How Many Objects Can Be Juggled (1997) | Hacker News Reader

How Many Objects Can Be Juggled (1997)

fermatslibrary.com

75 pointsmgdo51 comments

There's a fascinating video on this, which features the current world record holder, Alex Barron — who holds the records for most balls juggled (11) and most balls flashed (14).

See https://www.youtube.com/watch?v=7RDfNn7crqE

> I hate to break it to you aspiring numbers jugglers, but no human will ever juggle 100 balls.

I hate it when the in first sentence author makes a categorical sentence and then half a page later cites "a - acceleration of gravity = 9.81 m/s^2" as if it's a some sort of fundamental constant :) .

Here is an example of juggling under simulated different gravity: https://blog.ted.com/athletic-machines-raffaello-dandrea-at-...

Also, he wrote:

"g = acceleration due to gravity = 9.81 m/s^-2"

m/s^-2 is not the correct unit for acceleration.

Yes it is. That's the SI unit.

https://en.wikipedia.org/wiki/Acceleration

The text was "m/s^-2" (i.e., "m·s²") not "m/s^2".

No it isn't. That's not the SI unit.

https://en.wikipedia.org/wiki/Acceleration

TIP: Look closer =)

Oops, thanks, guess I have mental autocorrect turned on

Haha, no problem. It's easy to miss =)

> Here is an example of juggling under simulated different gravity

I've also seen juggling in VR where gravity can be modified.

If you're looking for a great long form article around juggling, I highly suggest Grantland's article[1] as one of my favorite "why did I just spend an hour reading about that?"

[1] http://grantland.com/features/anthony-gatto-juggling-cirque-...

One thing that needs to be discussed is the accuracy, which is independent of gravity. You can juggle at about 6 to 8 throws per second, so if you want to juggle, say, 50 objects, each one needs to be in flight for about 7 seconds.

That gives 7 seconds of time to drift from the intended landing location (time and space) and that's a long time.

Even if you don't have to throw things really high, their "beaten area" grows with time, and that's a quantity I'd be interested in seeing explored.

Also see (Claude) Shannon's Juggling Theorem: https://en.wikipedia.org/wiki/Juggling#Mathematics

Shannon also built juggling robots. (But not, to my knowledge, machines to measure human juggling.) Here's a video: https://www.youtube.com/watch?reload=9&v=sBHGzRxfeJY

> Shannon also built juggling robots. (

I'm disappointed this Kickstarter project never received funding:

https://www.kickstarter.com/projects/958028900/human-scale-p...

Here's the same paper without the maddening UX:

http://www.juggling.org/papers/limits/

I can't juggle. Not even in slow motion, with scarves. I've tried many times. I'm convinced some people just can't "get" it.

I found scarves harder to learn. You can learn to juggle three balls in 30 minutes:

1. Hold one bean bag in your dominant hand while standing, facing a wall, about a foot away.

2. Drop the bag. There, that's over with. (You'll be doing it a lot.)

3. Pick up the bag.

4. Toss the bag from one hand to the other, keeping your elbows loosely at your sides, until you consistently have it arcing at about eye level before descending to the other hand.

5. Do that 100 more times.

6. Hold one bag in each hand (drop them once, if you like).

7. Toss from your dominant hand and, when the bag is at the top of its arc in front of your eyes, toss the other bag. Catch them both.

8. Repeat without pausing another 100 times.

9. Hold two bags in your dominant hand and one in your other hand. Toss the first, wait for the top of the arc and toss the second, wait for the top of the arc and toss the third.

10. You're juggling. Drop all the bags to celebrate.

Definitely. I've had a hard time convincing people to do the required repetitions. They end early and say that they can't juggle. The few people that have, successfully juggle in a relatively short amount of time. Their form isn't great and they can't keep it up for a long time, but they have a great start.

A couple other exercises/tips worth mentioning:

- Start with your non-dominant hand for 2 balls as well (alternate which hand you start with).

- Stand over a couch or bed to make it less costly to drop a ball.

- Stand in front of a wall to notice when you're moving forward.

- For the advanced: try two in one hand (much harder than 3). Will make 3 ball juggling easier.

