# [r-t] Bobs only Stedman Triples

Robert Bennett rbennett at woosh.co.nz
Tue Jun 26 11:14:28 UTC 2012

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On Tue 26/06/12 7:21 PM , Stephen Beckingham bex280 at hotmail.com sent:

> From:
>
> > Each six of Erin is capable of being rung in 3 versions, 123...
231....
> > and 312...
> >
> > The Q sets apply regardless of what order the front bells are in.
>
> Does it? If one member of a q set is bobbed and another plained, then
the 3rd member cannot be obtained. but as we can start a six from any one
of 3 places, surely the following can apply:
> 1234567 is bobbed, and 1234675 is plained. This means that 1234765 can
not be had AT A SIX END, but surely 2314765 or 3124765 can be, thus making
the q set rule redundant in this case.

_Maybe. The above argument applies to Stedman as well._

_The peal of Stedman Triples without bobs rely on a set of 5 blocks of
two courses each, that are joined by 3 omits affecting 5,6,7._

_The first block starts from rounds 2.3.5.6.7.8.9... the sixes between the
omits at 1 and 4 being irregular._

_This is a completed Q set, even though some of the sixes are irregular._

_In principle, complete Q sets are not required for Stedman Triples
either, but no such bobs only peals have ever been found._

_My point was that it might be better to look for complete Q sets with
irregular sixes in Erin._

__

__

> > The peal of Erin requires 840 sixes.
> >
> > That makes 280 Q-sets. Now have to decide which ones are plained and
with
> > ones are bobbed.
>
> If it were that straightforward I would have thought someone's computer
>
> SJB

Sorry, clearly I meant:
1234567 is bobbed, and 1234675 is plained. This means that 1234756 can
not be had AT A SIX END, but surely 2314756 or 3124756 can be, thus making
the q set rule redundant in this case.

SJB

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