 _ [sigma] = 9
[symbol] 6 x 2 _ [sigma] = 13[bracket] 5 = AND.
These knots are not necessarily on /z.
Determination of z.
In [sigma] = 3, z = 1 except in the case of
2c, 2 + 6 (or a multiple of 6)c, in which
case z = 2. Every 3rd knot is on 2z.
(p 10 follows) 1st = 1 ([sigma]-1)c

[sigma] = 5. (4c z=1) [tick]
8c z = 3 8c z=3 [tick]
12c z = 1 12c z=1 [tick]
10c z = 16c z=1 [tick]
20c z = 1 20c z=1 [tick]
34c z = 24c z=1 [tick]
30c z = 28c z=3 [tick]
42c z = 32c
Every 5th knot is on 3z. 1st 3z = 2([sigma]-1)c

[sigma] = 7. (6c z = 1)[tick]
12c z = 1 [tick]
18c z = 4
24c z =
30c z =
36c z =
42c z =
48c z =
54c z =
60c z = 4
Every 7th is on 4z. 1st 4z = 3 ([sigma]-1)c

Examination of Sz 5b/5c.
[diagram][Caption: 1 2 3 4 5 6 7 8 9 10] Let B 10.1 be cut:
Join 1, 2. A twist is
formed [annabling?] a
crossing thus:-
[diagram] [Caption: 10 [2/1?] 3 4 5 6 7 8 9] If 10 & 3 are [symbol] no
new C is made to com-
pensate.^but I2 6/u results If one of
them is taken under & over the new B
2 new C result. Let 3 be taken under & over to 10.
[diagram] [Caption: 3 4 5 6 7 10] There are now 6C on
2 cords (S2 6b/6c) which
can be reduced only
by cutting 2 adjacent
B. Cut 4, 5, & 6, 7. &
join 5, 6, to make a twist.
[diagram] [Caption: 7 6 5 4] = [diagram] [Caption: 7 4] There are now 5C again. Join 4, 7.
Result: Original Twist.

[vertical line in left margin]The law for determining z is this:-
In any series [symbol], (where [symbol]=no.
of strands), the reversible having
([symbol]-1)2/2 c, and those having this
no. of c multiplied by [symbol] or any+ [symbol] ([symbol-1])c or + any multiple of
[symbol] ([symbol]-1) c multiple of [symbol] are on [symbol +1/2 z
and the others are on 1z.
[end of vertical line in left margin]Thus: Required reversibles on 1z, 8c.
8 is divisible by 2, 4
Values of [symbol]= 3, 5
[symbol]=3. (3-1)2/2 c = 2 c[symbol] x 2 = 6
[symbol] ([symbol]-1)c = 3 (3-1)c = 6c.
2c + 6c = 8c. But this then is not on
1z, and is no good.
[symbol] = 5 (5-1)2/2 = 8c. This then is no good.
Ans: No ^[symbol] 8c twist reversible on 1z.

Suppose however 6,7 & 8,9 had been cut
& 7 [symbol akin to > indicating ‘join to’or ‘joined to’ or ‘joining’] 8, 6 [>] 9. Then also the re-
sult is the same, viz Original Knot.
Next begin by [>]1 & 2 as before, &
join 10,4 & 3,5, keeping the C alter-
nate where the new C is made, thus:-
This knot results which
is a new i.k. of 5C,
viz the Timber Hitch,
which may be represented
It is found by experiment that this
same T.H. can be obtained by the
following processes:-
1 [>] 2, 3 [>] 5, 10 [>] 4
1 [>] 3, 2 [>] 8, 10 [>] 9
1 [>]5, 3 [>] 4, 10 [>] 2
1 [>]7, 6 [>] 8, 10 [>] 9
1 [>] 9, 7 [>] 8, 10 [>] 6,
1[>] 8, 4 [>] 6, 5[>] 7, 10 [>] 9.

