[r-t] Brunel

[Richard Smith]

Mike Ovenden wrote:

> With the lead-ends placed as given, each lead is one whole six and two half
> sixes.  When you look at what's needed to complete the half sixes, it forces
> a set of leads that are equivalent to a plain course of Stedman Triples.
> Therefore the extent would have to be partitionable into a true set of plain
> courses of Stedman Triples.  There's something on change-ringers from way
> back which says that's impossible - is that correct?

Yeah.  The extent doesn't partition into a true set of plain
courses, which is why you never see compositions of Stedman
Triples based around them.  Bob courses (of Stedman) work,
but obviously have two calls per twelve row "lead".  Hudson
blocks and the older Tebbs blocks also fully partition the
extent, but they also have bobs in consecutive sixes.

RAS