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Henry Bushby Transcription Pages

VM533B94v6_143.jpg

Revision as of Sep 18, 2014, 1:31:44 PM
created by GDavis
Revision as of Sep 18, 2014, 1:32:32 PM
edited by GDavis
Line 1: Line 1:
<p style="text-align:right;">141</p><br>  
+
<p style="text-align:right;">143</p> <br>
other side of 4 (in 4's division) and <br>
+
has just been placed in the vacancy <br>
"over" must be so interpreted through- <br>
+
as the left of a new pair con- <br>
out. As long as the movements <br>
+
sisting of itself & the adjacent right <br>
are consistent it does not matter. <br>
+
hand cord. So consider <i>all the old pairs</i> <s>the rest as</s> <br>
The 2nd <s>figure</s> <i>cord</i> (or one going under) <br>
+
<i>split up into</i> new pairs. Now take the left <br>
always goes immediately to the vacancy <br>
+
of each of the new pairs over,  <br>
left by the 1st (or one going over.) <br>
+
till the (N/2)th of the new pairs <br>
The scheme for 4 cords is <br>
+
is reached. Then treat the <s>right</s>  <br>
[to right of diagram] <br>
+
one most to the right of the <br>
(1)    2  1 <br>
+
(N/2)th new pair (when the crossing <br>
(2)   4  3 <br>
+
process is finished) as the left <br>
(3)   3  2 <br>
+
of a new set of pairs again, & so <br>
(4)    1  4 & repeat from (1) <br>
+
consider the right of each pair over <br>
[below figure] <br>
+
the left & so go on, (N/2) times <i>be-</i><s>al</s> <br>
In general, with N cords (even) <br>
+
<i>tween each [??]. </i><s>ternately con??? each (N/2)]th <br>
there are N/2 pairs: (Bung moves <u>with</u> watch) <br>
+
pair as it were with half of <br>
With these pairs taken consecutively, cross <br>
+
the next & so forming new pairs</s> <br>
the right hand cord of each pair over <br>
+
the left hand cord of the same and <br>
+
the left hand cord into the other's vacancy. <br>
+
This having been done to the (N/2)th <br>
+
pair, treat the cord of this pair which <br>
+

Revision as of Sep 18, 2014, 1:32:32 PM

143


has just been placed in the vacancy
as the left of a new pair con-
sisting of itself & the adjacent right
hand cord. So consider all the old pairs the rest as
split up into new pairs. Now take the left
of each of the new pairs over,
till the (N/2)th of the new pairs
is reached. Then treat the right
one most to the right of the
(N/2)th new pair (when the crossing
process is finished) as the left
of a new set of pairs again, & so
consider the right of each pair over
the left & so go on, (N/2) times be-al
tween each [??]. ternately con??? each (N/2)]th
pair as it were with half of
the next & so forming new pairs