# [r-t] semi-interesting observation

Robin Woolley robin at robinw.org.uk
Sun Oct 30 08:30:27 UTC 2005

```I remarked: "John David asks for a use for these courses. A 7 part a.t.w.
peal, possibly?"

Since no one has take up the challenge, I though I'd have a look myself.

Without loss of generality, select Group A for the example.

The first available calling position after the single at 5ths is, of course,
single at fourths. This leads to a part end 7823456 after 8 leads.

To obtain a 7-part peal of surprise, we need a minimum of 23 leads per part
so a simple solution to this seems to be to add three bobs 'home' at the end
of the part and include one plain lead between the pair of singles. This
also gives a part end cycling 2345678. This would give the 23 leads needed.

Unfortunately, in every group where the singles are not consecutive, this is
(trivially) false.

One solution is obtained by observing that, for group A methods, two
consecutive singles have the same net effect (on the c/o) as two consecutive

One way of doing this is: start Yorkshire, bob, Yorkshire, Rutland, bob and
we are in the course we want.

Applying this, and using the CC collection and Microsiril, I came up with
the following within 30 minutes:

5152 Spliced S Major (5m)
13578264 y -
18654327 y
13586742 r -
16452378 y
12748635 n
18375264 s
15634827 y
14267583 e
12783456 n -

13576248 y
16458327 y
18247635 y
17325864 y
15634782 y
14862573 y
18273456 y -

13526847 y
16457382 y
17842635 n
12385764 y
15634278 n
14768523 y
17823456 n -
7 part
n,r,s,y as usual; e = ebony

Not especially wonderful, I will admit but it's a start.

As far as I'm aware, this technique has not been used in this particular
way, but s5ths/s4ths also has the net effect of three bobs before and this
has been used, for example by MCW Sherwood.

Anyone prepared to offer a 5376 in the standard + 'Nottingham' 8's on this
plan?

Best wishes
Robin

PS - bearing in mind recent discussions, who composed the above?

```