- Yeh, so the 9th pass is the same as the first. This correlates with my

experience in the run-around side, as does your "B B B A A A A B" list at

the bottom, which shows the first passer doing exclusively self throws for

what feels like a long time after an initial burst. Yew Is A Genius.

Learning this with a pair of clubs in the passing slots which differ from

the other clubs (which always are self throws) seems necessary to work out

the cycle. This also means that you and Madison got about 1.5 times around

the cycle at the end of last night, when I counted a total of 12 three-count

passes as I was posting.

So the passing pattern is time-symetrical, right? An ongoing cycle would be

"B B B [A A A A] [B B B B} [A A A A] . . ."? Hmm. Some patterns in this

group are not time-symetrical in this fashion -- think, for example, of a

4-count with one side being two people half-juggling. There, if I am acting

the left hand, I never pass a club to anyone but my near-side partner. I

wonder if any time-asymetrical run-around patterns exist. I kind of think

not, but I'm not sure I could prove it.

Man, I wonder if anyone else in the world has done this stuff. We better

hustle up another person and try two facing 3-count run-arounds.

-- CHS

> -----Original Message-----

> From: Carl Raffa [SMTP:craffa@...]

> Sent: Friday, May 11, 2001 10:18 AM

> To: Charles Shapiro; mad@...

> Subject: 3 count run-around

>

> The 3 count run-around pattern repeats after 8 passes. The easiest way

> to analyze it is by looking at the throws. The throws in the basic

> run-around look like this:

>

> Start with A in front and B in back:

>

> B throws right to B's left

> A throws left to A's right

> A throws right to B's left

> B throws left to B's right

>

> B is now in front and A is in back:

>

> A throws right to A's left

> B throws left to B's right

> B throws right to A's left

> A throws left to A's right

>

> So, the basic run-around repeats after 8 throws.

>

> We can use the least common multiple to determine how a composite

> pattern repeats. With a 4 count, the solution is simple. The least

> common multiple of 4 and 8 is 8, so the whole pattern repeats every 8

> throws. We have to look at the run-around pattern to see who gets to

> pass. If you look at the 1st and 5th throws, you will see that B gets

> the first pass and A gets the second pass.

>

> Now, the 3 count is a bit more complex. The least common multiple of 3

> and 8 is 24! So, the whole pattern repeats every 24 throws. Again, the

> look at the run-around pattern to see who gets to pass. The pass

> sequence is B, B, B, A, A, A, A, B.

>

> Carl - Charles Shapiro wrote:
> So the passing pattern is time-symetrical, right? An ongoing cycle would be

You can also start the pattern anywhere in the sequence. Our usual fast

> "B B B [A A A A] [B B B B} [A A A A] . . ."?

start has been with B behind, slightly to the right, with a club in the

right hand. The first throw is B's right hand throw, which is a pass,

but leads to the BBBAAAAB pattern. Last night, we were able to slow

start directly into AAAABBBB by having B's first throw be a self throw,

and having A's first throw be a left hand pass. To fast start directly

into AAAABBBB we could start with B behind, slightly to the left, with a

club in the left hand, and have A throw the club in his left hand as the

first pass. In either case, though this requires A to start with a left

hand pass. I can't see a way to fast start into the AAAABBBB pattern

without a left hand pass. Unless we reverse the direction of the

run-around, of course.

As for a non-symmetrical pattern, an 8 count, or any multiple of an 8

count, would result in only one person getting to pass. But who wants

to wait around for an 8 count? A 7 count repeats every 56 throws with

pass pattern BABBABAA. A 6 count repeats every 24 throws with pass

pattern BBAA. A 5 count repeats every 40 throws with pass pattern

BBAAAABB. A 4 count repeats every 8 throws with pass pattern BA. A 3

count repeats every 24 throws with pass pattern BBBAAAAB. A 2 count

repeats every 8 throws with pass pattern BAAB. A 1 count repeats every

8 throws with pass pattern BAABABBA. That would be fun to try.

The end result is that anything other than a multiple of an 8 count will

always have an equal number of throws for each person. To get an

unequal number of passes, you'd have to have a mixed count. For

example, a 3 count followed by a 2 count followed by a 3 count would

only let B throw (assuming our standard start). You may be able to

accomplish the same thing by reversing the direction of the run-around,

but I haven't worked it out yet.

Carl

Hmm. Some patterns in this> group are not time-symetrical in this fashion -- think, for example, of a

> 4-count with one side being two people half-juggling. There, if I am acting

> the left hand, I never pass a club to anyone but my near-side partner. I

> wonder if any time-asymetrical run-around patterns exist. I kind of think

> not, but I'm not sure I could prove it.

>

> Man, I wonder if anyone else in the world has done this stuff. We better

> hustle up another person and try two facing 3-count run-arounds.

>

> -- CHS

>

> > -----Original Message-----

> > From: Carl Raffa [SMTP:craffa@...]

> > Sent: Friday, May 11, 2001 10:18 AM

> > To: Charles Shapiro; mad@...

> > Subject: 3 count run-around

> >

> > The 3 count run-around pattern repeats after 8 passes. The easiest way

> > to analyze it is by looking at the throws. The throws in the basic

> > run-around look like this:

> >

> > Start with A in front and B in back:

> >

> > B throws right to B's left

> > A throws left to A's right

> > A throws right to B's left

> > B throws left to B's right

> >

> > B is now in front and A is in back:

> >

> > A throws right to A's left

> > B throws left to B's right

> > B throws right to A's left

> > A throws left to A's right

> >

> > So, the basic run-around repeats after 8 throws.

> >

> > We can use the least common multiple to determine how a composite

> > pattern repeats. With a 4 count, the solution is simple. The least

> > common multiple of 4 and 8 is 8, so the whole pattern repeats every 8

> > throws. We have to look at the run-around pattern to see who gets to

> > pass. If you look at the 1st and 5th throws, you will see that B gets

> > the first pass and A gets the second pass.

> >

> > Now, the 3 count is a bit more complex. The least common multiple of 3

> > and 8 is 24! So, the whole pattern repeats every 24 throws. Again, the

> > look at the run-around pattern to see who gets to pass. The pass

> > sequence is B, B, B, A, A, A, A, B.

> >

> > Carl

>

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