[r-t] FCHs

Alan Reading alan.reading at googlemail.com
Thu Apr 23 11:13:03 UTC 2015

If you transpose a p.b. lead end by any in course FCH (doesn't have to be
tenors together), then 'pretend' what you end up with is a lead head, and
transpose it back so the tenor is at home you end up with an out of course
FCH that must also be present.
D FCH has component 32546
2345678 -> 3254678
8765432 -> 7856432
which is a lead head in the course 2435678 (24356 is an out of course
component of D).

Of course all we are really doing is observing that a course rung backwards
is of the opposite nature (but contains the same rows) to the course rung
forward for major + applying the technique Glint described.

Actually there is another thing you should do, thinking about the example
of E falseness which has 2 in course t.t. components 32465 and 43265.
You won't find 43265 from 32465 using just the technique with the lead
ends, but it does still follow thus:
2436587 -> 3425687 (applying 43265 to the p.b. lead end 2436587)
which ofcourse is a lead end of the course 3246578. So you also need to do
it for the lead ends as well as the lead heads.

I think I am correct in saying that if you start with any FCH and apply
these techniques for all plain bob lead heads and ends then you end up with
all other FCHs that *must* also be present (i.e. all the other ones that
fall under the same letter).
You can see how we are heavily relying on the assumptions that the method
is conventionally symmetric (so the inverse course contains the same rows)
and has plain bob lead heads.


On 23 April 2015 at 11:32, Andrew Rawlinson <andrewrawlinson at yahoo.co.uk>

> Alan Reading: "If you do the same thing for the with the lead ends as
> well as the lead
> heads then you get the out of course components"
> Is that transposing the lead ends by the in course tenors together or out of course? I assumed in course?
> Thanks,
> Andrew.
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