[r-t] Grandsire Triples

[edward martin]

I'm afraid that at first  glance, I don't see how your 2520 could have
a complementary 2520
If the halves are to be equal (on paper at least, ie before ever we
try to figure out how to join them up), it seems to me that if they
are to be equal in structure, each part of a 2520 would have to have a
total of six leads: one isolated lead  with 2 in the hunt  (between
two bobs) and then 5 members of another course ( however you can fit
them in)  . Yours has 7 leads per part ie 3 leads from one course,
then 3 leads from another plus the one lead for the PE. In the second
half, how are we going to get the two missing leads from the two
courses and have the whole thing be equal in structure?

Cheers
Eddie Martin


2009/9/19 Richard Pullin <grandsirerich at googlemail.com>:
> Hi all,
> On holiday I composed this 2520 of Grandsire Triples, a 15-part.
> Obviously it needs two Holt's Singles half a peal apart to join it to an
> equal and opposite half peal.
>
> Will the 2nd half of the peal be different and symmetrical in some way?
> (e.g: Holt's 10- part)
>
> Or
>
> Will the 2nd half of the peal be the same but started in a different place?
> (e.g Higgin's 10 part)
>
> It is beyond my capability to do this myself- I would really appreciate
> someone's help.
>
> Here is the 2520:
>
> 752634 1
> 347265 2
> 763452 3
> 247563 1
> 762435 3
> 357246 2
> 623457 1
> 576342 2
> 645723 3
> 376245 1
> 643752 3
> 526374 2
> 435726 1
> 264573 2
> 472635 3
> 564372 1
> 725436 2
> 647325 1
> 256734 2
> 5 part
>
> Thanks in advance, from rp