[r-t] Bobs only Stedman Triples

edward martin edward.w.martin at gmail.com
Tue Jun 26 08:09:08 UTC 2012

```On 26 June 2012 06:37, Robert Bennett <rbennett at woosh.co.nz> wrote:
>
>
> By a Q-set, I mean 3 bobs affecting the same 3 bells behind, or 3 omits
> affecting the same 3 bells.
>
> 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.

I'm afraid that that is NOT what most people understand to be a q-set
A q-set of calls at Home in Plain Bob Major involving the 2-3-4
requires that the other bells lie in the same order Thus:
12436587 either plain for 12345678 or bob for 14235678
If 12436587 bob for 14235678 then 14326587 cannot be plained without
running false

>
> 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.
>
> As you say, since there is no set of bob blocks true and complete, there
> needs to be a number of fragments joined together.
>  But I think that the bobs must still be used in threes.

Why?
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.

For this to work in Erin requires n number of compound blocks of plain
and bobbed sixes, equally structured not only to make provision for
all 5040 but to allow for at least two opportunities per block to say
plain an otherwise bobbed sixend involving the same 3 bells.

Cheers
Eddie

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