[r-t] Double Cambridge cyclic bob Minor

Alexander Holroyd holroyd at math.ubc.ca
Thu May 19 02:21:49 UTC 2011


With singles at every lead-end and half-lead it becomes easy to get a 720, 
similarly to the one below.  But I doubt if that is what you are after!

720 Unnamed Rotationally Symmetric Minor Method

   23456
   -----
s 53246 1
* 46325 2
s 35264 3
   -----
10 part, s at * in parts 5,10
single = 234

Method: -4-3-23-1-1-45, le 135264


On Wed, 18 May 2011, Alexander Holroyd wrote:

> If I've got it right, one can get a maximum of 40 mutually true leads, giving 
> a maximum possible length of 480 in whole leads (one would still have to join 
> them all together to achieve even this):
>
> 23465 24356 24653 25346 25634 26354 32456 32546 32645 34562 34625 34652 35642 
> 36254 42365 42536 43265 43562 45263 45362 46253 46325 52436 52634 53246 53462 
> 54236 54623 56234 56324 56432 62453 62534 62543 63245 63425 63524 64523 65342 
> 65423
>
> Ander
>
>
> On Wed, 18 May 2011, Richard Smith wrote:
>
>> First work out the false lead heads for the method.  For Double Cambridge 
>> they are:
>>
>>  123456
>>  135426
>>  136452
>>  152436
>>  162453
>>  164352
>
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