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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
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
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