Reversibility of Links
[Caption: Revolve top of 1 down ^ forwds, 2 steady.]
{Caption: Revolve top of 1 down forwds, 2 steady, revolve top of
3 up.]
And generally when thus regularly ^ placed at
linked first, revolve alternates.
If the links are thus:-
[Caption: simply revolve bottom
of 2 up.]
In all ^ these cases the links can only be
connected only in one way

Hence there are only 4 nos, to which
1 cannot be joined, viz itself 1; 10, re-
storing original B; 4, necessitating
a ring & [symbol for ‘therefore’] 2 cords; & 6, do. In
the case of 4 this is obvious, & it is
easy to demonstrate in the case of
6. For join 1,6.
Then if 3,4 are joined there is a ring.
If none of 2, 3, 4, & 5 are joined to one
another there will be 4 new crossings
& it is only possible to make 2 twists.
Therefore 2 of them must be joined.
It is immaterial which 2 are joined.
Join 2,3. If 4 & 5 are not now joined
there must be 2 new C &
[symbol for therefore] 2 twists. But if 2
twists are made by joining
7 8, 9 10, nothing is left to

As to Iz Reversibles
of the series 2, 8, 14 &c.

20,24, 28 appear in other series, but with
regard to 2 it is certain it cannot be
on Iz except as an unguarded twist. With
regard to 14c I know no it cannot occur
as a twist plait knot on any but 2z. With
regard to 8c it cannot appear as do on any
but 2z or 3z, but it does appear as I5 2b/8c.
Now of what series does this knot form a part, and can it include 14c ?
Values of x, with y = 14, are 2, 7. 3, 8.
We have ^ (z=1) I3 7b/14c, I8 2b/14c (z = 2)
(z = 1) II3 7b/14c (VII8 2b/14c reducible)
There are therefore 3 knots.
Note that I2 2b/2c, I3 2b/4c I5 2b/8c are
reversibles. I don’t feel sure that I4 2b/6c (2z) is
not so. But the only two regulars on 1z are
I3 7b/14c & II3 7b/14c
Examination shows conclusively that I3 7b/14c
is reversible.

join on to 4 & 5. Therefore join 4, 5.
Now if 10 is joined
to 9, 8 must be
joined to 7 & 2 C
are annulled. [sign for therefore]
10 must be joined
to 7 or 8. It can-
not be joined to 7 or a ring will be
formed. [sign for therefore] join 10 to 8 & 7 to 9. But
2 new C. will result. Therefore 1 cannot
be joined to 6 Q.E.D.
It appears from the foregoing that
provided a twist & a l. end are
formed, & the l. end joined to the
remaining end, the T.H. will always
be formed from S2 5b/5c.
Tait says there are only 2 1 cord
knots of 5c, & no doubt S2 5b/5c and 5C T.H. are those knots.

[The following 5 lines of text are crossed out by a bold diagonal line]
It looks as if with r. ks. When n
is even, an odd no. of bights indivisible
by n gives reversibles = none; and when
n is odd, an even ^ any no. of bights indivisible
by n gives reversibles. e.g. I3 7b/14c, I5 2b/8c
But try further.
I find all the r.ks up to I3 7b/14c are
reversible, & hence it may be safely inferred
that all I3 are reversible. The only knots
in I3 that coincides with plait-twists [greek cap sigma] = 3
are is I3 2b/4c. The others are different.
It is possible therefore that all in I5 are
reversible. And in I7. Also that all II3 are
reversible, & so forth. No: II3 4b/8c was not
reversible. But in I5, 2b8c, 3b12c, 4b16c, 6b24c
were all reversible, and probably it may be
safely said that all I5 are so. Further,
probably all with n odd of I are so.

