[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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