[r-t] Doubles methods with two hunts and a plain course of 120

[edward martin]

On 21 September 2010 14:47, Don Morrison <dfm at ringing.org> wrote:
> On Tue, Sep 21, 2010 at 6:04 AM, Robert Bennett <rbennett at woosh.co.nz> wrote:
>> There are some interesting doubles methods where there are two hunts, and
>> they (the hunts) cover all possible relative positions  (5 x 4 =20) in each
>> lead. If they do this once in each lead the plain course is 60, but if they
>> do this twice, the plain lead is 120.
>> An example of the shorter sort is:
>>
>> 5.1.3.5.3.5.3.125.3.5.1.5.3.125.3.5.3.5.3.1 (lead head 12453)
>
> I'm confused. By my reckoning this place notation leads to 35421, not
> 12453 as claimed:
> What am I missing?

According to my reckoning the above PN works if instead of both 125s
you have places in 145
ie with the hunt bells being 1-2 there are 20 different positional
relationships; in the above, the basic principle seems to be to work
this out via slow sixes on the back three bells and to cut these sixes
short where 1-2 would repeat positional relationship.

Eddie Martin