It is possible to represent S2 5b/5c
in long form thus:-
[Caption: Now join1, 2;
4,10; 3,5
The result
is obviously
the 5C. T.H.]
It might be thought that:-
obtained by [symbol resembling>] 1 8, 9 10, 4 6, 5 7
(turned upside down) was
a new k, but it is
clear that by untwisting
the bottom twist, a new
twist is added to the top, & the T.H. results.
Evidently 2 ^ or more unguarded twists at each
end are unstable & can be added on
to the reverse end from either until the
guard is reached. By a guard is meant
a cord passing through a bight above
the twist or below it, thus:-

[Blank except for the following lines in pencil]
And “A1” ii. 100 – 1.
I [wld?] I think [srihoh?] Reef & Granny

Examination of I. K. of 6C.
There is no r. k. on one cord of 6 C ex-
cept the Twist, because the factors of
6 are 1, 2, 3 & 6 giving n as 2, 3,4 & 7,
all of which except 7, the twist, have
measures in common with 6. (the same
applies to knots of 1 & 2 C.)
It is known that a [i?] k. of
T. H. form corresponds to every no. of C.
of 3 & upwards (that of 3 being an.
r. k. S2 3b/3c & that of 4 being S3 2b/4c)
Hence there is a T. H. of 6 C. Further
the Granny, the Reef & the Weaver
(or Bowline, or Crossed Running Knot)
are other knots of 6 C. on one cord. Tait
says there are only 4 such knots.
Problem: To derive these 4 ks. From
r. ks. of 6 C.
From S2 6b/6c the Reef is obtained
by joining 2,4; 1,3; 5,6; 12,7. In the
form of 2 dissimilar overhands on
and endless cord. From do. the Granny

No Text to transcrive on this page

on 1 cord is obtained by joining 1.b;
from S4 [2b/6c] the 6.c T.H arises by a
untwisting 1, 2, 3, 4; 1 & 2 being joined
and 3 & 4 joined through their bight.
from S3 [3b/6c] the Weaver G. can be
made by joining 1,4; 3, 5; 2, 6; of
which 3,4 are untwisted while 2,6
give the new C.
The T.H. is made thus

Compare p19.H for the
1st form of the T.H. here

But can S2 [6b/6c] supply the T.H. & the
Weaver as well as the Rief & Granny?

Page 24 - Only one Undocumented figure in center of page.

It is possible to represent S2 [6b/6c] in
long form thus: --
Proceed on analogous
lines with the plan
on p 19 & join 1,2.
There are now 5c & a new one must
be made. This can be done by
joining 3,5; 12,4 as before, when
the T.H. on 1 cord with 6C. results.
In both cases 1,2; 3,5; & 2,4 were
joined to produce the T.H.

[Therefore] 3 out of 4 possible [__ks??].
have been obtained from S2 [6b/6c]: can
the Weaver be got in some similar
It can be done as follows: Join 1,2;
3,4; 4, 7; 6, 10; 11,12. The stages are as

PAGE 26 - No Text to transcribe.

follows: --
In the above process 8,9 is the only
bight not cut. The operation is [________?] com-
plicated that it is going rather far to
say that the one knot is or can be
derived from the other. On the other
hand the T. H. is fairly derived both
on pp 19 & 25. It would be curious
if in the case of any knot in [__________???]
whether on 1 cord or more, a single
cord T.H. with the [_______??] no. of C as
the knot could always be produced by
joining 1,2; 3,5; 4 & [alpha??]. Try this. Proved
in case of Sz [5b/5c], Sz [6b/6c].
[under first fraction "z=1", under second fraction "z=2"]

Page 28 - Penciled note in lower right corner - "[2?] K regular knot."

No other text on page.

Join 1,2, 3,5; 2,4; in [J"S?"]2
There is still no 5 till we get to S2 [3b/3c]
& this is really a T.H. already. But
if the process is applied, result is the
reflection of the knot. (z=1)
Sz [4b/4c] (z=2) Result, Figure 8 on 1z
(a T.H.)
Sz [5b/5c], 5c T.H. (z=1) p 19.
Sz [6b/6c] (z=2): 6c T.H. (z=1)
Sz [7b/7c] (z=1): 7c T.H. (z=1)
Sz [8b/8c] (z=2): 8c T.H. (z=1)
Therefore this general rule may be
laid down: An irregular knot, the
Timber Hirch of the same no. of C. as
any knot in Sz with more than 2c, can
be obtained on 1 cord, [___________?] other methods by
joining 1,2; x,4; 3,5, and in the case
of Sz [3b/3c] it is merely another form of
the r.k. and in the case of Sz [4b/4c] it is

Page 30 - No text to transcribe.

another form of Sz [2b/4c], an r.k. so
that in these two cases the r.k. & i.k.
overlap. In other words there is no
i.k. of 3 or 4c on 1z, but these
two knots can be displayed as r.k. or
Query: Does the T.H. come within the
definition of a r.k.?
No: I think not, except of course
Sz [3b/3c], and S3 [2b/4c] re-arranged.

