[r-t] Bobs only Stedman Triples

[Robert Bennett]

 On Tue 26/06/12 8:09 PM , edward martin edward.w.martin at gmail.com sent:

On 26 June 2012 06:37, Robert Bennett  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

_Look at the 360 of Bob Minor. WHW,WHW,WHW. There are 3 bobbed Q sets:_ 

_e.g. 132546/123564 W,  152364/125346 H,  153246/135264 W_ 

_The second one is rung in the reverse order._ 

_This use of Q-sets is not my invention: it appears I think in the CC
Handbook of Composition? _ 

 >
 > 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? 

_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.._ 

 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. 

_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)._ 

 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.

_Tall order indeed! But I still think it would be worth looking._ 

__  

 Cheers
 Eddie

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