In this paper Jack Kalvan, a mechanical engineer and avid juggler, ...
Each hand has $b/h$ objects and so it would take $b/h \tau$ sec...
The equation for the velocity of a ball thrown in the air is $$ ...
In 1997, Kalvan didn't have smartphones or smartwatches equipped wi...
Calibration is a matter of getting some experimental points for a l...
To illustrate the importance of accuracy when juggling let's calcul...
How Many Objects Can Be Juggled
Jack Kalvan
Originally published in 1997
I hate to break it to you aspiring numbers jugglers,
but no human will ever juggle 100 balls. Only a handful
of people have reached a level to throw eleven or twelve
objects into the air, and so far, not for more than a few
seconds. No one has even come close to juggling 13 balls.
But is this within the realm of human possibility?
Hand speed is one of the main factors that limit the
number of objects one can juggle. (The other main fac-
tors being accuracy of throws and having long enough
arms and enough space in the air for the juggling pat-
tern.) I decided to find out if anyone has the hand speed
necessary to juggle 13 or more balls. So I designed an
experiment to measure the theoretical human juggling
limits - given the acceleration of the hands.
To write the necessary equations, I define the following
b = number of balls
h = number of hands
f = flight time of a ball from throw to catch
τ = time between throws from the same hand
= vertical throw velocity
g = acceleration due to gravity = 9.81 m/s
r = "dwell ratio" or fraction of time a hand is holding
a ball. My tests show r is usually about 2/3.
ω = average number of balls in flight per arc
One can also think of r as the average number of balls
in a hand while juggling. omega can be expressed as
the number of balls per hand minus the balls held in the
hand: ω = (b/h) r .
ω is also equal to the time that balls are in flight di-
vided by how often they are thrown: ω = f .
To simplify my analysis, I will assume balls are thrown
and caught at the same height. Newtonian physics tells
us the flight time of a ball, f = 2V
/g . Substituting this
equations for f into the second equation for omega gives
ω = 2V
Since g is a constant, we see that omega is proportional
to the throwing velocity of the hand divided by the time
between throws. This means the number of balls in the
air while juggling is closely related to the acceleration of
the hand. Although a juggler’s hands do not necessarily
accelerate smoothly, the number of balls one can get into
the air is approximately proportional to the maximum
acceleration of one’s hands .
I figured if I measure the maximum acceleration of
a juggler’s hands with a simple accelerometer, I could
roughly calculate the juggler’s maximum value for omega.
And substituting this value into the equation, b/h =
ω + r, gives an approximation of the maximum num-
ber of balls one can theoretically juggle. Remember, this
maximum number of balls is calculated only from the
speed a juggler can potentially throw balls into the air.
It does not take into account accuracy of throws or the
possibility of collisions.
Since the number of balls juggled is proportional to
hand acceleration, a corollary is that the height of your
juggling pattern is not related to your hand acceleration.
For example, if you juggle 5 balls high, you have about
the same hand acceleration as if you juggle them low.
The difference is that to juggle high, you accelerate for a
longer time and therefore have longer hand motion and
a higher throwing velocity.
I believe the following chart describes how the deriva-
tives of the vertical hand motion relate to juggling:
This simple device measures the hand acceleration.
A small mass is connected to a spring inside a tube.
When the hand accelerates the device (by shaking it
up and down), two opposing forces act on the mass:
the acceleration force (force = mass × acceleration) and
the spring force (force = spring stiffness × distance
stretched). These forces are in equilibrium when the
spring is stretched. A marker measures the maximum
distance the spring was stretched. Since the mass and
spring stiffness are constant, the maximum acceleration
is proportional to this distance. The distance the spring
stretches is therefore proportional to the number of balls
potentially juggled.
To calibrate the device, I attached it to the back of
a glove. This method worked well for low numbers of
balls but the glove and the weight of the jugglemeter
made juggling more difficult. Also imperfect juggling led
to artificially high readings because more acceleration is
needed to catch inaccurate throws.
To calibrate and test the jugglemeter for higher num-
bers of balls, I found it was easiest to first juggle a number
of balls for long enough to get the feel of it. Then, I’d
drop the balls, pick up the jugglemeter, and reproduce
the same hand motion while holding it. This process is
not extremely accurate since it relies on the ability to
reproduce the same hand motion as juggling but with
the balls removed. However, after tests with many jug-
glers and up to 10 balls, and extrapolating the results, I
feel confident that my calibration error is less than 10%.
Each juggler I tested repeated the calibration test with
an amount of balls that they could juggle comfortably,
to convince themselves that the meter calibration was
fairly accurate. Some people were unable to simulate
the proper hand motion without actually juggling the
balls. On the other hand, many were able to reproduce
the proper hand motion, and after pretending to juggle 5
balls, were shocked to see the meter reading "5". "How
does it know?!!" To find the maximum potential of each
subject, I told them to shake the jugglemeter as hard and
fast as they could, for a few seconds, in an up-and-down
motion. (This was the most interesting part for spec-
tators.) I measured the maximum with each hand and
allowed one retry if they thought they could do better.
I tested over 100 people ranging from non-jugglers to
some of the best jugglers in the world. Every one of the
test subjects had the hand speed to juggle 9 balls. The
average was about 16 balls, and the highest recorded was
about 24 balls. Hand speed did tend to increase slightly
with juggling ability. Those who said they had flashed 9
or more balls averaged about 18.3, while those who had
never flashed 5 balls averaged about 14.4. Comparing
subjects with comparable juggling ability: hand speed
averaged highest for people around 18 years old; there
was a small decrease in hand speed with age; and males
tended to get slightly higher readings than females. Often
there was a difference between the two hands and many
were surprised to find they did better with their "bad"
hand. I suspect that some of the test subjects were not
trying as hard as they could have. Also I suspect some
could have done better by relaxing their arms more it
helped me.