Twice the no. of knots in any S is
infinite, & Sz contains knots of all
possible C, [therefore] all the T.Hs. can be
derived from the S2 r.ks., & this disposes
of one class of i.ks., the rule above
laid down applying to all of them.
Query: Can the rule on p29 be
stated in more general terms on the lines
of p17.m? This can be said that
in any Sz knot if a twist adjacent parts
lying in two bights are joined, & 2 other parts

Page 32- No text to transcribe.

joined over a l. end, & the l. end
joined to the remaining free part, that
the rule holds good?
It has been shown so in S2 5b 5c -
Take S2 6b 6c where Z = 2.
Join 10,9; 4,6; if 5
is joined to 11, there still
remain 7 & 8 to be dealt
with. But they must not
be joined or a twist will
Can it be said that if
the twist is made, & the parts on each
side of it joined, & the parts on
each side of these again joined to
make a bight
through the new bt
a T.H. will result? No. There
remain uncut bights.
The rule however may be stated
more generally thus:- Any knot in S2
with 3 or more C. can be converted into
an i.k. in the form of a T.H. ^on 1 cord by cutting

Nothing to be transcribed on this page (34)

any 3 adjacent bights, joining 2 adjacent
parts, each in a different bight (thus
making a twist) joining the part farthest
away from the twist ^ [sorthing?] the half circumference to the part nearest
to it ^ the twist on the same side of it, and
joining the remaining ^ outside parts, ^ on either side of the uncut bight or bights. keeping
the crossings alternate.
(It will be noticed on p13 that
all the 3 bight methods deal with adjacent bights.)
With regard to the formation of the
granny (pp21-3) it will be noticed that
every alternate 2z knot in S2 has op-
posite bights on different cords. Such knots
are 2b 2c, 6b 6c, 10b10c &c. while the knots
3b 3c 4b 4c, 8b 8c, 12b 12c, have them on the same.
Hence it may be laid down as a general
rule that: An i.k. of the granny type
may be formed with the same no. of C. on
1z from every knot in S2 (except 2b 2c) with

Nothing to be transcribed on this page (36)

an even no. of C. not divisible by 4 by ^cutting
joining the nearest sides of ^ two opposite
bights. (In the case of 2b 2c an i. twist
results.) Also in those even C. knots
divisible by 4, an i.k. on (2+1) = 3z
by the same process, and on (2+N)z
for every pair when N pairs of opposite
bights are joined in such knots.
With regard to the formation of the
reef knot, it may also be laid down
that “ An [i?].k. may be formed with
the same no. of C. on 1z. from every
knot in S2 (except 2b 2c) with an even
no.of C. not divisible by 4 by joining
cutting 4 adjacent bights, joining 2
adjacent parts of adjacent bights, joining
an odd to an odd & and even to an even
part (in the same 1 2 circ. as the twist)
& joining the remaining ^2 outside parts on either side
of the uncut bight or bights.
For example, take S2 10b 10c :-

Nothing to be transcribed on this page (38)

so with regard to the weaver, it may
be said, From any S2 knot with even
C. indivisible by 4, an i.k. may be
formed by cutting 5 adjacent bights, say
x – 1 to 8 – 9, & joining 1,2; 3,4; 5,7; 6,9;
& x,8;
It appears probable that for every
additional bight cut, an additional i.k.
could be formed in the case of S2 knots
with even C. indivisible by 4, in which
case it might be said that the no.
of irregular knots capable of being
formed from such knots is B-2, giving
4 for S2 6b 6c, 8 for 10b 10c, 12 for 14b 14c & so on.

