Separation Anxiety

Submitted by simon on

How far apart do they get? That's often considered a factor in estimating the difficulty of a particular pair in a method. It explains the experience that ringing the slow work is the most difficult aspect of Kent on the tenors. Following on from Tina's recent article about the features that make a method handbell-friendly, we can look at the handbell-friendliness of each pair in standard methods, as measured by average separation. What do I mean by average separation? The separation of a pair (e.g. 7-8) in a change is the number of bells between them: 0 if they are in adjacent positions, 1 if they are in 2nd and 4th place (say), and so on. Adding up the separation of the pair in every change in the plain course, and then dividing by the length of the course, gives the average separation. Here are the results for the standard eight Surprise Major methods.

Average separation for each pair: Standard 8 Surprise Major

Method 1-2 3-4 5-6 7-8
Cambridge 2.00 2.11 2.27 1.63
Yorkshire 2.00 2.09 2.68 1.23
Lincolnshire 2.00 2.20 2.27 1.54
Rutland 2.00 2.21 2.27 1.52
Pudsey 2.00 2.05 2.57 1.38
Superlative 2.00 1.89 2.75 1.36
London 2.00 1.98 2.59 1.43
Bristol 2.00 2.21 2.82 0.96

The first interesting thing about this table is that the average separation of 1-2 is always 2. The second interesting thing is that for each method, adding up the average separations of all the pairs always gives a total of 8 (apart from rounding inaccuracies). In the comments I explain why this is.

For each method, the tenors stay closest together, then 3-4, then 5-6. This supports the usual perception that the tenors are the easiest inside pair, followed by 3-4, and 5-6 are the most difficult.

The smallest number in the table is 0.96 for the tenors in Bristol, but I'm not going to claim that Bristol on the tenors is easier than Yorkshire. However, if we look at the right-place methods, the smallest average separation for the tenors is in Yorkshire, and this supports the common approach of regarding the tenors to Yorkshire as the best starting point for Surprise Major.

The fact that the total of the average separations, for a given method, is always 8, means that if a method has one pair that stays close together, there must be other pairs that are further apart. For example, the very low 0.96 for 7-8 in Bristol is balanced by the high 2.82 for 5-6.

It's interesting to have some quantitative data to feed into the discussion about handbell-friendliness.

Method Strategies
Major (8 bells)

Comments

Here's why the average separation of the trebles is 2 for all the methods. When the treble is in 1st place, the total separation between it and all the other bells is 0+1+2+3+4+5+6 = 21. Because the 2nd rings all place bells during a plain course, all of these separations come into the calculation. When the treble is in 2nd place, the total separation between it and all the other bells is 15. When the treble is in 3rd place, the total is 11. When the treble is in 4th place, the total is 9. When the treble is between 5th and 8th place, the situation is symmetrical, so we get the same totals again. Putting all of this together, the total separation of the trebles in a whole course is 2*(21+15+11+9) = 112, multiplied by 4 because the treble rings in each position 4 times during a lead. So the total is 448. To get the average we divide by the length of a course, which is 224, giving the result 2 as seen in the table.

Submitted by simon on

Here's why the total of the average separation for each pair is always 8. Consider a change that occurs in one lead of a method, for example 13572468 (think of it as the treble's handstroke snap in Superlative). The separations between 1 and each other bell are achieved by 1-2 at the handstroke snap during the whole plain course, because 2 will be each place bell in turn. The separations between 3 & 4, 4 & 8, 5 & 8, 2 & 5, 2 & 6, 3 & 7 and 6 & 7 are achieved by 3-4 at the handstroke snap during the whole plain course, because those are the combinations of place bells that 3-4 ring. Similarly there are 7 pairs of bells whose separations are achieved by 5-6, and 7 pairs of bells whose separations are achieved by 7-8. The total of these 28 separations is the same as the total separation between 1st place and all other places, between 2nd place and all other places above it, between 3rd place and all other places above it, and so on. The total separation between 1st place and all other places is 1+2+3+4+5+6 = 21. The total between 2nd place and all other places above it is 1+2+3+4+5 = 15. (These are actually the so-called triangular numbers). So the total of the 28 separations is 21+15+10+6+3+1 = 56. The same calculation applies to each of the 32 rows in a lead, so we need 56 * 32, and then we are going to divide by the length of the course (224) to get the total of the average separations for each pair. The result is 8.

Submitted by simon on