> try two in one hand (much harder than 3). Will make 3 ball juggling easier.

What does this mean?

Not GP, but if I were to guess: Try juggling only two balls, but use only one hand. Here's a video for demonstration: https://www.youtube.com/watch?v=9uMui692JHU

Repetition is the key. It's very difficult, at first, to consistently throw a ball in the same way. It's also difficult to do it in the same way with both hands (right to left, left to right). Once you master the simple action of throwing the ball consistently from either hand, throwing 2, and then 3 balls at a time is much easier.

When you can juggle you won't ever look at your hands, you just look at the space in front of your eyes and your hands will be where they need to be.

I can juggle 3 balls quite well, but I can not for the life of me juggle 4 (different pattern) or 5 (same pattern as 3, more balls). 4 is extraordinarily difficult for me. I approach it by separately juggling 2 balls in each hand (no ball swaps hands), but my left hand betrays me!

Ahh, the Klutz method. I know it well. :)

One key thing that's always left out of these instructions is that you can't throw the same arc with both hands. This causes the balls to collide mid-air if your throws are too consistent. This was very frustrating for me when I was first learning. You have to move your hands side to side a little, throwing when your hand is closer to your center and catching when your hand is further from your center.

The pattern is like McDonald's Golden Arches but with the arches pushed inwards so that they mostly overlap, and as you say you always throw inside the ball coming down.

Two helpful things while learning are a) doing it standing in front of a wall, so you don't have to worry about keeping throws from going forward away from your body (common while starting), and b) doing it in front of a ledge, so you don't have to bend over so far when you drop balls (even more common while starting). Combine both if you can, makes it much less frustrating!

Have you ever tried with a good teacher there, in person, offering feedback, advice, and suggestions?

It's not something you can just pick up and try every once in a while. You have to seriously focus on it, like at least several hours a week. Once you get it, you'll never forget how, though. I learned in high school and only do it sporadically, and I can still pick up any random collection of 3 things and juggle 20 years later.

I've got a couple of other skills that took me a few months to pick up -- card tricks, and beatmatching records. They're all skills that look effortless but take a lot of practice to get right.

I suspect some people are significantly more strong hand dominate than other people.

I suspect that significantly adds to the learning curve.

---------

While I don't want to advocate a defeatist outlook (I'm sure if you truly plowed enough time into it you could brute force learn it), with juggling, the stakes are low enough, that maybe this is a nice opportunity to appreciate a little humility about how everyone's gifts are a little different,

and to be grateful for all the things you're able to accomplish w/o extraordinary brute force techniques

Claude Shannon authored Shannon's Juggling Theorem, and built juggling robots. I gave some links elsewhere in this thread.

My father-in-law knew worked with Shannon at Bell Labs, and then MIT. (Among other projects, they made useless machines[1].) One day Shannon rode in on a unicycle. "I don't know you could do that", Marvin said. "I just learned." "How long did it take you?" Shannon: "Only about twenty minutes of practice. But it took three months to get that twenty minutes."

Back to juggling: In addition to the other advice here, I recommend balls that don't bounce (or, balled-up socks), and stand over a bed so you don't have to chase the balls when you drop them. Then you can get your twenty (or thirty, or sixty) minutes of actual practice, in closer to twenty minutes (or thirty, or an hour).

Also, like many motor skills (sports, music), ten minutes of practice six days a week trumps an hour of practice once a week, especially if you're getting decent sleep in between.

[1] https://en.wikipedia.org/wiki/Useless_machine

My dad could juggle reasonably well, I'm fairly certain up to 6 balls. As a kid this was amazing, and it killed me that I couldn't do it at all. I was generally bad at physical things, but I tried this a lot, and never got anywhere.

Then, in my mid-twenties, my mother-in-law gave me a juggling book, apropos of nothing. I followed the instructions, and within a couple of days, could reasonably keep 3 balls in the air. I've never gotten any better than that, but by resetting my expectations and following instructions set out for me, I was able to cross that threshold so now my kids think I'm amazing. :)

I usually recommend juggling over a bed, because that limits the amount of exertion that happens during a drop. It also reduces the negative reinforcement that a drop is a failure. Because a drop is not a failure, it's just a drop.

