I must confess, which I think I have written before, that I have never been much of a Stedman conductor. Our recent campaign of ringing Stedman Triples regularly has given me the opportunity and motivation to try to work out how I can see more of what's going on. One approach which might become easier with practice is to watch how the bells are working together on the front, in the same way that in the tower we get used to working with two bells, one of which leaves and is replaced, and so on. Tina and Julia seem to be able to do this more easily than I can. But what I would really like is a view of the "big picture", which is where the coursing order comes in for treble-dominated methods. For non-Stedman experts such as myself, it's common to say (or assume) that Stedman doesn't have a coursing order, because the order in which bells reach the back is broken up by the fact that they alternately go in quick and slow, thus spending different periods of time away from the back. There is an order in which the bells reach the back throughout the whole course, which is a sequence of 14 numbers in which each bells occurs twice. David Parsons' little book on conducting Stedman says that you have to know this sequence. But then the difficulty is that although a bob affects three consecutive bells as usual, it also affects their other position in the overall sequence. And I don't think I can remember and transpose a 14-bell sequence.
So I started to look for something else. In our quarter this week I was having some success with seeing bells leading in turn, especially in the quick sixes. This led me to consider the order in which the bells go in quick, thinking that the quick bell leads in the middle of a quick six and could be a useful anchor:
1 6 5 2 3 4 7
Not only is this the order in which the bells go in quick, it's the order in which they go in slow or indeed do any fixed piece of work: first half turn, last whole turn, whatever. In fact this order is the reverse of the place bell order, thinking of place bell starts every 12 changes, i.e. with a block of two sixes as the repeating unit. We can see this in the diagram on the right (produced by Martin Bright's method printer at boojum.org.uk).
Another key point about this sequence, which we might as well call the coursing order, is that consecutive bells are in the 3-4 position relative to each other. This means that consecutive bells always do scissors work in 4-5 and 6-7 and cross in 5-6, and this is true whether they are going in quick or slow. They also dodge together in 4-5 in a slow six, between one and the other going in quick.
The only thing we don't immediately get from this coursing order is the order in which bells appear at the back. But this can be recovered by interleaving two copies of the coursing order, because bells at the back are alternately lying behind after quick work and lying behind after slow work, and each of those lying behinds is a piece of work that the bells reach in coursing order.
Knowing that the 7th lies behind first, followed by the 5th, we get the whole sequence of bells at the back as follows:
7 5 1 2 6 3 5 4 2 7 3 1 4 6
What might be more useful is the relationship between the quick bell and other bells on the front. For example, after the quick bell leads, the next full lead is the beginning of the last whole turn. This is done by the bell three positions after the quick bell, in the coursing order, and then the bell coming in for its last half turn is the next one after that in the coursing order. It remains to be seen how much of that I will be able to see while ringing.
If there's a coursing order to work with, a crucial question is whether it can be transposed in a reasonable way. Let's stick to twin-bob compositions, which means that the 7th is unaffected and we can leave it at the end of the coursing order, leaving just six bells to work with.
What happens is that a pair of bobs affects four bells: the two bells that dodge together at the back, and the two bells that make the bobs underneath them. For calls at S, H and L, these four bells are consecutive in the coursing order. The bells that make the bobs each move one position earlier, and the bells at the back each move one position later. Equivalently, the first pair swap and the last pair swap:
ABCD -> BADC
Here's how it works for S, H and L, writing a general coursing order as ABCDEF (omitting 7 at the end).
|S||ABCDEF -> BADCEF|
|H||ABCDEF -> ACBEDF|
|L||ABCDEF -> ABDCFE|
That's all straightforward, but what about Q? There isn't another set of four consecutive bells to affect. Actually Q is slightly different because the first bob is after a slow six, not after a quick six as in the other cases. Instead of being consecutive, the bells affected are the pairs before and after the 7th, or equivalently the first and last pairs in the coursing order:
|Q||ABCDEF -> BACDFE|
Let's check the transpositions for a Thurstans block to make sure it all works properly.
Thinking of the initial coursing order 165234 in relation to the first part head of 123456, we can see that the final coursing order 261345 has the same relationship to the second part head of 234516.
For a simpler example, we can see how the one-course touch S L Q works:
165234 -> 612534 -> 615243 -> 165234
Somehow I find this remarkable, although I suppose it's no more surprising than M B W as a one-course touch of Bristol.
We can also see that the use of H is crucial in general, because otherwise the bells stay as three pairs (16, 52, 34) that can only swap internally, limiting us to 4 courses (not 8 = 2³ because the coursing order remains in-course).
I'm sure this is all well-known to the Stedman experts, but it's been fun to work it out. Whether I can make use of it for conducting remains to be seen.