[r-t] Brunel

[Mike Ovenden]

With the lead-ends placed as given, each lead is one whole six and two half
sixes.  When you look at what's needed to complete the half sixes, it forces
a set of leads that are equivalent to a plain course of Stedman Triples.
Therefore the extent would have to be partitionable into a true set of plain
courses of Stedman Triples.  There's something on change-ringers from way
back which says that's impossible - is that correct?

Mike

----- Original Message ----- 
From: "Richard Smith" <richard at ex-parrot.com>
To: <ringing-theory at bellringers.net>
Sent: Monday, October 03, 2005 6:47 PM
Subject: Re: [r-t] Brunel


Rebecca Cox wrote:

> I have some straighforward compositions of Caters and
> Cinques but would be interested to know whether a peal of
> the triples is possible without using calls which make it
> effectively Stedman.

It appears that you can't get 420 mutually-true leads of the
triples method.  (The most I can find immediately is 367,
though it may well be possible to get slightly more than
this.)  This means that it will not be possible to get an
extent with just "lead-end" calls (where the lead-end is the
7 p.n. in the middle of a slow six).

RAS

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