Anyway, Rye told me about this double-star juggling pattern they used to do back in the day (when there were more than five people in juggling club). Of course, since then, the club has shrunk a bit, the rope they used to mark off where each person would stand has been appropriated for other uses, and the chart they drew that showed each person what angle and distance from their partners to stand at has disappeared.
President Rob and I managed to recreate the drawing for the pattern last night, but I don't have enough free time to sit down and figure out what the angles and distances are. It looks to me like there are two different angles in the pattern (one for the inside star and one for the outside star) and one standard passing distance. Maybe two--I could be wrong. I haven't really studied it too closely yet.
Anyway, what I want to know is, what are the missing angles, and how far apart should we mark ourselves off so that we're not hitting the person in front of us? (Note: We'd be throwing pins AROUND the person standing directly in front of us, so we'd also have to be far enough back to see around them a little bit too.)
5 comments:
The problem is missing a single critical piece of information. The problem cannot be solved until that one piece of information is provided.
The missing piece of information is: how far ahead of the outside-star people are the inside-star people standing?
That distance determines everything else in the problem.
The angles between the outside members are fixed at 108-degrees. The angles between the inside people are fixed at 108 degrees. That much is known. However, the arrows you have drawn are not between outside member of the group, they are between one outside member and two inside members.
As the relative size of the two stars **to each other** grows and shrinks, the angles between your arrows change. The size of the stars will combine with the angles to determine the exact measurement of the arrows.
So before you can specify either the angles, or the length of the arrows, you must provide the dimensions of the stars relative to each other.
This in turn, is determined by the answer to the question, "How far in front of an outside member must an inside member stand in order to have the club (or torch, or ball or whatever) clear the inside member's body and arms?"
This should be easily measurable based on the widest person's width and normal reach. Of course the clubs/balls/torches/etc. might be thrown OVER the inside person, in which case the distance might possibly be decreased considerably, but the visibility (along an arrow) will still depend on the widest person's width.
Once you know the widest person's clearance width (either visibility or physical clearance), you will know the distance across the base of a triangle formed by the arrows coming from an outside member and the inside member's clearance.
If you remember your trig, you will remember that on an isoceles triangle, if you know the length of the base and the top angle, you can calculate all other values, meaning you can calculate how far in front of the outside member the inside member must stand. Voila.
Once you know how far in front of the outside member the inside member must stand, it is a simple matter to calculate all other dimensions in the problem. And what you have done is find the MINIMIUM or smallest dimensions. You can easily make the stars larger by following the rules of proportionality.
Of course, if you change the PROPORTIONALITY, or the RELATIVE sizes of the stars (to each other) from their minimum you found, you will then need to recalculate the angles and distances. But as long as the proportions of the start to each other stay the same, you can SCALE the stars up in size without changing the angles.
Simple trig, eh?
The other problem and why it looks weird is you're not throwing at exact angles since you're throwing from right hands to left hands and not from the center of your body to the center of your body.
I'm so glad you've become more jugging fantastic then I have. I only do my diabolo now.
Ouch.
HaHa! What your dad said, though I'd say that you should at a minimum stay 8 ft from the person in front of you so you have the maximum clearance and will only bepaccing whatever about 20 ft or so. :)
There's my $.02.
er... that would be "passing". :)
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