Nothing to be transcribed on this page (40)

2 B must be cut to get the first
i.k., & one B must be left ap-
parently uncut.
Tait says there are 8 knots with
7c. Two of these are the r.k. S2 7b 7c
& the 7c. T.H. What are the other 5 on
1z? And why is it that only 1 i.k.
can be formed from S2 5b 5c & 7 i.ks from
7b 7c ?
According to Tait’s series the knots
are 5c 2k
6c 4k
7c 8k
The series 2,4,8 may be continued 16,
32, 64, & so on, or it may be the series
but it seems improbable that this should
be so. If however it is so it smashes
the B-2 theory above.

[Caption: What is this]
Cf V. 203.
If the ring in the knot opposite were
a metal ring & the cord cut at the
twist, this knot would form rather a
good mooring knot of the Larks Head
order, thus:-
B Either end will ^ may
[run?] be strained & I believe
this to be a new
knot ^ hitch. If a were
taken through the
left hand bight it would form an ordinary
running knot. If through the right hand
bight, a figure of 8. If instead of a a
bight were put down, it might serve as a
signal hitch &c. since on pulling a out
& straining b, the hitch runs out.

In ^ any knot of S2 the effect of a twist & join
over it (2 adjacent B cut) is to
reproduce the knot preceding it
in the series.
Try effect on S3 4b 8c
Here an i.k.of 7c on 2z is produced.
a circle interlaced with an [ounce?]
. By cutting the twisted bight
& the circle bight & joining the
following plait is obtained on 1z, 7c.
This cannot be joined up & made into

Nothing to be transcribed on this page (44)

an r.k.without obliterating
Of course this plait could be
obtained directly from S3 4b/8c thus :-
Here on a 1z k. 2 adjacent bights
have been cut & the alternate parts
x & 2, 1 & 3, joined without alternate
crossings, producing an i.k. of
c – 1 crossings = 7 on 1z.
Therefore among knots of 7c are
S2 7b/7c, the T.H; & this i.k. plait.
Presumably every 1z. k in S3
could be similarly made to produce
an i.k. on 1z. with 1c. less than
the number. (S3 5b/10c & S3 7b/14c can.)

Nothing to be transcribed on this page (46)

Question of rings. See p91.
A ring is formed in the following cases:-
2b/2c no 1 & no 4 (exists as such)
3b/3c 1,4.
4b/4c 1,4; - 1,8 (exists)
5b/5c 1,4; 1,8
6b/6c 1,4; 1,8; - 1-12 (exists)
7b/7c 1,4; 1,8; 1,12; & so on from
which it appears that any part with numeral x ^y joined to any other
part with numeral x ^y + (M2 X n-1)
gives a ring.
3b/3c 1 + 12 ^ 6 (exists)
4b/8c 1 + 6
5b/10c 1, 6;

Nothing in ink to be transcribed on this page (48) only some rough pencil jottings of formulae

(S3 cotd)
6b/6c 1,6; - 1,12 (exists)
7b/14c 1,6; 1,12;
8b/16c 1,6; 1,12;
9b/18c 1,6; 1,12; - 1,18 (exists)
10b/20c 1,6; 1,12; 1,18. & so on, from which
it appears that any part with a
numeral x^ y joined to any other part
with numeral x^ y - (M2 X S n) – 1 gives a
May the formula be stated generally
thus:- Rings are formed when x is
joined to (M2 X n) – 1? (M2 being any
multiple of 2)
Take S4 6b/18c . Here 1 + 4, 12 + 5 give
rings. According to the rule, 1 + (M2 X 4)
gives any rings, i.e. 1 + 8, 1 + 16 &c. Here 1 + 8 does
give a ring. 1 + 16 does not exist. The rule
holds good, but is not comprehensive enough.