People tend to have mental blocks around things, and I've found that when you break a mental block (this happened with me and Mill's Mess) the result is fantastic. It opens up levels of creativity you never believed you had.

It also teaches you patience with yourself. Good things are possible if you're patient with yourself. If you can't get something one day, give it a rest. Try it again the next day.

> Not even in slow motion, with scarves.

Scarves are harder than balls because their movement is unpredictable and your hands end up moving in all directions if you can't throw them well.

Find decently weighted balls (130g is standard) which will have enough weight that throws can be well controlled. Juggling is 90% throwing and 10% catching. Get the throws right and everything will quite literally fall in to place.

What I find fascinating is how there's a whole mathematical formalism for describing juggling patterns called siteswap which allows patterns to not only be described, but discovered. It's based on throw heights, with the basic 3-ball cascade being represented by 3 - a sequence of balls thrown so they land 3 beats later on, repeated indefinitely. 531 indicates one ball thrown high, one mid-height and then one passed directly across from one hand to the other.

Basic intro:

http://www.twjc.co.uk/siteswap.html

Much more in-depth treatment here including synchronous and multiplex patterns, ladder and causal diagrams and a bunch of proofs relating to the notation:

https://www.jugglingedge.com/pdf/BenBeeversGuidetoJugglingPa...

but you can always go here and try a few patterns such as 3, 441, 531, 504, (4,4) or in windmill/Mill's Mess mode, 423

http://www.gunswap.co/

Can more objects be juggled on the moon then on earth?

I don’t think so. The moon’s smaller gravitational acceleration helps, but on earth, you can cheat by juggling objects that have a low terminal velocity and/or that are only just heavier than the air they displace (as long as they are small and light enough so that you can push them upwards reasonably fast, the fact that they won’t go high once you let them go doesn’t matter much)

Yes, check the definitions of the author's theoretical max ball calculation on the first page gravitational acceleration determines the 'hang time' of a ball for a given value of hand acceleration. So holding everything else the same the same person with the same ability to throw 9 balls at g=-9.8m/s^2 would throw the balls higher on the moon giving them more room in their pattern for more balls.

I came here looking for a paper about OO and having too much state. But this was interesting.

I'm fascinated by these seemingly useless yet remarkably mind grasping problems.

I remember reading that Feynman had a profound insight while calculating the wobbling of plates being thrown on a ship.

You never know where a fertile mind can be taken by those aimless thoughts.

It was just the cafeteria at Cornell, slightly more prosaic than a ship.

I'm surprised that robots have not yet surpassed human jugglers.

https://en.wikipedia.org/wiki/Juggling_robot

Could a styrofoam or other ball fall more slowly (like a feather) and therefore allow more of them to be thrown?

The acceleration due to gravity is independent of mass. A 1cm ball made of lead and a 1cm ball made of wood would both fall at the same speed.

What does make a difference is shape, due to differing amounts of air resistance and lift.

The force from air resistance at a given velocity will be the same on the 1 cm lead ball and the 1 cm wood ball because they have the same shape and size. The acceleration from air resistance is the force divided by the mass, and so the 1 cm lead ball has much less acceleration from air resistance than does the 1 cm wood ball.

As you note, the acceleration due to gravity is the same on both balls.

The net acceleration, therefore, is larger for the 1 cm lead ball than for the 1 cm wood ball, and so the lead ball falls faster.

The difference in acceleration between the lead ball and the wood ball might not be large enough to easily see if you just drop them a couple meters.

Try a ping pong ball and a lead ball of the same diameter (4 cm). That lead ball would weigh 372 g. A ping pong ball is 2.7 g. At a given velocity, the acceleration from air resistance would be almost 140 times greater on the ping pong ball than on the lead ball. That should greatly reduce the height you need to drop from in order to see the difference.

This doesn't sound correct to me. The acceleration due to gravity is the same on both balls. However, since the lead ball weighs more than the wood ball, the force of gravity on the lead ball is stronger than the force on the wood ball (F_g = mg).

