Cylindrical Doubles

The basic premise of Cylindrical is that each bell keeps ringing at a constant speed. Of course, if those speeds are the same for every bell, you'd stay in rounds. However, like Plain Hunt, the rule is that odd numbered bells hunt out (i.e. always ringing slowly) and even numbered bells hunt in (i.e. always ringing quickly). So the treble starts off like this:

1 2 3 4 5
  1      
    1    
      1  
        1

Note that in rounds, ignoring the handstroke gap for now, the treble rings every five blows. But here, it rings one place later in each change, i.e. ringing every six blows, slower than it would ring in rounds. So where will it ring next? Six blows later again takes us past the next row and into the lead in the row after that.

1 2 3 4 5
  1      
    1    
      1  
        1
         
1        

And then it carries on at that speed, hunting up again.

1 2 3 4 5
  1      
    1    
      1  
        1
         
1        
  1      
    1    
      1  
        1

Now, what about the even bells? How about the 4th? It starts off normally:

1 2 3 4 5
    4    
  4      
4        

As we know, it's ringing more quickly than it would in rounds, in order to ring one place earlier in each row. It rings every 4 blows. So where does it ring next? 4 blows later would put it in 5ths place, i.e. ringing twice in the same row. And then it, too, carries on hunting in.

1 2 3 4 5
    4    
  4      
4       4
      4  
    4    
  4      

So, does this plan fit together? It seems very odd, some bells missing out rows, while others can ring twice in the same row. So far we've got:

1 2 3 4 5
  1 4    
  4 1    
4     1 4
      4 1
    4    
1 4      
4 1     4
    1 4  
    4 1  
  4     1
4       4
1     4  

The 2, being an even numbered bell, must, like the 4th, ring quickly, hunting in.

1 2 3 4 5
2 1 4   2
  4 1 2  
4   2 1 4
  2   4 1
2   4   2
1 4   2  
4 1 2   4
  2 1 4  
2   4 1 2
  4   2 1
4   2   4
1 2   4  

The 3rd is an odd numbered bell, so it hunts out, ringing slowly, every six blows:

1 2 3 4 5
2 1 4 3 2
  4 1 2 3
4   2 1 4
3 2   4 1
2 3 4   2
1 4 3 2  
4 1 2 3 4
  2 1 4 3
2   4 1 2
3 4   2 1
4 3 2   4
1 2 3 4  

And, finally, the 5th, also being odd-numbered, hunts out too. So in fact it doesn't ring in the first row!

1 2 3 4 5
2 1 4 3 2
5 4 1 2 3
4 5 2 1 4
3 2 5 4 1
2 3 4 5 2
1 4 3 2 5
4 1 2 3 4
5 2 1 4 3
2 5 4 1 2
3 4 5 2 1
4 3 2 5 4
1 2 3 4 5

So, yes, it does all fit together. And, after 12 rows, they're back in rounds. So that's the end of that. We just pause to note that the bells haven't all rung the same number of times. The treble, 3rd and 5th, at their slower rate, have each rung 10 times, while the 2nd and 4th have each rung 15 times, as they are each ringing more quickly.

But, hang on a minute... You've noticed, haven't you?

In the final rounds, the treble, 3rd and 5th must be at backstroke, having rung 10 times, while the other bells are at handstroke, as they rang 15 times. The strokes go strange right from the first row, in which the 2nd rings both at handstroke (in lead) and at backstroke (in 5ths place).

If we colour the handstroke blows as red, we can see more clearly, as well as be able to see some interesting patterns.

1 2 3 4 5
2 1 4 3 2
5 4 1 2 3
4 5 2 1 4
3 2 5 4 1
2 3 4 5 2
1 4 3 2 5
4 1 2 3 4
5 2 1 4 3
2 5 4 1 2
3 4 5 2 1
4 3 2 5 4
1 2 3 4 5

If you carry on, you should find that after 24 changes, we do get all the bells at backstroke.

But, you could argue that it comes round after at the end of the 4th row, where you get all the bells, in order, wrapping into the 5th row...

And why is it called Cylindrical? I think it's because bells don't confine themselves to one row, but go out of the front and back of the rows, joining up with other rows. So, rather than printing the diagram in 2 dimensions, as above, it perhaps makes more sense cut out, wrapped round and taped together, to make a cylinder. You can see the lines hunting up and down are much clearer:

Finally, does it work on 6? Does it work on 7? How long is the plain course this time?