Every person I tested demonstrated the hand acceler-
ation needed to juggle a smooth 9 ball cascade. Unfor-
tunately, catching wild throws in a messy, inaccurate 9
ball cascade requires much more acceleration than this.
And very few of the test subjects have the accuracy to
juggle 9 balls smoothly. By practicing, one’s accuracy im-
proves, the demands on one’s acceleration are lessened,
and the juggling actually requires less and less effort. It
requires much less effort to juggle 7 balls well than to
juggle 7 balls badly. Here are some illustrations of how
inaccuracy leads to higher acceleration. If one makes a
throw too far across and has to move the catching hand
twice the usual horizontal distance in the same amount of
time, this requires twice as much horizontal acceleration.
If one’s inaccuracy is in throw height, there is an even
more unfortunate relationship. Suppose one hand makes
a slightly high throw followed by a slightly low throw, so
that the two balls come down at almost the same time.
If a throw has to be made in half the normal time, this
requires four times the vertical acceleration. If juggling
7 balls, this correction would probably be beyond human
capabilities. I believe that eventually someone will juggle
13 balls, and flashing 15 doesn’t seem too unlikely. But
how can anyone get to juggling 13 balls smoothly without
wasting many years trying to juggle 13 balls badly? It
will probably require new training techniques; for exam-
ple a technique that allows the juggler to practice throw-
ing balls accurately at 15 ball juggling speed, without
worrying about catching them and picking them up ev-
ery few seconds. But that’s another paper.


The equation for the velocity of a ball thrown in the air is $$ v(t) = V_v - gt $$ At the point of maximum height, which happens at $t=f/2$, the velocity of the ball vanishes and so $f=2V_v/g$. Calibration is a matter of getting some experimental points for a low number of balls and fitting the function $y = ax + b$, where $y$ is the number of balls per hand, $x$ is the hand acceleration and $a$ and $b$ are constants that are independent from the number of balls being juggled (note that the "dwell ratio" seems to be a characteristic of the juggler and independent of the number of balls). In 1997, Kalvan didn't have smartphones or smartwatches equipped with powerful accelerometers but if you want to determine how many balls you could theoretically juggle based on your hands' acceleration you just need to download [an app like NCSU MyTech]( After you download the app simply grab your phone and move it as fast as you can in a juggle like movement. You can then use the data and calculate the maximum number of balls you could juggle - $b/h$ To illustrate the importance of accuracy when juggling let's calculate the horizontal displacement of a ball - $\Delta s$ - caused by a deviation of 1 degree ($\Delta \theta$) in the throwing angle. ![]( The equations of motion for a ball are $$ y(t) = v_{0y}t-\frac{1}{2}gt^2 \\ v_y = v_{0y}-gt $$ Now at the maximum height - $h$ - which happens at time $f/2$ $v_{0y}=g\frac{f}{2}$ and $h = \frac{gf^2}{4}-\frac{gf^2}{8} = \frac{1}{8}gf^2$ Now the horizontal position of the ball $s$ is $s = fv_{0x} = fv_{0y}\tan \theta = 4h \tan \theta$ since $v_{0x}/v_{0y} = \tan \theta$ and so $\theta = \arctan (s/4h)$ Now $\theta + \Delta \theta = \arctan [(s+\Delta s)/4h]$, giving $\theta = \arctan [(s+\Delta s)/4h] - \arctan (s/4h)$ Since $\arctan(x) = x + O(x^3)$ we have $$ \Delta \theta \approx \frac{s+\Delta s}{4h}- \frac{s}{4h} = \frac{\Delta s}{4h} $$ For $h=3m$, $\Delta \theta = \frac{2 \pi}{360}$, $\Delta s$ is 21 cm (!), which means that if you want to attempt to juggle a lot of balls hand acceleration is important but you also need an incredible throwing precision. In this paper Jack Kalvan, a mechanical engineer and avid juggler, tries to answer the following question: What is the maximum number of objects a human can juggle? The analysis is centered around the maximum velocity of the hands, which being a limiting factor, is only one piece of a more complex puzzle that involves throwing accuracy among others. Kalvan writes that flashing 15 balls (each ball is only thrown and caught once) "doesn't seem too unlikely" and his prediction appears to be not too far from reality - in April 2017, Alex Barron was able to flash 14 balls. [!['aw'](]( [For more information on Alex's record you can check this [article](] Each hand has $b/h$ objects and so it would take $b/h \tau$ seconds for a hand to go through a full cycle. Now, since each object will spend a fraction of time in the hand ($r \tau$), the total time each object spends in the air is simply the subtraction of these 2 times: $f = b/h \tau - r \tau$ or equivalently $f/\tau = b/h - r$. Now, at the point at which the objects reach the hand, the flux of incoming objects has to be equal to the flux of outgoing objects, otherwise there's going to be either a shortage or an accumulation of objects and the juggling pattern will break. ![]( Since objects are being thrown in the air at a frequency of $1/\tau$ and new objects are arriving at a frequency of $1/f$ times the average number of objects in the air ($\omega$) we get $$ \frac{\omega}{f} = \frac{1}{\tau} $$ or $$ \omega = \frac{f}{\tau} $$ Joining the two expressions we have $$ \omega= b/h - r $$