The effect of air resistance can be modeled by a different force. It is typically modeled as a linear function of velocity rather than mass, and it models the behaviour that the faster an object is traveling the more it is affected by air resistance which acts in the opposite direction as the direction the object is traveling in (F_a = -kv). By adding the two forces together, you get a second-order differential equation that describes how the object behaves (F_g + F_a = F_net = ma).

[1] https://oregonstate.edu/instruct/mth252h/Bogley/w02/resist.h...

> The effect of air resistance can be modeled by a different force. It is typically modeled as a linear function of velocity rather than mass [...]

That's the key. The air resistance force is a function of velocity [1], and it is not a function of mass. Two bodies that are aerodynamically identical (same shape, size, same boundary interactions with the air) experience the same air resistance force at a given velocity, regardless of their masses.

As you note, the falling body has two opposing forces. Gravity, which is proportional to the mass of the body, and air resistance, which depends on velocity and does not depend on mass. The motion of the body is determined by the net force.

The net force on the ball is mg + D(v), where m is the mass of the ball, g is 9.81 m/s^2, and D(v) is the function that gives air resistance of the ball at velocity v. (Note: I'm using a coordinate system where balls fall in the positive direction. In this system D(v) will be negative).

The net acceleration on the ball is net force divided by mass. This is g + D(v)/m.

Note the effect of varying the mass, leaving all else the same. Remember, D(v) is negative, so the effect of the D(v)/m term is to reduce the net acceleration the ball feels. In other words, it is to make the ball fall slower.

If we raise the mass, we reduce the magnitude of D(v)/m. We reduce the amount that air resistance slows down the ball. If we lower the mass, the opposite happens. Air resistance slows down the ball more.

For a given ball, as it falls and picks up speed, the air resistance goes up, becoming more and more effective at countering the gravitation force. This puts an upper limit on how fast the ball can fall--the so-called "terminal velocity". This is the velocity where D(v) = -gm. Note that for balls with larger mass, terminal velocity will be higher.

[1] a linear function at very low speed with no turbulence, a quadratic function in most situations we normally encounter in everyday life. The quadratic is 1/2 p Cd A v^2, where p is the density of the air, Cd is the drag coefficient (0.47 for a sphere), A is the cross sectional area, and v is the speed relative to the air.

You are correct only if all else is the same. However, I take issue with the assumption that you can leave all else the same. The net force on a heavier object is a higher for the lead ball than for the wood ball. The net force for the lead ball is m_l x a, but the net force on the wood ball is m_w x a where m_l>m_w, and the acceleration a depends on the shape of the object and can be thought of as the same.

We were given that the lead ball and the wood ball were the same size. I am assuming that they are both spherical.

The drag force is 1/2 p Cd A v^2 where p is the density of the air, Cd is the coefficient of drag, A is the cross sectional area, and v is the velocity.

If both balls are spheres, Cd is the same for them (0.47). The air density is the same for both. The cross sectional area is that same. Hence, the two balls have the same drag force at the same velocity. Hence, the deceleration from drag is lower for the heavier ball.

Here's an example with a basketball and a bowling ball showing what happens: https://www.youtube.com/watch?v=mGZLuaJ5MOc

Note that you need quite a long drop to see a noticeable difference.

Below is a simple simulator that drops two spherical balls of the same size but different mass, and prints how far they have fallen and how fast they are going each second for the first 10 seconds. I'll give some results first, and then the simulator code if anyone wants to play with it.

Here are the results for a 4 cm diameter lead ball and a 4 cm diameter ping pong ball:

  0.0 (0.0, 0.01) (0.0, 0.01)
  1.0 (4.9, 9.79) (4.11, 6.99)
  2.0 (19.51, 19.38) (12.02, 8.39)
  3.0 (43.54, 28.63) (20.51, 8.54)
  4.0 (76.58, 37.37) (29.06, 8.55)
  5.0 (118.06, 45.5) (37.62, 8.56)
  6.0 (167.34, 52.94) (46.17, 8.56)
  7.0 (223.7, 59.66) (54.73, 8.56)
  8.0 (286.41, 65.64) (63.29, 8.56)
  9.0 (354.74, 70.91) (71.84, 8.56)
  10.0 (428.0, 75.5) (80.4, 8.56)

Numbers in each row are: time in seconds since drop, distance first ball has fallen (in meters), velocity of first ball (meters/second), and the distance and velocity of the second ball.

