[r-t] Stedmonster triples

[Andrew Johnson]

> From: Alexander E Holroyd
> 
> This is the analogue of Stedman, but with alternate quick and slow 10s 
> and 4-pull dodges in 67:
> 
> 
http://www.boojum.org.uk/cgi-bin/line.pl?bells=7&pn=7.1.5.1.5.1.5.1.5.1.7.5.1.5.1.5.1.5.1.5&title=Stedmonster%20Triples&style=1&place-bellsx=on&rule=1%2C10&layout=1&action.x=1 

>
> The name "Stedmonster" seems to be in currency in certain places where 
> it is rung.
> 
> I think the method has been discussed before, but so far as I know it 
> was open whether an extent was possible.  It turns out that it is not so 

> hard to find one, although it would not be exactly easy to ring.

This method was one of the ideas I discussed in the Christmas issue of 
The Ringing World, issue 5669/5670, p.1246 
(subscriber link: https://bb.ringingworld.co.uk/issues/2019/1246) 

https://lists.ringingworld.co.uk/pipermail/ringing-theory_bellringers.org/2020-January/027060.html

I pointed that in Campanalogia (1677), p.168, Fabian Stedman says:‘ ‘Tis 
plainly demonstrable, that [t]he Principle upon five may go 420 triples 
upon seven, which is a twelfth part; 840 which is a sixth part; or 1260 
which is a fourth part of the whole, and the utmost period of triple 
changes. And then by making four extreams it may go 5040, the complete 
peal.’

From the article:

"Hunting on five, not three

Another idea is to vary the six structure such as plain hunt and reverse 
plain hunt to be on five, not three. This gives sections of ten rows 
instead of six. 
&5.1.5.1.7.5.1.5.1.5,1 
The line is like an extended Stedman front work with hunting and points on 
the front 5, with quadruple dodges at the back. This gives a 140 change 
plain course, so to extend to 420 needs a bob. Changing a 7 to a 5 will 
not work, as it is next to another 5. Changing a 1 to a 3, or a 5 to a 3 
or 5 to a 7 could work. For symmetry these should be in the middle of a 
slow or quick ten. It is quite easy to generate a 420, 840 or 1260 with 
calls changing a 5 to a 7 in the fourth or seventh slow ten of a course, 
e.g. the following three parts: 420: 7 or  820: 7;4,7 or 1260: 7;7;4,7. I 
cannot see an easy way of generating a 5040 with four extremes, and a peal 
in whole courses is unlikely as I have only found 19 mutual true courses, 
but 36 would be required. As a curiosity this extends to higher numbers, 
either with more dodging pairs behind, or with hunting on 7 or 9."

I didn't consider the bob you chose "- = 3 for 7" as I was trying to get a 
simple 420, 840 or 1260. Can you find a 5040 with 4 extremes?

Andrew Johnson








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