8 clubs 1-count (ultimates)
If we apply a basic mathematics theory, we can guess that for an 8 clubs ultimate,
both jugglers throw doubles, i.e. <4p | 4p>, just like in the following
scheme. Nevertheless there is a huge problem with collisions (every red circle).
Hence it is quite hard to do it this way (if you really wish to do it, you'll
have to define a "row" for each juggler and for each hand, and be
really precise -see page collisions).
NB : JoePass! is not concern about these collision problems.
8 ultimate, not so believable : <4p
8 ultimate, possible versions
A doable solution (but hard anyway) consists in having a juggler throwing very high (triples), and the other very low (singles). The juggler who starts with singles has 5 clubs (at least on the following scheme).
8 clubs ultimate, doable but hard version
: <5p | 3p>
A much easier solution is to have a juggler throwing floating doubles and the other one throwing floating singles, just under the doubles. On the following scheme, both jugglers starts with 4 clubs each.
8 clubs ultimate, "easier" version:
<4.5p | 3.5p>
Note : In the following schemes, I represented desynchronized rhythms (LH, then RH, LH...). Since in 8 ultimate you juggle 4 clubs separately on each side of the pattern (red lines & orange lines on one side, blue lines & black lines on the other), you can have the matching synchronized version, with RH and LH throws at the same time, in which it's easier to see what's going on in this pattern (even if it's harder to achieve).
There is also a last version, completely different from the firsts (cf. things to know about 8 clubs theory ). It consists in crossing every pass (RH->RH et LH->LH, see also avoid collisions ). You'll have to be desynchronized. The height of the passes is doubles' height, but you can also throw higher singles. Obviously it's much harder with singles.
8 clubs, crossed ultimate
: <4p | 4p>