Here's a pair of 12 cm diameter balls one weighing 16 pounds (maximum weight for a bowling ball), and one about the weight of a basketball:

  0.0 (0.0, 0.01) (0.0, 0.01)
  1.0 (4.9, 9.76) (4.75, 9.2)
  2.0 (19.4, 19.18) (17.42, 15.59)
  3.0 (43.03, 27.97) (34.92, 19.0)
  4.0 (75.04, 35.91) (54.81, 20.56)
  5.0 (114.53, 42.9) (75.76, 21.23)
  6.0 (160.5, 48.88) (97.15, 21.51)
  7.0 (211.96, 53.89) (118.72, 21.62)
  8.0 (267.99, 58.02) (140.37, 21.67)
  9.0 (327.74, 61.37) (162.05, 21.69)
  10.0 (390.51, 64.06) (183.74, 21.69)

The simulator is just doing a simple linear simulation that assumes constant velocity and acceleration during between simulation steps. That's not super accurate, but it is good enough to show the physics.

Simulator code below. Set r to the radius in meters of your spheres. Set m1 and m2 to the masses of your two spheres, in kilograms.

  #!/usr/bin/env python3
  import math
  
  p = 1.225   # kg/m^3
  Cd = 0.47   # for sphere
  r = .12     # m
  
  m1 = 7.2    # kg  mass of 4 cm lead ball
  m2 = .625    # kg mass of 4 cm ping pong ball
  
  def drag(v):    # m/s
      # 1/2 p Cd A v^2
      return 1/2 * p * Cd * math.pi * r**2 * v**2
  
  def sim(m1, m2):
      y1, y2 = 0.0, 0.0   # m
      v1, v2 = 0.0, 0.0   # m/s
      dt = 0.001          # s
  
      for ms in range(0, 10001):
          y1 += v1 * dt
          y2 += v2 * dt
  
          v1 -= drag(v1) * dt / m1
          v2 -= drag(v2) * dt / m2
  
          v1 += 9.81 * dt
          v2 += 9.81 * dt
  
          if ms % 1000 == 0:
              print(ms/1000, (round(y1,2), round(v1,2)), (round(y2,2), round(v2,2)))
  
  sim(m1, m2)

I appreciate your earnestness, but running your code gives:

  0 (0.0, 0.01) (0.0, 0.01)
  1 (4.91, 9.82) (4.91, 9.82)
  2 (19.63, 19.63) (19.63, 19.63)
  3 (44.16, 29.44) (44.16, 29.44)
  4 (78.5, 39.25) (78.5, 39.25)
  5 (122.65, 49.06) (122.65, 49.06)
  6 (176.61, 58.87) (176.61, 58.87)
  7 (240.38, 68.68) (240.38, 68.68)
  8 (313.96, 78.49) (313.96, 78.49)
  9 (397.35, 88.3) (397.35, 88.3)
  10 (490.55, 98.11) (490.55, 98.11)

It's Python 3. You ran it with Python 2. In Python 2, 1/2 == 0. In Python 3, 1/2 == 0.5. That means that in the drag function, this expression:

  1/2 * p * Cd * math.pi * r**2 * v**2

always gives 0 on Python 2 because of that 1/2 factor.

If you want to run it on Python 2, either change the 1/2 in the drag function to 1./2 or 0.5, or add

  from __future__ import division

at the top to tell Python 2 you want to use the Python 3 division behavior.

Then I stand corrected. Thank you for being patient.

Galileo says no.

Or what about say a sort of balloon that is partially filled with helium such that is it just shy of neutral buoyancy, that would be like low gravity. Then perhaps a person could do 100's. Maybe.

What about an object with a one-way parachute of sorts change that? That is an object that rises at normal speed but falls at a fraction of the speed due to inherent aerodynamics?

The effect would need to be super precise and air currents limited. And that may be impossible given all the motion going on.

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