
How Many Objects Can Be Juggled (1997) - mgdo
http://fermatslibrary.com/s/how-many-objects-can-be-juggled
======
jw1224
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](https://www.youtube.com/watch?v=7RDfNn7crqE)

------
avmich
> 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-...](https://blog.ted.com/athletic-machines-raffaello-dandrea-at-
tedglobal-2013/)

~~~
iiv
Also, he wrote:

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

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

~~~
mark_edward
Yes it is. That's the SI unit.

[https://en.wikipedia.org/wiki/Acceleration](https://en.wikipedia.org/wiki/Acceleration)

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

[https://en.wikipedia.org/wiki/Acceleration](https://en.wikipedia.org/wiki/Acceleration)

TIP: Look closer =)

~~~
mark_edward
Oops, thanks, guess I have mental autocorrect turned on

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

------
cridenour
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-...](http://grantland.com/features/anthony-gatto-juggling-cirque-du-
soleil-jason-fagone/)

------
ColinWright
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.

------
osteele
Also see (Claude) Shannon's Juggling Theorem:
[https://en.wikipedia.org/wiki/Juggling#Mathematics](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](https://www.youtube.com/watch?reload=9&v=sBHGzRxfeJY)

~~~
imron
> Shannon also built juggling robots. (

I'm disappointed this Kickstarter project never received funding:

[https://www.kickstarter.com/projects/958028900/human-
scale-p...](https://www.kickstarter.com/projects/958028900/human-scale-
prototype-for-bugjuggler-a-car-jugglin)

------
taco_emoji
Here's the same paper without the maddening UX:

[http://www.juggling.org/papers/limits/](http://www.juggling.org/papers/limits/)

------
nsxwolf
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.

~~~
patrickmay
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.

~~~
jay-anderson
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.

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

What does this mean?

~~~
scbrg
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](https://www.youtube.com/watch?v=9uMui692JHU)

------
spiralx
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](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...](https://www.jugglingedge.com/pdf/BenBeeversGuidetoJugglingPatterns.pdf)

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/](http://www.gunswap.co/)

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

~~~
rtkwe
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.

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

------
tambourine_man
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.

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

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

[https://en.wikipedia.org/wiki/Juggling_robot](https://en.wikipedia.org/wiki/Juggling_robot)

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

~~~
jcranmer
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.

~~~
tzs
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.

~~~
ctchocula
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...](https://oregonstate.edu/instruct/mth252h/Bogley/w02/resist.html)

~~~
tzs
> 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.

~~~
ctchocula
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.

~~~
tzs
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](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)

~~~
ctchocula
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)

~~~
tzs
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.

~~~
ctchocula
Then I stand corrected. Thank you for being patient.

