Ian.Fielding at nbt.nhs.uk
Thu Apr 23 09:36:06 UTC 2015
Or you can simply transpose the tenors together FCH by the rows of the plain course and then transposing the tenor back to home (although it won’t give you the incidence)
e.g. for B falseness (24365)
2345678 = 2436578
3527486 = 3254786 = 5372648
5738264 = 5372864 = 3256748 (which is 2365478 backwards)
I spent a large amount of time on this when I was young and keen and proudly showed Roger Baldwin the results only to be told that they were available in the back of composition 502.
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From: ringing-theory [mailto:ringing-theory-bounces at bellringers.net] On Behalf Of Chris02
Sent: 23 April 2015 00:21
To: ringing-theory at bellringers.net
Subject: Re: [r-t] FCHs
Robin Woolley wrote:
> The best I have found was by John Segar. It does not deal with
> split-tenors - but allows FCHs to be extracted with facility."
Andrew Rawlinson wrote:
> Segar's method can actually be used to extract split tenors falseness, as far
> as I know, providing you have the time!
> The only difference is that you look for two bells within the two proving rows that
> have the same amount of bells between them, eg. 7xx8xxx and xx7xx8x and
> then apply the same rules otherwise. In terms of the incidence, you take the
> leads from the top of the columns that the bell on furthest right in the proving row
> (making sure you're looking at them in the same way, for cyclical reasons!) and
> that is the lead that the falseness will be found in.
Or if you don't have the time (or inclination) the Method Master program will extract FCH's. This includes split tenors on eight bells and the incidence. It will also do inter-method falseness.
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