I had been using an interesting technique to splice Grandsire and St Clement's Triples together, without needing to stick to the usual hunt bell rule (click here and scroll down to see more about this project.) I realised that a similar technique could be used to splice together two Plain Major methods - Plain Bob and Hereward - that don't easily lend themselves to true touches.

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Hereward Bob Major is simply Double Norwich with a 2nds place lead end. It is difficult to compose true touches of spliced Hereward and Bob Major. Here is the first lead of Hereward divided into sections, each section running false against a particular course of Plain Bob, as shown:

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First lead of Hereward                          Runs false against this course of Plain Bob

12345678                                                 No falseness here

21436587

24135678                                                 False against 14235678

42316587

24361578                                                 False against 18576342

42635187

24365817                                                 No falseness here

42638571

46283751

64827315

46287135                                                 False against 13578264

64821753

46812735                                                 False against 14862735

64187253

61482735                                                 No falseness here

16847253

16482735

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The first false course head is 14235678. This means that if you rang a full plain course of Hereward, called a bob at the end of the course to produce 14235678, and then changed methods to ring a whole course of Plain Bob, the touch would be false. But the same thing would happen if we called a bob at any lead end in the first course of Hereward and then changed to Plain Bob - the touch would be false. As shown on the table, another false course is 13578264 - but that's the course you get to if you called a bob after six leads of Hereward.

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So this means that you are only ever one bob away from falseness - that is how the falseness between these two methods works. By symmetry, the courses of Bob Major which are only one bob away from getting back into the plain course are also false against the plain course of Hereward. These are courses 18576342 and 14862735 on the table.

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How on Earth can you splice these two methods together if calling a bob will always produce falseness between them? The usual solution is that all the in-course courses are rung to Bob Major and all the out-of-course courses are rung to Hereward (or vice versa).

In other words, the courses that are an even number of bell-swaps away from the plain course (such as 42356, 34256, 23645) are rung to one method, and the the courses that are an odd number of bell-swaps away from the plain course (such as 32456, 24356, 23546, 36425) are rung to the other. This will definitely be true if the tenors are kept together.

 

But the point of my project - as had been the point of my previous Triples project - was to experiment using Partner courses instead, so that a mixture of in-course and out-of-course courses could be rung to both methods. I wanted to see how many true courses I could attain, and what the longest length was.

 

What are the Partner Courses?

Every course has a partner, the two courses sharing a special relationship. The plain course, with its partner next to it, is:

 

12345678      12436578

 

The relationship between the plain course and its partner is that bells 1,2,3,4,5,6 ring exactly the same changes, but in reverse order, whilst bells 7,8 ring the same way round in both courses. This means that none of the changes are repeated, but it also means that bells 1 - 6 are in constant harmony with each other across both courses. For example, courses 12345678 and 12453678 are NOT partners with each other, so the difference between the two courses does not remain consistent in the way that it does for the married courses.

 

The relationship between all courses and partner courses is always the same as above, so the following are all partner couplets:

 

16452378      16543278

13265478      13624578

15246378      15423678

 

Because bells 1 - 6 ring the same combinations in a course and its partner course, this means that the difference between the two courses is always a 2-part transposition or transfigure, and never a 3-part difference. For example, the partner of 12345678 is 12436578, which is a 2-part transposition; pairs 3,4 and 5,6 are swapped over.

 

Because the difference between a course and its partner is always a 2-part transposition and never a 3-part, this means that a course of Bob Major will run true against its partner course being rung to Hereward. As the table shows, the falseness between the two methods is only ever via 3-part transpositions (the three bells that are affected by a bob.) The course 12345678 rung to Hereward is mutually true to the course 12436578 rung to Plain Bob (and v.v)

 

This means that we can use partner courses to build up a true touch of Spliced.

 

How many courses did I find?

I tend to work intuitively rather than deductively, so the best I could manage was 60 tenors-together courses out of 120. These form a five-part using Middleton's part ends (a group which works well for the falseness in question), with 12 courses per part. To get all 12 courses without falseness, I had to omit two of the partner courses. This entails that there are 7 courses of Hereward and 5 of Plain Bob in each course. Is it possible to get more than 60 tenors-together courses? Do let me know if you have an answer (email address above.) Interestingly, 60 is exactly half of all the 120 tenors-together courses, just as the number of true courses of spliced Grandsire and St Clement's Triples using this partner technique was also exactly half of the full number.

 

Here is a table of the mutually true courses (with the part end at the top of each section, and the imbalanced partner courses in brackets):

 

Courses of Hereward                    Courses of Plain Bob

  35264                                               32546

  32564                                               35246

  62534                                               65243

(65234                                              (25643)

  26534                                               45236

  52634)

  42563

 

  56342                                               53624

  53642                                               56324

  43652                                               46325

(46352                                              (36425)

  34652                                               26354

  63452)

  23645

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  64523                                               65432

  65423                                               64532

  25463                                               24536

(24563                                              (54236)

  52463                                               34562

  45263)

  35426

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  42635                                               46253

  46235                                               42653

  36245                                               32654

(32645                                              (62354)

  63245                                               52643

  26345)

  56234

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  23456                                               24365

  24356                                               23465

  54326                                               53462

(53426                                              (43562)

  45326                                               63425

  34526)

  64352

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Obviously the two methods can be swapped round, and the above table will still be true. The next task was to join the courses together. Lots of singles were necessary, in view of the problem with bobs as outlined above. Without much difficulty I managed to join all 60 courses together and produced this 6720. I used an irregular q-set to make the composition an exact five-part, a technique that has been explained in several other places on this website. This is why there are quite a lot of backstroke 87s.

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Split tenor courses

Splitting the tenors opens up a whole new can of worms. The partner relation is essentially the same (with bells 1-6 ringing backwards, with 7,8 the same way round), but it is still laborious working out how false three-part transpositions can creep in between the split tenor courses against the tenors-together courses, and the split tenor courses against the other types of split tenor courses.

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I managed to work out without too much difficulty a set of courses with the 7th in 5ths place (or 4ths if rung backwards) at the course ends, which ran true to my set of 60 tenors-together courses. The other type of split tenor courses with the 7th in 3rds place (or 2nds if rung backwards) was quite a head scratcher, so I left those for another time.

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I then did some work to join the 5ths courses to the tenors-together courses, and produced this 13440. This is where I left the project. It will be interesting to return at some point and see if the 3rds courses can be added in to make a 20160.

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Why these two methods?

The falseness between Plain Bob and Hereward is just the right type, and the right amount of it, to make this experiment the most effective. Having that 42356, 34256 falseness - but no more - is enough to provide a real challenge, but can still be closed into a logical group. Plain Bob and Double Mancroft Bob would have thrown in more difficulties (including out-of-course falseness), and Plain Bob and St Clement's would have been a nightmare!