No text in ink to be transcribed on this page (50) only some rough pencil jottings of formulae under pencilled sketches

1 + 4 points to S2, & it is to be
noted that S4 6b/18c is on 2z -
In S4 9b/27c, 1 + 4 also gives a ring
though this is on 1z.
It looks as though the rule
applying to S2 applied also to S4
and that the rule ought to be stated
Let P be any prime factor of n -
Thengreater than unity. ^ (It may equal n.) Then rings
are formed by joining x to
(M2 X P) – 1.
Test this with S6 11b/55c. A priori
since P may be 2 & 3, one would
expect that
+ 1 + (1 X 2 X 2) – 1 = 1 + 3 yes no
1 + (1 X 2 X 3) – 1 = 1 + 5 no
1 + (2 X 2 X 2) - 1 = 1 + 5 no
[Caption: Rule breaks down] 1 + (2 X 2 X 3) - 1 = 1 + 11 no
+ 1 + (4 X 2 X 2 ^ 16) – 1 = 1 + 15 yes no
1 + (4 X 2 X 3) – 1 = 1 + 23 non existent
1 + (6 X 2 X 2) – 1 =1 + 23 [ditto marks] [ditto marks]
1 + 12 ring = 1 +

No text to be transcribed on this page (52)

In S6 11c/22c the only rings that can
be formed is by adding 1 & ^ 1 + 6, 2, 2
& 13 &c. i.e. x to x + 2x.

Has it anything to do with the
number of times k 2 ^ or K2 will go into C?
Thus there are here 22c, & the added
number is (22/2 + 1) = 12.
But this would not suit S2 where
the C are often odd.
In S6 9b/45 it is obvious that
1 + 6 gives a ring.
It appears that in joining S6
1 + 6, 1 + 12, ^ 1+18, & so on give rings, i.e.
1 + Mn.
The same may be said of S2
when z = 1.
In all the series the part
numbered 1 regains the circumference
at 2n & 4n & 6n counting round from
the right & counting 1 for each part
gone by. The question is how to state

No text to be transcribed on this page (54)


this fact as a rule so as to em-
brace the cases where it emerges
e.g. a 2nd time in between its first
It depends on the no. of bights.
Where 4n is greater than the no.
of bights all is well

Take [I?]4 6b/18c 12 parts
Rings are formed at
1+16(4z)=1+, 4n-12
Take S6 11b/55c
Rings are formed at
non existent.
Therefore the rule may be stated
thus: Rings are formed when any
part numbered x ^y is joined to x ^y-- (M2×x)
-1, or, where M2 exceeds 2B, to

Usually y=1 to begin with, in which
case this practical rule is: Dont
join 1 to (M2×n) or to (M2×n)-2B.
& no M of 2 need be considered which
gives a value for the second part
higher than 2B, because such parts
do not exist.


x^y + (M2×n)-(1+2B).
But in the case of [S?]z with 1z, a
ring is formed by joining 1+2, and
not where z=2. This apparent
exception is met by applying the
2nd part of the rule.
The rule of
B=C/n-1, ∴ 2B=2C/n-1
Hence the Rule of Rings may be
stated thus:-
[Blue vertical line crossed out down to end of paragraph] A ring is formed when any
part numbered x^y is joined to
any other part the number of
which = x^y+(M2×n)-1, or to
any other part the number of which
=x^y+(M2×n)-(1+2C/n-1), and
for 2C/n-1 may be substituted 2B.
Therefore in forming i.ks. with 1z
it is necessary to set out all the
parts existing comprehended in this double
formula as prohibited joins.

Since x=2B=2C/n-1, more simply
thus: Don't join 1 as forbidden on
p 56 nor (when the joined no. is even)
must it equal x+(M2×n)-2 or
(M2×n)-2; nor when odd, to
x+(M2×n) or M2×n
But x is always even: ∴ these last 2 ex-
pressions can never be odd. x+(M2×n ^-2 can
not exist. But (M2×n)-2=M2 ? No. no.


