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](https://projects.delta.ncsu.edu/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. [](https://www.youtube.com/watch?v=WK6AlhTjUbY) [For more information on Alex's record you can check this [article](https://www.wired.com/story/the-physicsand-physicalityof-extreme-juggling/)] 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 $$ \omega= b/h - r $$