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Henry Bushby Transcription pages

VM533B94v4-p017.jpg

Revision as of Jun 18, 2014, 9:31:12 AM
created by DPepper
Revision as of Jun 18, 2014, 9:32:27 AM
edited by DPepper
Line 1: Line 1:
17<br>join on to 4 & 5. Therefore join 4, 5. <br>Now if 10 is joined<br>to 9, 8 must be<br>joined to 7 & 2 C<br>are annulled. [sign for therefore]<br>10 must be joined<br>to 7 or 8. It can-<br>not be joined to 7 or a ring will be<br>formed. [sign for therefore] join 10 to 8 & 7 to 9. But<br>2 new C. will result. Therefore 1 cannot<br>be joined to 6 Q.E.D.<br>It appears from the foregoing that<br>provided a twist & a l. end are<br>formed, & the l. end joined to the<br>remaining end, the T.H. will always<br>be formed from S<sub>2</sub> <sup>5b</sup>/<sub>5c</sub>.<br><u>Tait</u>says there are only 2 1 cord<br>knots of 5c, & no doubt S<sub>2</sub> <sup>5b</sup>/<sub>5c</sub> and 5C T.H. are those knots.<br>
+
17<br>join on to 4 & 5. Therefore join 4, 5. <br>Now if 10 is joined<br>to 9, 8 must be<br>joined to 7 & 2 C<br>are annulled. [sign for therefore]<br>10 must be joined<br>to 7 or 8. It can-<br>not be joined to 7 or a ring will be<br>formed. [sign for therefore] join 10 to 8 & 7 to 9. But<br>2 new C. will result. Therefore 1 cannot<br>be joined to 6 Q.E.D.<br>It appears from the foregoing that<br>provided a twist & a l. end are<br>formed, & the l. end joined to the<br>remaining end, the T.H. will always<br>be formed from S<sub>2</sub> <sup>5b</sup>/<sub>5c</sub>.<br><u>Tait</u> says there are only 2 1 cord<br>knots of 5c, & no doubt S<sub>2</sub> <sup>5b</sup>/<sub>5c</sub> and 5C T.H. are those knots.<br>

Revision as of Jun 18, 2014, 9:32:27 AM

17
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.