With regard to this rule it is
to be noted that 2 parts cannot be
joined without 2 other parts being
joined. Therefore if only 4 parts are
to be joined (the minimum number) i.e.
if only 2 bights are cut, the joining
of the complements of these parts
would necessitate the rule being
broken in the 2nd join. In such
a case therefore, where only 2
bights are cut & y=1, to the
working rule on p.56 must be
added this:-
The no. to which 1 is joined must
not come within the formula on p.
56, nor must it, when even, be such
that it equals x+(M2×n)-2 or to
x+(M2×n)-(2+2C/n-1); nor, when odd,
to x+(M2×n) or to x+(M2×n)-2C/n-1.

[vertical line]otherwise [vertical line] the [vertical line] primary formula will be
broken in the case of x, where x
=the part numbered 2B.


The complete working rule for
joining 1 where 2B are cut (which
may or may not be adjacent is):-
Don't join 1 to
M2 being any multiple of 2, which it
is no good to consider when any of
these 3 expressions represent a number
equal to or higher than 2B(=2C/n-1).
If such joins are made, rings will
M2×n may be written "M2n"
when the 3 prohibitions are
2Mn or M2n
2(Mn-B) or M2n-2B
2(Mn-1) or M2n-2
Unfortunately the rule breaks down
since S6 12b/60c shows that there is
only 1 exclusion. But it may perhaps
hold when z=1 to begin with.


When z exceeds 1
Subject to the case of z=1, a
rule may be stated thus:-
The no. of ring-giving joins is
B/Z-1 (By ring-giving is meant
that the no. of Z is increased.)
Cases where Z exceeds unity
S2 4b/4c 1 exclusion 4b/2-1=1
viz 2n
S2 6b/6c 2 ex. 6b/2 -1=2, ^4 2n, ^8 4n
& so on.
S3 3b/6c 0 ex 3/3-1=0
S3 6b/12c 1 ex 6/3-1=1, ^6 2n
S3 9b/18c 2 ex 9/3-1=2 ^6 2n, ^12 4n
& so on -
S4 6b/18c 2 ex 6/4 2 -1=2 ^4n-2B no 4, ^2n no 8


Rule of Rings

In any [insk?]. r. k. the number of
ways in which 2 any 2 bights may
be cut and have their cords rejoined
so as to increase the existing value
of Z by forming an additional Z
(or "ring") is limited to ^cannot exceed B/Z -1,
(or C/Z(n-1) -1),
A ring is formed when any part
in such a case numbered y is joined
to any other part the number of
which is y+M2n -1 or y+M2n-(1+2B)
(= y+M2n-(1+2C/n-1)) or y+M2n-3
(as in this case the complementary part
of y's bight would have to break ^come within the
first part of the rule.), provided always
that the number of cases is limited
to B/Z -1, and that the values of
the part-numerals obtained by this
formula are the first (B/Z -1) one or
more in order of arithmetical greatness.

Where 1 = y, exclusions under
this rule are:-
M2n - 2B
M2n - 2
where M is any multiple: but subject
to the rule that there are no
more such values than B/2 - 1, and
further that these are taken in order
of arithmetical magnitude; and
of course, since x = 2B, no value
can equal or exceed 2B.
Test this rule with S5 12b/48c, z = 1.
B/z - 1 = 12/1 - 1 = 11.
M2x= Mx10 = 10, 20, exceeds
10, 20, [30 exceeds 2B]
M2n - 2B = M2 M x 10 - 24
= 6, 16, [26 exceeds 2B]
M2n = 2 = 8, 18, [28 exceeds 2B]
[therefore] the prohibitions are
6, 8, 10, 16, 18, 20

This test is found to confirm the
Next take S6 8b/40c. z = 2
B/z - 1 = 8/2 - 1 = 3
It is obvious that
there are 3 cases = 4, 8, 12.
M2n = M12 = 12, [24 exceeds 2B]
M2n - 2B = 8, [20 exceeds 2B]
M2n - 2 = 10, [22 exceeds 2B]
Of these 12 & 8 are right; 10 is wrong.
Therefore strike out the (M2n - 2)
part of he rule & state the exclusions thus
M2n - M'B
where M & M' stand for any multi-
pliers which may be the same or
The rule thus stated meets the
cases of both S5 12b/48c & S6 8b/40c. Test
it further.

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