[r-t] Bobs only Stedman Triples

[edward martin]

On 26 June 2012 12:51, Robert Bennett <rbennett at woosh.co.nz> wrote:
>

>  >  But I think that the bobs must still be used in threes.
>
> _I suspect that this is the case. These things are hard to prove. It might
> be interesting to count up the bobs for Colin Wyld's peal and assign the
> bobs and omits into complete, incomplete and pseudo q-sets..

I really hope that you will spend time doing that
As I recall there are 705 bobs (which number IS divisible by 3!)
I'd be interested if you find any "pseudo q-sets" whatever you mean by that
As for being hard to prove... I've told you :

>
>  The Magic Block works in Stedman because in a bobbed block from
>  2314567 there are two places where a bob may be omitted when say the
>  567 would be involved:
>  A = quick six end where the front four bells lie in the order 2314 and
>  B = slow six end where the front four lie in the order 4321.
>  If you always omit bobs at A you successfully link a true block of 60
>  with 75 in 6-7, then another with 56 in 6-7 and finally another with
>  67 in 6-7.
>  If you always omit bobs at B you get the same true touch.
>  If you always omit bobs at both A & B you still get the same thing. In
>  each case the q-set of 3 omits (or of 3 bobs) is completed at A and at
>  B.
>
>  However, if you first omit at A then at B but do nor omit until the
>  next B, you certainly have 3 omits involving 567 but you do NOT have
>  the complete q-set of omits (or of bobs) at both A & at B
>  At A you have 2314567 plained to give 3426175 but at the next A you
>  have 2314756 bobbed to give 3427156, therefore 2314675 cannot occur as
>  a sixend. Similarly at B you have 4132675 plained to give 4127356, the
>  next B 4132756 is also plained to give 4125367 but the row 4132567
>  never occurs as a six end. Both q-sets involving 5-6-7 are therefore
>  incomplete.
>

But you say:

> _Having two calling positions that do the same thing doesn't always help
> much, if the leads are not changeable. For instance, I found with
> methods like Grandsire Triples, Oxford Bob Triples etc. that having two
> bobs at different places still doesn't allow a peal, because the law of Q
> sets still seems to prevent it. Two different bobs at the same place are
> effective, because a composite Q set is then possible. E.g. my peals of
> Grandsire Triples with bobs and two hics (ringing.org)._
>

I despair!